1 Functions of Several Variables Some Examples Level Curves / Contours Functions of More Variables... 6

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1 Contents 1 Functions of Several Variables Some Examples Level Curves / Contours Functions of More Variables Partial Derivatives First Derivatives Second Derivatives Partial Differential Equations Tangent Planes and Linear Approximations Equation of a Tangent Plane Differentials The Chain Rule Multivariable Chain Rules Implicit Differentiation Directional Derivatives and the Gradient Vector The Directional Derivative Properties of the Gradient Vector Tangent Planes of Level Surfaces Maximum Rate of Change Minimum and Maximum Values Local Extrema and Critical Points Absolute Minimum and Maximum Values Functions of Several Variables In Calculus I and II, you learned how to differentiate and integrate functions of a single variable of the form y = f(x). This type of function is represented geometrically as a curve in the xy plane where the x axis is the domain of the function (the set of input values) and the y axis is the range (the set of output values) of the function f. We now want to extend this idea to a multivariable function, i.e., functions that have more than one variable. A function of two variables has the general form z = f(x, y) and would appear geometrically as a surface in xyz space. The input to this function is now an ordered pair of real numbers (x, y) and the output is a real number z. For any point in the xy plane, the function would produce a z value via the relation z = f(x, y). By varying x and y over all possible values, this would produce a surface in xyz space, which is the set of all points (x, y, z) such that z = f(x, y). For functions of a single variable, the domain is the set of all x values for which the function is defined and is represented geometrically as an interval of the x axis. The domain of a two variable function is the set of all ordered pairs (x, y) for which the function is defined and is represented geometrically as a region in the xy plane. For many functions, the domain is the whole xy plane but there are certain functions such as logarithms, square roots and rational functions that may restrict the domain to some subset of the xy plane.

2 1.1 Some Examples Example 1.1. Consider the function f(x, y) = x 2 + y 2. The surface created by this function is called a paraboloid. This surface can obtained by drawing the parabola z = x 2 in the xz plane and then revolving it around the z axis. Its vertex is at the origin and it opens upward, forming a bowl shape. The domain of this function is all of the xy plane, since it is defined for all values for x and y. Figure 1: A surface plot of the paraboloid z = x 2 + y 2. The surface is colored according to its z values where blue is the lowest z value and red is the highest z value. Example 1.2. Consider the function f(x, y) = x 2 y 2. The surface created by this function is called a hyperboloid and is the 3D analog of a hyperbola. The domain of this function is all of the xy plane, since it is defined for all values for x and y. Figure 2: A surface plot of the hyperboloid z = x 2 y 2.

3 Example 1.3. Consider the function f(x, y) = x 2 + y 2. The surface created by this function is a cone. The domain of this function is all of the xy plane, since it is defined for all values for x and y. Figure 3: A surface plot of the cone z = x 2 + y 2. Example 1.4. Find the domains of the following functions: a) f(x, y) = 2 x y b) f(x, y) = ln(x 2 + y 2 1) c) f(x, y) = 1/ xy Solution: a) Since this function involves a square root, it is limited to values of x and y that will make the argument of the square root non-negative, i.e., we must have 2 x y 0 y 2 x. The domain D can then be stated in set notation as D = {(x, y) y 2 x}. Geometrically, the domain is a portion of the xy that is either above or below the line y = 2 x (include the line itself). To determine which side of the line the inequality is satisfied, we can just pick any test point and if the inequality is satisfied for that point, then we know to choose the side that contains that point. If we plug in (0, 0) as our test point, we get 0 2, which is true so the side of the line containing (0, 0) is the domain. b) Since this function involves a logarithm, it is limited to values of x and y that will make the argument of the logarithm strictly positive, i.e., we must have x 2 + y 2 1 > 0 x 2 + y 2 > 1.

4 The domain D can then be stated in set notation as D = {(x, y) x 2 + y 2 > 1}. Geometrically, the domain is the region outside of the unit circle, not including the circle itself. c) There now two things restricting the domain of the function: the argument of the square root must be non-negative and the denominator can not be zero. Therefore the product xy must be strictly positive, i.e., xy > 0 x, y > 0 or x, y < 0. The domain D can then be stated in set notation as D = {(x, y) x, y > 0 or x, y < 0}. Geometrically, the domain is the first and third quadrants of the xy plane but does not include the x or y axes, since this would result in division by zero. 1.2 Level Curves / Contours The level curves, or contours of a function of two variables are the curves in the xy plane where the function z = f(x, y) remains at a constant height. The contour corresponding to the elevation c is the implicit curve f(x, y) = c. For example, suppose a the height z of a mountain above the point (x, y) is represented by the surface z = f(x, y). Then the contours of this function would create an elevation map of the terrain. If your were to walk along any of these curves, you would be following a path of constant elevation, i.e., you would not be moving up or down as you walked the path. Note that contours can never cross or touch each other, since this would imply that the function is double-valued at the intersection point. Regions where the contours are close together indicate that the surface is very steep there since the elevation changes rapidly due to small movements in the xy plane. Figure 4: An elevation map (contour plot) of Camelback Mountain. Each line shows a path of constant elevation. The peak of the mountain is marked with a triangle. Notice that the peak is surrounded by closed contours that get smaller as they get closer to the peak. Example 1.5. The contours of the paraboloid z = x 2 + y 2 are circles. If we set z = c, we get the equation for the contour at height c s z = x 2 + y 2 = c,

5 Figure 5: The surface and contour plots of the paraboloid z = x 2 + y 2. which is a circle with radius r = c. All of the contours are thus concentric circles around the origin but they are not evenly spaced. The paraboloid gets steeper as you move away from the origin so the contours get closer together. Example 1.6. The contours of the cone z = x 2 + y 2 are also circles. If we set z = c, we get the equation for the contour at height c s z = x 2 + y 2 = c x 2 + y 2 = c 2, which is a circle with radius r = c. All of the contours are thus concentric circles around the origin that are evenly spaced. The steepness of the cone never changes so the contours always remain a fixed distance from each other. Figure 6: The surface and contour plots of the cone z = x 2 + y 2.

6 Example 1.7. The contours of the hyperboloid z = x 2 y 2 are hyperbolas. If we set z = c, we get the equation for the contour at height c s z = x 2 y 2 = c, which is the equation of a hyperbola centered at the origin. Figure 7: The surface and contour plots of the hyperboloid z = x 2 y Functions of More Variables Until now, we have used the specific case of two-variable functions to demonstrate the idea of a multivariable function. However, a function can have any number of variables and the concepts we discuss in this class can usually be extended to deal with functions of any number of variables. A function of n variables has the form f(x) = f(x 1, x 2,..., x n ) where x = x 1, x 2,..., x n is an n-dimensional vector of variables and f(x) is a scalar-valued quantity. Although our geometric intuition is restricted to three dimensions or less, there are many situations where a function may have a large number of variables. For example, consider if you wanted to model a company s profit with a multivariable function. Each component of the vector of variables would represent some quantity or aspect of the company that is within their ability to change, which may number in the hundreds or thousands. Example 1.8. Consider the function f(x) = x for x R n. If n = 1, then we get f(x 1 ) = x 1. For n = 2, we would get f(x 1, x 2 ) = x x 2 2, which is a cone. For n > 2, we would call this a hypercone. This tells us that the absolute value function is really the 1D version of a cone. Example 1.9. Consider the function f(x) = c x for c, x R n. If n = 1, then we get f(x 1 ) = c 1 x 1, which represents a line through the origin with slope c 1. For n = 2, we would get f(x 1, x 2 ) = c 1 x 1 + c 2 x 2, which is a plane that passes through the origin. For n > 2, we would call this a hyperplane. This demonstrates that a line is really the 1D version of a plane.

