x = 1 n (x 1 + x 2 + +x n )

Transcription

1 3 PARTIAL DERIVATIVES Science Photo Librar Three-dimensional surfaces have high points and low points that are analogous to the peaks and valles of a mountain range. In this chapter we will use derivatives to locate these points and to stud other features of such surfaces. In this chapter we will etend man of the basic concepts of calculus to functions of two or more variables, commonl called functions of several variables. We will begin b discussing limits and continuit for functions of two and three variables, then we will define derivatives of such functions, and then we will use these derivatives to stud tangent planes, rates of change, slopes of surfaces, and maimiation and minimiation problems. Although man of the basic ideas that we developed for functions of one variable will carr over in a natural wa, functions of several variables are intrinsicall more complicated than functions of one variable, so we will need to develop new tools and new ideas to deal with such functions. 3. FUNCTIONS OF TWO OR MORE VARIABLES In previous sections we studied real-valued functions of a real variable and vector-valued functions of a real variable. In this section we will consider real-valued functions of two or more real variables. NOTATION AND TERMINOLOGY There are man familiar formulas in which a given variable depends on two or more other variables. For eample, the area A of a triangle depends on the base length b and height h b the formula A = bh; the volume V of a rectangular bo depends on the length l, the width w, and the height h b the formula V = lwh; and the arithmetic average of n real numbers,,,..., n, depends on those numbers b the formula Thus, we sa that = n ( n ) 96 A is a function of the two variables b and h; V is a function of the three variables l, w, and h; is a function of the n variables,,..., n. The terminolog and notation for functions of two or more variables is similar to that for functions of one variable. For eample, the epression = f(,)

2 3. Functions of Two or More Variables 97 means that is a function of and in the sense that a unique value of the dependent variable is determined b specifing values for the independent variables and. Similarl, epresses w as a function of,, and, and w = f(,,) u = f(,,..., n ) epresses u as a function of,,..., n. As with functions of one variable, the independent variables of a function of two or more variables ma be restricted to lie in some set D, which we call the domain of f. Sometimes the domain will be determined b phsical restrictions on the variables. If the function is defined b a formula and if there are no phsical restrictions or other restrictions stated eplicitl, then it is understood that the domain consists of all points for which the formula ields a real value for the dependent variable. We call this the natural domain of the function. The following definitions summarie this discussion. B etension, one can define the notion of n-dimensional space in which a point is a sequence of n real numbers (,,..., n ), and a function of n real variables is a rule that assigns a unique real number f(,,..., n ) to each point in some set in this space. 3.. definition A function f of two variables, and, is a rule that assigns a unique real number f(,) to each point (, ) in some set D in the -plane. 3.. definition A function f of three variables,,, and, is a rule that assigns a unique real number f(,,) to each point (,,) in some set D in threedimensional space. = Eample domain of f. Let f(,) = + + ln( ). Find f(e,) and sketch the natural Solution. B substitution, f(e,) = + + ln(e ) = + ln(e ) = + = 3 = The solid boundar line is included in the domain, while the dashed boundar is not included in the domain. To find the natural domain of f, we note that + isdefined onl when, while ln( ) is defined onl when < or <. Thus, the natural domain of f consists of all points in the -plane for which <. To sketch the natural domain, we first sketch the parabola = as a dashed curve and the line = as a solid curve. The natural domain of f is then the region ling above or on the line = and below the parabola = (Figure 3...) Figure 3.. Eample Let f(,,) = Find f (,, ) and the natural domain of f. Solution. B substitution, f (,, ) = () ( ) ( ) = Because of the square root sign, we must have in order to have a real

3 98 Chapter3 / Partial Derivatives value for f(,,). Rewriting this inequalit in the form + + we see that the natural domain of f consists of all points on or within the sphere + + = The wind chill inde is that temperature (in F) which would produce the same sensation on eposed skin at a wind speed of 3 mi/h as the temperature and wind speed combination in current weather conditions. wind speed v (mi/h) Table 3.. temperature T ( F) FUNCTIONS DESCRIBED BY TABLES Sometimes it is either desirable or necessar to represent a function of two variables in table form, rather than as an eplicit formula. For eample, the U.S. National Weather Service uses the formula W = T + (.475T 35.75)v.6 () to model the wind chill inde W (in F) as a function of the temperature T (in F) and the wind speed v (in mi/h) for wind speeds greater than 3 mi/h. This formula is sufficientl comple that it is difficult to get an intuitive feel for the relationship between the variables. One can get a clearer sense of the relationship b selecting sample values of T and v and constructing a table, such as Table 3.., in which we have rounded the values of W to the nearest integer. For eample, if the temperature is 3 F and the wind speed is 5 mi/h, it feels as if the temperature is 5 F. If the wind speed increases to 5 mi/h, the temperature then feels as if it has dropped to 9 F. Note that in this case, an increase in wind speed of mi/h causes a 6 F decrease in the wind chill inde. To estimate wind chill values not displaed in the table, we can use linear interpolation. For eample, suppose that the temperature is 3 F and the wind speed is 7 mi/h. A reasonable estimate for the drop in the wind chill inde from its value when the wind speed is 5 mi/h would be 6 F =. F. (Wh?) The resulting estimate in wind chill would then be 5. = 3.8 F. In some cases, tables for functions of two variables arise directl from eperimental data, in which case one must either work directl with the table or else use some technique to construct a formula that models the data in the table. Such modeling techniques are developed in statistics and numerical analsis tets. GRAPHS OF FUNCTIONS OF TWO VARIABLES Recall that for a function f of one variable, the graph of f() in the -plane was defined to be the graph of the equation = f(). Similarl, if f is a function of two variables, we define the graph of f(,) in -space to be the graph of the equation = f(,). In general, such a graph will be a surface in 3-space. Eample 3 In each part, describe the graph of the function in an -coordinate sstem. Solution (a). (a) f(,) = (b) f(,) = (c) f(,) = + B definition, the graph of the given function is the graph of the equation = which is a plane. A triangular portion of the plane can be sketched b plotting the intersections with the coordinate aes and joining them with line segments (Figure 3..a). Solution (b). B definition, the graph of the given function is the graph of the equation = ()

4 3. Functions of Two or More Variables 99 (,, ) = (,, ) = (a) (b) (,, ) = + After squaring both sides, this can be rewritten as + + = which represents a sphere of radius, centered at the origin. Since () imposes the added condition that, the graph is just the upper hemisphere (Figure 3..b). Solution (c). After squaring, we obtain The graph of the given function is the graph of the equation = + (3) = + which is the equation of a circular cone (see Table.7.). Since (3) imposes the condition that, the graph is just the lower nappe of the cone (Figure 3..c). LEVEL CURVES We are all familiar with the topographic (or contour) maps in which a three-dimensional landscape, such as a mountain range, is represented b two-dimensional contour lines or curves of constant elevation. Consider, for eample, the model hill and its contour map shown in Figure The contour map is constructed b passing planes of constant elevation through the hill, projecting the resulting contours onto a flat surface, and labeling the contours with their elevations. In Figure 3..3, note how the two gullies appear as indentations in the contour lines and how the curves are close together on the contour map where the hill has a steep slope and become more widel spaced where the slope is gradual. Figure 3.. (c) Hundreds of feet = f(, ) = k Figure 3..3 A perspective view of a model hill with two gullies A contour map of the model hill Level curve of height k Figure 3..4 f(, ) = k Contour maps are also useful for studing functions of two variables. If the surface = f(,) is cut b the horiontal plane = k, then at all points on the intersection we have f(,) = k. The projection of this intersection onto the -plane is called the level curve of height k or the level curve with constant k (Figure 3..4). A set of level curves for = f(,) is called a contour plot or contour map of f. Eample 4 The graph of the function f(,) = in -space is the hperbolic paraboloid (saddle surface) shown in Figure 3..5a. The level curves have equations of the form = k. For k>these curves are hperbolas opening along lines parallel to the -ais; for k< the are hperbolas opening along lines parallel to the -ais; and for k = the level curve consists of the intersecting lines + = and = (Figure 3..5b).

5 9 Chapter3 / Partial Derivatives Figure 3..5 (a) (b) Eample 5 Sketch the contour plot of f(,) = 4 + using level curves of height k =,,, 3, 4, 5. Solution. The graph of the surface = 4 + is the paraboloid shown in the left part of Figure 3..6, so we can reasonabl epect the contour plot to be a famil of ellipses centered at the origin. The level curve of height k has the equation 4 + = k. Ifk =, then the graph is the single point (, ). For k>we can rewrite the equation as k/4 + k = which represents a famil of ellipses with -intercepts ± k/ and -intercepts ± k. The contour plot for the specified values of k is shown in the right part of Figure = 4 + k = 5 k = 4 k = 3 k = = k k = Figure 3..6 Eample 6 Sketch the contour plot of f(,) = using level curves of height k = 6, 4,,,, 4, 6. Solution. The graph of the surface = is the plane shown in the left part of Figure 3..7, so we can reasonabl epect the contour plot to be a famil of parallel lines. The level curve of height k has the equation = k, which we can rewrite as = + ( k) This represents a famil of parallel lines of slope. The contour plot for the specified values of k is shown in the right part of Figure 3..7.

6 3. Functions of Two or More Variables 9 = = k 5 Figure k = 4 k = 6 k = 6 k = 4 k = k = k = CONTOUR PLOTS USING TECHNOLOGY Ecept in the simplest cases, contour plots can be difficult to produce without the help of a graphing utilit. Figure 3..8 illustrates how graphing technolog can be used to displa level curves. The table shows two graphical representations of the level curves of the function f(,) = sin sin produced with a CAS over the domain π, π. Figure >.98 <.98 <.833 <.737 <.64 <.547 <.45 <.356 <.6 <.65 < The term level surface is standard but confusing, since a level surface need not be level in the sense of being horiontal it is simpl a surface on which all values of f are the same. LEVEL SURFACES Observe that the graph of = f() is a curve in -space, and the graph of = f(,) is a surface in 3-space, so the number of dimensions required for these graphs is one greater than the number of independent variables. Accordingl, there is no direct wa to graph a function of three variables since four dimensions are required. However, if k is a constant, then the graph of the equation f(,,) = k will generall be a surface in 3-space (e.g., the graph of + + = is a sphere), which we call the level surface with constant k. Some geometric insight into the behavior of the function f can sometimes be obtained b graphing these level surfaces for various values of k. Eample 7 Describe the level surfaces of (a) f(,,) = + + (b) f(,,) = Solution (a). The level surfaces have equations of the form + + = k

7 9 Chapter3 / Partial Derivatives For k> the graph of this equation is a sphere of radius k, centered at the origin; for k = the graph is the single point (,, ); and for k< there is no level surface (Figure 3..9). Level surfaces of f (,, ) = + + Figure 3..9 k < k = k > Solution (b). The level surfaces have equations of the form = k As discussed in Section.7, this equation represents a cone if k =, a hperboloid of two sheets if k>, and a hperboloid of one sheet if k< (Figure 3..). GRAPHING FUNCTIONS OF TWO VARIABLES USING TECHNOLOGY Generating surfaces with a graphing utilit is more complicated than generating plane curves because there are more factors that must be taken into account. We can onl touch on the ideas here, so if ou want to use a graphing utilit, its documentation will be our main source of information. Graphing utilities can onl show a portion of -space in a viewing screen, so the first step in graphing a surface is to determine which portion of -space ou want to displa. This region is called the viewing bo or viewing window. For eample, Figure 3.. shows the effect of graphing the paraboloid = + in three different viewing windows. However, within a fied viewing bo, the appearance of the surface is also affected b the viewpoint, that is, the direction from which the surface is viewed, and the distance from the viewer to the surface. For eample, Figure 3.. shows the graph of the paraboloid = + from three different viewpoints using the first viewing bo in Figure Level surfaces of f (,, ) = Figure 3.. TECHNOLOGY MASTERY 4 5 Figure 3.. Varing the viewing bo. If ou have a graphing utilit that can generate surfaces in 3-space, read the documentation and tr to duplicate some of the surfaces in Figures 3.. and 3.. and Table Figure 3.. Varing the viewpoint. 8 4 Table 3.. shows si surfaces in 3-space along with their associated contour plots. Note that the mesh lines on the surface are traces in vertical planes, whereas the level curves correspond to traces in horiontal planes. In these contour plots the color gradation varies from purple through green to red as increases.

8 3. Functions of Two or More Variables 93 Table 3.. surface contour plot surface contour plot = cos = 5e sin = sin + = e ( + ) = cos() = QUICK CHECK EXERCISES 3. (See page 97 for answers.). The domain of f(,) = ln is and the domain of g(,) = ln + ln is.. Let f(,) = + +. (a) f(, ) = (b) f(, ) = (c) f (a,a) = (d) f(+,)= 3. Let f(,) = e +. (a) For what values of k will the graph of the level curve f(,) = k be nonempt? (b) Describe the level curves f(,) = k for the values of k obtained in part (a). 4. Let f(,,) = (a) For what values of k will the graph of the level surface f(,,) = k be nonempt? (b) Describe the level surfaces f(,,) = k for the values of k obtained in part (a).

9 94 Chapter3 / Partial Derivatives EXERCISE SET 3. Graphing Utilit C CAS 8 These eercises are concerned with functions of two variables.. Let f(,) = +. Find (a) f(, ) (b) f(, ) (c) f(, ) (d) f(, 3) (e) f(3a,a) (f ) f(ab,a b).. Let f(,) = + 3. Find (a) f(t,t ) (b) f(, ) (c) f(, 4). 3. Let f(,) = + 3. Find (a) f( +, ) (b) f(,3 3 ). 4. Let g() = sin. Find (a) g(/) (b) g() (c) g( ). 5. Find F(g(), h()) if F(,) = e, g() = 3, and h() = Find g(u(, ), v(, )) if g(,) = sin( ), u(, ) = 3, and v(,) = π. 7. Let f(,) = + 3, (t) = t, and (t) = t 3. Find (a) f((t), (t)) (b) f((), ()) (c) f((), ()). 8. Let g(,) = e 3, (t) = ln(t + ), and (t) = t. Find g((t), (t)). 9. Refer to Table 3.. to estimate the wind chill inde when (a) the temperature is 5 F and the wind speed is 7 mi/h (b) the temperature is 8 F and the wind speed is 5 mi/h.. Refer to Table 3.. to estimate the wind chill inde when (a) the temperature is 35 F and the wind speed is 4 mi/h (b) the temperature is 3 F and the wind speed is 5 mi/h.. One method for determining relative humidit is to wet the bulb of a thermometer, whirl it through the air, and then compare the thermometer reading with the actual air temperature. If the relative humidit is less than %, the reading on the thermometer will be less than the temperature of the air. This difference in temperature is known as the wet-bulb depression. The accompaning table gives the relative humidit as a function of the air temperature and the wet-bulb depression. Use the table to complete parts (a) (c). (a) What is the relative humidit if the air temperature is C and the wet-bulb thermometer reads 6 C? (b) Estimate the relative humidit if the air temperature is 5 C and the wet-bulb depression is 3.5 C. (c) Estimate the relative humidit if the air temperature is C and the wet-bulb depression is 5 C. wet-bulb depression ( C) Table E- air temperature ( C) Use the table in Eercise to complete parts (a) (c). (a) What is the wet-bulb depression if the air temperature is 3 C and the relative humidit is 73%? (b) Estimate the relative humidit if the air temperature is 5 C and the wet-bulb depression is 4.5 C. (c) Estimate the relative humidit if the air temperature is 6 C and the wet-bulb depression is 3 C. 3 6 These eercises involve functions of three variables. 3. Let f(,,) = Find (a) f(,, ) (b) f( 3,, ) (c) f(,, ) (d) f(a,a,a) (e) f(t,t, t) (f ) f(a + b, a b, b). 4. Let f(,,) = +. Find (a) f( +,, ) (b) f(,/,). 5. Find F(f(), g(), h()) if F(,,) = e, f() =, g() = +, and h() =. 6. Find g(u(,, ), v(,, ), w(,, )) if g(,,) = sin, u(,,)= 3, v(,,) = π, and w(,,) = /. 7 8 These eercises are concerned with functions of four or more variables. 7. (a) Let f(,,,t) = 3 + t. Find f( 5,,π,3π). n (b) Let f(,,..., n ) = k k. k= Find f(,,...,). 8. (a) Let f(u, v, λ, φ) = e u+v cos λ tan φ. Find f(,,,π/4). (b) Let f(,,..., n ) = n. Find f(,,...,n). 9 Sketch the domain of f. Use solid lines for portions of the boundar included in the domain and dashed lines for portions not included. 9. f(,) = ln( ). f(,) = + 4. f(,) =. f(,) = ln 3 4 Describe the domain of f in words. 3. (a) f(,) = e + (b) f(,,) = 5 (c) f(,,) = e 4 4. (a) f(,) = (b) f(,) = ln( ) + 3 (c) f(,,) = + +

10 3. Functions of Two or More Variables True False Determine whether the statement is true or false. Eplain our answer. 5. If the domain of f(,) is the -plane, then the domain of f(sin t, t)is the interval [, ]. 6. If f(,) = /, then a contour f(,) = m is the straight line = m. 7. The natural domain of f(,,) = is a disk of radius centered at the origin in the -plane. 8. Ever level surface of f(,,) = is a plane. (I) (c) (d) (II) 9 38 Sketch the graph of f. 9. f(,) = 3 3. f(,) = 9 3. f(,) = + 3. f(,) = f(,) = 34. f(,) = f(,) = f(,) = f(,) = f(,) = (III) (IV) FOCUS ON CONCEPTS 39. In each part, match the contour plot with one of the functions f(,) = +, f(, ) = +, f(,) = b inspection, and eplain our reasoning. Larger values of are indicated b lighter colors in the contour plot, and the concentric contours correspond to equall spaced values of. (a) (b) (c) 4. In each part, match the contour plot with one of the surfaces in the accompaning figure b inspection, and eplain our reasoning. The larger the value of, the lighter the color in the contour plot. Figure E-4 4. In each part, the questions refer to the contour map in the accompaning figure. (a) Is A or B the higher point? Eplain our reasoning. (b) Is the slope steeper at point A or at point B? Eplain our reasoning. (c) Starting at A and moving so that remains constant and increases, will the elevation begin to increase or decrease? (d) Starting at B and moving so that remains constant and increases, will the elevation begin to increase or decrease? (e) Starting at A and moving so that remains constant and decreases, will the elevation begin to increase or decrease? (f ) Starting at B and moving so that remains constant and decreases, will the elevation begin to increase or decrease? (a) (b) 3 4 A B Elevations in hundreds of feet 3 Figure E-4

11 96 Chapter3 / Partial Derivatives 4. A curve connecting points of equal atmospheric pressure on a weather map is called an isobar. On a tpical weather map the isobars refer to pressure at mean sea level and are given in units of millibars (mb). Mathematicall, isobars are level curves for the pressure function p(, ) defined at the geographic points (, ) represented on the map. Tightl packed isobars correspond to steep slopes on the graph of the pressure function, and these are usuall associated with strong winds the steeper the slope, the greater the speed of the wind. (a) Referring to the accompaning weather map, is the wind speed greater in Medicine Hat, Alberta or in Chicago? Eplain our reasoning. (b) Estimate the average rate of change in atmospheric pressure (in mb/mi) from Medicine Hat to Chicago, given that the distance between the two cities is approimatel 4 mi. 6 Medicine Hat 8 6 Pressure in millibars (mb) Figure E-4 Chicago Sketch the level curve = k for the specified values of k. 43. = + ; k =,,, 3, = /; k =,,,, 45. = + ; k =,,,, 46. = + 9 ; k =,,, 3, = ; k =,,,, 48. = csc ; k =,,,, 49 5 Sketch the level surface f(,,) = k. 49. f(,,) = ; k = 6 5. f(,,) = + ; k = 5. f(,,) = + 4; k = 7 5. f(,,) = 4 + ; k = Describe the level surfaces in words. 53. f(,,) = ( ) f(,,) = f(,,) = f(,,) = 57. Let f(,) = Find an equation of the level curve that passes through the point (a) (, ) (b) (, ) (c) (, ). 58. Let f(,) = e. Find an equation of the level curve that passes through the point (a) (ln, ) (b) (, 3) (c) (, ). 59. Let f(,,) = +. Find an equation of the level surface that passes through the point (a) (,, ) (b) (,, 3) (c) (,, ). 6. Let f(,,) = + 3. Find an equation of the level surface that passes through the point (a) (,, ) (b) (, 4, ) (c) (,, ). 6. If T (,) is the temperature at a point (, ) on a thin metal plate in the -plane, then the level curves of T are called isothermal curves. All points on such a curve are at the same temperature. Suppose that a plate occupies the first quadrant and T (,) =. (a) Sketch the isothermal curves on which T =, T =, and T = 3. (b) An ant, initiall at (, 4), wants to walk on the plate so that the temperature along its path remains constant. What path should the ant take and what is the temperature along that path? 6. If V(,) is the voltage or potential at a point (, ) in the -plane, then the level curves of V are called equipotential curves. Along such a curve, the voltage remains constant. Given that 8 V(,) = sketch the equipotential curves at which V =., V =., and V = Let f(,) = + 3. (a) Use a graphing utilit to generate the level curve that passes through the point (, ). (b) Generate the level curve of height. 64. Let f(,) =. (a) Use a graphing utilit to generate the level curve that passes through the point (, ). (b) Generate the level curve of height 8. C 65. Let f(,) = e ( + ). (a) Use a CAS to generate the graph of f for and. (b) Generate a contour plot for the surface, and confirm visuall that it is consistent with the surface obtained in part (a). (c) Read the appropriate documentation and eplore the effect of generating the graph of f from various viewpoints. C 66. Let f(,) = e sin. (a) Use a CAS to generate the graph of f for 4 and π. (b) Generate a contour plot for the surface, and confirm visuall that it is consistent with the surface obtained in part (a). (cont.)

