TOPICS IN DIFFERENTIATION

Transcription

1 3 TOPICS IN DIFFERENTIATION Craig Lovell/Corbis Images The growth an ecline of animal populations an natural resources can be moele using basic functions stuie in calculus. We begin this chapter b etening the process of ifferentiation to functions that are either ifficult or impossible to ifferentiate irectl. We will iscuss a combination of irect an inirect methos of ifferentiation that will allow us to evelop a number of new erivative formulas that inclue the erivatives of logarithmic, eponential, an inverse trigonometric functions. Later in the chapter, we will consier some applications of the erivative. These will inclue was in which ifferent rates of change can be relate as well as the use of linear functions to approimate nonlinear functions. Finall, we will iscuss L Hôpital s rule, a powerful tool for evaluating its. 3. IMPLICIT DIFFERENTIATION Up to now we have been concerne with ifferentiating functions that are given b equations of the form = f(). In this section we will consier methos for ifferentiating functions for which it is inconvenient or impossible to epress them in this form. FUNCTIONS DEFINED EXPLICITLY AND IMPLICITLY An equation of the form = f() is sai to efine eplicitl as a function of because the variable appears alone on one sie of the equation an oes not appear at all on the other sie. However, sometimes functions are efine b equations in which is not alone on one sie; for eample, the equation + + = () is not of the form = f(), but it still efines as a function of since it can be rewritten as = + Thus, we sa that () efines implicitl as a function of, the function being f() = + 85

2 86 Chapter 3 / Topics in Differentiation + = An equation in an can implicitl efine more than one function of. This can occur when the graph of the equation fails the vertical line test, so it is not the graph of a function of. For eample, if we solve the equation of the circle + = () for in terms of, we obtain =±, so we have foun two functions that are efine implicitl b (), namel, f () = an f () = (3) The graphs of these functions are the upper an lower semicircles of the circle + = (Figure 3..). This leas us to the following efinition. = 3.. efinition We will sa that a given equation in an efines the function f implicitl if the graph of = f()coincies with a portion of the graph of the equation. Eample The graph of = is not the graph of a function of, since it oes not pass the vertical line test (Figure 3..). However, if we solve this equation for in terms of, we obtain the equations = an =, whose graphs pass the vertical line test an are portions of the graph of = (Figure 3..). Thus, the equation = implicitl efines the functions f () = an f () = Figure 3.. = = = Figure 3.. The graph of = oes not pass the vertical line test, but the graphs of = an = o. Although it was a trivial matter in the last eample to solve the equation = for in terms of, it is ifficult or impossible to o this for some equations. For eample, the equation = 3 (4) can be solve for in terms of, but the resulting formulas are too complicate to be practical. Other equations, such as sin() =, cannot be solve for b an elementar metho. Thus, even though an equation ma efine one or more functions of, it ma not be possible or practical to fin eplicit formulas for those functions. Fortunatel, CAS programs, such as Mathematica an Maple, have implicit plotting capabilities that can graph equations such as (4). The graph of this equation, which is calle the Folium of Descartes, is shown in Figure 3..3a. Parts (b) an (c) of the figure show the graphs (in blue) of two functions that are efine implicitl b (4) Figure = (a) (b) (c) 3 4

3 3. Implicit Differentiation 87 IMPLICIT DIFFERENTIATION In general, it is not necessar to solve an equation for in terms of in orer to ifferentiate the functions efine implicitl b the equation. To illustrate this, let us consier the simple equation = (5) One wa to fin / is to rewrite this equation as = (6) from which it follows that = (7) Another wa to obtain this erivative is to ifferentiate both sies of (5) before solving for in terms of, treating as a (temporaril unspecifie) ifferentiable function of. With this approach we obtain [] = [] []+ [] =0 + = 0 = If we now substitute (6) into the last epression, we obtain = which agrees with Equation (7). This metho of obtaining erivatives is calle implicit ifferentiation. Eample Use implicit ifferentiation to fin / if 5 + sin =. [5 + sin ] = [ ] 5 [ ]+ [sin ] = ( 5 ) + (cos ) = The chain rule was use here because is a function of. 0 + (cos ) = René Descartes (596 l650) Descartes, a French aristocrat, was the son of a government official. He grauate from the Universit of Poitiers with a law egree at age 0. After a brief probe into the pleasures of Paris he became a militar engineer, first for the Dutch Prince of Nassau an then for the German Duke of Bavaria. It was uring his service as a solier that Descartes began to pursue mathematics seriousl an evelop his analtic geometr. After the wars, he returne to Paris where he stalke the cit as an eccentric, wearing a swor in his belt an a plume hat. He live in leisure, selom arose before a.m., an abble in the stu of human phsiolog, philosoph, glaciers, meteors, an rainbows. He eventuall move to Hollan, where he publishe his Discourse on the Metho, an finall to Sween where he ie while serving as tutor to Queen Christina. Descartes is regare as a genius of the first magnitue. In aition to major contributions in mathematics an philosoph he is consiere, along with William Harve, to be a founer of moern phsiolog.