7 Functions of three variables are still somewhat in our ability to grasp geometrically. Although we would need 4 dimensions to view such a function, we can still make sense of it through its level surfaces, or isosurfaces. A level surface of a function of three variables f(x, y, z) is a surface in xyz space where the function remains constant. Example Suppose the temperature at any point in a room is given by the function T (x, y, z) = 1 x 2 + y 2 + z The level surfaces of this function would represent the surfaces along which the temperature in the room is at the same value (isotherms). For example, the set of all points where T = 0.1 would be T (x, y, z) = 1 x 2 + y 2 + z = 1 10 x2 + y 2 + z = 10 x 2 + y 2 + z 2 = 9, which is a sphere of radius 3 centered at the origin. Any particle traveling on the surface of this sphere would always stay at the temperature T = 0.1. The set of all level surfaces for this function would be concentric spheres that get further apart from each other as their radius increases. The general level surfaces where T = c are given by 1 x 2 + y 2 + z = c x2 + y 2 + z = 1 c x2 + y 2 + z 2 = 1 c 1, 1 which is a sphere with radius r = 1. Note that this function always has values between c 0 and 1, so these are the only valid values for c.

8 2 Partial Derivatives 2.1 First Derivatives Consider a function of two variables, z = f(x, y). We can extend the idea of a derivative to a multivariable function with the concept of a partial derivative. A partial derivative of f is an ordinary derivative with respect to one of the variables while all other variables are treated as constants. For a function of two variables, there are two possible partial derivatives: x and y. Note that this is similar to the differential notation df from single-variable calculus, but we dx use the symbol instead of d to emphasize that it is a partial derivative, as opposed to an ordinary derivative. Another common notation for partial derivatives is to just write the function symbol f and use subscripts of x or y to denote partial differentiation, such as f x = x, f y = y. Partial derivatives are also functions of x and y and when we evaluate a partial derivative at the point (x, y), we would write this as f x (x, y) or f y (x, y). The technical definition of a partial derivative is very similar to the definition of an ordinary derivative: x = lim h 0 f(x + h, y) f(x, y), h y = lim f(x, y + h) f(x, y) h 0 h Note that for each partial derivative, we apply the definition of an ordinary derivative to one variable and hold the other variable constant. Partial derivatives can be interpreted geometrically as the slope of a surface z = f(x, y) in the directions corresponding to the x or y axes. For example, imagine standing on the surface above some point (x, y). There are infinitely many directions in which you could step, and the surface may have a different slope in each direction. The partial derivative f x (x, y) is the slope of the surface if you were to step in the direction of the +x axis, and f y (x, y) is the slope in the direction of the +y axis. As we ll see later in the course, partial derivatives are special cases of a more general concept, called directional derivatives. Partial derivatives can be defined for functions with any number of variables by differentiating with respect to the indicated variable, while treating all other variables as constants. A function with n variables will have n partial derivatives since we can differentiate with respect to each of the n variables. Example 2.1. Compute the partial derivatives of the function f(x, y) = x/ y at the point (3, 1). Solution: First we write the function as f(x, y) = xy 1/2. To compute the partial derivative with respect to x, we treat y as a constant and get Plugging in the point (3, 1) gives us x = y 1/2. x (3, 1) = 1 1/2 = 1.

9 Likewise, for the partial derivative with respect to y, we treat x as a constant and get Plugging in the point (3, 1) gives us y = 1 2 xy 3/2. y (3, 1) = /2 = 3 2. Example 2.2. Compute the partial derivatives of the function f(x, y) = x 2 y 3 sin(2y)+y x. Solution: The partial derivative with respect to x is x = 2xy3 sin(2y) 1. For the partial derivative with respect to y, we have to use the product rule: y = 3x2 y 2 sin(2y) + 2x 2 y 3 cos(2y) + 1. Example 2.3. Compute the partial derivatives of the function f(x, y) = xye x2 y 2. Solution: Using the product rule, the partial derivative with respect to x is y2 = ye x2 2x 2 ye x2 y 2 = (1 2x 2 )ye x2 y 2. x Likewise, the partial derivative with respect to y is y = xe x2 y2 2xy 2 e x2 y 2 = (1 2y 2 )xe x2 y 2. Example 2.4. Compute the partial derivatives of the function f(x, y) = (x 2 + 3y 2 ) 1. Solution: The partial derivative with respect to x is x = (x2 + 3y 2 ) 2 (2x) = 2x(x 2 + 3y 2 ) 2. Likewise, the partial derivative with respect to y is y = 2(x2 + 3y 2 ) 2 (6y) = 12y(x 2 + 3y 2 ) 2. Example 2.5. Compute the partial derivatives of the function f(x, y, z) = z 2 e xz/y. Solution: This is a function of 3 variables so we need to compute f x, f y and f z. The partial derivative with respect to x is ( ) z x = z2 e xz/y = z3 e xz/y. y y The partial derivative with respect to y is y = z2 e xz/y The partial derivative with respect to z is y = 2zexz/y + z 2 e xz/y ( xz y 2 ( ) x = y ) = xz3 e xz/y y 2. ) (2z + xz2 e xz/y. y

10 Example 2.6. Use implicit differentiation to compute x x y2 9 + z2 = 1. and y Solution: Differentiating the equation with respect to x gives us 1 x + 2z 2 x = 0. Solving for gives us x x = x 4z. Differentiating the equation with respect to y gives us 2 y + 2z 9 y = 0. for the ellipsoid defined by Solving for y gives us y = y 9z. 2.2 Second Derivatives We can also define second partial derivatives, although there become more possibilities. For a function of two variables, there are four possible second partial derivatives: f xx = ( ), f yy = ( ) x x y y f xy = ( ), f yx = ( ). x y y x These can also be stated using the notation, although it is a bit clumsy: f xx = 2 f x 2, f xy = 2 f x y, f yx = 2 f y x, f yy = 2 f y 2. In general, for a function of n variables, there will be n 2 possible second derivatives. These can be arranged into an n n matrix called the Hessian matrix of f, which we will learn more about later. One important property of second partial derivatives is given by Clairaut s Theorem, which says that partial derivatives can be applied in any order. In other words, all mixed partial derivatives are equal, i.e., f xy = f yx. This also applies for higher derivatives, for example, f xxy = f xyx = f yxx. All that matters is the number of times each variable is differentiated, not the order in which the differentiation is done. Example 2.7. Compute all second partial derivatives of the function f(x, y) = x 3 y 2 e 2x and verify that f xy = f yx. Solution: The first partial derivatives are f x = 3x 2 y 2 e 2x + x 3 y 2 (2e 2x ) = (3x 2 y 2 + 2x 3 y 2 )e 2x,