12 3. Limits and Continuit 97 (c) Read the appropriate documentation and eplore the effect of generating the graph of f from various viewpoints. 67. In each part, describe in words how the graph of g is related to the graph of f. (a) g(,) = f(,) (b) g(,) = + f(,) (c) g(,) = f(, + ) 68. (a) Sketch the graph of f(,) = e ( + ). (b) Describe in words how the graph of the function g(,) = e a( + ) is related to the graph of f for positive values of a. 69. Writing Find a few practical eamples of functions of two and three variables, and discuss how phsical considerations affect their domains. 7. Writing Describe two different was in which a function f(,) can be represented geometricall. Discuss some of the advantages and disadvantages of each representation. QUICK CHECK ANSWERS 3.. points (, ) in the first or third quadrants; points (, ) in the first quadrant. (a) 4 (b) 4 (c) (d) /( + ) 3. (a) k> (b) the lines + = ln k 4. (a) <k (b) spheres of radius ( k)/k for <k<, the single point (,, ) for k = 3. LIMITS AND CONTINUITY In this section we will introduce the notions of limit and continuit for functions of two or more variables. We will not go into great detail our objective is to develop the basic concepts accuratel and to obtain results needed in later sections. A more etensive stud of these topics is usuall given in advanced calculus. Figure 3.. (, ) (, ) In words, Formulas () and () state that a limit of a function f along a parametric curve can be obtained b substituting the parametric equations for the curve into the formula for the function and then computing the limit of the resulting function of one variable at the appropriate point. LIMITS ALONG CURVES For a function of one variable there are two one-sided limits at a point, namel, lim f() and lim f() + reflecting the fact that there are onl two directions from which can approach, the right or the left. For functions of two or three variables the situation is more complicated because there are infinitel man different curves along which one point can approach another (Figure 3..). Our first objective in this section is to define the limit of f(,) as (, ) approaches a point (, ) along a curve C (and similarl for functions of three variables). If C is a smooth parametric curve in -space or 3-space that is represented b the equations = (t), = (t) or = (t), = (t), = (t) and if = (t ), = (t ), and = (t ), then the limits are defined b lim (, ) (, ) (along C) f(,) and lim f(,,) (,,) (,, ) (along C) lim f(,) = lim f((t), (t)) () (, ) (, ) t t (along C) lim f(,,) = lim f((t), (t), (t)) () (,,) (,, ) t t (along C)

13 98 Chapter3 / Partial Derivatives L ((t), (t), f((t), (t))) (, ) = f(, ) ((t), (t)) lim f(, ) = L (, ) (, ) (along C ) Figure 3.. C In these formulas the limit of the function of t must be treated as a one-sided limit if (, ) or (,, ) is an endpoint of C. A geometric interpretation of the limit along a curve for a function of two variables is shown in Figure 3..: As the point ((t), (t)) moves along the curve C in the plane toward (, ), the point ((t), (t), f((t), (t))) moves directl above it along the graph of = f(,) with f((t), (t)) approaching the limiting value L. In the figure we followed a common practice of omitting the ero -coordinate for points in the -plane. Eample Figure 3..3a shows a computer-generated graph of the function f(,) = + The graph reveals that the surface has a ridge above the line =, which is to be epected since f(,) has a constant value of for =, ecept at (, ) where f is undefined (verif). Moreover, the graph suggests that the limit of f(,) as (, ) (, ) along a line through the origin varies with the direction of the line. Find this limit along (a) the -ais (b) the -ais (c) the line = (d) the line = (e) the parabola =.5.5 = =.5.5 (a).5 = ( = ) = = (b) = ( = ).5 = = + (c) Figure 3..3 Solution (a). The -ais has parametric equations = t, =, with (, ) corresponding to t =, so ( t ) lim f(,) = lim f(t,) = lim (, ) (, ) t t (along = ) which is consistent with Figure 3..3b. = lim t = Solution (b). The -ais has parametric equations =, = t, with (, ) corresponding to t =, so ( t ) lim f(,) = lim f(,t)= lim (, ) (, ) t t (along = ) which is consistent with Figure 3..3b. = lim t =

14 3. Limits and Continuit 99 Solution (c). The line = has parametric equations = t, = t, with (, ) corresponding to t =, so lim f(,) = lim f(t,t) = lim ( t ) ( = lim ) = (, ) (, ) t t t t (along = ) which is consistent with Figure 3..3b. Solution (d). The line = has parametric equations = t, = t, with (, ) corresponding to t =, so t lim f(,) = lim f(t, t) = lim (, ) (, ) t t t = lim t = (along = ) which is consistent with Figure 3..3b. For uniformit, we have chosen the same parameter t in each part of Eample. We could have used or as the parameter, according to the contet. For eample, part (b) could be computed using lim f(,) and part (e) could be computed using lim f(, ) Solution (e). The parabola = has parametric equations = t, = t, with (, ) corresponding to t =, so lim f(,) = lim f(t,t ) = lim ( t 3 ) ( = lim t ) = (, ) (, ) t t t + t 4 t + t (along = ) This is consistent with Figure 3..3c, which shows the parametric curve = t, = t, = t + t superimposed on the surface. A closed disk includes all of the points on its bounding circle. Figure 3..4 An interior point Figure 3..5 A boundar point An interior point An open disk contains none of the points on its bounding circle. A boundar point OPEN AND CLOSED SETS Although limits along specific curves are useful for man purposes, the do not alwas tell the complete stor about the limiting behavior of a function at a point; what is required is a limit concept that accounts for the behavior of the function in an entire vicinit of a point, not just along smooth curves passing through the point. For this purpose, we start b introducing some terminolog. Let C be a circle in -space that is centered at (, ) and has positive radius δ. The set of points that are enclosed b the circle, but do not lie on the circle, is called the open disk of radius δ centered at (, ), and the set of points that lie on the circle together with those enclosed b the circle is called the closed disk of radius δ centered at (, ) (Figure 3..4). Analogousl, if S is a sphere in 3-space that is centered at (,, ) and has positive radius δ, then the set of points that are enclosed b the sphere, but do not lie on the sphere, is called the open ball of radius δ centered at (,, ), and the set of points that lie on the sphere together with those enclosed b the sphere is called the closed ball of radius δ centered at (,, ). Disks and balls are the two-dimensional and three-dimensional analogs of intervals on a line. The notions of open and closed can be etended to more general sets in -space and 3-space. If D is a set of points in -space, then a point (, ) is called an interior point of D if there is some open disk centered at (, ) that contains onl points of D, and (, ) is called a boundar point of D if ever open disk centered at (, ) contains both points in D and points not in D. The same terminolog applies to sets in 3-space, but in that case the definitions use balls rather than disks (Figure 3..5). For a set D in either -space or 3-space, the set of all interior points is called the interior of D and the set of all boundar points is called the boundar of D. Moreover, just as for disks, we sa that D is closed if it contains all of its boundar points and open if it contains none of its boundar points. The set of all points in -space and the set of all points in

15 9 Chapter3 / Partial Derivatives 3-space have no boundar points (wh?), so b agreement the are regarded to be both open and closed. GENERAL LIMITS OF FUNCTIONS OF TWO VARIABLES The statement lim (,) (, ) f(,) = L is intended to conve the idea that the value of f(,) can be made as close as we like to the number L b restricting the point (, ) to be sufficientl close to (but different from) the point (, ). This idea has a formal epression in the following definition and is illustrated in Figure When convenient, (3) can also be written as lim f(,) = L or as f(,) L as (, ) (, ) 3.. definition Let f be a function of two variables, and assume that f is defined at all points of some open disk centered at (, ), ecept possibl at (, ). We will write lim f(,) = L (3) (,) (, ) if given an number ɛ>, we can find a number δ> such that f(,) satisfies f(,) L <ɛ whenever the distance between (, ) and (, ) satisfies < ( ) + ( ) <δ L + e f(, ) L L e = L + e = L e In Figure 3..6, the condition f(,) L <ɛ is satisfied at each point (, ) within the circular region. However, the fact that this condition is satisfied at the center of the circular region is not relevant to the limit. This circular region with the center removed consists of all points (, ) that satisf < ( ) + ( ) < d. Figure 3..6 d (, ) (, ) = f (, ) lim f(, ) = L (, ) (, ) Another illustration of Definition 3.. is shown in the arrow diagram of Figure As in Figure 3..6, this figure is intended to conve the idea that the values of f(,) can be forced within ɛ units of L on the -ais b restricting (, ) to lie within δ units of (, ) in the -plane. We used a white dot at (, ) to suggest that the epsilon condition need not hold at this point. We note without proof that the standard properties of limits hold for limits along curves and for general limits of functions of two variables, so that computations involving such limits can be performed in the usual wa.

16 3. Limits and Continuit 9 (, ) d (, ) ( L e L f (, ) L + e ( Figure 3..7 Eample lim (,) (,4) [53 9] = lim (,) (,4) [53 ] [ = 5 lim (,) (,4) ] 3 [ = 5() 3 (4) 9 = 7 lim 9 (,) (,4) ] lim 9 (,) (,4) RELATIONSHIPS BETWEEN GENERAL LIMITS AND LIMITS ALONG SMOOTH CURVES Stated informall, if f(,) has limit L as (, ) approaches (, ), then the value of f(,) gets closer and closer to L as the distance between (, ) and (, ) approaches ero. Since this statement imposes no restrictions on the direction in which (, ) approaches (, ), it is plausible that the function f(,) will also have the limit L as (, ) approaches (, ) along an smooth curve C. This is the implication of the following theorem, which we state without proof. WARNING In general, one cannot show that lim f(,) = L (,) (, ) b showing that this limit holds along a specific curve, or even some specific famil of curves. The problem is there ma be some other curve along which the limit does not eist or has a value different from L (see Eercise 34, for eample). 3.. theorem (a) (b) If f(,) L as (, ) (, ), then f(,) L as (, ) (, ) along an smooth curve. If the limit of f(,) fails to eist as (, ) (, ) along some smooth curve, or if f(,) has different limits as (, ) (, ) along two different smooth curves, then the limit of f(,) does not eist as (, ) (, ). Eample 3 The limit lim (,) (,) + does not eist because in Eample we found two different smooth curves along which this limit had different values. Specificall, lim = and lim (, ) (, ) + (, ) (, ) + = (along = ) (along = ) CONTINUITY Stated informall, a function of one variable is continuous if its graph is an unbroken curve without jumps or holes. To etend this idea to functions of two variables, imagine that the graph of = f(,) is formed from a thin sheet of cla that has been molded into peaks

17 9 Chapter3 / Partial Derivatives and valles. We will regard f as being continuous if the cla surface has no tears or holes. The functions graphed in Figure 3..8 fail to be continuous because of their behavior at (, ). The precise definition of continuit at a point for functions of two variables is similar to that for functions of one variable we require the limit of the function and the value of the function to be the same at the point. Hole at the origin 3..3 definition Afunction f(,)is said to be continuous at (, ) if f(, ) is defined and if lim f(,) = f(, ) (,) (, ) In addition, if f is continuous at ever point in an open set D, then we sa that f is continuous on D, and if f is continuous at ever point in the -plane, then we sa that f is continuous everwhere. Infinite at the origin The following theorem, which we state without proof, illustrates some of the was in which continuous functions can be combined to produce new continuous functions. Figure 3..8 Vertical jump at the origin 3..4 theorem (a) (b) (c) If g() is continuous at and h() is continuous at, then f(,) = g()h() is continuous at (, ). If h(, ) is continuous at (, ) and g(u) is continuous at u = h(, ), then the composition f(,) = g(h(, )) is continuous at (, ). If f(,) is continuous at (, ), and if (t) and (t) are continuous at t with (t ) = and (t ) =, then the composition f ((t), (t)) is continuous at t. Eample 4 Use Theorem 3..4 to show that the functions f(,) = 3 5 and f(,) = sin(3 5 ) are continuous everwhere. Solution. The polnomials g() = 3 and h() = 5 are continuous at ever real number, and therefore b part (a) of Theorem 3..4, the function f(,) = 3 5 is continuous at ever point (, ) in the -plane. Since 3 5 is continuous at ever point in the -plane and sin u is continuous at ever real number u, it follows from part (b) of Theorem 3..4 that the composition f(,) = sin(3 5 ) is continuous everwhere. Theorem 3..4 is one of a whole class of theorems about continuit of functions in two or more variables. The content of these theorems can be summaried informall with three basic principles: Recogniing Continuous Functions A composition of continuous functions is continuous. A sum, difference, or product of continuous functions is continuous. A quotient of continuous functions is continuous, ecept where the denominator is ero.

18 3. Limits and Continuit 93 B using these principles and Theorem 3..4, ou should be able to confirm that the following functions are all continuous everwhere: e + /3, cosh( 3 ), + + Eample 5 Evaluate lim (,) (,) +. Solution. Since f(,) = /( + ) is continuous at (, ) (wh?), it follows from the definition of continuit for functions of two variables that lim (,) (,) + = ( )() ( ) + () = 5 Eample 6 Since the function f(,) = 3 is a quotient of continuous functions, it is continuous ecept where =. Thus, f(,) is continuous everwhere ecept on the hperbola =. LIMITS AT DISCONTINUITIES Sometimes it is eas to recognie when a limit does not eist. For eample, it is evident that lim (,) (,) + =+ which implies that the values of the function approach + as (, ) (, ) along an smooth curve (Figure 3..9). However, it is not evident whether the limit lim (,) (,) ( + ) ln( + ) = + eists because it is an indeterminate form of tpe. Although L Hôpital s rule cannot be applied directl, the following eample illustrates a method for finding this limit b converting to polar coordinates. Figure 3..9 Eample 7 Find lim (,) (,) ( + ) ln( + ). Solution. Let (r, θ) be polar coordinates of the point (, ) with r. Then we have = r cos θ, = r sin θ, r = + Moreover, since r we have r = +, so that r + if and onl if (, ) (, ). Thus, we can rewrite the given limit as lim (,) (,) ( + ) ln( + ) = lim r r ln r + = lim r + lnr /r This converts the limit to an indeterminate form of tpe /. = lim r + /r /r 3 = lim r + ( r ) = L Hôpital s rule

19 94 Chapter3 / Partial Derivatives REMARK The graph of f(,) = ( + ) ln( + ) in Eample 7 is a surface with a hole (sometimes called a puncture) at the origin (Figure 3..). We can remove this discontinuit b defining f(, ) to be. (See Eercises 39 and 4, which also deal with the notion of a removable discontinuit.) = ( + )ln( + ) Figure 3.. CONTINUITY AT BOUNDARY POINTS Recall that in our stud of continuit for functions of one variable, we first defined continuit at a point, then continuit on an open interval, and then, b using one-sided limits, we etended the notion of continuit to include the boundar points of the interval. Similarl, for functions of two variables one can etend the notion of continuit of f(,) to the boundar of its domain b modifing Definition 3.. appropriatel so that (, ) is restricted to approach (, ) through points ling wholl in the domain of f. We will omit the details. = Figure 3.. Eample 8 The graph of the function f(,) = is the upper hemisphere shown in Figure 3.., and the natural domain of f is the closed unit disk + The graph of f has no tears or holes, so it passes our intuitive test of continuit. In this case the continuit at a point (, ) on the boundar reflects the fact that lim = (,) (, ) = when (, ) is restricted to points on the closed unit disk +. It follows that f is continuous on its domain. EXTENSIONS TO THREE VARIABLES All of the results in this section can be etended to functions of three or more variables. For eample, the distance between the points (,,)and (,, ) in 3-space is ( ) + ( ) + ( ) so the natural etension of Definition 3.. to 3-space is as follows: 3..5 definition Let f be a function of three variables, and assume that f is defined at all points within a ball centered at (,, ), ecept possibl at (,, ). We will write lim f(,,) = L (4) (,,) (,, ) if given an number ɛ>, we can find a number δ>such that f(,,) satisfies f(,,) L <ɛ whenever the distance between (,,)and (,, ) satisfies < ( ) + ( ) + ( ) <δ As with functions of one and two variables, we define a function f(,,) of three variables to be continuous at a point (,, ) if the limit of the function and the value of the function are the same at this point; that is, lim f(,,) = f(,, ) (,,) (,, ) Although we will omit the details, the properties of limits and continuit that we discussed for functions of two variables, including the notion of continuit at boundar points, carr over to functions of three variables.

20 3. Limits and Continuit 95 QUICK CHECK EXERCISES 3. (See page 97 for answers.). Let f(,) = + Determine the limit of f(,) as (, ) approaches (, ) along the curve C. (a) C: = (b) C: = (c) C: = (d) C: =. (a) lim cos π = (,) (3,) (b) lim (,) (,) e = ( ) (c) lim (,) (,) ( + ) sin = + 3. A function f(,) is continuous at (, ) provided f(, ) eists and provided f(,) has limit as (, ) approaches. 4. Determine all values of the constant a such that the function f(,) = a + is continuous everwhere. EXERCISE SET 3. 6 Use limit laws and continuit properties to evaluate the limit.. lim (,) (,3) (4 ). lim ( sin ) (,) (/,π) 3 3. lim (,) (,) + 4. lim (,) (, 3) e 5. lim (,) (,) ln( + 3 ) 6. lim (,) (4, ) Show that the limit does not eist b considering the limits as (, ) (, ) along the coordinate aes (a) lim (b) lim (,) (,) + (,) (,) + cos 8. (a) lim (b) lim (,) (,) + (,) (,) + 9 Evaluate the limit using the substitution = + and observing that + if and onl if (, ) (, ). sin( + ) cos( + ) 9. lim. lim (,) (,) + (,) (,) +. lim (,) (,) e /( + ). lim e / + (,) (,) + 3 Determine whether the limit eists. If so, find its value lim 4. lim (,) (,) + (,) (,) lim 6. lim (,) (,) 3 + (,) (,) + 7. lim (,,) (,,) lim ln( + ) (,,) (,, ) sin( + + ) 9. lim (,,) (,,) + + sin. lim + + (,,) (,,) + +. lim (,,) (,,). lim (,,) (,,) tan + + e + + [ ] Evaluate the limits b converting to polar coordinates, as in Eample lim + ln( + ) (,) (,) 4. lim (,) (,) ln( + ) 5. lim (,) (,) + 6. lim (,) (,) Evaluate the limits b converting to spherical coordinates (ρ,θ,φ) and b observing that ρ + if and onl if (,,) (,, ). 7. lim (,,) (,,) + + sin sin 8. lim (,,) (,,) True False Determine whether the statement is true or false. Eplain our answer. 9. If D is an open set in -space or in 3-space, then ever point in D is an interior point of D. 3. If f(,) L as (, ) approaches (, ) along the -ais, and if f(,) L as (, ) approaches (, ) along the -ais, then lim (,) (,) f(,) = L. 3. If f and g are functions of two variables such that f + g and fg are both continuous, then f and g are themselves continuous.

21 96 Chapter3 / Partial Derivatives 3. If lim + f() = L =, then FOCUS ON CONCEPTS lim (,) (,) + f( + ) = 33. The accompaning figure shows a portion of the graph of f(,) = 4 + (a) Based on the graph in the figure, does f(,) have a limit as (, ) (, )? Eplain our reasoning. (b) Show that f(,) as(, ) (, ) along an line = m. Does this impl that f(,) as (, ) (, )? Eplain. (c) Show that f(,) as (, ) (, ) along the parabola =, and confirm visuall that this is consistent with the graph of f(,). (d) Based on parts (b) and (c), does f(,) have a limit as (, ) (, )? Is this consistent with our answer to part (a)? Figure E (a) Show that the value of approaches as (, ) (, ) along an straight line = m, or along an parabola = k. (b) Show that 3 lim (,) (,) 6 + does not eist b letting (, ) (, ) along the curve = (a) Show that the value of approaches as (,,) (,, ) along an line = at, = bt, = ct. (b) Show that the limit lim (,,) (,,) does not eist b letting (,,) (,, ) along the curve = t, = t, = t. [ 36. Find lim (,) (,) tan [ + + ( ) + ( ) 37. Find lim (,) (,) tan sin( + ), (,) = (, ) 38. Let f(,) = +, (,) = (, ). Show that f is continuous at (, ) A function f(,) is said to have a removable discontinuit at (, ) if lim (,) (, ) f(,) eists but f is not continuous at (, ), either because f is not defined at (, ) or because f(, ) differs from the value of the limit. Determine whether f(, ) has a removable discontinuit at (, ). 39. f(,) = + 4. f(,) = ln( + ) 4 48 Sketch the largest region on which the function f is continuous. 4. f(,) = ln( + ) 4. f(,) = 43. f(,) = f(,) = ln( + ) ( ) 45. f(,) = cos f(,) = e 47. f(,) = sin () 48. f(,) = tan ( ) 49 5 Describe the largest region on which the function f is continuous. 49. f(,,) = 3 e cos() 5. f(,,) = ln(4 ) + 5. f(,,) = + 5. f(,,) = sin Writing Describe the procedure ou would use to determine whether or not the limit eists. ]. ]. lim (,) (, ) f(,) 54. Writing In our own words, state the geometric interpretations of ɛ and δ in the definition of given in Definition 3... lim (,) (, ) f(,) = L

22 3.3 Partial Derivatives 97 QUICK CHECK ANSWERS 3.. (a) (b) (c) (d). (a) 3 (b) (c) 3. f(, ); (, ) 4. a 3.3 PARTIAL DERIVATIVES In this section we will develop the mathematical tools for studing rates of change that involve two or more independent variables. PARTIAL DERIVATIVES OF FUNCTIONS OF TWO VARIABLES If = f(,), then one can inquire how the value of changes if is held fied and is allowed to var, or if is held fied and is allowed to var. For eample, the ideal gas law in phsics states that under appropriate conditions the pressure eerted b a gas is a function of the volume of the gas and its temperature. Thus, a phsicist studing gases might be interested in the rate of change of the pressure if the volume is held fied and the temperature is allowed to var, or if the temperature is held fied and the volume is allowed to var. We now define a derivative that describes such rates of change. Suppose that (, ) is a point in the domain of a function f(,). Ifwefi =, then f(, ) is a function of the variable alone. The value of the derivative d d [f(, )] at then gives us a measure of the instantaneous rate of change of f with respect to at the point (, ). Similarl, the value of the derivative d d [f(,)] at gives us a measure of the instantaneous rate of change of f with respect to at the point (, ). These derivatives are so basic to the stud of differential calculus of multivariable functions that the have their own name and notation definition If = f(,) and (, ) is a point in the domain of f, then the partial derivative of f with respect to at (, ) [also called the partial derivative of with respect to at (, )] is the derivative at of the function that results when = is held fied and is allowed to var. This partial derivative is denoted b f (, ) and is given b f (, ) = d d [f(, )] = f( +, ) f(, ) = lim () The limits in () and () show the relationship between partial derivatives and derivatives of functions of one variable. In practice, our usual method for computing partial derivatives is to hold one variable fied and then differentiate the resulting function using the derivative rules for functions of one variable. Similarl, the partial derivative of f with respect to at (, ) [also called the partial derivative of with respect to at (, )] is the derivative at of the function that results when = is held fied and is allowed to var. This partial derivative is denoted b f (, ) and is given b f (, ) = d d [f(,)] = f(, + ) f(, ) = lim ()

23 98 Chapter3 / Partial Derivatives Eample Find f (, 3) and f (, 3) for the function f(,) = Solution. Since f (, 3) = d d [f(,3)] = d d [ ] = we have f (, 3) = = 58. Also, since f (,)= d d [f(,)]= d d [ + + 4] =4 + we have f (, 3) = 4(3) + = 4. THE PARTIAL DERIVATIVE FUNCTIONS Formulas () and () define the partial derivatives of a function at a specific point (, ). However, often it will be desirable to omit the subscripts and think of the partial derivatives as functions of the variables and. These functions are f( +, ) f(,) f (, ) = lim f(, + ) f(,) f (, ) = lim The following eample gives an alternative wa of performing the computations in Eample. Eample Find f (, ) and f (, ) for f(,) = , and use those partial derivatives to compute f (, 3) and f (, 3). TECHNOLOGY MASTERY Computer algebra sstems have specific commands for calculating partial derivatives. If ou have a CAS, use it to find the partial derivatives f (, ) and f (, ) in Eample. Solution. Keeping fied and differentiating with respect to ields f (, ) = d d [ ] =6 + 4 and keeping fied and differentiating with respect to ields Thus, f (, ) = d d [ ] =4 3 + f (, 3) = 6( )(3 ) + 4 = 58 and f (, 3) = 4( 3 )3 + = 4 which agree with the results in Eample. The smbol is called a partial derivative sign. It is derived from the Crillic alphabet. PARTIAL DERIVATIVE NOTATION If = f(,), then the partial derivatives f and f are also denoted b the smbols f, and f, Some tpical notations for the partial derivatives of = f(,) at a point (, ) are f, f, f, =,= (, ), (, ) (, ) (, )