11 The second partial derivatives are f y = 2x 3 ye 2x. f xx = (6xy 2 + 6x 2 y 2 )e 2x + (3x 2 y 2 + 2x 3 y 2 )(2e 2x ) = (6xy x 2 y 2 + 4x 3 y 2 )e 2x, f yy = 2x 3 e 2x, f xy = (6x 2 y + 4x 3 y)e 2x, f yx = 2y(3x 2 e 2x + x 3 (2e 2x )) = (6x 2 y + 4x 3 y)e 2x. 2.3 Partial Differential Equations An equation that involves a multi-variable function and some of its partial derivatives is called a partial differential equation (PDE). These equations are of utmost importance in mathematical applications since most physical laws take the form of a PDE. The goal in solving a PDE is to find a function that satisfies the relations of its partial derivatives specified by the PDE. However, PDEs allow for infinitely many solutions, and certain extra information, called initial and/or boundary conditions, must be known in order to determine a unique solution. Example 2.8. The heat equation describes how heat spreads through a conducting medium. Let u(x, t) represent the temperature of a metal bar at the point x and at time t. The heat equation is u t = Du xx where D is the thermal conductivity of the bar, which is an experimentally determined parameter that quantifies how well the bar conducts heat. Example 2.9. The wave equation describes the motion of a wave through a medium and it applies to all different kinds of waves such as water waves, sound waves, light waves, etc. Imagine a guitar string that is fastened down at both ends. If the string is plucked, it creates a wave that propagates along the string, causing it to vibrate. Let u(x, t) represent the height of the string at the point x and at time t. The wave equation is u tt = c 2 u xx where c is the speed at which the wave propagates, which depends on properties of the medium of propagation. Example Show that u(x, t) = e Dt sin(x) satisfies the heat equation. Solution: We just need to verify that u t = Du xx. Calculating these partial derivatives gives us u t = De Dt sin(x), u x = e Dt cos(x), u xx = e Dt sin(x). Now compare u t and u xx and notice that u xx = ut D u t = Du xx. Note that this is certainly not the only solution to the heat equation. A unique solution would require us to know the solution everywhere in space at one particular time (initial condition) and at two particular points in space for all time (boundary conditions). For example, if we had the initial condition u(x, 0) = sin(x) and the boundary conditions u(0, t) = 0 and u(π, t) = 0, then the solution given above is the only solution.

12 3 Tangent Planes and Linear Approximations A function f of a single variable can be visualized as a curve in the xy plane where y = f(x). At each point on the curve, we can compute the tangent line with the point-slope formula y = y 0 + m(x x 0 ) where (x 0, y 0 ) is a point on the line and m is the slope. At the point (x 0, f(x 0 )), the slope of the tangent line is just f (x 0 ). Plugging these values into the point-slope formula gives us the equation of the tangent line y = f(x 0 ) + f (x 0 )(x x 0 ) L(x) x0 This formula is also referred to as the linearization of f about the point x 0. The function L(x) represents the linear function that is the best linear approximation of f for values of x near x 0. Linearizations are an important tool in mathematics because they allow us to approximate a nonlinear problem as a linear one, so long as the approximation is only used in a small interval around x 0. The accuracy of the linearization of a function around x 0 depends on the curvature at x 0 : the smaller the curvature, the better the approximation. 3.1 Equation of a Tangent Plane Functions of two variables are represented as surfaces in R 3 consisting of all points (x, y, z) such that z = f(x, y). Let P (x 0, y 0, z 0 ) be a point on this surface. There are actually infinitely many tangent lines to the surface at the point because there are infinitely many directions by which we can approach the point P. However, all of the possible tangent lines lie in the same plane, which is defined as the tangent plane of the surface at the point P. Just like the tangent line represents the linearization of a single variable function, the tangent plane represents the linearization for a two variable function. The equation for the tangent plane at the point (x 0, y 0, z 0 ) will have the general form A(x x 0 ) + B(y y 0 ) + C(z z 0 ) = 0. Dividing the equation through by C and defining a = A/C and b = B/C and solving for z gives us z = z 0 + a(x x 0 ) + b(y y 0 ). To find a and b, we make the following observations. Along the plane y = y 0, the tangent plane equation reduces to z = z 0 + a(x x0) which is actually the linearization of the single variable function g(x) = f(x, y 0 ). Comparing this with the single variable linearization, we see that a = g (x 0 ) = f x (x 0, y 0 ). Similarly, along the plane x = x 0, the tangent plane equation reduces to z = z 0 + b(y y0) which is the linearization of the single variable function h(y) = f(x 0, y) which tells us that b = h (y 0 ) = f y (x 0, y 0 ). Therefore the equation of the tangent plane at the point (x 0, y 0, z 0 ) is z = z 0 + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 )

13 which is the linearization of f at the point (x 0, y 0, z 0 ). The linearization is a linear function L(x, y) that represents the best linear approximation to the f at the point (x 0, y 0, f(x 0, y 0 )) and is denoted as L(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) (x0,y 0 ) Example 3.1. Find the equation of the tangent plane of the function f(x, y) = 3x 2 +y 2 2xy at the point (3, 1). Solution: Using (x 0, y 0 ) = (3, 1), the equation of the tangent plane will be The partial derivatives of f are Evaluating these at (3, 1) gives z = f(3, 1) + f x (3, 1)(x 3) + f y (3, 1)(y + 1). f x = 6x 2y, f y = 2y 2x. f x (3, 1) = 6(3) 2( 1) = 20, f y (3, 1) = 2( 1) 2(3) = 8. Evaluating f at (3, 1) gives us f(3, 1) = 3(3) 2 + ( 1) 2 2(3)( 1) = = 34. Plugging in all of these values into the tangent plane formula gives us z = (x 3) 8(y + 1). In other words, the linearization of f at (3, 1) is defined by the linear function L(x, y) = (x 3) 8(y + 1). (3, 1) 3.2 Differentials For a single variables function y = f(x), the differential of the function is defined as dy = f (x)dx. Intuitively, we think of dy as the change in the height of the function between x and x + dx. For a function of two variables z = f(x, y), the differential is defined as dz = f x (x, y)dx + f y (x, y)dy = dx + x y dy which represents the change in the height of the function when moving from the point (x, y) to the point (x + dx, y + dy). We can use this idea to approximate how much a function changes between the points (x, y) and (x + x, y + y) with the formula z = f(x + x, y + y) f(x, y) x + x y y

14 where x and y are small increments. This approximation is equivalent to using the linearization of f at the point (x, y) to approximate f(x + x, y + y) and is thus only a good approximation when x and y are small in magnitude. The concept of a differential can be easily extended for functions with nay number of variables. For a function of n variables of the form f(x 1, x 2,..., x n ), the differential is given by df = x 1 dx 1 + x 2 dx x n dx n = n k=1 x k dx k. Example 3.2. Suppose that a function has the properties f(1, 2) = 3, f x (1, 2) = 1 and f y (1, 2) = 6. Approximate the value of f(0.9, 2.2). Solution: Since x goes from 1 to 0.9, x = = 0.1. Likewise, y = = 0.2. Then the change in f can approximation as f = f(0.9, 2.2) f(1, 2) f x (1, 2) x + f y (1, 2) y = ( 1)( 0.1) + (6)(0.2) = 1.3. Therefore f(0.9, 2.2) f(1, 2) + f = = 4.3. Example 3.3. A cylinder has a radius r = 5 and a height h = 2. Approximate the change in the volume of the cylinder if the radius is increased by 0.1 and the height is decreased by Solution: The volume of a cylinder is a function of its radius r and height h given by V (r, h) = πr 2 h. The exact change in volume can be calculated as V = V ( , ) V (5, 2) Now we want to compute the approximate differential of this function with the formula V V V (r, h) r + (r, h) h r h where r = 5, h = 2 and r = 0.1 is the change in the radius and h = 0.05 is the change in height. Note that h is negative since the height is decreased. The partial derivatives of the volume function are V r = 2πrh, V h = πr2. Evaluating these for the given radius and height gives V (5, 2) = 2π(5)(2) = 20π, r V h (5, 2) = π(5)2 = 25π. The change in V is then approximated as V (20π)(0.1) + (25π)( 0.05) = (2 1.25)π = 0.75π Since the answer is positive, it means that the overall volume was increased.