24 3.3 Partial Derivatives 99 Eample 3 Find / and / if = 4 sin( 3 ). Solution. = [4 sin( 3 )]= 4 [sin(3 )]+sin( 3 ) (4 ) = 4 cos( 3 ) 3 + sin( 3 ) 4 3 = 4 3 cos( 3 ) sin( 3 ) = [4 sin( 3 )]= 4 [sin(3 )]+sin( 3 ) (4 ) = 4 cos( 3 ) 3 + sin( 3 ) = 3 5 cos( 3 ) PARTIAL DERIVATIVES VIEWED AS RATES OF CHANGE AND SLOPES Recall that if = f(), then the value of f ( ) can be interpreted either as the rate of change of with respect to at or as the slope of the tangent line to the graph of f at. Partial derivatives have analogous interpretations. To see that this is so, suppose that C is the intersection of the surface = f(,) with the plane = and that C is its intersection with the plane = (Figure 3.3.). Thus, f (, ) can be interpreted as the rate of change of with respect to along the curve C, and f (,)can be interpreted as the rate of change of with respect to along the curve C. In particular, f (, ) is the rate of change of with respect to along the curve C at the point (, ), and f (, ) is the rate of change of with respect to along the curve C at the point (, ). Slope = f (, ) Slope = f (, ) = f(, ) C C = f (, ) = = Figure 3.3. (, ) (, ) In an applied problem, the interpretations of f (, ) and f (, ) must be accompanied b the proper units. See Eample 4. Eample 4 Recall that the wind chill temperature inde is given b the formula W = T + (.475T 35.75)v.6 Compute the partial derivative of W with respect to v at the point (T, v) = (5, ) and interpret this partial derivative as a rate of change. Solution. Holding T fied and differentiating with respect to v ields W v (T, v) = + + (.475T 35.75)(.6)v.6 = (.475T 35.75)(.6)v.84 Since W is in degrees Fahrenheit and v is in miles per hour, a rate of change of W with respect to v will have units F/(mi/h) (which ma also be written as F h/mi). Substituting

25 93 Chapter3 / Partial Derivatives Confirm the conclusion of Eample 4 b calculating W(5, + v) W(5, ) v for values of v near. T = 5 and v = gives W v (5, ) = F ( 4.) mi/h as the instantaneous rate of change of W with respect to v at (T, v) = (5, ). We conclude that if the air temperature is a constant 5 F and the wind speed changes b a small amount from an initial speed of mi/h, then the ratio of the change in the wind chill inde to the change in wind speed should be about.58 F/(mi/h). Geometricall, f (, ) can be viewed as the slope of the tangent line to the curve C at the point (, ), and f (, ) can be viewed as the slope of the tangent line to the curve C at the point (, ) (Figure 3.3.). We will call f (, ) the slope of the surface in the -direction at (, ) and f (, ) the slope of the surface in the -direction at (, ). Eample 5 Let f(,) = (a) Find the slope of the surface = f(,) in the -direction at the point (, ). (b) Find the slope of the surface = f(,) in the -direction at the point (, ). Solution (a). Differentiating f with respect to with held fied ields f (, ) = Thus, the slope in the -direction is f (, ) = 4; that is, is decreasing at the rate of 4 units per unit increase in. Solution (b). Differentiating f with respect to with held fied ields f (, ) = + 5 Thus, the slope in the -direction is f (, ) = 6; that is, is increasing at the rate of 6 units per unit increase in. ESTIMATING PARTIAL DERIVATIVES FROM TABULAR DATA For functions that are presented in tabular form, we can estimate partial derivatives b using adjacent entries within the table. Table 3.3. temperature T ( F) Eample 6 Use the values of the wind chill inde function W(T,v) displaed in Table 3.3. to estimate the partial derivative of W with respect to v at (T, v) = (5, ). Compare this estimate with the value of the partial derivative obtained in Eample 4. wind speed v (mi/h) Solution. Since W W(5, + v) W(5, ) W(5, + v) 5 (5, ) = lim = lim v v v v v we can approimate the partial derivative b W W(5, + v) 5 (5, ) v v With v = 5 this approimation is W W(5, + 5) 5 W(5, 5) (5, ) = = = v F mi/h

26 and with v = 5 this approimation is 3.3 Partial Derivatives 93 W W(5, 5) 5 W(5, 5) (5, ) = = = 4 v F mi/h We will take the average, 3 5 =.6 F/(mi/h), of these two approimations as our estimate of ( W / v)(5, ). This is close to the value found in Eample 4. W v (5, ) = F ( 4.) mi/h IMPLICIT PARTIAL DIFFERENTIATION Figure 3.3.,, 3 3 3,, Check the results in Eample 7 b differentiating the functions = and directl. = Eample 7 Find the slope of the sphere + + = inthe-direction at the points ( 3, 3, ( 3) and 3, 3, 3) (Figure 3.3.). Solution. The point ( 3, 3, 3) lies on the upper hemisphere =, and the point ( 3, 3, 3) lies on the lower hemisphere =. We could find the slopes b differentiating each epression for separatel with respect to and then evaluating the derivatives at = 3 and =. However, it is more efficient to differentiate the 3 given equation + + = implicitl with respect to, since this will give us both slopes with one differentiation. To perform the implicit differentiation, we view as a function of and and differentiate both sides with respect to, taking to be fied. The computations are as follows: [ + + ]= [] + + = = Substituting the - and -coordinates of the points ( 3, 3, ( 3) and 3, 3, 3) in this epression, we find that the slope at the point ( 3, 3, ) 3 is and the slope at ( 3, 3, ) 3 is. Eample 8 Suppose that D = + is the length of the diagonal of a rectangle whose sides have lengths and that are allowed to var. Find a formula for the rate of change of D with respect to if varies with held constant, and use this formula to find the rate of change of D with respect to at the point where = 3 and = 4. Solution. Differentiating both sides of the equation D = + with respect to ields D D D = and thus D = Since D = 5 when = 3 and = 4, it follows that 5 D D = 3 or =3,=4 = 3 =3,=4 5 Thus, D is increasing at a rate of 3 unit per unit increase in at the point (3, 4). 5

27 93 Chapter3 / Partial Derivatives PARTIAL DERIVATIVES AND CONTINUITY In contrast to the case of functions of a single variable, the eistence of partial derivatives for a multivariable function does not guarantee the continuit of the function. This fact is shown in the following eample. Eample 9 Let, (,) = (, ) f(,) = +, (,) = (, ) (3) (a) Show that f (, ) and f (, ) eist at all points (, ). (b) Eplain wh f is not continuous at (, ). Solution (a). Figure shows the graph of f. Note that f is similar to the function considered in Eample of Section 3., ecept that here we have assigned f a value of at (, ). Ecept at this point, the partial derivatives of f are Figure f (, ) = ( + ) () ( + ) = 3 ( + ) (4) f (, ) = ( + ) () = 3 (5) ( + ) ( + ) It is not evident from Formula (3) whether f has partial derivatives at (, ), and if so, what the values of those derivatives are. To answer that question we will have to use the definitions of the partial derivatives (Definition 3.3.). Appling Formulas () and () to (3) we obtain f(, ) f(, ) f (, ) = lim = lim = f (, ) = lim f(, ) f(, ) = lim = This shows that f has partial derivatives at (, ) and the values of both partial derivatives are at that point. Solution (b). We saw in Eample 3 of Section 3. that lim (,) (,) + does not eist. Thus, f is not continuous at (, ). We will stud the relationship between the continuit of a function and the properties of its partial derivatives in the net section. PARTIAL DERIVATIVES OF FUNCTIONS WITH MORE THAN TWO VARIABLES For a function f(,,) of three variables, there are three partial derivatives: f (,, ), f (,, ), f (,,) The partial derivative f is calculated b holding and constant and differentiating with respect to. For f the variables and are held constant, and for f the variables and are held constant. If a dependent variable w = f(,,)

28 is used, then the three partial derivatives of f can be denoted b w, w, and w 3.3 Partial Derivatives 933 Eample If f(,,) = , then f (,,)= f (,,)= f (,,)= f (,, ) = 4( ) 3 () () 3 + = 3 Eample If f(ρ,θ,φ) = ρ cos φ sin θ, then f ρ (ρ,θ,φ)= ρ cos φ sin θ f θ (ρ,θ,φ)= ρ cos φ cos θ f φ (ρ,θ,φ)= ρ sin φ sin θ In general, if f(v,v,...,v n ) is a function of n variables, there are n partial derivatives of f, each of which is obtained b holding n of the variables fied and differentiating the function f with respect to the remaining variable. If w = f(v,v,...,v n ), then these partial derivatives are denoted b w v, w v,..., w v n where w/ v i is obtained b holding all variables ecept v i fied and differentiating with respect to v i. HIGHER-ORDER PARTIAL DERIVATIVES Suppose that f is a function of two variables and. Since the partial derivatives f / and f / are also functions of and, these functions ma themselves have partial derivatives. This gives rise to four possible second-order partial derivatives of f, which are defined b f = ( ) f f = f = ( ) f = f Differentiate twice with respect to. Differentiate twice with respect to. f = ( ) f f = f = ( ) f = f Differentiate first with respect to and then with respect to. Differentiate first with respect to and then with respect to. The last two cases are called the mied second-order partial derivatives or the mied second partials. Also, the derivatives f / and f / are often called the first-order partial derivatives when it is necessar to distinguish them from higher-order partial derivatives. Similar conventions appl to the second-order partial derivatives of a function of three variables.

29 934 Chapter3 / Partial Derivatives WARNING Observe that the two notations for the mied second partials have opposite conventions for the order of differentiation. In the notation the derivatives are taken right to left, and in the subscript notation the are taken left to right. The conventions are logical if ou insert parentheses: f = ( ) f Right to left. Differentiate Left to right. Differentiate f inside the parentheses first. = (f ) inside the parentheses first. Eample Find the second-order partial derivatives of f(,) = Solution. so that We have f = and f = f = ( ) f = ( ) = 3 + f = ( ) f = (3 + 4 ) = 6 f = ( ) f = (3 + 4 ) = f = ( f ) = ( ) = Third-order, fourth-order, and higher-order partial derivatives can be obtained b successive differentiation. Some possibilities are 3 f = ( ) f 4 f = f 3 = ( 3 ) f = f f = ( ) f 4 f = f = ( 3 ) f = f Eample 3 Let f(,) = e +. Find f. Solution. f = ( ) 3 f = f = ( e ) = (e ) = e If f is a function of three variables, then the analog of Theorem 3.3. holds for each pair of mied second-order partials if we replace open disk b open ball. How man second-order partials does f(,,) have? EQUALITY OF MIXED PARTIALS For a function f(,) it might be epected that there would be four distinct second-order partial derivatives: f,f,f, and f. However, observe that the mied second-order partial derivatives in Eample are equal. The following theorem (proved in Appendi D) eplains wh this is so theorem Let f be a function of two variables. If f and f are continuous on some open disk, then f = f on that disk.

30 3.3 Partial Derivatives 935 It follows from this theorem that if f (, ) and f (, ) are continuous everwhere, then f (, ) = f (, ) for all values of and. Since polnomials are continuous everwhere, this eplains wh the mied second-order partials in Eample are equal. THE WAVE EQUATION Consider a string of length L that is stretched taut between = and = L on an -ais, and suppose that the string is set into vibrator motion b plucking it at time t = (Figure 3.3.4a). The displacement of a point on the string depends both on its coordinate and the elapsed time t, and hence is described b a function u(, t) of two variables. For a fied value t, the function u(, t) depends on alone, and the graph of u versus describes the shape of the string think of it as a snapshot of the string at time t (Figure 3.3.4b). It follows that at a fied time t, the partial derivative u/ represents the slope of the string at, and the sign of the second partial derivative u/ tells us whether the string is concave up or concave down at (Figure 3.3.4c). u u u Slope = u u < (concave down) L (a) L (b) L (c) Figure Leverett Bradle/Gett Images The vibration of a plucked string is governed b the wave equation. For a fied value of, the function u(, t) depends on t alone, and the graph of u versus t is the position versus time curve of the point on the string with coordinate. Thus, for a fied value of, the partial derivative u/ t is the velocit of the point with coordinate, and u/ t is the acceleration of that point. It can be proved that under appropriate conditions the function u(, t) satisfies an equation of the form u t = u c (6) where c is a positive constant that depends on the phsical characteristics of the string. This equation, which is called the one-dimensional wave equation, involves partial derivatives of the unknown function u(, t) and hence is classified as a partial differential equation. Techniques for solving partial differential equations are studied in advanced courses and will not be discussed in this tet. Show that the function u(, t) = sin( ct) is a solution of Equa- Eample 4 tion (6). Solution. We have u = cos( ct), u = ccos( ct), t Thus, u(, t) satisfies (6). u = sin( ct) u t = c sin( ct)

31 936 Chapter3 / Partial Derivatives QUICK CHECK EXERCISES 3.3 (See page 94 for answers.). Let f(,) = sin. Then f (, ) = and f (, ) =.. The slope of the surface = in the -direction at the point (, 3) is, and the slope of this surface in the -direction at the point (, 3) is. 3. The volume V of a right circular cone of radius r and height h is given b V = 3 πr h. (a) Find a formula for the instantaneous rate of change of V with respect to r if r changes and h remains constant. (b) Find a formula for the instantaneous rate of change of V with respect to h if h changes and r remains constant. 4. Find all second-order partial derivatives for the function f(,) = 3. EXERCISE SET 3.3 Graphing Utilit. Let f(,) = 3 3. Find (a) f (, ) (b) f (, ) (c) f (,) (d) f (, ) (e) f (,) (f ) f (, ) (g) f (, ) (h) f (, ).. Let = e sin. Find (a) / (b) / (c) / (,) (d) / (,) (e) / (,) (f ) / (,) (g) / (ln,) (h) / (ln,). 3. Let f(,) = 3 +. (a) Find the slope of the surface = f(,) in the - direction at the point (4, ). (b) Find the slope of the surface = f(,) in the - direction at the point (4, ). 4. Let f(,) = e + 5. (a) Find the slope of the surface = f(,) in the - direction at the point (3, ). (b) Find the slope of the surface = f(,) in the - direction at the point (3, ). 5. Let = sin( 4). (a) Find the rate of change of with respect to at the point (, ) with held fied. (b) Find the rate of change of with respect to at the point (, ) with held fied. 6. Let = ( + ). (a) Find the rate of change of with respect to at the point (, 4) with held fied. (b) Find the rate of change of with respect to at the point (, 4) with held fied. FOCUS ON CONCEPTS 7. Use the information in the accompaning figure to find the values of the first-order partial derivatives of f at the point (, ). = f(, ) (,, 3) (,, 4) (,, ) (,, ) Figure E-7 8. The accompaning figure shows a contour plot for an unspecified function f(,). Make a conjecture about the signs of the partial derivatives f (, ) and f (, ), and eplain our reasoning Figure E-8 9. Suppose that Nolan throws a baseball to Ran and that the baseball leaves Nolan s hand at the same height at which it is caught b Ran. It we ignore air resistance, the horiontal range r of the baseball is a function of the initial speed v of the ball when it leaves Nolan s hand and the angle θ above the horiontal at which it is thrown. Use the accompaning table and the method of Eample 6 to estimate (a) the partial derivative of r with respect to v when v = 8 ft/s and θ = 4 (b) the partial derivative of r with respect to θ when v = 8 ft/s and θ = 4. speed v (ft/s) angle u (degrees) Table E-9. Use the table in Eercise 9 and the method of Eample 6 to estimate (a) the partial derivative of r with respect to v when v = 85 ft/s and θ = 45 (b) the partial derivative of r with respect to θ when v = 85 ft/s and θ = 45.

32 3.3 Partial Derivatives 937. The accompaning figure shows the graphs of an unspecified function f(,) and its partial derivatives f (, ) and f (, ). Determine which is which, and eplain our reasoning. 5 I II III Figure E-. What can ou sa about the signs of /, /, /, and / at the point P in the accompaning figure? Eplain our reasoning. P = f (, ) Figure E- 3 6 True False Determine whether the statement is true or false. Eplain our answer. 3. If the line = is a contour of f(,)through (4, ), then f (4, ) =. 4. If the plane = 3 intersects the surface = f(,) in a curve that passes through (3, 4, 6) and satisfies =, then f (3, 4) = If the graph of = f(,) is a plane in 3-space, then both f and f are constant functions. 6. There eists a polnomial f(,) that satisfies the equations f (, ) = and f (, ) = +. 7 Find / and /. 7. = 4e 3 8. = cos( 5 4 ) 9. = 3 ln( + 3/5 ). = e sin 4. = +. = Find f (, ) and f (, ). 3. f(,) = f(,) = + 5. f(,) = 3/ tan (/) 6. f(,) = 3 e + 3 sec 7. f(,) = ( tan ) 4/3 8. f(,) = cosh( )sinh ( ) Evaluate the indicated partial derivatives. 9. f(,) = ; f (3, ), f (3, ) 3. f(,) = e ; f / (, ), f / (, ) 3. = + 4 ; / (, ), / (, ) 3. w = cos ; w/ (,π), w/ (,π) 33. Let f(,,) = Find (a) f (,,) (b) f (,,) (c) f (,,) (d) f (,,) (e) f (,,) (f ) f (,, 3). 34. Let w = cos. Find (a) w/ (,,) (b) w/ (,,) (c) w/ (,,) (d) w/ (,,) (e) w/ (,,) (f ) w/ (,, ) Find f, f, and f. 35. f(,,) = ln( cos ) ( ) 36. f(,,) = 3/ sec ( ) 37. f(,,) = tan f(,,) = cosh( )sinh ( ) 39 4 Find w/, w/, and w/. 39. w = e sin 4. w = + 4. w = w = 3 e Let f(,,) = e. Find (a) f / (,,) (b) f / (,,) (c) f / (,,). 44. Let w = + 4. Find (a) w/ (,, ) (b) w/ (,, ) (c) w/ (,, ). 45. Let f(,) = e cos. Use a graphing utilit to graph the functions f (,)and f (, π/). 46. Let f(,) = e sin. Use a graphing utilit to graph the functions f (,)and f (, ). 47. A point moves along the intersection of the elliptic paraboloid = + 3 and the plane =. At what rate is changing with respect to when the point is at (,, 7)? 48. A point moves along the intersection of the elliptic paraboloid = + 3 and the plane =. At what rate is changing with respect to when the point is at (,, 7)? 49. Apoint moves along the intersection of the plane = 3 and the surface = 9. At what rate is changing with respect to when the point is at (4, 3, )? 5. Find the slope of the tangent line at (,, 5) to the curve of intersection of the surface = + 4 and (a) the plane = (b) the plane =. 5. The volume V of a right circular clinder is given b the formula V = πr h, where r is the radius and h is the height. (a) Find a formula for the instantaneous rate of change of V with respect to r if r changes and h remains constant. (cont.)

33 938 Chapter3 / Partial Derivatives (b) Find a formula for the instantaneous rate of change of V with respect to h if h changes and r remains constant. (c) Suppose that h has a constant value of 4 in, but r varies. Find the rate of change of V with respect to r at the point where r = 6 in. (d) Suppose that r has a constant value of 8 in, but h varies. Find the instantaneous rate of change of V with respect to h at the point where h = in. 5. The volume V of a right circular cone is given b V = π 4 d 4s d where s is the slant height and d is the diameter of the base. (a) Find a formula for the instantaneous rate of change of V with respect to s if d remains constant. (b) Find a formula for the instantaneous rate of change of V with respect to d if s remains constant. (c) Suppose that d has a constant value of 6 cm, but s varies. Find the rate of change of V with respect to s when s = cm. (d) Suppose that s has a constant value of cm, but d varies. Find the rate of change of V with respect to d when d = 6 cm. 53. According to the ideal gas law, the pressure, temperature, and volume of a gas are related b P = kt /V, where k is a constant of proportionalit. Suppose that V is measured in cubic inches (in 3 ), T is measured in kelvins (K), and that for a certain gas the constant of proportionalit is k = in lb/k. (a) Find the instantaneous rate of change of pressure with respect to temperature if the temperature is 8 K and the volume remains fied at 5 in 3. (b) Find the instantaneous rate of change of volume with respect to pressure if the volume is 5 in 3 and the temperature remains fied at 8 K. 54. The temperature at a point (, ) on a metal plate in the -plane is T (,) = degrees Celsius. Assume that distance is measured in centimeters and find the rate at which temperature changes with respect to distance if we start at the point (, ) and move (a) to the right and parallel to the -ais (b) upward and parallel to the -ais. 55. The length, width, and height of a rectangular bo are l = 5,w =, and h = 3, respectivel. (a) Find the instantaneous rate of change of the volume of the bo with respect to the length if w and h are held constant. (b) Find the instantaneous rate of change of the volume of the bo with respect to the width if l and h are held constant. (c) Find the instantaneous rate of change of the volume of the bo with respect to the height if l and w are held constant. 56. The area A of a triangle is given b A = ab sin θ, where a and b are the lengths of two sides and θ is the angle between these sides. Suppose that a = 5, b =, and θ = π/3. (a) Find the rate at which A changes with respect to a if b and θ are held constant. (b) Find the rate at which A changes with respect to θ if a and b are held constant. (c) Find the rate at which b changes with respect to a if A and θ are held constant. 57. The volume of a right circular cone of radius r and height h is V = 3 πr h. Show that if the height remains constant while the radius changes, then the volume satisfies V r = V r 58. Find parametric equations for the tangent line at (, 3, 3) to the curve of intersection of the surface = and (a) the plane = (b) the plane = (a) B differentiating implicitl, find the slope of the hperboloid + = inthe-direction at the points (3, 4, 6) and (3, 4, 6). (b) Check the results in part (a) b solving for and differentiating the resulting functions directl. 6. (a) B differentiating implicitl, find the slope of the hperboloid + = inthe-direction at the points (3, 4, 6 ) and (3, 4, 6 ). (b) Check the results in part (a) b solving for and differentiating the resulting functions directl Calculate / and / using implicit differentiation. Leave our answers in terms of,, and. 6. ( + + ) 3/ = 6. ln( + 3 ) = sin = 64. e sinh + = Find w/, w/, and w/ using implicit differentiation. Leave our answers in terms of,,, and w. 65. ( w ) 3/ = ln( w) = 67. w + w sin = 68. e sinh w w + = 69 7 Find f and f. 69. f(,) = 7. f(,) = 3 e t dt 7. f(,) = sin t 3 dt 7. f(,) = + e t dt sin t 3 dt 73. Let = cos. Find (a) / (b) / (c) / (d) /. 74. Let f(,) = Find (a) f (b) f (c) f (d) f.

34 3.3 Partial Derivatives Confirm that the mied second-order partial derivatives of f are the same. 75. f(,) = f(,) = f(,) = e cos 78. f(,) = e 79. f(,) = ln(4 5) 8. f(,) = ln( + ) 8. f(,) = ( )/( + ) 8. f(,) = ( )/( + ) 83. Epress the following derivatives in notation. (a) f (b) f (c) f (d) f 84. Epress the derivatives in subscript notation. 3 f (a) (b) 4 f 4 f 5 f (c) (d) Given f(,) = 3 5 +, find (a) f (b) f (c) f. 86. Given = ( ) 5, find 3 3 (a) (b) (c) Given f(,) = 3 e 5, find (a) f (, ) (b) f (, ) (c) f (, ). 88. Given w = e cos, find 3 w (a) (π/4,) (b) 3 w (π/4,) 89. Let f(,,) = Find (a) f (b) f (c) f (d) f (e) f (f ) f (g) f (h) f. 9. Let w = (4 3 + ) 5. Find w 3 w 4 w (a) (b) (c). 9. Show that the function satisfies Laplace s equation + = (a) = + (b) = e sin + e cos (c) = ln( + ) + tan (/) 9. Show that the function satisfies the heat equation t = c (a) = e t sin(/c) (c >, constant) (b) = e t cos(/c) 93. Show that the function u(, t) = sin cωt sin ω satisfies the wave equation [Equation (6)] for all real values of ω. 94. In each part, show that u(, ) and v(,) satisf the Cauch Riemann equations u = v u and = v (a) u =, (b) u = e cos, (c) u = ln( + ), v = v = e sin v = tan (/) 95. Show that if u(, ) and v(,) each have equal mied second partials, and if u and v satisf the Cauch Riemann equations (Eercise 94), then u, v, and u + v satisf Laplace s equation (Eercise 9). 96. When two resistors having resistances R ohms and R ohms are connected in parallel, their combined resistance R in ohms is R = R R /(R + R ). Show that R R = R R 4R (R + R ) 4 97 Find the indicated partial derivatives. 97. f(v,w,,) = 4v w ; f / v, f / w, f /, f / 98. w = r cos st + e u sin ur; w/ r, w/ s, w/ t, w/ u 99. f(v,v,v 3,v 4 ) = v v v3 + ; v 4 f / v, f / v, f / v 3, f / v 4. V = e + we w + w; V /, V /, V /, V / w. Let u(w,,,)= e w sin. Find (a) u u (,,,π) (b) (,,,π) u u (c) (,,,π) (d) (,,,π) w 4 u 4 u (e) (f ) w w.. Let f(v,w,,) = v / w 4 / /3. Find f v (,, 4, 8), f w (,, 4, 8), f (,, 4, 8), and f (,, 4, 8). 3 4 Find w/ i for i =,,...,n. 3. w = cos( + + +n n ) ( n ) /n 4. w = k k= 5 6 Describe the largest set on which Theorem 3.3. can be used to prove that f and f are equal on that set. Then confirm b direct computation that f = f on the given set. 5. (a) f(,) = (b) f(,) = 3 / 6. (a) f(,) = + (b) f(,) = sin( + 3 ) 7. Let f(,) = 3 +. Find f (, ) and f (, ) b evaluating the limits in Definition Then check our work b calculating the derivative in the usual wa.