15 Example 3.4. Approximate the change in the surface area S of a rectangular box if the dimesions are perturbed by 0.05, 0.06 and Solution: The surface area of the box is the sum of the surface areas for each face, which is S(x, y, z) = 2xy + 2yz + 2xz. We want to compute the approximate differential of this function given by S S x (x, y, z) x + S y S (x, y, z) y + (x, y, z) z where x = 1, y = 2, z = 4, x = 0.05, y = 0.06 and z = The partial derivatives of the surface area function are S x = 2y + 2z, S y = 2x + 2z, S Evaluating these for the given dimensions of the box gives us S (1, 2, 4) = 2(2 + 4) = 12, x S (1, 2, 4) = 2(1 + 4) = 10, y S (1, 2, 4) = 2(1 + 2) = 6. The change in S is then approximated as = 2x + 2y. S 12(0.05) + 10( 0.06) + 6(0.01) = Since the answer is positive, it means that the overall surface area was increased.

16 4 The Chain Rule 4.1 Multivariable Chain Rules Recall from single variable calculus that if y = f(x) and x = g(t), then we can compute the derivative of y with respect to t using the standard chain rule: dy dt = dy dx dx dt. We now want to extend the chain rule for multi-variable functions. We will start with the simplest case. Let z = f(x, y) be a two variable function where x = g(t) and y = h(t) are functions of the variable t. This means that ultimately z is just a function of t and we would like to know what the rate of change of z with respect to t, i.e., we want to compute dz/dt. From the last section, we know that the differential of z is given by dz = dx + x y dy. Dividing this equation through by dt gives us the chain rule for this case: dz dt = dx x dt + dy y dt. We can interpret dz/dt geometrically as follows. The functions x(t) and y(t) describe some parameterized curve in the xy plane. Evaluating z = f(x, y) at every point along this curve produces the space curve r(t) = x(t), y(t), z(t) that lies entirely on the surface z = f(x, y). Then dz/dt tells us how fast this curve is rising along the z axis as we move along the curve. In other words, it gives us the z-component of the velocity vector for r(t). Example 4.1. The height of Mount Doom is given by the function z = 10 2x 2 y 2. If Gollum is skulking along a path that is always directly above the point (3t, sin t) at time t, then at what rate is his elevation changing at t = π/3? Solution: We need to evaluate dz/dt at t = π/3. By the chain rule, The partial derivatives of z are The derivatives of x(t) and y(t) are dz dt = dx x dt + dy y dt. x = 4x, dx dt = 3, dy dt Plugging these into the chain rule formula gives us dz dt y = 2y. = cos t. = ( 4x)(3) + ( 2y)(cos t) = 12x 2y cos t. At t = π/3, x = 3(π/3) = π and y = sin(π/3) = 3/2. We then get dz 3 3 dt = 12π 2 cos(π/3) = 12π t=π/

17 Example 4.2. Suppose a cone has a radius r(t) that is increasing by 0.1 cm/s and a height h(t) that is decreasing at 0.2 cm/s. Find the rate at which the volume is changing when r = 2 and h = 6. Solution: The volume of a cone is a function of its radius and height: V (r, h) = 1 3 πr2 h. Then differentiating V with respect to t using the chain rule, we get The partial derivatives of V are V t = V r dr dt + V dh h ht. V r = 2π 3 rh, Evaluating these when r = 2 and h = 6, we get V V (2, 6) = 8π, r Based on the rates we are given, we know that dr dt = 0.1, V h = 1 3 πr2. h (2, 6) = 4π 3. dh dt = 0.2. Plugging these values into the chain rule formula, we get V t (2, 6) = (8π)(0.1) + 4π 3 (0.2) = 16π Now we consider a more complicated case. Suppose that we again have a two variable function z = f(x, y) but now x and y are also functions of two variables s and t, i.e., x = g(s, t) and y = h(s, t). This means that z is ultimately a function of s and t and we would like to know how to compute / s and / t. For this case, the chain rule has the form s = x x s + y y s, t = x x t + y y t. Example 4.3. Let z = x 2 y 5y 2 where x(s, t) = s 2 2t and y(s, t) = se 2t. Compute / s and / t when s = 1 and t = 0. Solution: The partial derivatives of z with respect to x and y are x = 2xy, y = x2 10y. The partial derivatives of x and y with respect to s and t are Using the chain rule, x s = 2s, x t = 2, y s = e2t y t = 2se2t s = x x s + y y s = (2xy)(2s) + (x2 10y)(e 2t ).

18 Evaluating x and y when s = 1 and t = 0 gives x(1, 0) = 1 and y(1, 0) = 1 so (1, 0) = 4 + (1 10)(1) = 4 9 = 5. s Now we use the chain rule again to compute t = x x t + y y t = (2xy)( 2) + (x2 10y)(2se 2t ). Evaluating this when s = 1 and t = 0 gives (1, 0) = 4 + (1 10)(2) = 4 18 = 22. t The chain rule can be generalized to work for functions with any number of variables. Suppose f(x 1, x 2,..., x n ) is a function of n variables where each x i is a function of m variables x i (t 1, t 2,..., t m ). Then the derivative of f with respect to t i = x 1 + x x n n x j = t i x 1 t i x 2 t i x n t i x j t i for each i = 1, 2,..., m. Example 4.4. Let where w(x, y, z) = xy + yz + xz j=1 x(u, v) = u cos(v), y(u, v) = u sin(v), z(u, v) = uv. Compute w/ u and w/ v when u = 2 and v = π/2. Solution: Using the chain rule, we get w u = w x x u + w y y u + w u. The partial derivatives of w with respect to x, y and z are w x = y + z, w y = x + z, w = x + y. The partial derivatives of x, y and z with respect to u are x u = cos(v), y u = sin(v), u = v. Plugging this into the chain rule formula gives w = (y + z) cos(v) + (x + z) sin(v) + (x + y)v. u When u = 2 and v = π/2, we get x = 0, y = 2 and z = π. Plugging in the values gives us w (2, π/2) = (2 + π) cos(π/2) + (0 + π) sin(π/2) + (0 + 2)π/2 = 0 + π + π = 2π. u The partial derivatives of x, y and z with respect to v are x y = u sin(v), v v = u cos(v), v = u. Plugging this into the chain rule formula gives w = (y + z)u sin(v) + (x + z)u cos(v) + (x + y)u. u When u = 2 and v = π/2, we get x = 0, y = 2 and z = π. Plugging in the values gives us w (2, π/2) = (2 + π)( 2) + (0 + π) 0 + (2)(2) = 4 2π + 4 = 2π. v