35 94 Chapter3 / Partial Derivatives 8. Let f(,) = ( + ) /3. Show that 4, (,) = (, ) f (, ) = 3( + ) /3, (,) = (, ) Source: This problem, due to Don Cohen, appeared in Mathematics and Computer Education, Vol. 5, No., 99, p Writing Eplain how one might use the graph of the equation = f(,) to determine the signs of f (, ) and f (, ) b inspection.. Writing Eplain how one might use the graphs of some appropriate contours of = f(,) to determine the signs of f (, ) and f (, ) b inspection. 9. Let f(,) = ( ) /3. (a) Show that f (, ) =. (b) At what points, if an, does f (, ) fail to eist? QUICK CHECK ANSWERS 3.3. sin + cos ; cos. 9; 3. (a) 3 πrh (b) 3 πr s 4. f (, ) = 3, f (, ) = 6, f (, ) = f (, ) = DIFFERENTIABILITY, DIFFERENTIALS, AND LOCAL LINEARITY In this section we will etend the notion of differentiabilit to functions of two or three variables. Our definition of differentiabilit will be based on the idea that a function is differentiable at a point provided it can be ver closel approimated b a linear function near that point. In the process, we will epand the concept of a differential to functions of more than one variable and define the local linear approimation of a function. DIFFERENTIABILITY Recall that a function f of one variable is called differentiable at if it has a derivative at, that is, if the limit f f( + ) f( ) ( ) = lim () eists. As a consequence of () a differentiable function enjos a number of other important properties: (,, f(, )) The graph of = f() has a nonvertical tangent line at the point (,f( )); f ma be closel approimated b a linear function near (Section 3.5); f is continuous at. Our primar objective in this section is to etend the notion of differentiabilit to functions of two or three variables in such a wa that the natural analogs of these properties hold. For eample, if a function f(,) of two variables is differentiable at a point (, ), we want it to be the case that Figure 3.4. (, ) = f(, ) the surface = f(,) has a nonvertical tangent plane at the point (,,f(, )) (Figure 3.4.); the values of f at points near (, ) can be ver closel approimated b the values of a linear function; f is continuous at (, ).

36 3.4 Differentiabilit, Differentials, and Local Linearit 94 One could reasonabl conjecture that a function f of two or three variables should be called differentiable at a point if all the first-order partial derivatives of the function eist at that point. Unfortunatel, this condition is not strong enough to guarantee that the properties above hold. For instance, we saw in Eample 9 of Section 3.3 that the mere eistence of both first-order partial derivatives for a function is not sufficient to guarantee the continuit of the function. To determine what else we should include in our definition, it will be helpful to reeamine one of the consequences of differentiabilit for a single-variable function f(). Suppose that f()is differentiable at = and let f = f( + ) f( ) denote the change in f that corresponds to the change in from to +. We saw in Section 3.5 that f f ( ) provided is close to. In fact, for close to the error f f ( ) in this approimation will have magnitude much smaller than that of because f f ( ) ( ) f( + ) f( ) lim = lim f ( ) = f ( ) f ( ) = Since the magnitude of is just the distance between the points and +, we see that when the two points are close together, the magnitude of the error in the approimation will be much smaller than the distance between the two points (Figure 3.4.). The etension of this idea to functions of two or three variables is the etra ingredient needed in our definition of differentiabilit for multivariable functions. For a function f(,), the smbol f, called the increment of f, denotes the change in the value of f(,) that results when (, ) varies from some initial position (, ) to some new position ( +, + ); thus f = f( +, + ) f(, ) () (see Figure 3.4.3). [If a dependent variable = f(,) is used, then we will sometimes write rather than f.] Let us assume that both f (, ) and f (, ) eist and (b analog with the one-variable case) make the approimation Show that if f(,)is a linear function, then (3) becomes an equalit. f f (, ) + f (, ) (3) For and close to, we would like the error f f (, ) f (, ) in this approimation to be much smaller than the distance ( ) + ( ) between (, ) and ( +, + ). We can guarantee this b requiring that f f (, ) f (, ) lim = (, ) (,) ( ) + ( ) f( + ) f( ) Figure 3.4. f f ( ) f ( ) + f f(, ) (, ) Figure f f( +, + ) ( +, + )

37 94 Chapter3 / Partial Derivatives Based on these ideas, we can now give our definition of differentiabilit for functions of two variables definition A function f of two variables is said to be differentiable at (, ) provided f (, ) and f (, ) both eist and f f (, ) f (, ) lim = (4) (, ) (,) ( ) + ( ) As with the one-variable case, verification of differentiabilit using this definition involves the computation of a limit. Eample (, ). Use Definition 3.4. to prove that f(,) = + is differentiable at Solution. The increment is f = f( +, + ) f(, ) = ( ) + ( ) Since f (, ) = and f (, ) =, we have f (, ) = f (, ) =, and (4) becomes lim (, ) (,) ( ) + ( ) ( ) + ( ) = Therefore, f is differentiable at (, ). lim ( ) + ( ) = (, ) (,) We now derive an important consequence of limit (4). Define a function ɛ = ɛ(, ) = f f (, ) f (, ) ( ) + ( ) for (, ) = (, ) and define ɛ(, ) to be. Equation (4) then implies that lim ɛ(, ) = (, ) (,) Furthermore, it immediatel follows from the definition of ɛ that f = f (, ) + f (, ) + ɛ ( ) + ( ) (5) In other words, if f is differentiable at (, ), then f ma be epressed as shown in (5), where ɛ as(, ) (, ) and where ɛ = if(, ) = (, ). For functions of three variables we have an analogous definition of differentiabilit in terms of the increment f = f( +, +, + ) f(,, ) 3.4. definition A function f of three variables is said to be differentiable at (,, ) provided f (,, ), f (,, ), and f (,, ) eist and lim (,, ) (,,) f f (,, ) f (,, ) f (,, ) ( ) + ( ) + ( ) = (6)

38 3.4 Differentiabilit, Differentials, and Local Linearit 943 In a manner similar to the two-variable case, we can epress the limit (6) in terms of a function ɛ(,, ) that vanishes at (,, ) = (,, ) and is continuous there. The details are left as an eercise for the reader. If a function f of two variables is differentiable at each point of a region R in the plane, then we sa that f is differentiable on R; and if f is differentiable at ever point in the -plane, then we sa that f is differentiable everwhere. For a function f of three variables we have corresponding conventions. DIFFERENTIABILITY AND CONTINUITY Recall that we want a function to be continuous at ever point at which it is differentiable. The net result shows this to be the case theorem If a function is differentiable at a point, then it is continuous at that point. The converse of Theorem is false. For eample, the function f(,) = + is continuous at (, ) but is not differentiable at (, ). Wh not? proof We will give the proof for f(,), a function of two variables, since that will reveal the essential ideas. Assume that f is differentiable at (, ). To prove that f is continuous at (, ) we must show that lim f(,) = f(, ) (,) (, ) which, on letting = + and = +, is equivalent to B Equation () this is equivalent to lim f( +, + ) = f(, ) (, ) (,) lim f = (, ) (,) However, from Equation (5) [ lim f = lim f (, ) + f (, ) (, ) (,) (, ) (,) + ɛ(, ) ] ( ) + ( ) = + + = It can be difficult to verif that a function is differentiable at a point directl from the definition. The net theorem, whose proof is usuall studied in more advanced courses, provides simple conditions for a function to be differentiable at a point theorem If all first-order partial derivatives of f eist and are continuous at a point, then f is differentiable at that point. For eample, consider the function f(,,) = + Since f (,,)=,f (,,)=, and f (,,)= are defined and continuous everwhere, we conclude from Theorem that f is differentiable everwhere.

39 944 Chapter3 / Partial Derivatives DIFFERENTIALS As with the one-variable case, the approimations f f (, ) + f (, ) for a function of two variables and the approimation f f (,, ) + f (,, ) + f (,, ) (7) for a function of three variables have a convenient formulation in the language of differentials. If = f(,) is differentiable at a point (, ),welet d = f (, )d+ f (, )d (8) denote a new function with dependent variable d and independent variables d and d. We refer to this function (also denoted df )asthetotal differential of at (, ) or as the total differential of f at (, ). Similarl, for a function w = f(,,) of three variables we have the total differential of w at (,, ), dw = f (,, )d+ f (,, )d+ f (,, )d (9) which is also referred to as the total differential of f at (,, ). It is common practice to omit the subscripts and write Equations (8) and (9) as d = f (,)d+ f (,)d () and dw = f (,,)d+ f (,,)d+ f (,,)d () In the two-variable case, the approimation f f (, ) + f (, ) can be written in the form f df () for d = and d =. Equivalentl, we can write approimation () as d (3) In other words, we can estimate the change in b the value of the differential d where d is the change in and d is the change in. Furthermore, it follows from (4) that if and are close to, then the magnitude of the error in approimation (3) will be much smaller than the distance ( ) + ( ) between (, ) and ( +, + ). Eample Use (3) to approimate the change in = from its value at (.5,.) to its value at (.53,.4). Compare the magnitude of the error in this approimation with the distance between the points (.5,.) and (.53,.4). Solution. For = we have d = d + d. Evaluating this differential at (, ) = (.5,.), d = =.53.5 =.3, and d = =.4. =.4 ields d =. (.3) + (.5)(.)(.4) =.7 Since =.5 at(, ) = (.5,.) and = at (, ) = (.53,.4), we have = =.7348 and the error in approimating b d has magnitude d = =.348

40 3.4 Differentiabilit, Differentials, and Local Linearit 945 Since the distance between (.5,.) and (.53,.4) = (.5 +,. + ) is ( ) + ( ) = (.3) + (.4) =.5 =.5 we have d ( ) + ( ) =.348 =.6496 <.5 5 Thus, the magnitude of the error in our approimation is less than of the distance between 5 the two points. With the appropriate changes in notation, the preceding analsis can be etended to functions of three or more variables. Eample 3 The length, width, and height of a rectangular bo are measured with an error of at most 5%. Use a total differential to estimate the maimum percentage error that results if these quantities are used to calculate the diagonal of the bo. Solution. The diagonal D of a bo with length, width, and height is given b D = + + Let,,, and D = + + denote the actual values of the length, width, height, and diagonal of the bo. The total differential dd of D at (,, ) is given b dd = + + d d If,,, and D = + + are the measured and computed values of the length, width, height, and diagonal, respectivel, then d and =, =, =.5,.5,.5 We are seeking an estimate for the maimum sie of D/D. With the aid of Equation () we have D dd = D D + + [ + + ] [ ] = Since dd D = ( ) ( + + (.5) + (.5) + (.5)) =.5 we estimate the maimum percentage error in D to be 5%.

41 946 Chapter3 / Partial Derivatives LOCAL LINEAR APPROXIMATIONS We now show that if a function f is differentiable at a point, then it can be ver closel approimated b a linear function near that point. For eample, suppose that f(,) is differentiable at the point (, ). Then approimation (3) can be written in the form f( +, + ) f(, ) + f (, ) + f (, ) If we let = + and = +, this approimation becomes Show that if f(,)is a linear function, then (4) becomes an equalit. f(,) f(, ) + f (, )( ) + f (, )( ) (4) which ields a linear approimation of f(,). Since the error in this approimation is equal to the error in approimation (3), we conclude that for (, ) close to (, ), the error in (4) will be much smaller than the distance between these two points. When f(,) is differentiable at (, ) we get Eplain wh the error in approimation (4) is the same as the error in approimation (3). L(, ) = f(, ) + f (, )( ) + f (, )( ) (5) and refer to L(, ) as the local linear approimation to f at (, ). Eample 4 Let L(, ) denote the local linear approimation to f(,) = + at the point (3, 4). Compare the error in approimating f(3.4, 3.98) = (3.4) + (3.98) b L(3.4, 3.98) with the distance between the points (3, 4) and (3.4, 3.98). Solution. We have f (, ) = and f (, ) = + + with f (3, 4) = 3 5 and f (3, 4) = 4. Therefore, the local linear approimation to f at (3, 4) 5 is given b L(, ) = ( 3) + 4 ( 4) 5 Consequentl, Since f(3.4, 3.98) L(3.4, 3.98) = (.4) + 4 (.) = f(3.4, 3.98) = (3.4) + (3.98) 5.89 the error in the approimation is about =.9. This is less than of the distance (3.4 3) + (3.98 4).45 between the points (3, 4) and (3.4, 3.98). For a function f(,,) that is differentiable at (,, ), the local linear approimation is L(,,)= f(,, ) + f (,, )( ) (6) + f (,, )( ) + f (,, )( ) We have formulated our definitions in this section in such a wa that continuit and local linearit are consequences of differentiabilit. In Section 3.7 we will show that

42 3.4 Differentiabilit, Differentials, and Local Linearit 947 if a function f(,) is differentiable at a point (, ), then the graph of L(, ) is a nonvertical tangent plane to the graph of f at the point (,,f(, )). QUICK CHECK EXERCISES 3.4 (See page 949 for answers.). Assume that f(,) is differentiable at (, ) and let f denote the change in f from its value at (, ) to its value at ( +, + ). (a) f (b) The limit that guarantees the error in the approimation in part (a) is ver small when both and are close tois.. Compute the differential of each function. (a) = e (b) w = sin() 3. If f is differentiable at (, ), then the local linear approimation to f at (, ) is L() =. 4. Assume that f(, ) = 4 and f(,) is differentiable at (, ) with f (, ) = and f (, ) = 3. Estimate the value of f(.9,.95). EXERCISE SET 3.4 FOCUS ON CONCEPTS. Suppose that a function f(,) is differentiable at the point (3, 4) with f (3, 4) = and f (3, 4) =. If f(3, 4) = 5, estimate the value of f(3., 3.98).. Suppose that a function f(,) is differentiable at the point (, ) with f (, ) = and f (, ) = 3. If f(, ) =, estimate the value of f(.99,.). 3. Suppose that a function f(,,) is differentiable at the point (,, 3) with f (,, 3) =, f (,, 3) =, and f (,, 3) = 3. If f(,, 3) = 4, estimate the value of f(.,., 3.3). 4. Suppose that a function f(,,) is differentiable at the point (,, ), f (,, ) =, f (,, ) =, and f (,, ) =. If f(,, ) =, estimate the value of f(.98,.99,.97). 5. Use Definitions 3.4. and 3.4. to prove that a constant function of two or three variables is differentiable everwhere. 6. Use Definitions 3.4. and 3.4. to prove that a linear function of two or three variables is differentiable everwhere. 7. Use Definition 3.4. to prove that f(,,) = + + is differentiable at (,, ). 8. Use Definition 3.4. to determine all values of r such that f(,,) = ( + + ) r is differentiable at (,, ). 9 Compute the differential d or dw of the function. 9. = 7. = e. = 3. = = tan 4. = sec ( 3) 5. w = w = e 7. w = 3 8. w = w = tan (). w = Use a total differential to approimate the change in the values of f from P to Q. Compare our estimate with the actual change in f.. f(,) = + 4; P(, ), Q(.,.4). f(,) = /3 / ; P(8, 9), Q(7.78, 9.3) 3. f(,) = + ; P(, ), Q(.,.4) 4. f(,) = ln + ; P(, ), Q(.9,.98) 5. f(,,) = 3 ; P(,, ), Q(.99,.,.) 6. f(,,) = ; P(,, 4), + + Q(.4,.98, 3.97) 7 3 True False Determine whether the statement is true or false. Eplain our answer. 7. B definition, a function f(,) is differentiable at (, ) provided both f (, ) and f (, ) are defined. 8. For an point (, ) in the domain of a function f(,), we have lim f = (, ) (,) where f = f( +, + ) f(, ) 9. If f and f are both continuous at (, ), then so is f. 3. The graph of a local linear approimation to a function f(,) is a plane.

43 948 Chapter3 / Partial Derivatives 3. In the accompaning figure a rectangle with initial length and initial width has been enlarged, resulting in a rectangle with length + and width +. What portion of the figure represents the increase in the area of the rectangle? What portion of the figure represents an approimation of the increase in area b a total differential? + Figure E The volume V of a right circular cone of radius r and height h is given b V = 3 πr h. Suppose that the height decreases from in to 9.95 in and the radius increases from 4 in to 4.5 in. Compare the change in volume of the cone with an approimation of this change using a total differential (a) Find the local linear approimation L to the specified function f at the designated point P. (b) Compare the error in approimating f b L at the specified point Q with the distance between P and Q. 33. f(,) = ; P(4, 3), Q(3.9, 3.) f(,) =.5.3 ; P(, ), Q(.5,.97) 35. f(,) = sin ; P(, ), Q(.3,.4) 36. f(,) = ln ; P(, ), Q(.,.) 37. f(,,) = ; P(,, 3), Q(.,., 3.3) 38. f(,,) = + ; P(,, ), Q(.99,.99,.) f(,,) = e ; P(,, ), Q(.99,.,.99) 4. f(,,) = ln( + ); P(,, ), Q(.,.97,.) 4. In each part, confirm that the stated formula is the local linear approimation at (, ). (a) e sin (b) Show that the local linear approimation of the function f(,) = α β at (, ) is α β + α( ) + β( ) 43. In each part, confirm that the stated formula is the local linear approimation at (,, ). (a) (b) Based on Eercise 4, what would ou conjecture is the local linear approimation to α β γ at (,, )? Verif our conjecture b finding this local linear approimation. 45. Suppose that a function f(,) is differentiable at the point (, ) with f (, ) = and f(, ) = 3. Let L(, ) denote the local linear approimation of f at (, ). If L(.,.9) = 3.5, find the value of f (, ). 46. Suppose that a function f(,) is differentiable at the point (, ) with f (, ) = and f(, ) = 3. Let L(, ) denote the local linear approimation of f at (, ). IfL(.,.) = 3.3, find the value of f (, ). 47. Suppose that a function f(,,) is differentiable at the point (3,, ) and L(,,)= + is the local linear approimation to f at (3,, ). Find f(3,, ), f (3,, ), f (3,, ), and f (3,, ). 48. Suppose that a function f(,,) is differentiable at the point (,, ) and L(,,)= is the local linear approimation to f at (,, ). Find f(,, ), f (,, ), f (,, ), and f (,, ) A function f is given along with a local linear approimation L to f at a point P. Use the information given to determine point P. 49. f(,) = + ; L(, ) = 5. f(,) = ; L(, ) = f(,,) = + ; L(,,)= + 5. f(,,) = ; L(,,)= 53. The length and width of a rectangle are measured with errors of at most 3% and 5%, respectivel. Use differentials to approimate the maimum percentage error in the calculated area. 54. The radius and height of a right circular cone are measured with errors of at most % and 4%, respectivel. Use differentials to approimate the maimum percentage error in the calculated volume. 55. The length and width of a rectangle are measured with errors of at most r%, where r is small. Use differentials to approimate the maimum percentage error in the calculated length of the diagonal. 56. The legs of a right triangle are measured to be 3 cm and 4 cm, with a maimum error of.5 cm in each measurement. Use differentials to approimate the maimum possible error in the calculated value of (a) the hpotenuse and (b) the area of the triangle. 57. The period T of a simple pendulum with small oscillations is calculated from the formula T = π L/g, where L is the length of the pendulum and g is the acceleration due to gravit. Suppose that measured values of L and g have errors of at most.5% and.%, respectivel. Use differentials to approimate the maimum percentage error in the calculated value of T. 58. According to the ideal gas law, the pressure, temperature, and volume of a confined gas are related b P = kt /V, where k is a constant. Use differentials to approimate the

44 3.5 The Chain Rule 949 percentage change in pressure if the temperature of a gas is increased 3% and the volume is increased 5%. 59. Suppose that certain measured quantities and have errors of at most r% and s%, respectivel. For each of the following formulas in and, use differentials to approimate the maimum possible error in the calculated result. (a) (b) / (c) 3 (d) 3 6. The total resistance R of three resistances R, R, and R 3, connected in parallel, is given b R = + + R R R 3 Suppose that R, R, and R 3 are measured to be ohms, ohms, and 5 ohms, respectivel, with a maimum error of % in each. Use differentials to approimate the maimum percentage error in the calculated value of R. 6. The area of a triangle is to be computed from the formula A = ab sin θ, where a and b are the lengths of two sides and θ is the included angle. Suppose that a, b, and θ are measured to be 4 ft, 5 ft, and 3, respectivel. Use differentials to approimate the maimum error in the calculated value of A if the maimum errors in a, b, and θ are ft, 4 ft, and, respectivel. 6. The length, width, and height of a rectangular bo are measured with errors of at most r% (where r is small). Use differentials to approimate the maimum percentage error in the computed value of the volume. 63. Use Theorem to prove that f(,) = sin is differentiable everwhere. 64. Use Theorem to prove that f(,,) = sin is differentiable everwhere. 65. Suppose that f(,) is differentiable at the point (, ) and let = f(, ). Prove that g(,,) = f(,) is differentiable at (,, ). 66. Suppose that f satisfies an equation in the form of (5), where ɛ(, ) is continuous at (, ) = (, ) with ɛ(, ) =. Prove that f is differentiable at (, ). 67. Writing Discuss the similarities and differences between the definition of differentiabilit for a function of a single variable and the definition of differentiabilit for a function of two variables. 68. Writing Discuss the use of differentials in the approimation of increments and in the estimation of errors. QUICK CHECK ANSWERS 3.4 f f (, ) f (, ). (a) f (, ) + f (, ) (b) lim =. (a) d = e d + e d (, ) (,) ( ) + ( ) (b) dw = sin() d + cos() d + cos() d 3. f(, ) + f (, )( ) + f (, )( ) THE CHAIN RULE In this section we will derive versions of the chain rule for functions of two or three variables. These new versions will allow us to generate useful relationships among the derivatives and partial derivatives of various functions. CHAIN RULES FOR DERIVATIVES If is a differentiable function of and is a differentiable function of t, then the chain rule for functions of one variable states that, under composition, becomes a differentiable function of t with d dt = d d d dt We will now derive a version of the chain rule for functions of two variables. Assume that = f(,) is a function of and, and suppose that and are in turn functions of a single variable t, sa = (t), = (t) The composition = f((t), (t)) then epresses as a function of the single variable t. Thus, we can ask for the derivative d/dt and we can inquire about its relationship to the