19 4.2 Implicit Differentiation The chain rule can be used to relate partial derivatives of a two-variable function F (x, y) to the implicit derivatives of its contours. Consider the in the xy plane F (x, y) = C. If we treat y as an implicit function of x, then we can write the equation of the curve as F (x, y(x)) = C. Now if we differentiate this expression with respect to x using the chain rule, we get Solving for dy/dx gives us F x + F dy y dx = 0. F dy dx = x F y = F x F y. This gives us a convenient way of computing implicit derivatives of plane curves using partial derivatives of a multi-variable function. Notice that the above results is independent of the value of C. This is because this ratio of partial derivatives actually contains the information about the slopes for every contour of F. To make this more clear, consider a point (x 0, y 0 ). There is only one contour of F that runs through this point and the slope of the this contour at the point (x 0, y 0 ) is dy = F x(x 0, y 0 ) dx F y (x 0, y 0 ). (x0,y 0 ) Example 4.5. Use( implicit differentiation to find the equation of the tangent line of the unit 1 circle at the point, ) Solution: The unit circle has the equation x 2 + y 2 = 1 and is therefore a contour of the two variable function F (x, y) = x 2 + y 2. Then the slope of any contour is dy dx = F x = 2x F y 2y = x y. ( 1 Evaluating this at the point (x 0, y 0 ) =, ) 3, we get the slope of the tangent line is 2 2 dy dx = (x0,y 0 ) = 1 3. Then by the point-slope formula, the equation of the tangent line is 3 y = 2 1 ( x 1 ). 3 2 This idea can also be applied to find partial derivatives of implicit surfaces. For example, an arbitrary surface in R 3 can be stated implicitly as F (x, y, z) = C. If we treat z as being an implicit function of x and y, then F (x, y, z(x, y)) = C. We can apply the chain rule to this equation to calculate the implicit partial derivatives / x and / y. Differentiating F (x, y, z(x, y)) = 0 with respect to x gives us F x + F y y x + F x = 0. However, since x and y are independent variables, y x F x + F z x = 0 = 0 and we just get

20 which gives us x = F x F z. This exact process can applied to also find / y which results in y = F y F z. Example 4.6. Compute the partial derivatives / x and / y for the unit sphere and find the equation of its tangent plane at the point (1/2, 1/2, 1/ 2). Solution: The unit sphere is defined implicitly using the function The partial derivatives of F are Then and F (x, y, z) = x 2 + y 2 + z 2 = 1. F x = 2x, F y = 2y, F z = 2z. x = F x F z = 2x 2z = x z y = F y F z = 2y 2z = y z. Evaluating the partial derivatives at the given point, we get z x (x 0, y 0 ) = x 0 = = 1, z z y (x 0, y 0 ) = y 0 = = 1. z The equation of the tangent plane at the point (x 0, y 0, z 0 ) is L(x, y) = z 0 + z x (x 0, y 0, z 0 )(x x 0 ) + z y (x 0, y 0, z 0 )(y y 0 ). Plugging in the partial derivatives of z evaluated at (x 0, y 0, z 0 ) gives L(x, y) = (x 1 2 ) 1 2 (y 1 2 )

21 5 Directional Derivatives and the Gradient Vector 5.1 The Directional Derivative As we ve already seen, partial derivatives tell us how a function is changing as we move along a direction that is parallel to the x or y axes. We now want to generalize this concept which leads us to the idea of a directional derivative. Let f(x, y) be a function of two variables. If we define the vector r = x, y, then we can use the notation f(r), instead of f(x, y). Let u be a unit vector in some direction. Then the directional derivative of f in the direction of u is defined as f(r + hu) f(r) D u f(r) = lim. h 0 h In other words, this tells us the rate of change of the function in the direction along the unit vector u. The partial derivatives f x and f y are just the directional derivatives of f along the unit vectors i and j, respectively. For example, if u = i, then r + hu = x, y + h 1, 0 = x + h, y and the formula for the directional derivative reduces to Likewise, if u = j, then f(r + hi) f(r) lim h 0 h = lim h 0 f(x + h, y) f(x, y) h r + hu = x, y + h 0, 1 = x, y + h and the formula for the directional derivative reduces to f(r + hj) f(r) lim h 0 h = lim h 0 f(x, y + h) f(x, y) h = f x. = f y. Here is how we find a formula for the directional derivative. Let u = u 1, u 2 be a unit vector and r 0 = x 0, y 0 be some fixed point. Define the function Then by definition of the derivative g(h) = f(r 0 + hu) = f(x 0 + hu 1, y 0 + hu 2 ). g (0) = lim h 0 g(h) g(0) h = lim h 0 f(r 0 + hu) f(r 0 ) h = D u f(r 0 ). If we differentiate g respect to h and use the chain rule for the right hand side, we get g (h) = dx x dh + dy y dh = f x(x 0 + hu 1, y 0 + hu 2 )u 1 + f y (x 0 + hu 1, y 0 + hu 2 )u 2. Evaluating this at h = 0 gives g (0) = f x (x 0, y 0 )u 1 + f y (x 0, y 0 )u 1 = f x (x 0, y 0 ), f y (x 0, y 0 ) u 1, u 2. Equating each of the expressions for g (0) gives us the formula We now define the del operator as D u f(r) = f x (r)u 1 + f y (r)u 2 = f x (r), f y (r) u 1, u 2. = x, = x, y. y

22 Note that the del operator is a vector, but only in an abstract sense. Its components are operators, not numbers, so there is no arrow we could draw to represent this vector. Therefore the del operator doesn t really mean anything by itself but when it operates on a function f, we get the well-defined gradient vector of f, given by f = x, y f = x f, y f = f x, f y. The expression f is usually pronounced as del f or grad f. Using the gradient vector of f gives us a convenient notation for the directional derivative: D u f(r) = f(r) u. Example 5.1. Find the directional derivative of f(x, y) = y 2 /x along the vector v = 2, 3 at the point (1, 2). Solution: We first need to find the unit vector in the direction of v, which is u = v v = 1 1 2, 3 = 2, Now we compute the partial derivatives of f to get the gradient vector: f x = y2 x, f 2 y = 2y x f = y2 x, 2y. 2 x Evaluating the gradient at the point (1, 2) gives us The directional derivative at (1, 2) is then D u f(1, 2) = f(1, 2) u = 4, 4 f(1, 2) = 2 2 /1 2, 4/1 = 4, , 3 = 1 (( 4)(2) + (4)(3)) = Example 5.2. Find the directional derivative of f(x, y) = x 3 y 4 x 2 y at the point (1, 1) in the direction of the unit vector v that makes an angle of θ = π/3 with the +x-axis. Solution: We first need to find the unit vector in the direction of v, which is u = cos(θ), sin(θ) = 1 2 1, 3. Now we compute the partial derivatives of f to get the gradient vector: f x = 3x 2 y 4 2xy, f y = 4x 3 y 3 x 2 f = 3x 2 y 4 2xy, 4x 3 y 3 x 2. Evaluating the gradient at the point (1, 1) gives us The directional derivative at (1, 1) is then f(1, 1) = 3 2, 4 1 = 1, 3. D u f(1, 1) = f(1, 1) u = 1, , 3 = 1 2 ( ).