45 95 Chapter3 / Partial Derivatives derivatives /, /, d/dt, and d/dt. Letting,, and denote the changes in,, and, respectivel, that correspond to a change of t in t, we have d dt = lim t t, d dt It follows from (3) of Section 3.4 that = lim t t, and d dt = lim t t + () where the partial derivatives are evaluated at ((t), (t)). Dividing both sides of () b t ields t t + () t Similarl, we can produce the analog of () for functions of three variables as follows: assume that w = f(,,) is a function of,, and, and suppose that,, and are functions of a single variable t. As above we define w,,, and to be the changes in w,,, and that correspond to a change of t in t. Then (7) in Section 3.4 implies that w w and dividing both sides of (3) b t ields w t + w w t + w t + w t w + (3) Taking the limit as t of both sides of () and (4) suggests the following results. (A complete proof of the two-variable case can be found in Appendi D.) (4) 3.5. theorem (Chain Rules for Derivatives) If = (t) and = (t) are differentiable at t, and if = f(,) is differentiable at the point (, ) = ((t), (t)), then = f((t), (t)) is differentiable at t and d dt = d dt + d dt (5) where the ordinar derivatives are evaluated at t and the partial derivatives are evaluated at (, ). If each of the functions = (t), = (t), and = (t) is differentiable at t, and if w = f(,,) is differentiable at the point (,,)= ((t), (t), (t)), then the function w = f((t), (t), (t)) is differentiable at t and dw dt = w d dt + w d dt + w d dt (6) d dt d dt where the ordinar derivatives are evaluated at t and the partial derivatives are evaluated at (,,). t d dt Figure 3.5. = d dt + d dt t Formula (5) can be represented schematicall b a tree diagram that is constructed as follows (Figure 3.5.). Starting with at the top of the tree and moving downward, join each variable b lines (or branches) to those variables on which it depends directl. Thus, is joined to and and these in turn are joined to t. Net, label each branch with a

46 3.5 The Chain Rule 95 derivative whose numerator contains the variable at the top end of that branch and whose denominator contains the variable at the bottom end of that branch. This completes the tree. To find the formula for d/dt, follow the two paths through the tree that start with Create a tree diagram for Formula (6). and end with t. Each such path corresponds to a term in Formula (5). Eample Suppose that =, = t, = t 3 Use the chain rule to find d/dt, and check the result b epressing as a function of t and differentiating directl. Solution. B the chain rule [Formula (5)], d dt = d dt + d dt = (t 5 )(t) + (t 4 )(3t ) = 7t 6 = ()(t) + ( )(3t ) Alternativel, we can epress directl as a function of t, = = (t ) (t 3 ) = t 7 and then differentiate to obtain d/dt = 7t 6. However, this procedure ma not alwas be convenient. Eample Suppose that w = + +, = cos θ, = sin θ, = tan θ Use the chain rule to find dw/dθ when θ = π/4. Solution. From Formula (6) with θ in the place of t, we obtain dw dθ = w d dθ + w d dθ + w d dθ = ( + + ) / ()( sin θ)+ ( + + ) / ()(cos θ) When θ = π/4, we have + ( + + ) / ()(sec θ) = cos π 4 =, = sin π 4 =, = tan π 4 = Confirm the result of Eample b epressing w directl as a function of θ. Substituting = /, = /, =, θ = π/4 in the formula for dw/dθ ields dw dθ = ( ) ( ( ) ) + ( ) ( ( ) ) + ( ) ()() θ=π/4 =

47 95 Chapter3 / Partial Derivatives REMARK There are man variations in derivative notations, each of which gives the chain rule a different look. If = f(,), where and are functions of t, then some possibilities are d dt = f d dt + f d dt df dt df dt = f d dt + f d dt = f (t) + f (t) CHAIN RULES FOR PARTIAL DERIVATIVES In Formula (5) the variables and are each functions of a single variable t. We now consider the case where and are each functions of two variables. Let = f(,) and suppose that and are functions of u and v, sa = (u, v), = (u, v) The composition = f((u, v), (u, v)) epresses as a function of the two variables u and v. Thus, we can ask for the partial derivatives / uand / v; and we can inquire about the relationship between these derivatives and the derivatives /, /, / u, / v, / u, and / v. Similarl, if w = f(,,) and,, and are each functions of u and v, then the composition w = f((u, v), (u, v), (u, v)) epresses w as a function of u and v. Thus we can also ask for the derivatives w/ u and w/ v; and we can investigate the relationship between these derivatives, the partial derivatives w/, w/, and w/, and the partial derivatives of,, and with respect to u and v theorem (Chain Rules for Partial Derivatives) If = (u, v) and = (u, v) have first-order partial derivatives at the point (u, v), and if = f(,) is differentiable at the point (, ) = ((u, v), (u, v)), then = f((u, v), (u, v)) has firstorder partial derivatives at the point (u, v) given b u = u + u and v = v + v (7 8) If each function = (u, v), = (u, v), and = (u, v) has first-order partial derivatives at the point (u, v), and if the function w = f(,,) is differentiable at the point (,,)= ((u, v), (u, v), (u, v)), then w = f((u, v), (u, v), (u, v)) has firstorder partial derivatives at the point (u, v) given b w u = w u + w u + w u and w v = w v + w v + w v (9 ) proof We will prove Formula (7); the other formulas are derived similarl. If v is held fied, then = (u, v) and = (u, v) become functions of u alone. Thus, we are back to the case of Theorem If we appl that theorem with u in place of t, and if we use rather than d to indicate that the variable v is fied, we obtain u = u + u

48 3.5 The Chain Rule 953 u v u u v u v u u v = + u u u u v u v v Figure 3.5. u w v = + v v u w w w u v v u v u v u v w u w v v = w w u + u + = w w v + v + Figure w u w v v Figures 3.5. and show tree diagrams for the formulas in Theorem As illustrated in Figure 3.5., the formula for / u can be obtained b tracing all paths through the tree that start with and end with u, and the formula for / v can be obtained b tracing all paths through the tree that start with and end with v. Figure displas analogous results for w/ u and w/ v. Eample 3 Given that find / u and / v using the chain rule. Solution. u = Eample 4 = v = = u + [ u v + u + v v v + [ = e, = u + v, = u/v ( ) u = (e )() + (e ) = v ] ] e (u+v)(u/v) = [ 4u v + e (u+v)(u/v) v = (e )() + (e ) ( u ) v ( u v )] e = = u v e(u+v)(u/v) Suppose that [ + ] e v [ u v (u + v) ( u v )] e (u+v)(u/v) w = e, = 3u + v, = 3u v, = u v Use appropriate forms of the chain rule to find w/ u and w/ v. Solution. and From the tree diagram and corresponding formulas in Figure we obtain w u = e (3) + e (3) + e (uv) = e ( uv) w v = e () + e ( ) + e (u ) = e ( + u ) If desired, we can epress w/ u and w/ v in terms of u and v alone b replacing,, and b their epressions in terms of u and v. OTHER VERSIONS OF THE CHAIN RULE Although we will not prove it, the chain rule etends to functions w = f(v,v,...,v n ) of n variables. For eample, if each v i is a function of t, i =,,...,n, the relevant formula is dw dt = w dv + w dv + + w dv n v dt v dt v n dt () Note that () is a natural etension of Formulas (5) and (6) in Theorem There are infinitel man variations of the chain rule, depending on the number of variables and the choice of independent and dependent variables. Agood working procedure is to use tree diagrams to derive new versions of the chain rule as needed.

49 954 Chapter3 / Partial Derivatives Eample 5 Suppose that w = + and = ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ Use appropriate forms of the chain rule to find w/ ρ and w/ θ. w w w w r u r u r f f f r f u r f u r f w r w u = w w r + r + = w u + Figure w u w r Solution. From the tree diagram and corresponding formulas in Figure we obtain w ρ w θ = sin φ cos θ + sin φ sin θ cos φ = ρ sin φ cos θ + ρ sin φ sin θ ρ cos φ = ρ sin φ(cos θ + sin θ) ρ cos φ = ρ(sin φ cos φ) = ρ cos φ = ()( ρ sin φ sin θ)+ ()ρ sin φ cos θ = ρ sin φ sin θ cos θ + ρ sin φ sin θ cos θ = This result is eplained b the fact that w does not var with θ. You can see this directl b epressing the variables,, and in terms of ρ,φ, and θ in the formula for w. (Verif that w = ρ cos φ.) dw d w Figure w w d d = w w d + d + w d d w d d Eample 6 Suppose that w = +, = sin, = e Use an appropriate form of the chain rule to find dw/d. Solution. From the tree diagram and corresponding formulas in Figure we obtain dw = + ( + ) cos + e d = sin + ( + e ) cos + e sin This result can also be obtained b first epressing w eplicitl in terms of as w = sin + e sin and then differentiating with respect to ; however, such direct substitution is not alwas possible. WARNING The smbol, unlike the differential d, has no meaning of its own. For eample, if we were to cancel partial smbols in the chain-rule formula u = u + u we would obtain u = u + u which is false in cases where / u =. One of the principal uses of the chain rule for functions of a single variable was to compute formulas for the derivatives of compositions of functions. Theorems 3.5. and 3.5. are important not so much for the computation of formulas but because the allow us

50 3.5 The Chain Rule 955 to epress relationships among various derivatives. As an illustration, we revisit the topic of implicit differentiation. IMPLICIT DIFFERENTIATION Consider the special case where = f(,)is a function of and and is a differentiable function of. Equation (5) then becomes d d = f d d + f d d = f + f d d This result can be used to find derivatives of functions that are defined implicitl. For eample, suppose that the equation () f(,) = c (3) defines implicitl as a differentiable function of and we are interested in finding d/d. Differentiating both sides of (3) with respect to and appling () ields Thus, if f / =, we obtain In summar, we have the following result. f + f d d = d d = f / f / Show that the function = is defined implicitl b the equation theorem If the equation f(,) = c defines implicitl as a differentiable function of, and if f / =, then + = but that Theorem is not applicable for finding d/d. d d = f / f / (4) Eample 7 Given that = find d/d using (4), and check the result using implicit differentiation. Solution. B (4) with f(,) = 3 + 3, d d = f / + f / = 3 Alternativel, differentiating implicitl ields ( d ) = d which agrees with the result obtained b (4). or d + d = 3 The chain rule also applies to implicit partial differentiation. Consider the case where w = f(,,) is a function of,, and and is a differentiable function of and. It follows from Theorem 3.5. that w = f + f (5)

51 956 Chapter3 / Partial Derivatives If the equation f(,,) = c (6) defines implicitl as a differentiable function of and, then taking the partial derivative of each side of (6) with respect to and appling (5) gives If f / =, then A similar result holds for /. f + f = = f / f / theorem If the equation f(,,) = c defines implicitl as a differentiable function of and, and if f / =, then = f / f / and = f / f / ( Eample 8 3, 3, ) 3. Consider the sphere + + =. Find / and / at the point Note the similarit between the epression for / found in Eample 8 and that found in Eample 7 of Section 3.3. Solution. B Theorem with f(,,) = + +, = f / f / = = and = f / f / = = At the point ( 3, 3, 3), evaluating these derivatives gives / = and / =. QUICK CHECK EXERCISES 3.5 (See page 959 for answers.). Suppose that = and and are differentiable functions of t with =, =, d/dt =, and d/dt = 3 when t =. Then d/dt = when t =.. Suppose that C is the graph of the equation f(,) = and that this equation defines implicitl as a differentiable function of. If the point (, ) belongs to C with f (, ) = 3 and f (, ) =, then the tangent line to C at the point (, ) has slope. 3. A rectangle is growing in such a wa that when its length is 5 ft and its width is ft, the length is increasing at a rate of3ft/s and its width is increasing at a rate of 4 ft/s. At this instant the area of the rectangle is growing at a rate of. 4. Suppose that = /, where and are differentiable functions of u and v such that = 3, =, / u = 4, / v =, / u =, and / v = when u = and v =. When u = and v =, / u = and / v =. EXERCISE SET Use an appropriate form of the chain rule to find d/dt.. = 3 3 ; = t 4, = t. = ln( + ); = t, = t /3 3. = 3 cos sin ; = /t, = 3t 4. = + 4 ; = ln t, = t 5. = e ; = t /3, = t 3

52 3.5 The Chain Rule = cosh ; = t/, = e t 7 Use an appropriate form of the chain rule to find dw/dt. 7. w = ; = t, = t 3, = t 5 8. w = ln( ); = t /, = t /3, = t 9. w = 5 cos sin ; = /t, = t, = t 3. w = + 4 ; = ln t, = t, = 4t FOCUS ON CONCEPTS. Suppose that w = 3 4 ; = t, = t +, = t 4 Find the rate of change of w with respect to t at t = b using the chain rule, and then check our work b epressing w as a function of t and differentiating.. Suppose that w = sin ; = cos t, = t, = e t Find the rate of change of w with respect to t at t = b using the chain rule, and then check our work b epressing w as a function of t and differentiating. 3. Suppose that = f(,) is differentiable at the point (4, 8) with f (4, 8) = 3 and f (4, 8) =. If = t and = t 3, find d/dt when t =. 4. Suppose that w = f(,,) is differentiable at the point (,, ) with f (,, ) =, f (,, ) =, and f (,, ) = 3. If = t, = sin(πt), and = t +, find dw/dt when t =. 5. Eplain how the product rule for functions of a single variable ma be viewed as a consequence of the chain rule applied to a particular function of two variables. 6. A student attempts to differentiate the function using the power rule, mistakenl getting. A second student attempts to differentiate b treating it as an eponential function, mistakenl getting (ln ). Use the chain rule to eplain wh the correct derivative is the sum of these two incorrect results. 7 Use appropriate forms of the chain rule to find / u and / v. 7. = 8 + 3; = uv, = u v 8. = tan ; = u/v, = u v 9. = /; = cos u, = 3 sin v. = 3 ; = u + v ln u, = u v ln v. = e ; = uv, = /v. = cos sin ; = u v, = u + v 3 3 Use appropriate forms of the chain rule to find the derivatives. 3. Let T = 3 + ; = r cos θ, = r sin θ. Find T/ r and T/ θ. 4. Let R = e s t ; s = 3φ, t = φ /. Find dr/dφ. 5. Let t = u/v; u =, v = 4 3. Find t/ and t/. 6. Let w = rs/(r + s ); r = uv, s = u v. Find w/ u and w/ v. 7. Let = ln( + ), where = r cos θ. Find / r and / θ. 8. Let u = rs ln t, r =, s = 4 +, t = 3. Find u/ and u/. 9. Let w = , = ρ sin φ cos θ, = ρ sin φ sin θ, = ρ cos φ. Find w/ ρ, w/ φ, and w/ θ. 3. Let w = 3 3, = 3 +, =. Find dw/d. 3. Use a chain rule to find the value of dw ds if s=/4 w = r r tan θ; r = s, θ = πs. 3. Use a chain rule to find the values of f f and u v u=,v= u=,v= if f(,) = + ; = u, = uv Use a chain rule to find the values of and r θ r=,θ=π/6 r=,θ=π/6 if = e / ; = r cos θ, = r sin θ. 34. Use a chain rule to find d dt if = ; = t,= t + 7. t= True False Determine whether the statement is true or false. Eplain our answer. 35. The smbols and are defined in such a wa that the partial derivative / can be interpreted as a ratio. 36. If is a differentiable function of,, and 3, and if i is a differentiable function of t for i =,, 3, then is a differentiable function of t and d dt = 3 i= d i i dt 37. If is a differentiable function of and, and if and are twice differentiable functions of t, then is a twice differentiable function of t and d dt = d dt + d dt 38. If is a differentiable function of and such that = when = = and such that = then =.

53 958 Chapter3 / Partial Derivatives 39 4 Use Theorem to find d/d and check our result using implicit differentiation cos = = 5 4. e + e = = Find / and / b implicit differentiation, and confirm that the results obtained agree with those predicted b the formulas in Theorem = 44. ln( + ) + + = 45. e 5 sin 3 = e cos e sin + = 47. (a) Suppose that = f(u) and u = g(,). Draw a tree diagram, and use it to construct chain rules that epress / and / in terms of d/du, u/, and u/. (b) Show that = d ( ) u du + d u du = d ( ) u du + d u du = d u du + d u u du 48. (a) Let = f( ). Use the result in Eercise 47(a) to show that + = (b) Let = f(). Use the result in Eercise 47(a) to show that = (c) Confirm the result in part (a) in the case where = sin( ). (d) Confirm the result in part (b) in the case where = e. 49. Let f be a differentiable function of one variable, and let = f( + ). Show that = 5. Let f be a differentiable function of one variable, and let = f( + ). Show that = 5. Let f be a differentiable function of one variable, and let w = f(u), where u = Show that w + w + w = 6 dw du 5. Let f be a differentiable function of one variable, and let w = f(ρ), where ρ = ( + + /. Show that ( ) w ( ) w ( ) w ( ) dw + + = dρ 53. Let = f(, ). Show that / + / =. 54. Let f be a differentiable function of three variables and suppose that w = f(,, ). Show that w + w + w = 55. Suppose that the equation = f(,) is epressed in the polar form = g(r, θ) b making the substitution = r cos θ and = r sin θ. (a) View r and θ as functions of and and use implicit differentiation to show that r = cos θ and θ = sin θ r (b) View r and θ as functions of and and use implicit differentiation to show that r = sin θ and θ = cos θ r (c) Use the results in parts (a) and (b) to show that = r cos θ r θ sin θ = r sin θ + r θ cos θ (d) Use the result in part (c) to show that ( ) ( ) ( ) + = + r r ( ) θ (e) Use the result in part (c) to show that if = f(,) satisfies Laplace s equation + = then = g(r, θ) satisfies the equation r + r θ + r r = and conversel. The latter equation is called the polar form of Laplace s equation. 56. Show that the function = tan satisfies Laplace s equation; then make the substitution = r cos θ, = r sin θ, and show that the resulting function of r and θ satisfies the polar form of Laplace s equation given in part (e) of Eercise (a) Show that if u(, ) and v(, ) satisf the Cauch Riemann equations (Eercise 94, Section 3.3), and if = r cos θ and = r sin θ, then u r = v r θ and v r = u r θ This is called the polar form of the Cauch Riemann equations. (b) Show that the functions u = ln( + ), v = tan (/) satisf the Cauch Riemann equations; then make the substitution = r cos θ, = r sin θ, and show that the resulting functions of r and θ satisf the polar form of the Cauch Riemann equations.

54 3.5 The Chain Rule Recall from Formula (6) of Section 3.3 that under appropriate conditions a plucked string satisfies the wave equation u t = u c where c is a positive constant. (a) Show that a function of the form u(, t) = f( + ct) satisfies the wave equation. (b) Show that a function of the form u(, t) = g( ct) satisfies the wave equation. (c) Show that a function of the form u(, t) = f( + ct) + g( ct) satisfies the wave equation. (d) It can be proved that ever solution of the wave equation is epressible in the form stated in part (c). Confirm that u(, t) = sin t sin satisfies the wave equation in which c =, and then use appropriate trigonometric identities to epress this function in the form f( + t) + g( t). 59. Let f be a differentiable function of three variables, and let w = f(,,), = ρ sin φ cos θ, = ρ sin φ sin θ, and = ρ cos φ. Epress w/ ρ, w/ φ, and w/ θ in terms of w/, w/, and w/. 6. Let w = f(,,) be differentiable, where = g(,). Taking and as the independent variables, epress each of the following in terms of f /, f /, f /, /, and /. (a) w/ (b) w/ 6. Let w = ln(e r + e s + e t + e u ). Show that w rstu = 6e r+s+t+u 4w [Hint: Take advantage of the relationship e w = e r + e s + e t + e u.] 6. Suppose that w is a differentiable function of,, and 3, and = a + b = a + b 3 = a 3 + b 3 where the a s and b s are constants. Epress w/ and w/ in terms of w/, w/, and w/ (a) Let w be a differentiable function of,, 3, and 4, and let each i be a differentiable function of t. Find a chain-rule formula for dw/dt. (b) Let w be a differentiable function of,, 3, and 4, and let each i be a differentiable function of v, v, and v 3. Find chain-rule formulas for w/ v, w/ v, and w/ v Let w = ( n )k, where n. For what values of k does w + w + + w n = hold? 65. We showed in Eercise 8 of Section 5. that d g() f(t)dt = f(g())g () f(h())h () d h() Derive this same result b letting u = g() and v = h() and then differentiating the function F(u, v) = u v f(t)dt with respect to. 66. Prove: If f, f, and f are continuous on a circular region containing A(, ) and B(, ), then there is a point (, ) on the line segment joining A and B such that f(, ) f(, ) = f (, )( ) + f (, )( ) This result is the two-dimensional version of the Mean- Value Theorem. [Hint: Epress the line segment joining A and B in parametric form and use the Mean-Value Theorem for functions of one variable.] 67. Prove: If f (, ) = and f (, ) = throughout a circular region, then f(,) is constant on that region. [Hint: Use the result of Eercise 66.] 68. Writing Use differentials to give an informal justification for the chain rules for derivatives. 69. Writing Compare the use of the formula d d = f / f / with the process of implicit differentiation. QUICK CHECK ANSWERS ft /s 4. ;

55 96 Chapter3 / Partial Derivatives 3.6 DIRECTIONAL DERIVATIVES AND GRADIENTS The partial derivatives f (, ) and f (, ) represent the rates of change of f(,) in directions parallel to the - and -aes. In this section we will investigate rates of change of f(,) in other directions. Figure 3.6. (, ) u DIRECTIONAL DERIVATIVES In this section we etend the concept of a partial derivative to the more general notion of a directional derivative. We have seen that the partial derivatives of a function give the instantaneous rates of change of that function in directions parallel to the coordinate aes. Directional derivatives allow us to compute the rates of change of a function with respect to distance in an direction. Suppose that we wish to compute the instantaneous rate of change of a function f(,) with respect to distance from a point (, ) in some direction. Since there are infinitel man different directions from (, ) in which we could move, we need a convenient method for describing a specific direction starting at (, ). One wa to do this is to use a unit vector u = u i + u j that has its initial point at (, ) and points in the desired direction (Figure 3.6.). This vector determines a line l in the -plane that can be epressed parametricall as = + su, = + su () Since u is a unit vector, s is the arc length parameter that has its reference point at (, ) and has positive values in the direction of u. For s =, the point (, ) is at the reference point (, ), and as s increases, the point (, ) moves along l in the direction of u. On the line l the variable = f( + su, + su ) is a function of the parameter s. The value of the derivative d/ds at s = then gives an instantaneous rate of change of f(,) with respect to distance from (, ) in the direction of u definition If f(,) is a function of and, and if u = u i + u j is a unit vector, then the directional derivative of f in the direction of u at (, ) is denoted b D u f(, ) and is defined b Slope in u direction = rate of change of with respect to s D u f(, ) = d [ f( + su, + su ) ] () ds s= Q = f(, ) C provided this derivative eists. s (, ) u Figure 3.6. (, ) l Geometricall, D u f(, ) can be interpreted as the slope of the surface = f (, ) in the direction of u at the point (,,f(, )) (Figure 3.6.). Usuall the value of D u f(, ) will depend on both the point (, ) and the direction u. Thus, at a fied point the slope of the surface ma var with the direction (Figure 3.6.3). Analticall, the directional derivative represents the instantaneous rate of change of f (, ) with respect to distance in the direction of u at the point (, ). Eample Let f(,) =. Find and interpret D u f(, ) for the unit vector u = 3 i + j

56 = f (, ) (, ) The slope of the surface varies with the direction of u. Figure u Solution. Since f we have ( + It follows from Equation () that [ ( D u f(, ) = d ds ) ( 3s, + s = Directional Derivatives and Gradients 96 f 3s + )] 3s, + s s= ) ( + s ) ( 3 = 4 s + + ) 3 s + [ D u f(, ) = d ( 3 ds 4 s + + ) ] 3 s + [ 3 = s + + ] 3 = + 3 Since + 3.3, we conclude that if we move a small distance from the point (, ) in the direction of u, the function f(,) = will increase b about.3 times the distance moved. s= s= The definition of a directional derivative for a function f(,,) of three variables is similar to Definition definition If u = u i + u j + u 3 k is a unit vector, and if f(,,) is a function of,, and, then the directional derivative of f in the direction of u at (,, ) is denoted b D u f(,, ) and is defined b D u f(,, ) = d [ f( + su, + su, + su 3 ) ] (3) ds s= provided this derivative eists. Although Equation (3) does not have a convenient geometric interpretation, we can still interpret directional derivatives for functions of three variables in terms of instantaneous rates of change in a specified direction. For a function that is differentiable at a point, directional derivatives eist in ever direction from the point and can be computed directl in terms of the first-order partial derivatives of the function theorem (a) If f(,) is differentiable at (, ), and if u = u i + u j is a unit vector, then the directional derivative D u f(, ) eists and is given b D u f(, ) = f (, )u + f (, )u (4) (b) If f(,,) is differentiable at (,, ), and if u = u i + u j + u 3 k is a unit vector, then the directional derivative D u f(,, ) eists and is given b D u f(,, ) = f (,, )u + f (,, )u + f (,, )u 3 (5)

57 96 Chapter3 / Partial Derivatives proof We will give the proof of part (a); the proof of part (b) is similar and will be omitted. The function = f( + su, + su ) is the composition of the function = f(,) with the functions = (s) = + su and = (s) = + su As such, the chain rule in Formula (5) of Section 3.5 immediatel gives D u f(, ) = d [ f( + su, + su ) ] ds s= = d ds = f (, )u + f (, )u s= We can use Theorem to confirm the result of Eample. For f(,) = we have f (, ) = and f (, ) = (verif). With Equation (4) becomes u = 3 i + j ( ) 3 D u f(, ) = + = 3 + which agrees with our solution in Eample. Recall from Formula (3) of Section. that a unit vector u in the -plane can be epressed as u = cos φ i + sin φ j (6) where φ is the angle from the positive -ais to u. Thus, Formula (4) can also be epressed as D u f(, ) = f (, ) cos φ + f (, ) sin φ (7) Eample Find the directional derivative of f(,) = e at (, ) in the direction of the unit vector that makes an angle of π/3 with the positive -ais. Solution. The partial derivatives of f are f (, ) = e, f (, ) = e f (, ) =, f (, ) = The unit vector u that makes an angle of π/3 with the positive -ais is Note that in Eample 3 we used a unit vector to specif the direction of the directional derivative. This is required in order to appl either Formula (4) or Formula (5). Thus, from (7) u = cos(π/3)i + sin(π/3)j = i + 3 j D u f(, ) = f (, ) cos(π/3) + f (, ) sin(π/3) = (/) + ( )( 3/) = 3 Eample 3 Find the directional derivative of f(,,) = 3 + at the point (,, ) in the direction of the vector a = i + j k.