23 Note that the gradient vector can be defined for functions with any number of variables. For example, if f(x, y, z) is a function of 3 variables then f = f x, f y, f z. The directional derivative for functions of three variables has the exact same formula as for two variables, the vectors would just have 3 components. Example 5.3. Find the directional derivative of f(x, y, z) = yze x at the point (0, 3, 1) in the direction of the vector v = 1, 2, 1. Solution: We first need to find the unit vector in the direction of v, which is u = v v = 1 1 1, 2, 1 = 1, 2, Now we compute the partial derivatives of f to get the gradient vector: f x = yze x, f y = ze x, f z = ye x f = yze x, ze x, ye x. Evaluating the gradient at the point (0, 3, 1) gives us The directional derivative at (0, 3, 1) is then D u f(0, 3, 1) = f(0, 3, 1) u = 3, 1, 3 f(0, 3, 1) = 3e 0, 1e 0, 3e 0 = 3, 1, , 2, 1 = 1 6 ( ) = Properties of the Gradient Vector The gradient vector of a function has a special geometric significance. Let z = f(x, y) be a function of two variables. Then f(x, y) = c is the equation of the contour for which z = c. The contour is a curve in the xy plane that can be parameterized as r(t) = x(t), y(t). The equation of for the contour then becomes f(x(t), y(t)) = c. Differentiating each side of this equation with respect to t with the chain rule gives which can be written in vector notation as dx x dt + dy y dt = 0 f x, f y x (t), y (t) = 0 f r (t) = 0. This tells us that the gradient vector f is always perpendicular to the tangent vector of the contour r (t). In other words, the gradient of a function is always perpendicular to the contours of that function. Example 5.4. Verify that the gradient vector of f(x, y) = x 2 + y 2 is orthogonal to its contours at every point. Solution: The gradient vector of this function at an arbitrary point (x, y) is f = 2x, 2y.

24 A general contour for this function has the form x 2 + y 2 = c which is a circle of radius c, which can be parameterized as The tangent vector of this curve is r(t) = c cos(t), sin(t). r (t) = c sin(t), cos(t). Evaluating the gradient vector along this circle gives Then so f is orthogonal to r. f = 2x, 2y = 2 c cos(t), sin(t). f r = 2c( cos(t) sin(t) + sin(t) cos(t)) = Tangent Planes of Level Surfaces The idea that the gradient vector is orthogonal to the contours of a function of two variables can also be extended to functions of three or more variables. In particular, if F (x, y, z) is a function of three variables, then the gradient vector F = F x, F y, F z is a vector in R 3 that is orthogonal to the level surfaces F (x, y, z) = c. This gives us a convenient method for finding the normal vector for the tangent plane of any surface. If (x 0, y 0, z 0 ) is a point on the surface, then F (x 0, y 0, z 0 ) is orthogonal to the surface at this point, i.e., it is the normal vector for the tangent plane at this point. Therefore, the equation of the tangent plane is F (x 0, y 0, z 0 ) x x 0, y y 0, z z 0 = 0. The normal line of a level surface F (x, y, z) = c at the point (x 0, y 0, z 0 ) is defined as the line that is orthogonal to the surface at that point. Note that since this line is orthogonal to the surface, its direction vector is the normal vector of the tangent plane, which is the gradient vector. The equation of the normal line is thus r(t) = x 0, y 0, z 0 + t F (x 0, y 0, z 0 ). Note that these results are also valid for surfaces of the form z = f(x, y), since we can define a function of three variables as F (x, y, z) = f(x, y) z. Then the surface z = f(x, y) is the level surface of F defined by F (x, y, z) = 0. In this case, the gradient vector of F reduces to F = F x, F y, F z = f x, f y, 1. The equation of the tangent plane at the point (x 0, y 0, f(x 0, y 0 )) can then be stated as f x (x 0, y 0 ), f y (x 0, y 0 ), 1 x x 0, y y 0, z f(x 0, y 0 ) = 0. Solving this equation for z would give the linearization of f at the point (x 0, y 0 ).

25 Example 5.5. Find the equations of the tangent plane and the normal line of the implicit surface defined by F (x, y, z) = xyz 2 = 6 at the point (3, 2, 1). Solution: We first find the normal vector of the tangent plane, which is given by the gradient vector of F, evaluated at the point (3, 2, 1). The gradient vector of F is Evaluating this at the point (3, 2, 1) gives F = F x, F y, F z = yz 2, xz 2, 2xyz. F (3, 2, 1) = 2, 3, 12, which is the normal vector of the tangent plane. Since the plane contains the point (3, 2, 1), we get the equation of the plane is 2, 3, 12 x 3, y 2, z 1 = 0 2(x 3) + 3(y 2) + 12(z 1) = 0. The normal line is then r(t) = 3, 2, 1 + t 2, 3, 12 = 3 + 2t, 2 + 3t, t. 5.4 Maximum Rate of Change The directional derivative tells us how fast a function is changing at a given point and in a given direction. It is natural to then ask in which direction is the function changing the fastest, i.e., in which direction is the directional derivative at a maximum. The directional derivative of f(x, y) in the direction of the unit vector u is given by the dot product Let θ be the angle between f and u. Then Since u is a unit vector, this reduces to D u f(x, y) = f(x, y) u. D u f(x, y) = f(x, y) u = f(x, y) u cos(θ). D u f(x, y) = f(x, y) cos(θ). Since 1 cos(θ) 1, the directional derivative is at a maximum when cos(θ) = 1 θ = 0, which implies that u is in the same direction as f(x, y). In other words, the function is increasing the fastest in the direction of f(x, y) and the directional derivative in this direction is f(x, y). This is referred to as the direction of steepest ascent. On the other hand, if cos(θ) = 1 then θ = π and u and f(x, y) point in opposite directions. In this case, the direction derivative is f(x, y), which is the minimum value of the directional derivative. This means that the function is decreasing the fastest in the direction opposite to f, which makes intuitive sense. This is referred to as the direction of steepest descent. In the case where cos(θ) = 0 then θ = π/2 which means that u is orthogonal to f(x, y). The directional derivative in this case is zero, which means the function is not changing at all in the direction of u, i.e., u is tangent to the contour of f at this point. This should be expected since we already know that f(x, y) is orthogonal to the contours of f at every point. Since the direction of steepest ascent at the point (x, y) is f(x, y) and f(x, y) is orthogonal to the contours of f, then we must conclude that the direction of steepest ascent for a function f is always orthogonal to the contours of f.

26 Example 5.6. In which direction is the function f(x, y) = xye x y increasing the fastest at the point (2, 2) and what is the maximum value of the directional derivative there? Also, find a vector v that is tangent to the contour of f at this point. Solution: The gradient vector of this function at an arbitrary point (x, y) is f(x, y) = ye x y + xye x y, xe x y xye x y = e x y y + xy, x xy. The direction of maximum increase at (2, 2) is then f(2, 2) = e , 2 4 = 6, 2 and the maximum value of the directional derivative is f(2, 2) = ( 2) 2 = 40. The tangent vector of the contour at (2, 2) must be orthogonal to the gradient vector there and there are two possibilities: v = 2, 6 or v = 2, 6. Example 5.7. Let u = f be the direction of steepest ascent for a surface f(x, y). Suppose the tangent vector of the surface in the direction of u makes an angle of θ with the xy plane. Find the angle φ that the tangent vector would make with the xy plane in a direction that deviates from u by the angle α. Solution: The key to the solution is to recognize that the directional derivative along u represents the slope of the surface in the direction of u, i.e., it is the rise/run ratio of the tangent vector to the surface in that direction. This ratio should be the tangent of the angle between the tangent vector and the xy plane, so tan(θ) = D u f = f, since u is in the direction of steepest ascent. Now let v be a unit vector that makes an angle of α with respect to u. Then the directional derivative in the direction of v is D v f = f v = f cos(α) = tan(θ) cos(α). As mentioned before, this must be tangent of the angle φ between the tangent vector of the surface in the direction of v and the xy plane. Therefore tan(φ) = D v f = tan(θ) cos(α) φ = tan 1 (tan(θ) cos(α)).

27 6 Minimum and Maximum Values Just as with functions of a single variables, function of multiple variables have minimum and maximum values. As we will see, it is possible to find these values using partial derivatives, although the situation becomes more complicated. Let z = f(x, y) be a function of two variables. Geometrically speaking, the local extrema of this function correspond to the peaks and troughs of the surface z = f(x, y). More precisely, we define a local minimum of f to be a point (x 0, y 0 ) such that f(x 0, y 0 ) f(x, y) for all points (x, y) in a disc with a sufficiently small radius, centered at (x 0, y 0 ). In other words, the value of the function at the local minimum is less than all of the nearby points. If the above inequality holds for any radius, then f(x 0, y 0 ) is the absolute minimum of f. Likewise, we define a local maximum of f to be a point (x 0, y 0 ) such that f(x 0, y 0 ) f(x, y) for all points (x, y) in a disc with a sufficiently small radius, centered at (x 0, y 0 ). In other words, the value of the function at the local maximum is greater than all of the nearby points. If the above inequality holds for any radius then f(x 0, y 0 ) is the absolute maximum of f. At any local extrema, the tangent plane of the surface there must be parallel to the xy plane. Therefore the normal vector of the tangent plane must be parallel (or anti-parallel) to the normal vector of the xy plane, which is k = 0, 0, 1. We saw in the last section that f x (x 0, y 0 ), f y (x 0, y 0 ), 1 is a normal vector for the tangent plane at the point (x 0, y 0 ). In order for this vector to be parallel to k, we must have f x (x 0, y 0 ) = 0 and f y (x 0, y 0 ) = 0. In other words, all components of the gradient vector at a local minimum or maximum must be zero: f(x 0, y 0 ) = 0. Note that this statement applies to functions with any number of variables. One consequence of the gradient vector being all zeros is that the directional derivative is zero in every possible direction, which in plain English, just means that the surface is flat there. 6.1 Local Extrema and Critical Points We define a critical point, or stationary point as any point (x 0, y 0 ) such that f(x 0, y 0 ) = 0. Local minima and maxima automatically satisfy this property but there is also a third kind of critical point that is not encountered in single variable calculus, called a saddle point, which is neither a local minimum or maximum. This terminology comes from the fact that surfaces near these points have a saddle-like shape, with the saddle point being at the center. The contours around a local minima or maxima are closed curves centered around the critical point. However, the contours around a saddle point look like little hyperbolas with the saddle point at the center. Just as in single variable calculus, there is a second derivative test that can be used to classify critical points as either a local maximum, local minimum or saddle point. To this end, we define the Hessian matrix of the function f(x, y) as [ ] fxx f H = xy. f yx All of the local curvature properties of the surface are contained within this matrix. In particular, the type of critical point is classified in terms of the determinant of the Hessian matrix D = f xx f xy = f xxf yy f xy f yx = f xx f yy fxy. 2 f yx f yy For any two variable function, there are 3 possible types of critical points and the second derivative test tells us how to classify them. f yy

28 Two-Variable Second Derivative Test: 1. If D(x 0, y 0 ) > 0 and f xx (x 0, y 0 ) > 0, then (x 0, y 0 ) is a local minimum. 2. If D(x 0, y 0 ) > 0 and f xx (x 0, y 0 ) < 0, then (x 0, y 0 ) is a local maximum. 3. If D(x 0, y 0 ) < 0, then (x 0, y 0 ) is a saddle point. Note that if D = 0, then the second derivative test is inconclusive and we can not classify the critical point. Figure 8: Surface and contour plots for a two-variable function with several critical points. There are 3 local maxima, 2 local minima and 3 saddle points. Can you spot them all? Example 6.1. Find and classify the critical points of f(x, y) = x 2 + y 2 + 2x + 2y xy. Solution: We set each of the partial derivatives equal to zero, which gives us f x = 2x + 2 y = 0, f y = 2y + 2 x = 0. This gives us a system of two equations for the two unknowns, x and y. Solving for y in the first equation gives us y = 2x + 2. Inserting this into the second equation gives 2(2x + 2) + 2 x = 0 3x + 6 = 0 x = 2, y = 2( 2) + 2 = 2 so ( 2, 2) is the only critical point. The second derivatives for this function are f xx = 2, f yy = 2, f xy = 1 so the Hessian determinant evaluated at the critical point is D( 2, 2) = f xx f yy f 2 xy = (2)(2) ( 1) 2 = 3. Since D > 0 and f xx > 0, then by the second derivative test, the point ( 2, 2) is a local minimum.

29 Example 6.2. Find and classify the critical points of f(x, y) = 3x 2 y + y 3 3x 2 3y Solution: We set each of the partial derivatives equal to zero, which gives us f x = 6xy 6x = 0 6x(y 1) = 0, f y = 3x 2 + 3y 2 6y = 0. From the first equation, we get that either x = 0 or y = 1 and we must consier each of these cases separately. If x = 0, then the second equation reduces to 3y 2 6y = 0 3y(y 2) = 0 y = 0, 2. Therefore (0, 0) and (0, 2) are both critical points. If y = 1, then the second equation reduces to 3x = 0 3x 2 = 3 x = 1, 1. Therefore ( 1, 1) and (1, 1) are also critical points. Now we must classify each of the 4 critical points with the second derivative test. The second partial derivatives of f are The Hessian determinant is thus f xx = 6y 6, f yy = 6y 6, f xy = 6x. D = (6y 6) 2 (6x) 2. We now apply the second derivative test for each critical point: Test for (0, 0): D(0, 0) = 36 > 0, f xx (0, 0) = 6 < 0 (0, 0) is a local maximum. Test for (0, 2): D(0, 2) = 36 > 0, f xx (0, 2) = 6 > 0 (0, 2) is a local minimum. Test for ( 1, 1): D( 1, 1) = 36 < 0 ( 1, 1) is a saddle point. Test for (1, 1): D(1, 1) = 36 < 0 (1, 1) is a saddle point. Example 6.3. Find the maximum volume of a rectangular box that has a surface area of 8 ft 2. Solution: Let x, y and z denote the dimensions of the box. Then the volume of the box is V (x, y, z) = xyz. We need to now reduce this to a function of two variables, using the constraint on the surface area. Since the surface area of the box is 8, we know that Solving for z, we get S = 2(xy + yz + xz) = 8 xy + yz + xz = 4. z = 4 xy x + y. The volume can then be stated as a function of two varaibles The partial derivatives of V are V (x, y) = xyz = xy 4 xy x + y = 4xy x2 y 2. x + y V x = (4y 2xy2 )(x + y) (4xy x 2 y 2 ) (x + y) 2,