58 Solution. The partial derivatives of f are 3.6 Directional Derivatives and Gradients 963 f (,,)=, f (,,)= 3, f (,,)= 3 + f (,, ) = 4, f (,, ) =, f (,, ) = Since a is not a unit vector, we normalie it, getting u = Formula (5) then ields a a = 9 (i + j k) = 3 i + 3 j 3 k D u f(,, ) = ( 4) ( ) = 3 THE GRADIENT Formula (4) can be epressed in the form of a dot product as D u f(, ) = (f (, )i + f (, )j) (u i + u j) Similarl, Formula (5) can be epressed as = (f (, )i + f (, )j) u D u f(,, ) = (f (,, )i + f (,, )j + f (,, )k) u In both cases the directional derivative is obtained b dotting the direction vector u with a new vector constructed from the first-order partial derivatives of f. Remember that f is not a product of and f. Think of as an operator that acts on a function f to produce the gradient f definition (a) If f is a function of and, then the gradient of f is defined b f(,) = f (, )i + f (, )j (8) (b) If f is a function of,, and, then the gradient of f is defined b f(,,) = f (,,)i + f (,,)j + f (,,)k (9) The smbol (read del ) is an inverted delta. (It is sometimes called a nabla because of its similarit in form to an ancient Hebrew ten-stringed harp of that name.) Formulas (4) and (5) can now be written as D u f(, ) = f(, ) u () Slope = f. u and D u f(,, ) = f(,, ) u () (, ) Figure f u = f (, ) respectivel. For eample, using Formula () our solution to Eample 3 would take the form D u f(,, ) = f(,, ) u = ( 4i + j + k) ( 3 i + 3 j 3 k) = ( 4) ( ) = 3 Formula () can be interpreted to mean that the slope of the surface = f(,) at the point (, ) in the direction of u is the dot product of the gradient with u (Figure 3.6.4).

59 964 Chapter3 / Partial Derivatives PROPERTIES OF THE GRADIENT The gradient is not merel a notational device to simplif the formula for the directional derivative; we will see that the length and direction of the gradient f provide important information about the function f and the surface = f(,). For eample, suppose that f(,) =, and let us use Formula (4) of Section.3 to rewrite () as D u f(,) = f(,) u = f(,) u cos θ = f(,) cos θ () where θ is the angle between f(,) and u. Equation () tells us that the maimum value of D u f at the point (, ) is f(,), and this maimum occurs when θ =, that is, when u is in the direction of f(,). Geometricall, this means: At (, ), the surface = f(,) has its maimum slope in the direction of the gradient, and the maimum slope is f(,). Increasing most rapidl That is, the function f(,) increases most rapidl in the direction of its gradient (Figure 3.6.5). Similarl, () tells us that the minimum value of D u f at the point (, ) is f(,), and this minimum occurs when θ = π, that is, when u is oppositel directed to f(,). Geometricall, this means: Decreasing most rapidl At (, ), the surface = f(,) has its minimum slope in the direction that is opposite to the gradient, and the minimum slope is f(,). f (, ) f Figure That is, the function f(,) decreases most rapidl in the direction opposite to its gradient (Figure 3.6.5). Finall, in the case where f(,) =, it follows from () that D u f(,) = inall directions at the point (, ). This tpicall occurs where the surface = f(,) has a relative maimum, a relative minimum, or a saddle point. A similar analsis applies to functions of three variables. As a consequence, we have the following result theorem Let f be a function of either two variables or three variables, and let P denote the point P(, ) or P(,, ), respectivel. Assume that f is differentiable at P. (a) (b) (c) If f = at P, then all directional derivatives of f at P are ero. If f = at P, then among all possible directional derivatives of f at P, the derivative in the direction of f at P has the largest value. The value of this largest directional derivative is f at P. If f = at P, then among all possible directional derivatives of f at P, the derivative in the direction opposite to that of f at P has the smallest value. The value of this smallest directional derivative is f at P. Eample 4 Let f(,) = e. Find the maimum value of a directional derivative at (, ), and find the unit vector in the direction in which the maimum value occurs.

60 3.6 Directional Derivatives and Gradients 965 What would be the minimum value of a directional derivative of f(,) = e at (, )? Solution. Since the gradient of f at (, ) is f(,) = f (, )i + f (, )j = e i + e j f(, ) = 4i + 4j B Theorem 3.6.5, the maimum value of the directional derivative is f(, ) = ( 4) + 4 = 3 = 4 This maimum occurs in the direction of f(, ). The unit vector in this direction is u = f(, ) f(, ) = 4 ( 4i + 4j) = i + j GRADIENTS ARE NORMAL TO LEVEL CURVES We have seen that the gradient points in the direction in which a function increases most rapidl. For a function f(,) of two variables, we will now consider how this direction of maimum rate of increase can be determined from a contour map of the function. Suppose that (, ) is a point on a level curve f(,) = c of f, and assume that this curve can be smoothl parametried as = (s), = (s) (3) where s is an arc length parameter. Recall from Formula (6) of Section.4 that the unit tangent vector to (3) is T = T(s) = ( ) d i + ds ( ) d j ds Since T gives a direction along which f is nearl constant, we would epect the instantaneous rate of change of f with respect to distance in the direction of T to be. That is, we would epect that D T f(,) = f(,) T(s) = To show this to be the case, we differentiate both sides of the equation f(,) = c with respect to s. Assuming that f is differentiable at (, ), we can use the chain rule to obtain f d ds + f d ds = which we can rewrite as ( f i + f ) ( d j ds i + d ) ds j = or, alternativel, as f(,) T = Therefore, if f(,) =, then f(,) should be normal to the level curve f(,) = c at an point (, ) on the curve. It is proved in advanced courses that if f(,) has continuous first-order partial derivatives, and if f(, ) =, then near (, ) the graph of f(,) = c is indeed a smooth curve through (, ). Furthermore, we also know from Theorem that f will be differentiable at (, ). We therefore have the following result. Show that the level curves for f(,) = + are circles and verif Theorem at (, ) = (3, 4) theorem Assume that f(,) has continuous first-order partial derivatives in an open disk centered at (, ) and that f(, ) =. Then f(, ) is normal to the level curve of f through (, ). When we eamine a contour map, we instinctivel regard the distance between adjacent contours to be measured in a normal direction. If the contours correspond to equall spaced

61 966 Chapter3 / Partial Derivatives values of f, then the closer together the contours appear to be, the more rapidl the values of f will be changing in that normal direction. It follows from Theorems and that this rate of change of f is given b f(,). Thus, the closer together the contours appear to be, the greater the length of the gradient of f. Eample 5 A contour plot of a function f is given in Figure 3.6.6a. Sketch the directions of the gradient of f at the points P, Q, and R. At which of these three points does the gradient vector have maimum length? Minimum length? 3.5 P R 3.5 P R Q.5 Q Figure (a) Vectors not to scale (b) Solution. It follows from Theorems and that the directions of the gradient vectors will be as given in Figure 3.6.6b. Based on the densit of the contour lines, we would guess that the gradient of f has maimum length at R and minimum length at P, with the length at Q somewhere in between. REMARK f (, ) = c u f (, ) (, ) f (, ) Figure If (, ) is a point on the level curve f(,) = c, then the slope of the surface = f(,) at that point in the direction of u is D u f(, ) = f(, ) u If u is tangent to the level curve at (, ), then f(,) is neither increasing nor decreasing in that direction, so D u f(, ) =. Thus, f(, ), f(, ), and the tangent vector u mark the directions of maimum slope, minimum slope, and ero slope at a point (, ) on a level curve (Figure 3.6.7). Good skiers use these facts intuitivel to control their speed b igagging down ski slopes the ski across the slope with their skis tangential to a level curve to stop their downhill motion, and the point their skis down the slope and normal to the level curve to obtain the most rapid descent. AN APPLICATION OF GRADIENTS There are numerous applications in which the motion of an object must be controlled so that it moves toward a heat source. For eample, in medical applications the operation of certain diagnostic equipment is designed to locate heat sources generated b tumors or infections, and in militar applications the trajectories of heat-seeking missiles are controlled to seek and destro enem aircraft. The following eample illustrates how gradients are used to solve such problems. UPI Photo/Michael Ammons/Air Force/Digital Railroad, Inc Heat-seeking missiles such as "Stinger" and "Sidewinder" use infrared sensors to measure gradients. Eample 6 A heat-seeking particle is located at the point (, 3) on a flat metal plate whose temperature at a point (, ) is T (,) = 8

62 3.6 Directional Derivatives and Gradients 967 Find an equation for the trajector of the particle if it moves continuousl in the direction of maimum temperature increase Figure Solution. Assume that the trajector is represented parametricall b the equations = (t), = (t) where the particle is at the point (, 3) at time t =. Because the particle moves in the direction of maimum temperature increase, its direction of motion at time t is in the direction of the gradient of T (,), and hence its velocit vector v(t) at time t points in the direction of the gradient. Thus, there is a scalar k that depends on t such that from which we obtain Equating components ields d dt and dividing to eliminate k ields v(t) = k T (,) d dt i + d j = k( 6i 4 j) dt = 6k, d dt = 4k d d = 4k 6k = 4 Thus, we can obtain the trajector b solving the initial-value problem d d =, () = 3 4 The differential equation is a separable first-order equation and hence can be solved b the method of separation of variables discussed in Section 8.. We leave it for ou to show that the solution of the initial-value problem is = 4 3 /4 The graph of the trajector and a contour plot of the temperature function are shown in Figure QUICK CHECK EXERCISES 3.6 (See page 97 for answers.). The gradient of f(,,) = 3 at the point (,, ) is.. Suppose that the differentiable function f(,) has the propert that ( f + s 3, + s ) = 3se s The directional derivative of f in the direction of 3 u = i + j at (, ) is. 3. If the gradient of f(,) at the origin is 6i + 8j, then the directional derivative of f in the direction of a = 3i + 4j at the origin is. The slope of the tangent line to the level curve of f through the origin at (, ) is. 4. If the gradient of f(,,) at (,, 3) is i j + k, then the maimum value for a directional derivative of f at (,, 3) is and the minimum value for a directional derivative at this point is.

63 968 Chapter3 / Partial Derivatives EXERCISE SET 3.6 Graphing Utilit C CAS 8 Find D u f at P.. f(,) = ( + ) 3/ ; P(3, ); u = i + j. f(,) = e ; P(4, ); u = 3 5 i j 3. f(,) = ln( + + ); P(, ); u = i 3 j 4. f(,) = c + d ; P(3, 4); u = 4 5 i j 5. f(,,) = ; P(,, ); u = 3 i + 3 j 3 k 6. f(,,) = e + ; P(,, 3); u = 7 i 3 7 j k 7. f(,,) = ln( ); P(,, 4); u = 3 3 i 4 3 j 3 k 8. f(,,) = sin ; P (, 3,π) ; u = i j + k Find the directional derivative of f at P in the direction of a. 9. f(,) = 4 3 ; P(, ); a = 4i 3j. f(,) = ; P(, ); a = i + j. f(,) = ln ; P(, 4); a = 3i + 3j. f(,) = e cos ; P(,π/4); a = 5i j 3. f(,) = tan (/); P(, ); a = i j 4. f(,) = e e ; P(, ); a = 5i j 5. f(,,) = 3 + ; P(,, ); a = 3i j + k 6. f(,,) = + ; P( 3,, 4); a = i j k 7. f(,,) = ; P(,, 3); a = 6i + 3j k + 8. f(,,) = e ++3 ; P(,, ); a = i 4j + 5k 9 Find the directional derivative of f at P in the direction of a vector making the counterclockwise angle θ with the positive -ais. 9. f(,) = ; P(, 4); θ = π/3. f(,) = + ; P(, ); θ = π /. f(,) = tan( + ); P(π/6,π/3); θ = 7π/4. f(,) = sinh cosh ; P(, ); θ = π 3. Find the directional derivative of f(,) = + at P(, ) in the direction of Q(, ). 4. Find the directional derivative of f(,) = e sec at P(,π/4) in the direction of the origin. 5. Find the directional derivative of f(,) = e at P(, ) in the direction of the negative -ais. 6. Let f(,) = + Find a unit vector u for which D u f(, 3) =. 7. Find the directional derivative of f(,,) = + at P(,, ) in the direction from P to Q(,, ). 8. Find the directional derivative of the function f(,,) = at P(,, ) in the direction of the negative -ais. FOCUS ON CONCEPTS 9. Suppose that D u f(, ) = 5 and D v f(, ) =, where u = 3 5 i 4 5 j and v = 4 5 i + 3 j. Find 5 (a) f (, ) (b) f (, ) (c) the directional derivative of f at (, ) in the direction of the origin. 3. Given that f ( 5, ) = 3 and f ( 5, ) =, find the directional derivative of f at P( 5, ) in the direction of the vector from P to Q( 4, 3). 3. The accompaning figure shows some level curves of an unspecified function f(,). Which of the three vectors shown in the figure is most likel to be f? Eplain. 3. The accompaning figure shows some level curves of an unspecified function f(,). Of the gradients at P and Q, which probabl has the greater length? Eplain. I Figure E-3 III II Find or w. P Figure E = = e 3 cos w = ln w = e 5 sec 37 4 Find the gradient of f at the indicated point. 37. f(,) = ( + ) 3 ; (, ) 38. f(,) = ( + ) / ; (3, 4) 39. f(,,) = ln( + + ); ( 3, 4, ) 4. f(,,) = tan 3 ; (π/4, 3, ) Q 3

64 3.6 Directional Derivatives and Gradients Sketch the level curve of f(,) that passes through P and draw the gradient vector at P. 4. f(,) = 4 + 3; P(, ) 4. f(,) = / ; P(, ) 43. f(,) = + 4 ; P(, ) 44. f(,) = ; P(, ) 45. Find a unit vector u that is normal at P(, ) to the level curve of f(,) = 4 through P. 46. Find a unit vector u that is normal at P(, 3) to the level curve of f(,) = 3 through P Find a unit vector in the direction in which f increases most rapidl at P, and find the rate of change of f at P in that direction. 47. f(,) = 4 3 ; P(, ) 48. f(,) = 3 ln ; P(, 4) 49. f(,) = + ; P(4, 3) 5. f(,) = ; P(, ) + 5. f(,,) = ; P(,, ) 5. f(,,) = 3 + 4; P(, 3, ) 53. f(,,) = + ; P(,, ) 54. f(,,) = tan ( + ) ; P(4,, ) 55 6 Find a unit vector in the direction in which f decreases most rapidl at P, and find the rate of change of f at P in that direction. 55. f(,) = ; P(, 3) 56. f(,) = e ; P(, 3) 57. f(,) = cos(3 ); P(π/6,π/4) 58. f(,) = ; P(3, ) f(,,) = + ; P(5, 7, 6) 6. f(,,) = 4e cos ; P(,,π/4) 6 64 True False Determine whether the statement is true or false. Eplain our answer. In each eercise, assume that f denotes a differentiable function of two variables whose domain is the -plane. 6. If v = u, then the directional derivative of f in the direction of v at a point (, ) is twice the directional derivative of f in the direction of u at the point (, ), 6. If = is a contour of f, then f (, ) =. 63. If u is a fied unit vector and D u f(,) = for all points (, ), then f is a constant function. 64. If the displacement vector from (, ) to (, ) is a positive multiple of f(, ), then f(, ) f(, ). FOCUS ON CONCEPTS 65. Given that f(4, 5) = i j, find the directional derivative of the function f at the point (4, 5) in the direction of a = 5i + j. 66. Given that f(, ) = i j and D u f(, ) =, find u (two answers). 67. The accompaning figure shows some level curves of an unspecified function f(,). (a) Use the available information to approimate the length of the vector f(, ), and sketch the approimation. Eplain how ou approimated the length and determined the direction of the vector. (b) Sketch an approimation of the vector f(4, 4) Figure E The accompaning figure shows a topographic map of a hill and a point P at the bottom of the hill. Suppose that ou want to climb from the point P toward the top of the hill in such a wa that ou are alwas ascending in the direction of steepest slope. Sketch the projection of our path on the contour map. This is called the path of steepest ascent. Eplain how ou determined the path. ft P Figure E Let = 3. Find all points at which =6. 7. Given that = 3 +, find at the point (5, ). 7. A particle moves along a path C given b the equations = t and = t.if = +, find d/ds along C at the instant when the particle is at the point (, 4). 7. The temperature (in degrees Celsius) at a point (, ) on a metal plate in the -plane is T (,) = + + (cont.)

65 97 Chapter3 / Partial Derivatives (a) Find the rate of change of temperature at (, ) in the direction of a = i j. (b) An ant at (, ) wants to walk in the direction in which the temperature drops most rapidl. Find a unit vector in that direction. 73. If the electric potential at a point (, ) in the -plane is V(, ), then the electric intensit vector at the point (, ) is E = V(,). Suppose that V(,) = e cos. (a) Find the electric intensit vector at (π/4, ). (b) Show that at each point in the plane, the electric potential decreases most rapidl in the direction of the vector E. 74. On a certain mountain, the elevation above a point (, ) in an -plane at sea level is =..4, where,, and are in meters. The positive -ais points east, and the positive -ais north. A climber is at the point (, 5, 99). (a) If the climber uses a compass reading to walk due west, will she begin to ascend or descend? (b) If the climber uses a compass reading to walk northeast, will she ascend or descend? At what rate? (c) In what compass direction should the climber begin walking to travel a level path (two answers)? 75. Given that the directional derivative of f(,,)at the point (3,, ) in the direction of a = i j k is 5 and that f(3,, ) =5, find f(3,, ). 76. The temperature (in degrees Celsius) at a point (,,)in a metal solid is T (,,) = (a) Find the rate of change of temperature with respect to distance at (,, ) in the direction of the origin. (b) Find the direction in which the temperature rises most rapidl at the point (,, ). (Epress our answer as a unit vector.) (c) Find the rate at which the temperature rises moving from (,, ) in the direction obtained in part (b). 77. Let r = +. (a) Show that r = r, where r = i + j. r (b) Show that f(r) = f (r) r = f (r) r. r 78. Use the formula in part (b) of Eercise 77 to find (a) f(r) if f(r) = re 3r (b) f(r) if f(r) = 3r r and f() =. 79. Let u r be a unit vector whose counterclockwise angle from the positive -ais is θ, and let u θ be a unit vector 9 counterclockwise from u r. Show that if = f(,), = r cos θ, and = r sin θ, then = r u r + r θ u θ [Hint: Use part (c) of Eercise 55, Section 3.5.] 8. Prove: If f and g are differentiable, then (a) (f + g) = f + g (b) (cf ) = c f (c constant) (c) (fg) ( ) = f g + g f f g f f g (d) = g g (e) (f n ) = nf n f. 8 8 A heat-seeking particle is located at the point P on a flat metal plate whose temperature at a point (, ) is T (,). Find parametric equations for the trajector of the particle if it moves continuousl in the direction of maimum temperature increase. 8. T (,) = 5 4 ; P(, 4) 8. T (,) = ; P(5, 3) 83. Use a graphing utilit to generate the trajector of the particle together with some representative level curves of the temperature function in Eercise Use a graphing utilit to generate the trajector of the particle together with some representative level curves of the temperature function in Eercise 8. C 85. (a) Use a CAS to graph f(,) = ( + 3 )e ( + ). (b) At how man points do ou think it is true that D u f(,) = for all unit vectors u? (c) Use a CAS to find f. (d) Use a CAS to solve the equation f(,) = for and. (e) Use the result in part (d) together with Theorem to check our conjecture in part (b). 86. Prove: If = (t) and = (t) are differentiable at t, and if = f(,) is differentiable at the point ((t), (t)), then d dt = r (t) where r(t) = (t)i + (t)j. 87. Prove: If f, f, and f are continuous on a circular region, and if f(,) = throughout the region, then f(,) is constant on the region. [Hint: See Eercise 67, Section 3.5.] 88. Prove: If the function f is differentiable at the point (, ) and if D u f(,) = in two nonparallel directions, then D u f(,) = in all directions. 89. Given that the functions u = u(,,), v = v(,,), w = w(,,), and f(u, v, w) are all differentiable, show that f(u, v, w) = f f f u + v + u v w w 9. Writing Let f denote a differentiable function of two variables. Write a short paragraph that discusses the connections between directional derivatives of f and slopes of tangent lines to the graph of f. 9. Writing Let f denote a differentiable function of two variables. Although we have defined what it means to sa that f is differentiable, we have not defined the derivative of f. Write a short paragraph that discusses the merits of defining the derivative of f to be the gradient f.