30 V x = (4x 2x2 y)(x + y) (4xy x 2 y 2 ) (x + y) 2, Simplifying the numerators and setting them equal to zero gives Since x, y 0, we get Subtracting these two equations, we get 4y 2 x 2 y 2 2xy 3 = 0 y 2 (4 x 2 2xy) = 0, 4x 2 x 2 y 2 2x 3 y = 0 x 2 (4 y 2 2xy) = 0. 4 x 2 2xy = 0, 4 y 2 2xy = 0. x 2 + y 2 = 0 y = ±x. However, since x and y must both be positive, we must have y = x. Plugging this into one of the previous equation gives us 4 x 2 2x 2 = 0 4 3x 2 = 0 x = 4/3 = 2/ 3. Therefore x = y = 2/ 3 are the optimal dimensions. The corresponding z value is z = 4 xy x + y = 4 4/3 4/ 3 = 2 3. Notice that this is the same value we got for x and y, so the volume of the box is maximized when all three dimensions are the same, i.e., the box is a perfect cube. The maximum volume is ( ) 3 2 V max = 3. Example 6.4. Find the point(s) on the plane x + y + z = 0 that are closest to the point P (1, 1, 0). Solution: The function that we need to minimize is the distance between P and a point on the plane. For convenience, we use the square of the distance as the function to be minimized. For any point (x, y, z), the square of the distance to P is F (x, y, z) = (x 1) 2 + (y 1) 2 + z 2. However, since z = x y, we can reduce this to a function of 2 variables f(x, y) = (x 1) 2 + (y 1) 2 + ( x y) 2. Setting the partial derivatives of f equal to 0 gives us f x = 2(x 1) 2( x y) = 0 4x + 2y = 2, f y = 2(y 1) 2( x y) = 0 2x + 4y = 2. The solution to this system is x = 1/3 and y = 1/3. At this point, z = 1/3 1/3 = 2/3 so the closes point on the plane to P is (1/3, 1/3, 2/3).

31 6.2 Absolute Minimum and Maximum Values Most functions do not have absolute minimums or maximums if the domain of the function is infinite in extent. However, when we restrict the domain of a function to a bounded set (a set with finite extent), then there must be an absolute minimum and absolute maximum on this set. Recall from single variable calculus the following problem: find the absolute minimum and absolute maximum of f(x) on the interval [a, b]. For this problem, the absolute extrema either occur at a critical point of f in the interval [a, b] or at one of the boundaries (endpoints) of the interval. To find the absolute extrema, we evaluate f at every critical point in [a, b] and at the endpoints and select the largest/smallest value for the absolute maximum/minimum. For functions of two variables, this same idea applies, but becomes more complicated. The complication comes from the fact that the boundary of a region in R 2 is a curve with infinitely many points, whereas the boundary of a region in R only has two points (the endpoints of the interval). This fact will require a significant modification in the process of finding absolute extrema for functions of two variables, which we will now discuss. Let f(x, y) be a function defined on some bounded (finite) set D in R 2. The absolute maximum/minimum of f could possible occur at a critical point inside D, but we must also consider the fact that the absolute minimum/maximum may occur on the boundary of D. The boundary of D is a closed plane curve C that can be parameterized as r(t) = x(t), y(t). When f is evaluated on the boundary of D, it reduces to a single-variable function g(t) = f(x(t), y(t)) and we must consider the critical points of g(t) as as candidates for the absolute extrema of f. If the boundary curve C is not smooth, then we must parameterize and find the critical points for each smooth segment separately. In this case, we must also consider the corner points as candidates for the boundary extrema. To summarize, there are three types of points that could possibly be the absolute extrema: the critical points of f(x, y) in the region D, the critical points of g(t) on the boundary curve C, and the corner points (if any) of C. The process of finding absolute extrema for f(x, y) on the region D can be summarized as follows: 1. Find all critical points of f(x, y) that are inside D. 2. Parameterize each smooth segment of the boundary curve C and find the critical points of g(t) = f(x(t), y(t)) for each segment. 3. Evaluate f(x, y) for every point found in steps 1 and If C contains any corners, evaluate f(x, y) at the corner points. 5. Using the values computed in steps 3 and 4, select the largest value as the absolute maximum and the smallest value as the absolute minimum. Example 6.5. Find the absolute minimum and maximum of the function f(x, y) = x 2 y 2 on the unit disc: D = {(x, y) x 2 + y 2 1}. Solution: First we find the critical points of f in D. Setting each partial derivative to zero gives us f x = 2x = 0 x = 0, f y = 2y = 0 y = 0, so (0, 0) is the only critical point of f. The boundary curve for the unit disc is the unit circle, which we parameterize as r(t) = cos(t), sin(t), 0 t 2π.

32 Evaluating f on the boundary then gives us The derivative is thus g(t) = x(t) 2 y(t) 2 = cos 2 (t) sin 2 (t) = cos(2t). g (t) = 2 sin(2t). Setting this equal to zero gives us the critical t-values: These four t-values correspond to the points g (t) = 2 sin(2t) = 0 t = 0, π/2, π, 3π/2. r(0) = 1, 0, r(π/2) = 0, 1, r(π) = 1, 0, r(3π/2) = 0, 1, which are all candidates points, in addition to (0, 0). Note that since the boundary curve is smooth, we can skip step 4. Evaluating f at each of the candidate points gives f(0, 0) = 0, f(1, 0) = 1, f(0, 1) = 1, f( 1, 0) = 1, f(0, 1) = 1. The largest of these values is 1 and the smallest is 1. Therefore the absolute maximum of f on D is 1, which occurs at the points (±1, 0) and the absolute minimum of f on D is 1, which occurs at the points (0, ±1). A plot of this function over the region D is shown below, which we can use to verify our answers. Example 6.6. Find the absolute minimum and maximum of the function f(x, y) = x 2 + y 2 xy on the unit disc. Solution: First we find the critical points of f in D. zero gives us f x = 2x y = 0, f y = 2y x = 0. Setting each partial derivative to The only solution to this system is x = 0, y = 0 so (0, 0) is the only critical point of f and it is inside the region D. The boundary curve for the unit disc is the unit circle, which we parameterize as r(t) = cos(t), sin(t), 0 t 2π. Evaluating f on the boundary then gives us g(t) = x(t) 2 + y(t) 2 x(t)y(t) = cos 2 (t) + sin 2 (t) sin(t) cos(t) = sin(2t).

33 The derivative is thus g (t) = cos(2t). Setting this equal to zero gives us the critical t-values: g (t) = 2 sin(2t) = 0 t = π/4, 3π/4, 5π/4, 7π/4. These four t-values correspond to the points r(π/4) = 1/ 2, 1/ 2, r(3π/4) = 1/ 2, 1/ 2, r(5π/4) = 1/ 2, 1/ 2, r(7π/4) = 1/ 2, 1/ 2, which are all candidates points, in addition to the interior critical point (0, 0). Evaluating f at each of the candidate points gives f(0, 0) = 0, f(1/ 2, 1/ 2) = 1/2, f( 1/ 2, 1/ 2) = 3/2, f( 1/ 2, 1/ 2) = 1/2, f(1/ 2, 1/ 2) = 3/2. The largest of these values is 3/2 and the smallest is 0. Therefore the absolute maximum of f on D is 3/2, which occurs at the points (1/ 2, 1/ 2) and ( 1/ 2, 1/ 2). The absolute minimum of f on D is 0, which occurs at the point (0, 0). A plot of this function over the region D is shown below, which we can use to verify our answers.

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