66 3.7 Tangent Planes and Normal Vectors 97 QUICK CHECK ANSWERS 3.6.,, ; ; TANGENT PLANES AND NORMAL VECTORS In this section we will discuss tangent planes to surfaces in three-dimensional space. We will be concerned with three main questions: What is a tangent plane? When do tangent planes eist? How do we find equations of tangent planes? TANGENT PLANES AND NORMAL VECTORS TO LEVEL SURFACES F(,, ) =c We begin b considering the problem of finding tangent planes to level surfaces of a function F(,,). These surfaces are represented b equations of the form F(,,) = c. T We will assume that F has continuous first-order partial derivatives, since this has an important geometric consequence. Fi c, and suppose that P (,, ) satisfies the equation P C F(,,) = c. In advanced courses it is proved that if F has continuous first-order partial derivatives, and if F(,, ) =, then near P the graph of F(,,) = c is indeed a surface rather than some possibl eotic-looking set of points in 3-space. We will base our concept of a tangent plane to a level surface S: F(,,) = c on the more elementar notion of a tangent line to a curve C in 3-space (Figure 3.7.). Intuitivel, Figure 3.7. we would epect a tangent plane to S at a point P to be composed of the tangent lines at P of all curves on S that pass through P (Figure 3.7.). Suppose C is a curve on S through P that is parametried b = (t), = (t), = (t) with = (t ), = (t ), and = (t ). The tangent line l to C through P is then parallel to the vector Tangent lines P (, ) Tangent line F(,, ) = c All tangent lines at P lie in the tangent plane. Figure 3.7. S Tangent plane r = (t )i + (t )j + (t )k where we assume that r = (Definition..7). Since C is on the surface F(,,) = c, we have c = F ((t), (t), (t)) () Computing the derivative at t of both sides of (), we have b the chain rule that = F (,, ) (t ) + F (,, ) (t ) + F (,, ) (t ) We can write this equation in vector form as = (F (,, )i + F (,, )j + F (,, )k) ( (t )i + (t )j + (t )k) or = F(,, ) r () F(,, ) S (,, ) F(,, ) = c It follows that if F(,, ) =, then F(,, ) is normal to line l. Therefore, the tangent line l to C at P is contained in the plane through P with normal vector F(,, ). Since C was arbitrar, we conclude that the same is true for an curve on S through P (Figure 3.7.3). Thus, it makes sense to define the tangent plane to S at P to be the plane through P whose normal vector is n = F(,, ) = F (,, ), F (,, ), F (,, ) Figure Using the point-normal form [see Formula (3) in Section.6], we have the following definition.

67 97 Chapter3 / Partial Derivatives Definition 3.7. can be viewed as an etension of Theorem from curves to surfaces definition Assume that F(,,)has continuous first-order partial derivatives and that P (,, ) is a point on the level surface S: F(,,) = c. If F(,, ) =, then n = F(,, ) is a normal vector to S at P and the tangent plane to S at P is the plane with equation F (,, )( ) + F (,, )( ) + F (,, )( ) = (3) The line through the point P parallel to the normal vector n is perpendicular to the tangent plane (3). We will call this the normal line, or sometimes more simpl the normal to the surface F(,,) = c at P. It follows that this line can be epressed parametricall as = + F (,, )t, = + F (,, )t, = + F (,, )t (4) Eample Consider the ellipsoid = 8. (a) Find an equation of the tangent plane to the ellipsoid at the point (,, ). (b) Find parametric equations of the line that is normal to the ellipsoid at the point (,, ). (c) Find the acute angle that the tangent plane at the point (,, ) makes with the -plane. Solution (a). We appl Definition 3.7. with F(,,) = and (,, ) = (,, ). Since F(,,) = F (,, ), F (,, ), F (,,) =,8, we have n = F(,, ) =, 6, Hence, from (3) the equation of the tangent plane is ( ) + 6( ) + ( ) = or = 8 Solution (b). Since n =, 6, at the point (,, ), it follows from (4) that parametric equations for the normal line to the ellipsoid at the point (,, ) are = + t, = + 6t, = + t Figure (,, ) Solution (c). To find the acute angle θ between the tangent plane and the -plane, we will appl Formula (9) of Section.6 with n = n =, 6, and n =,,. This ields, 6,,, cos θ =, 6,,, = ( 66 )() = 66 Thus, ( ) θ = cos (Figure 3.7.4). TANGENT PLANES TO SURFACES OF THE FORM = f (, ) To find a tangent plane to a surface of the form = f(,), we can use Equation (3) with the function F(,,) = f(,).

68 3.7 Tangent Planes and Normal Vectors 973 Eample Find an equation for the tangent plane and parametric equations for the normal line to the surface = at the point (,, 4). Solution. Let F(,,) =. Then F(,,) = on the surface, so we can find the find the gradient of F at the point (,, 4): F(,,) = i j + k F(,, 4) = 4i 4j + k From (3) the tangent plane has equation 4( ) 4( ) + ( 4) = or = 8 and the normal line has equations = 4t, = 4t, = 4 + t Suppose that f(,)is differentiable at a point (, ) and that = f(, ). It can be shown that the procedure of Eample can be used to find the tangent plane to the surface = f(,) at the point (,, ). This ields an alternative equation for a tangent plane to the graph of a differentiable function theorem If f(,) is differentiable at the point (, ), then the tangent plane to the surface = f(,) at the point P (,,f(, )) [or (, )] is the plane = f(, ) + f (, )( ) + f (, )( ) (5) proof Consider the function F(,,) = f(,). Since F(,,) = onthe surface, we will appl (3) to this function. The partial derivatives of F are F (,,)= f (, ), F (,,)= f (, ), F (,,)= Since the point at which we evaluate these derivatives lies on the surface, it will have the form (,,f(, )). Thus, (3) gives = F (,, )( ) + F (,, )( ) + F (,, )( f(, )) = f (, )( ) f (, )( ) + ( f(, )) which is equivalent to (5). Recall from Section 3.4 that if a function f(,) is differentiable at a point (, ), then the local linear approimation L(, ) to f at (, ) has the equation L(, ) = f(, ) + f (, )( ) + f (, )( ) Notice that the equation = L(, ) is identical to that of the tangent plane to f(,) at the point (, ). Thus, the graph of the local linear approimation to f(,) at the point (, ) is the tangent plane to the surface = f(,) at the point (, ). TANGENT PLANES AND TOTAL DIFFERENTIALS Recall that for a function = f(,) of two variables, the approimation b differentials is = f = f(,) f(, ) d = f (, )( ) + f (, )( )

69 974 Chapter3 / Partial Derivatives Note that the tangent plane in Figure is analogous to the tangent line in Figure The tangent plane provides a geometric interpretation of this approimation. We see in Figure that is the change in along the surface = f(,) from the point P (,,f(, )) to the point P(,,f(,)), and d is the change in along the tangent plane from P to Q(,, L(, )). The small vertical displacement at (, ) between the surface and the plane represents the error in the local linear approimation to f at (, ). We have seen that near (, ) this error term has magnitude much smaller than the distance between (, ) and (, ). = f(, ) d Figure P (,, f(, )) (, ) Tangent plane at P = L(, ) (, ) d = = d = = G(,, ) = Figure G F(,, ) = T F (,, ) USING GRADIENTS TO FIND TANGENT LINES TO INTERSECTIONS OF SURFACES In general, the intersection of two surfaces F(,,) = and G(,,)= will be a curve in 3-space. If (,, ) is a point on this curve, then F(,, ) will be normal to the surface F(,,) = at(,, ) and G(,, ) will be normal to the surface G(,,)= at(,, ). Thus, if the curve of intersection can be smoothl parametried, then its unit tangent vector T at (,, ) will be orthogonal to both F(,, ) and G(,, ) (Figure 3.7.6). Consequentl, if F(,, ) G(,, ) = then this cross product will be parallel to T and hence will be tangent to the curve of intersection. This tangent vector can be used to determine the direction of the tangent line to the curve of intersection at the point (,, ). Eample 3 Find parametric equations of the tangent line to the curve of intersection of the paraboloid = + and the ellipsoid = 9 at the point (,, ) (Figure 3.7.7). Figure (,, ) Solution. We begin b rewriting the equations of the surfaces as + = and = and we take F(,,) = + and G(,,)= We will need the gradients of these functions at the point (,, ). The computations are F(,,) = i + j k, G(,,)= 6i + 4 j + k F(,, ) = i + j k, G(,, ) = 6i + 4j + 4k Thus, a tangent vector at (,, ) to the curve of intersection is i j k F(,, ) G(,, ) = = i 4j 4k 6 4 4

70 3.7 Tangent Planes and Normal Vectors 975 Since an scalar multiple of this vector will do just as well, we can multipl b to reduce the sie of the coefficients and use the vector of 6i 7j k to determine the direction of the tangent line. This vector and the point (,, ) ield the parametric equations = + 6t, = 7t, = t QUICK CHECK EXERCISES 3.7 (See page 977 for answers.). Suppose that f(,, ) =, and f(,,) is differentiable at (,, ) with f(,, ) =,,. An equation for the tangent plane to the level surface f(,,) = at the point (,, ) is, and parametric equations for the normal line to the level surface through the point (,, ) are =, =, =. Suppose that f(, ) is differentiable at the point (3, ) with f(3, ) = 4,f (3, ) =, and f (3, ) = 3. An equation for the tangent plane to the graph of f at the point (3,, 4) is, and parametric equations for the normal line to the graph of f through the point (3,, 4) are =, =, = 3. An equation for the tangent plane to the graph of = at the point (, 4, 8) is, and parametric equations for the normal line to the graph of = through the point (, 4, 8) are =, =, = 4. The sphere + + = 9 and the plane + + = 5 intersect in a circle that passes through the point (,, ). Parametric equations for the tangent line to this circle at (,, ) are =, =, = EXERCISE SET 3.7 C CAS. Consider the ellipsoid =. (a) Find an equation of the tangent plane to the ellipsoid at the point (,, ). (b) Find parametric equations of the line that is normal to the ellipsoid at the point (,, ). (c) Find the acute angle that the tangent plane at the point (,, ) makes with the -plane.. Consider the surface 3 + =. (a) Find an equation of the tangent plane to the surface at the point (,, ). (b) Find parametric equations of the line that is normal to the surface at the point (,, ). (c) Find the acute angle that the tangent plane at the point (,, ) makes with the -plane. 3 Find an equation for the tangent plane and parametric equations for the normal line to the surface at the point P = 5; P( 3,, 4) 4. 4 = 7; P( 3,, ) 5. = ; P(,, ) 6. = 7 ; P(, 4, 4) 7. = e ; P(,, ) 8. = ln + ; P(,, ) 9. = e 3 sin 3; P(π/6,, ). = / + / ; P(4, 9, 5) FOCUS ON CONCEPTS. Find all points on the surface at which the tangent plane is horiontal. (a) = 3 (b) = Find a point on the surface = 3 at which the tangent plane is parallel to the plane = Find a point on the surface = 8 3 at which the tangent plane is perpendicular to the line = 3t, = 7 + 8t, = 5 t. 4. Show that the surfaces = + and = ( + ) + 5 intersect at (3, 4, 5) and have a common tangent plane at that point. 5. (a) Find all points of intersection of the line = + t, = + t, = t + 7 and the surface = + (b) At each point of intersection, find the cosine of the acute angle between the given line and the line normal to the surface. 6. Show that if f is differentiable and = f(/), then all tangent planes to the graph of this equation pass through the origin.

71 C 976 Chapter3 / Partial Derivatives 7 True False Determine whether the statement is true or false. Eplain our answer. 7. If the tangent plane to the level surface of F(,,) at the point P (,, ) is also tangent to a level surface of G(,,)at P, then F(,, ) = G(,, ). 8. If the tangent plane to the graph of = f(,) at the point (,, ) has equation + = 4, then f (, ) = and f (, ) =. 9. If the tangent plane to the graph of = f(,) at the point (,, ) has equation + = 3, then the local linear approimation to f at (, ) is given b the function L(, ) = + ( ) + ( ).. The normal line to the surface = f(,) at the point P (,,f(, )) has a direction vector given b f (, )i + f (, )j k. Find two unit vectors that are normal to the given surface at the point P. +. = ; P(3, 5, ). sin 4 cos = 4; P (π, π, ) 3. Show that ever line that is normal to the sphere passes through the origin. + + = 4. Find all points on the ellipsoid = 9at which the plane tangent to the ellipsoid is parallel to the plane + 3 = Find all points on the surface + = at which the normal line is parallel to the line through P(,, ) and Q(4,, ). 6. Show that the ellipsoid = 9 and the sphere = have a common tangent plane at the point (,, ). 7. Find parametric equations for the tangent line to the curve of intersection of the paraboloid = + and the ellipsoid = 9 at the point (,, ). 8. Find parametric equations for the tangent line to the curve of intersection of the cone = + and the plane + + = at the point (4, 3, 5). 9. Find parametric equations for the tangent line to the curve of intersection of the clinders + = 5 and + = 5 at the point (3, 3, 4). 3. The accompaning figure shows the intersection of the surfaces = 8 and 4 + =. (a) Find parametric equations for the tangent line to the curve of intersection at the point (,, 4). (b) Use a CAS to generate a reasonable facsimile of the figure. You need not generate the colors, but tr to obtain a similar viewpoint Figure E-3 3. Show that the equation of the plane that is tangent to the ellipsoid a + b + c = at (,, ) can be written in the form a + b + c = 3. Show that the equation of the plane that is tangent to the paraboloid = a + b at (,, ) can be written in the form + = + a b 33. Prove: If the surfaces = f(,)and = g(,)intersect at P(,, ), and if f and g are differentiable at (, ), then the normal lines at P are perpendicular if and onl if f (, )g (, ) + f (, )g (, ) = 34. Use the result in Eercise 33 to show that the normal lines to the cones = + and = + are perpendicular to the normal lines to the sphere + + = a at ever point of intersection (see Figure E-36). 35. Two surfaces f(,,) = and g(,,) = are said to be orthogonal at a point P of intersection if f and g are nonero at P and the normal lines to the surfaces are perpendicular at P. Show that if f(,, ) = and g(,, ) =, then the surfaces f(,,) = and g(,,) = are orthogonal at the point (,, ) if and onl if f g + f g + f g = at this point. [Note: This is a more general version of the result in Eercise 33.] 36. Use the result of Eercise 35 to show that the sphere + + = a and the cone = + are orthogonal at ever point of intersection (see the accompaning figure). Figure E-36

72 3.8 Maima and Minima of Functions of Two Variables Show that the volume of the solid bounded b the coordinate planes and a plane tangent to the portion of the surface = k, k>, in the first octant does not depend on the point of tangenc. 38. Writing Discuss the role of the chain rule in defining a tangent plane to a level surface. 39. Writing Discuss the relationship between tangent planes and local linear approimations for functions of two variables. QUICK CHECK ANSWERS 3.7. ( ) + + ( + ) = ; = + t; = t; = + t. = 4 + ( 3) 3( ); = 3 + t; = 3t; = 4 t 3. = 8 + 8( ) + ( 4); = + 8t; = 4 + t; = 8 t 4. = + t; = ; = t 3.8 MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES Earlier in this tet we learned how to find maimum and minimum values of a function of one variable. In this section we will develop similar techniques for functions of two variables. Figure 3.8. = f(, ) EXTREMA If we imagine the graph of a function f of two variables to be a mountain range (Figure 3.8.), then the mountaintops, which are the high points in their immediate vicinit, are called relative maima of f, and the valle bottoms, which are the low points in their immediate vicinit, are called relative minima of f. Just as a geologist might be interested in finding the highest mountain and deepest valle in an entire mountain range, so a mathematician might be interested in finding the largest and smallest values of f(,) over the entire domain of f. These are called the absolute maimum and absolute minimum values of f. The following definitions make these informal ideas precise definition A function f of two variables is said to have a relative maimum at a point (, ) if there is a disk centered at (, ) such that f(, ) f(,) for all points (, ) that lie inside the disk, and f is said to have an absolute maimum at (, ) if f(, ) f(,) for all points (, ) in the domain of f. Relative maimum Absolute maimum = f (, ) B C A Relative minimum Absolute minimum Figure 3.8. D 3.8. definition A function f of two variables is said to have a relative minimum at a point (, ) if there is a disk centered at (, ) such that f(, ) f(,) for all points (, ) that lie inside the disk, and f is said to have an absolute minimum at (, ) if f(, ) f(,) for all points (, ) in the domain of f. If f has a relative maimum or a relative minimum at (, ), then we sa that f has a relative etremum at (, ), and if f has an absolute maimum or absolute minimum at (, ), then we sa that f has an absolute etremum at (, ). Figure 3.8. shows the graph of a function f whose domain is the square region in the -plane whose points satisf the inequalities,. The function f has

73 978 Chapter3 / Partial Derivatives relative minima at the points A and C and a relative maimum at B. There is an absolute minimum at A and an absolute maimum at D. For functions of two variables we will be concerned with two important questions: Are there an relative or absolute etrema? If so, where are the located? Eplain wh an subset of a bounded set is also bounded. BOUNDED SETS Just as we distinguished between finite intervals and infinite intervals on the real line, so we will want to distinguish between regions of finite etent and regions of infinite etent in -space and 3-space. A set of points in -space is called bounded if the entire set can be contained within some rectangle, and is called unbounded if there is no rectangle that contains all the points of the set. Similarl, a set of points in 3-space is bounded if the entire set can be contained within some bo, and is unbounded otherwise (Figure 3.8.3). Figure A bounded set in -space An unbounded set in -space (the first quadrant) A bounded set in 3-space THE EXTREME-VALUE THEOREM For functions of one variable that are continuous on a closed interval, the Etreme-Value Theorem (Theorem 4.4.) answered the eistence question for absolute etrema. The following theorem, which we state without proof, is the corresponding result for functions of two variables theorem (Etreme-Value Theorem) If f(,) is continuous on a closed and bounded set R, then f has both an absolute maimum and an absolute minimum on R. Eample The square region R whose points satisf the inequalities and is a closed and bounded set in the -plane. The function f whose graph is shown in Figure 3.8. is continuous on R; thus, it is guaranteed to have an absolute maimum and minimum on R b the last theorem. These occur at points D and A that are shown in the figure. REMARK If an of the conditions in the Etreme-Value Theorem fail to hold, then there is no guarantee that an absolute maimum or absolute minimum eists on the region R. Thus, a discontinuous function on a closed and bounded set need not have an absolute etrema, and a continuous function on a set that is not closed and bounded also need not have an absolute etrema.

74 3.8 Maima and Minima of Functions of Two Variables 979 Relative maimum = = FINDING RELATIVE EXTREMA Recall that if a function g of one variable has a relative etremum at a point where g is differentiable, then g ( ) =. To obtain the analog of this result for functions of two variables, suppose that f(,) has a relative maimum at a point (, ) and that the partial derivatives of f eist at (, ). It seems plausible geometricall that the traces of the surface = f(,) on the planes = and = have horiontal tangent lines at (, ) (Figure 3.8.4), so f (, ) = and f (, ) = The same conclusion holds if f has a relative minimum at (, ), all of which suggests the following result, which we state without formal proof. (, ) = f(, ) = f(, ) Figure theorem If f has a relative etremum at a point (, ), and if the firstorder partial derivatives of f eist at this point, then f (, ) = and f (, ) = Recall that the critical points of a function f of one variable are those values of in the domain of f at which f () = orf is not differentiable. The following definition is the analog for functions of two variables. Eplain wh D u f(, ) = for all u if (, ) is a critical point of f and f is differentiable at (, ). (, ) = The function f (, ) = has neither a relative maimum nor a relative minimum at the critical point (, ). Figure definition A point (, ) in the domain of a function f(,) is called a critical point of the function if f (, ) = and f (, ) = or if one or both partial derivatives do not eist at (, ). It follows from this definition and Theorem that relative etrema occur at critical points, just as for a function of one variable. However, recall that for a function of one variable a relative etremum need not occur at ever critical point. For eample, the function might have an inflection point with a horiontal tangent line at the critical point (see Figure 4..6). Similarl, a function of two variables need not have a relative etremum at ever critical point. For eample, consider the function f(,) = This function, whose graph is the hperbolic paraboloid shown in Figure 3.8.5, has a critical point at (, ), since from which it follows that f (, ) = and f (, ) = f (, ) = and f (, ) = However, the function f has neither a relative maimum nor a relative minimum at (, ). For obvious reasons, the point (, ) is called a saddle point of f. In general, we will sa that a surface = f(,) has a saddle point at (, ) if there are two distinct vertical planes through this point such that the trace of the surface in one of the planes has a relative maimum at (, ) and the trace in the other has a relative minimum at (, ). Eample The three functions graphed in Figure all have critical points at (, ). For the paraboloids, the partial derivatives at the origin are ero. You can check this

75 98 Chapter3 / Partial Derivatives algebraicall b evaluating the partial derivatives at (, ), but ou can see it geometricall b observing that the traces in the -plane and -plane have horiontal tangent lines at (, ). For the cone neither partial derivative eists at the origin because the traces in the -plane and the -plane have corners there. The paraboloid in part (a) and the cone in part (c) have a relative minimum and absolute minimum at the origin, and the paraboloid in part (b) has a relative maimum and an absolute maimum at the origin. = + = = + f (, ) = f (, ) = relative and absolute min at (, ) f (, ) = f (, ) = relative and absolute ma at (, ) f (, ) and f (, ) do not eist relative and absolute min at (, ) Figure (a) (b) (c) THE SECOND PARTIALS TEST For functions of one variable the second derivative test (Theorem 4..4) was used to determine the behavior of a function at a critical point. The following theorem, which is usuall proved in advanced calculus, is the analog of that theorem for functions of two variables theorem (The Second Partials Test) Let f be a function of two variables with continuous second-order partial derivatives in some disk centered at a critical point (, ), and let D = f (, )f (, ) f (, ) With the notation of Theorem 3.8.6, show that if D>, then f (, ) and f (, ) have the same sign. Thus, we can replace f (, ) b f (, ) in parts (a) and (b) ofthe theorem. (a) If D>and f (, )>, then f has a relative minimum at (, ). (b) If D>and f (, )<, then f has a relative maimum at (, ). (c) If D<, then f has a saddle point at (, ). (d ) If D =, then no conclusion can be drawn. Eample 3 Locate all relative etrema and saddle points of f(,) = Solution. Since f (, ) = 6 and f (, ) = + 8, the critical points of f satisf the equations 6 = + 8 = Solving these for and ields =, = 6 (verif), so (, 6) is the onl critical point. To appl Theorem we need the second-order partial derivatives f (, ) = 6, f (, ) =, f (, ) =

76 3.8 Maima and Minima of Functions of Two Variables At the point (, 6) we have D = f (, 6)f (, 6) f (, 6) = (6)() ( ) = 8 > and f (, 6) = 6 > so f has a relative minimum at (, 6) b part (a) of the second partials test. Figure shows a graph of f in the vicinit of the relative minimum. f (, ) = Figure Eample 4 Locate all relative etrema and saddle points of f(,) = f (, ) = Solution. Since f (, ) = f (, ) = () the critical points of f have coordinates satisfing the equations = = or = 3 = 3 () Substituting the top equation in the bottom ields = ( 3 ) 3 or, equivalentl, 9 = or ( 8 ) =, which has solutions =, =, =. Substituting these values in the top equation of (), we obtain the corresponding -values =, =, =. Thus, the critical points of f are (, ), (, ), and (, ). From (), f (, ) =, f (, ) =, f (, ) = 4 which ields the following table: Figure critical point (, ) f (, ) f (, ) f (, ) D = f f f (, ) (, ) (, ) At the points (, ) and (, ), we have D> and f <, so relative maima occur at these critical points. At (, ) there is a saddle point since D<. The surface and a contour plot are shown in Figure The figure eight pattern at (, ) in the contour plot for the surface in Figure is tpical for level curves that pass through a saddle point. If a bug starts at the point (,, ) on the surface, in how man directions can it walk and remain in the -plane? The following theorem, which is the analog for functions of two variables of Theorem 4.4.3, will lead to an important method for finding absolute etrema theorem If a function f of two variables has an absolute etremum (either an absolute maimum or an absolute minimum) at an interior point of its domain, then this etremum occurs at a critical point.

77 98 Chapter3 / Partial Derivatives proof If f has an absolute maimum at the point (, ) in the interior of the domain of f, then f has a relative maimum at (, ). If both partial derivatives eist at (, ), then f (, ) = and f (, ) = b Theorem 3.8.4, so (, ) is a critical point of f. If either partial derivative does not eist, then again (, ) is a critical point, so (, ) is a critical point in all cases. The proof for an absolute minimum is similar. FINDING ABSOLUTE EXTREMA ON CLOSED AND BOUNDED SETS If f(,) is continuous on a closed and bounded set R, then the Etreme-Value Theorem (Theorem 3.8.3) guarantees the eistence of an absolute maimum and an absolute minimum of f on R. These absolute etrema can occur either on the boundar of R or in the interior of R, but if an absolute etremum occurs in the interior, then it occurs at a critical point b Theorem Thus, we are led to the following procedure for finding absolute etrema: Compare this procedure with that in Section 4.4 for finding the etreme values of f()on a closed interval. How to Find the Absolute Etrema of a Continuous Function f of Two Variables on a Closed and Bounded Set R Step. Find the critical points of f that lie in the interior of R. Step. Find all boundar points at which the absolute etrema can occur. Step 3. Evaluate f(,) at the points obtained in the preceding steps. The largest of these values is the absolute maimum and the smallest the absolute minimum. (, 5) R (, ) (3, ) Figure Eample 5 Find the absolute maimum and minimum values of f(,) = (3) on the closed triangular region R with vertices (, ), (3, ), and (, 5). Solution. The region R is shown in Figure We have f = 3 6 and f = 3 3 so all critical points occur where 3 6 = and 3 3 = Solving these equations ields = and =, so (, ) is the onl critical point. As shown in Figure 3.8.9, this critical point is in the interior of R. Net we want to determine the locations of the points on the boundar of R at which the absolute etrema might occur. The boundar of R consists of three line segments, each of which we will treat separatel: The line segment between (, ) and (3, ): On this line segment we have =, so (3) simplifies to a function of the single variable, u() = f(,) = 6 + 7, 3 This function has no critical points because u () = 6 is nonero for all. Thus the etreme values of u() occur at the endpoints = and = 3, which correspond to the points (, ) and (3, ) of R.

78 3.8 Maima and Minima of Functions of Two Variables 983 The line segment between (, ) and (, 5): On this line segment we have =, so (3) simplifies to a function of the single variable, v() = f(,)= 3 + 7, 5 This function has no critical points because v () = 3 is nonero for all. Thus, the etreme values of v() occur at the endpoints = and = 5, which correspond to the points (, ) and (, 5) of R. The line segment between (3, ) and (, 5): In the -plane, an equation for this line segment is = 5 + 5, 3 (4) 3 so (3) simplifies to a function of the single variable, w() = f (, ) = 3 ( ) 6 3 ( ) + 7 = , 3 Since w () = + 4, the equation w () = ields = 7 as the onl critical point of 5 w. Thus, the etreme values of w occur either at the critical point = 7 or at the endpoints 5 = and = 3. The endpoints correspond to the points (, 5) and (3, ) of R, and from (4) the critical point corresponds to ( 7 5, 3) 8. Finall, Table 3.8. lists the values of f(,) at the interior critical point and at the points on the boundar where an absolute etremum can occur. From the table we conclude that the absolute maimum value of f is f(, ) = 7 and the absolute minimum value is f(3, ) =. Table 3.8. (, ) f(, ) (, ) 7 (3, ) (, 5) 8 7 8, (, ) Eample 6 Determine the dimensions of a rectangular bo, open at the top, having a volume of 3 ft 3, and requiring the least amount of material for its construction. Solution. Let = length of the bo (in feet) = width of the bo (in feet) = height of the bo (in feet) S = surface area of the bo (in square feet) We ma reasonabl assume that the bo with least surface area requires the least amount of material, so our objective is to minimie the surface area Two sides each have area. Two sides each have area. The base has area. Figure 3.8. (Figure 3.8.) subject to the volume requirement From (6) we obtain = 3/, so (5) can be rewritten as S = + + (5) = 3 (6) S = (7)

79 984 Chapter3 / Partial Derivatives 4, R 4 4 4, = Figure , 3 which epresses S as a function of two variables. The dimensions and in this formula must be positive, but otherwise have no limitation, so our problem reduces to finding the absolute minimum value of S over the open first quadrant: >, >. Because this region is neither closed nor bounded, we have no mathematical guarantee at this stage that an absolute minimum eists. However, if S has an absolute minimum value in the open first quadrant, then it must occur at a critical point of S. Thus, our net step is to find the critical points of S. Differentiating (7) we obtain S = 64, so the coordinates of the critical points of S satisf =, = Solving the first equation for ields S = 64 (8) = 64 (9) and substituting this epression in the second equation ields 64 (64/ ) = which can be rewritten as ) ( 3 = 64 The solutions of this equation are = and = 4. Since we require >, the onl solution of significance is = 4. Substituting this value into (9) ields = 4. We conclude that the point (, ) = (4, 4) is the onl critical point of S in the first quadrant. Since S = 48 if = = 4, this suggests we tr to show that the minimum value of S on the open first quadrant is 48. It immediatel follows from Equation (7) that 48 <Sat an point in the first quadrant for which at least one of the inequalities 64 > 48, > 48, 64 > 48 is satisfied. Therefore, to prove that 48 S, we can restrict attention to the set of points in the first quadrant that satisf the three inequalities 64 48, 48, These inequalities can be rewritten as 48, 4 3, 4 3 and the define a closed and bounded region R within the first quadrant (Figure 3.8.). The function S is continuous on R, so Theorem guarantees that S has an absolute minimum value somewhere on R. Since the point (4, 4) lies within R, and 48 <Son the boundar of R (wh?), the minimum value of S on R must occur at an interior point. It then follows from Theorem that the mimimum value of S on R must occur at a critical point of S. Hence, the absolute minimum of S on R (and therefore on the entire open first quadrant) is S = 48 at the point (4, 4). Substituting = 4 and = 4 into (6) ields =, so the bo using the least material has a height of ft and a square base whose edges are 4 ft long.

80 3.8 Maima and Minima of Functions of Two Variables 985 S 75 REMARK Fortunatel, in our solution to Eample 6 we were able to prove the eistence of an absolute minimum of S in the first quadrant. The general problem of finding the absolute etrema of a function on an unbounded region, or on a region that is not closed, can be difficult and will not be considered in this tet. However, in applied problems we can sometimes use phsical considerations to deduce that an absolute etremum has been found. For eample, the graph of Equation (7) in Figure 3.8. strongl suggests that the relative minimum at = 4 and = 4 is also an absolute minimum. 5 4 Figure QUICK CHECK EXERCISES 3.8 (See page 989 for answers.). The critical points of the function f(,) = are.. Suppose that f(,) has continuous second-order partial derivatives everwhere and that the origin is a critical point for f. State what information (if an) is provided b the second partials test if (a) f (, ) =, f (, ) =, f (, ) = (b) f (, ) =, f (, ) =, f (, ) = (c) f (, ) = 3, f (, ) =, f (, ) = (d) f (, ) = 3, f (, ) =, f (, ) =. 3. For the function f(,) = , state what information (if an) is provided b the second partials test at the point (a) (, ) (b) (, ) (c) (, ). 4. A rectangular bo has total surface area of ft. Epress the volume of the bo as a function of the dimensions and of the base of the bo. EXERCISE SET 3.8 Graphing Utilit C CAS Locate all absolute maima and minima, if an, b inspection. Then check our answers using calculus.. (a) f(,) = ( ) + ( + ) (b) f(,) = (c) f(,) = + 5. (a) f(,) = ( + ) ( 5) (b) f(,) = e (c) f(,) = 3 4 Complete the squares and locate all absolute maima and minima, if an, b inspection. Then check our answers using calculus. 3. f(,) = f(,) = + 4 FOCUS ON CONCEPTS 5 8 The contour plots show all significant features of the function. Make a conjecture about the number and the location of all relative etrema and saddle points, and then use calculus to check our conjecture f(, ) = f(, ) = f(, ) = f(, ) =

81 C C 986 Chapter3 / Partial Derivatives 9 Locate all relative maima, relative minima, and saddle points, if an. 9. f(,) = f(,) = + +. f(,) = f(,) = 3 3. f(,) = f(,) = e 5. f(,) = + e 6. f(,) = f(,) = e sin 8. f(,) = sin 9. f(,) = e ( + +). f(,) = + a3 + b3 (a =,b = ). Use a CAS to generate a contour plot of f(,) = for and, and use the plot to approimate the locations of all relative etrema and saddle points in the region. Check our answer using calculus, and identif the relative etrema as relative maima or minima.. Use a CAS to generate a contour plot of f(,) = + 4 for 5 5 and 5 5, and use the plot to approimate the locations of all relative etrema and saddle points in the region. Check our answer using calculus, and identif the relative etrema as relative maima or minima. 3 6 True False Determine whether the statement is true or false. Eplain our answer. In these eercises, assume that f(,) has continuous second-order partial derivatives and that D(,) = f (, )f (, ) f (, ) 3. If the function f is defined on the disk +, then f has a critical point somewhere on this disk. 4. If the function f is defined on the disk +, and if f is not a constant function, then f has a finite number of critical points on this disk. 5. If P(, ) is a critical point of f, and if f is defined on a disk centered at P with D(, )>, then f has a relative etremum at P. 6. If P(, ) is a critical point of f with f(, ) =, and if f is defined on a disk centered at P with D(, )<, then f has both positive and negative values on this disk. FOCUS ON CONCEPTS 7. (a) Show that the second partials test provides no information about the critical points of the function f(,) = (b) Classif all critical points of f as relative maima, relative minima, or saddle points. 8. (a) Show that the second partials test provides no information about the critical points of the function f(,) = 4 4. (b) Classif all critical points of f as relative maima, relative minima, or saddle points. 9. Recall from Theorem that if a continuous function of one variable has eactl one relative etremum on an interval, then that relative etremum is an absolute etremum on the interval. This eercise shows that this result does not etend to functions of two variables. (a) Show that f(,) = 3e 3 e 3 has onl one critical point and that a relative maimum occurs there. (See the accompaning figure.) (b) Show that f does not have an absolute maimum. Source: This eercise is based on the article The Onl Critical Point in Town Test b Ira Rosenholt and Lowell Smlie, Mathematics Magaine, Vol. 58, No. 3, Ma 985, pp = 3e 3 e 3 Figure E-9 3. If f is a continuous function of one variable with two relative maima on an interval, then there must be a relative minimum between the relative maima. (Convince ourself of this b drawing some pictures.) The purpose of this eercise is to show that this result does not etend to functions of two variables. Show that f(,) = 4 e 4 e 4 has two relative maima but no other critical points (see Figure E-3). Source: This eercise is based on the problem Two Mountains Without a Valle proposed and solved b Ira Rosenholt, Mathematics Magaine, Vol. 6, No., Februar 987, p. 48. = 4 e 4 e 4 Figure E Find the absolute etrema of the given function on the indicated closed and bounded set R. 3. f(,) = 3; R is the triangular region with vertices (, ), (, 4), and (5, ).

82 3.8 Maima and Minima of Functions of Two Variables f(,) = ; R is the triangular region with vertices (, ), (, 4), and (4, ). 33. f(,) = 3 + 6; R is the region bounded b the square with vertices (, ), (, ), (, ), and (, ). 34. f(,) = e e ; R is the rectangular region with vertices (, ), (, ), (, ), and (, ). 35. f(,) = + ; R is the disk f(,) = ; R is the region that satisfies the inequalities,, and Find three positive numbers whose sum is 48 and such that their product is as large as possible. 38. Find three positive numbers whose sum is 7 and such that the sum of their squares is as small as possible. 39. Find all points on the portion of the plane + + = 5in the first octant at which f(,,) = has a maimum value. 4. Find the points on the surface = 5 that are closest to the origin. 4. Find the dimensions of the rectangular bo of maimum volume that can be inscribed in a sphere of radius a. 4. Find the maimum volume of a rectangular bo with three faces in the coordinate planes and a verte in the first octant on the plane + + =. 43. A closed rectangular bo with a volume of 6 ft 3 is made from two kinds of materials. The top and bottom are made of material costing per square foot and the sides from material costing 5 per square foot. Find the dimensions of the bo so that the cost of materials is minimied. 44. A manufacturer makes two models of an item, standard and delue. It costs $4 to manufacture the standard model and $6 for the delue. A market research firm estimates that if the standard model is priced at dollars and the delue at dollars, then the manufacturer will sell 5( ) of the standard items and 45, + 5( ) of the delue each ear. How should the items be priced to maimie the profit? 45. Consider the function f(,) = over the unit square,. (a) Find the maimum and minimum values of f on each edge of the square. (b) Find the maimum and minimum values of f on each diagonal of the square. (c) Find the maimum and minimum values of f on the entire square. 46. Show that among all parallelograms with perimeter l, a square with sides of length l/4 has maimum area. [Hint: The area of a parallelogram is given b the formula A = ab sin α, where a and b are the lengths of two adjacent sides and α is the angle between them.] 47. Determine the dimensions of a rectangular bo, open at the top, having volume V, and requiring the least amount of material for its construction. 48. A length of sheet metal 7 inches wide is to be made into a water trough b bending up two sides as shown in the accompaning figure. Find and φ so that the trapeoidshaped cross section has a maimum area. f 7 Figure E A common problem in eperimental work is to obtain a mathematical relationship = f() between two variables and b fitting a curve to points in the plane that correspond to eperimentall determined values of and, sa (, ), (, ),...,( n, n ) The curve = f()is called a mathematical model of the data. The general form of the function f is commonl determined b some underling phsical principle, but sometimes it is just determined b the pattern of the data. We are concerned with fitting a straight line = m + b to data. Usuall, the data will not lie on a line (possibl due to eperimental error or variations in eperimental conditions), so the problem is to find a line that fits the data best according to some criterion. One criterion for selecting the line of best fit is to choose m and b to minimie the function n g(m, b) = (m i + b i ) i= This is called the method of least squares, and the resulting line is called the regression line or the least squares line of best fit. Geometricall, m i + b i is the vertical distance between the data point ( i, i ) and the line = m + b. = m + b m i + b i ( i, i ) i These vertical distances are called the residuals of the data points, so the effect of minimiing g(m, b) is to minimie the sum of the squares of the residuals. In these eercises, we will derive a formula for the regression line. 49. The purpose of this eercise is to find the values of m and b that produce the regression line. (a) To minimie g(m, b), we start b finding values of m and b such that g/ m = and g/ b =. Show

83 988 Chapter3 / Partial Derivatives that these equations are satisfied if m and b satisf the conditions ( n ) ( n ) n m + i b = i i i= i i= ( n ) i m + nb = i= (b) Let = ( n )/n denote the arithmetic average of,,..., n. Use the fact that n ( i ) to show that n ( n i= i= i n i= i i= ) ( n ) i with equalit if and onl if all the i s are the same. (c) Assuming that not all the i s are the same, prove that the equations in part (a) have the unique solution n n n n i i m = i= n i= i= i i= ( n n ) i i i= i= i= ( n ) b = n i m i n [Note: We have shown that g has a critical point at these values of m and b. In the net eercise we will show that g has an absolute minimum at this critical point. Accepting this to be so, we have shown that the line = m + b is the regression line for these values of m and b.] 5. Assume that not all the i s are the same, so that g(m, b) has a unique critical point at the values of m and b obtained in Eercise 49(c). The purpose of this eercise is to show that g has an absolute minimum value at this point. (a) Find the partial derivatives g mm (m, b), g bb (m, b), and g mb (m, b), and then appl the second partials test to show that g has a relative minimum at the critical point obtained in Eercise 49. (b) Show that the graph of the equation = g(m, b) is a quadric surface. [Hint: See Formula (4) of Section.7.] (c) It can be proved that the graph of = g(m, b) is an elliptic paraboloid. Accepting this to be so, show that this paraboloid opens in the positive -direction, and eplain how this shows that g has an absolute minimum at the critical point obtained in Eercise Use the formulas obtained in Eercise 49 to find and draw the regression line. If ou have a calculating utilit that can calculate regression lines, use it to check our work. i= i (, ) (, ) (3, 4) (, 3) (, ) (, ) 3 4 (4, ) 55. The following table shows the life epectanc b ear of birth of females in the United States: ear of birth life epectanc (a) Take t = to be the ear 93, and let be the life epectanc for birth ear t. Use the regression capabilit of a calculating utilit to find the regression line of as a function of t. (b) Use a graphing utilit to make a graph that shows the data points and the regression line. (c) Use the regression line to make a conjecture about the life epectanc of females born in the ear. 56. A compan manager wants to establish a relationship between the sales of a certain product and the price. The compan research department provides the following data: price () in dollars $35. $4. $45. $48. $5. dail sales volume () in units (a) Use a calculating utilit to find the regression line of as a function of. (b) Use a graphing utilit to make a graph that shows the data points and the regression line. (c) Use the regression line to make a conjecture about the number of units that would be sold at a price of $ If a gas is cooled with its volume held constant, then it follows from the ideal gas law in phsics that its pressure drops proportionall to the drop in temperature. The temperature that, in theor, corresponds to a pressure of ero is called absolute ero. Suppose that an eperiment produces the following data for pressure P versus temperature T with the volume held constant: (cont.)

84 3.9 Lagrange Multipliers 989 P (kilopascals) T ( celsius) (a) Use a calculating utilit to find the regression line of P as a function of T. (b) Use a graphing utilit to make a graph that shows the data points and the regression line. (c) Use the regression line to estimate the value of absolute ero in degrees Celsius. 58. Find (a) a continuous function f(,) that is defined on the entire -plane and has no absolute etrema on the -plane; (b) a function f(,) that is defined everwhere on the rectangle, and has no absolute etrema on the rectangle. 59. Show that if f has a relative maimum at (, ), then G() = f(, ) has a relative maimum at = and H() = f(,)has a relative maimum at =. 6. Writing Eplain how to determine the location of relative etrema or saddle points of f(,) b eamining the contours of f. 6. Writing Suppose that the second partials test gives no information about a certain critical point (, ) because D(, ) =. Discuss what other steps ou might take to determine whether there is a relative etremum at that critical point. QUICK CHECK ANSWERS 3.8. (, ) and ( 6, ). (a) no information (b) a saddle point at (, ) (c) a relative minimum at (, ) (d) a relative maimum at (, ) 3. (a) a saddle point at (, ) (b) no information, since (, ) is not a critical point ( ) (c) a relative minimum at (, ) 4. V = LAGRANGE MULTIPLIERS In this section we will stud a powerful new method for maimiing or minimiing a function subject to constraints on the variables. This method will help us to solve certain optimiation problems that are difficult or impossible to solve using the methods studied in the last section. EXTREMUM PROBLEMS WITH CONSTRAINTS In Eample 6 of the last section, we solved the problem of minimiing subject to the constraint This is a special case of the following general problem: S = + + () 3 = () 3.9. Three-Variable Etremum Problem with One Constraint Maimie or minimie the function f(,,) subject to the constraint g(,,) =. We will also be interested in the following two-variable version of this problem: 3.9. Two-Variable Etremum Problem with One Constraint Maimie or minimie the function f(,) subject to the constraint g(,) =.

85 99 Chapter3 / Partial Derivatives g(, ) = Maimum of f (, ) is g(, ) = (a) Minimum of f (, ) is Figure 3.9. (b) (, ) LAGRANGE MULTIPLIERS One wa to attack problems of these tpes is to solve the constraint equation for one of the variables in terms of the others and substitute the result into f. This produces a new function of one or two variables that incorporates the constraint and can be maimied or minimied b appling standard methods. For eample, to solve the problem in Eample 6 of the last section we substituted () into () to obtain S = which we then minimied b finding the critical points and appling the second partials test. However, this approach hinges on our abilit to solve the constraint equation for one of the variables in terms of the others. If this cannot be done, then other methods must be used. One such method, called the method of Lagrange multipliers, will be discussed in this section. To motivate the method of Lagrange multipliers, suppose that we are tring to maimie a function f(,) subject to the constraint g(,) =. Geometricall, this means that we are looking for a point (, ) on the graph of the constraint curve at which f(,) is as large as possible. To help locate such a point, let us construct a contour plot of f(,) in the same coordinate sstem as the graph of g(,) =. For eample, Figure 3.9.a shows some tpical level curves of f(,) = c, which we have labeled c =,, 3, 4, and 5 for purposes of illustration. In this figure, each point of intersection of g(,) = with a level curve is a candidate for a solution, since these points lie on the constraint curve. Among the seven such intersections shown in the figure, the maimum value of f(,) occurs at the intersection (, ) where f(,) has a value of 4. Note that at (, ) the constraint curve and the level curve just touch and thus have a common tangent line at this point. Since f(, ) is normal to the level curve f(,) = 4 at (, ), and since g(, ) is normal to the constraint curve g(,) = at(, ), we conclude that the vectors f(, ) and g(, ) must be parallel. That is, f(, ) = λ g(, ) (3) for some scalar λ. The same condition holds at points on the constraint curve where f(,) has a minimum. For eample, if the level curves are as shown in Figure 3.9.b, then the minimum value of f(,) occurs where the constraint curve just touches a level curve. Joseph Louis Lagrange (736 83) French Italian mathematician and astronomer. Lagrange, the son of a public official, was born in Turin, Ital. (Baptismal records list his name as Giuseppe Lodovico Lagrangia.) Although his father wanted him to be a lawer, Lagrange was attracted to mathematics and astronom after reading a memoir b the astronomer Halle. At age 6 he began to stud mathematics on his own and b age 9 was appointed to a professorship at the Roal Artiller School in Turin. The following ear Lagrange sent Euler solutions to some famous problems using new methods that eventuall blossomed into a branch of mathematics called calculus of variations. These methods and Lagrange s applications of them to problems in celestial mechanics were so monumental that b age 5 he was regarded b man of his contemporaries as the greatest living mathematician. In 776, on the recommendations of Euler, he was chosen to succeed Euler as the director of the Berlin Academ. During his sta in Berlin, Lagrange distinguished himself not onl in celestial mechanics, but also in algebraic equations and the theor of numbers. After twent ears in Berlin, he moved to Paris at the invitation of Louis XVI. He was given apartments in the Louvre and treated with great honor, even during the revolution. Napoleon was a great admirer of Lagrange and showered him with honors count, senator, and Legion of Honor. The ears Lagrange spent in Paris were devoted primaril to didactic treatises summariing his mathematical conceptions. One of Lagrange s most famous works is a memoir, Mécanique Analtique, in which he reduced the theor of mechanics to a few general formulas from which all other necessar equations could be derived. It is an interesting historical fact that Lagrange s father speculated unsuccessfull in several financial ventures, so his famil was forced to live quite modestl. Lagrange himself stated that if his famil had mone, he would not have made mathematics his vocation. In spite of his fame, Lagrange was alwas a sh and modest man. On his death, he was buried with honor in the Pantheon.