4 88 Chapter 3 / Topics in Differentiation Solving for / we obtain = (8) 0 + cos Note that this formula involves both an. In orer to obtain a formula for / that involves alone, we woul have to solve the original equation for in terms of an then substitute in (8). However, it is impossible to o this, so we are force to leave the formula for / in terms of an. Eample 3 Use implicit ifferentiation to fin / if 4 = 9. Solution. Differentiating both sies of 4 = 9 with respect to iels 8 4 = 0 from which we obtain = (9) Differentiating both sies of (9) iels ()() ()(/) = (0) Substituting (9) into (0) an simplifing using the original equation, we obtain = ( /) = 4 = In Eamples an 3, the resulting formulas for / involve both an. Although it is usuall more esirable to have the formula for / epresse in terms of alone, having the formula in terms of an is not an impeiment to fining slopes an equations of tangent lines provie the - an -coorinates of the point of tangenc are known. This is illustrate in the following eample. Eample 4 Fin the slopes of the tangent lines to the curve + = 0atthe points (, ) an (, ). = (, ) (, ) = Figure 3..4 Solution. We coul procee b solving the equation for in terms of, an then evaluating the erivative of = at (, ) an the erivative of = at(, ) (Figure 3..4). However, implicit ifferentiation is more efficient since it can be use for the slopes of both tangent lines. Differentiating implicitl iels [ + ] = [0] [ ] []+ [] = [0] = 0 = At (, ) we have =, an at (, ) we have =, so the slopes of the tangent lines to the curve at those points are = an = = = = =

5 3. Implicit Differentiation 89 Eample 5 (a) Use implicit ifferentiation to fin / for the Folium of Descartes = 3. (b) Fin an equation for the tangent line to the Folium of Descartes at the point ( 3, ) 3. (c) At what point(s) in the first quarant is the tangent line to the Folium of Descartes horizontal? Formula () cannot be evaluate at (0, 0) an hence provies no information about the nature of the Folium of Descartes at the origin. Base on the graphs in Figure 3..3, what can ou sa about the ifferentiabilit of the implicitl efine functions graphe in blue in parts (b) an (c) of the figure? Solution (a). Differentiating implicitl iels [3 + 3 ]= [3] = = + ( ) = = () 3 3 Figure , Solution (b). At the point ( 3, 3 of the tangent line at this point is m tan = ), we have = 3 an = 3, so from () the slope m tan =3/ =3/ = (3 /) (3/) (3/) (3/) = Thus, the equation of the tangent line at the point ( 3, 3 ) is 3 = ( ) 3 or + = 3 which is consistent with Figure Solution (c). The tangent line is horizontal at the points where / = 0, an from () this occurs onl where = 0or = () Substituting this epression for in the equation = 3 for the curve iels 3 + ( ) 3 = = 0 3 ( 3 ) = 0 whose solutions are = 0 an = /3. From (), the solutions = 0 an = /3 iel the points (0, 0) an ( /3, /3 ), respectivel. Of these two, onl ( /3, /3 ) is in the first quarant. Substituting = /3, = /3 into () iels = = /3 = /3 0 4/3 /3 = 0 4 We conclue that ( /3, /3 ) (.6,.59) is the onl point on the Folium of Descartes Figure 3..6 in the first quarant at which the tangent line is horizontal (Figure 3..6).

6 90 Chapter 3 / Topics in Differentiation DIFFERENTIABILITY OF FUNCTIONS DEFINED IMPLICITLY When ifferentiating implicitl, it is assume that represents a ifferentiable function of. If this is not so, then the resulting calculations ma be nonsense. For eample, if we ifferentiate the equation + + = 0 (3) we obtain + = 0 or = However, this erivative is meaningless because there are no real values of an that satisf (3) (wh?); an hence (3) oes not efine an real functions implicitl. The nonsensical conclusion of these computations conves the importance of knowing whether an equation in an that is to be ifferentiate implicitl actuall efines some ifferentiable function of implicitl. Unfortunatel, this can be a ifficult problem, so we will leave the iscussion of such matters for more avance courses in analsis. QUICK CHECK EXERCISES 3. (See page 9 for answers.). The equation + = efines implicitl the function =.. Use implicit ifferentiation to fin / for 3 =. 3. The slope of the tangent line to the graph of + + = 3 at (, ) is. 4. Use implicit ifferentiation to fin / for sin =. EXERCISE SET 3. C CAS (a) Fin / b ifferentiating implicitl. (b) Solve the equation for as a function of, an fin / from that equation. (c) Confirm that the two results are consistent b epressing the erivative in part (a) as a function of alone =. sin = 3 Fin / b implicit ifferentiation = = = = 7. + = 8. = + 9. sin( ) = 0. cos( ) =. tan 3 ( + ) =. 3 + sec = Fin / b implicit ifferentiation = = = = 7. + sin = 8. cos = 9 0 Fin the slope of the tangent line to the curve at the given points in two was: first b solving for in terms of an ifferentiating an then b implicit ifferentiation = ; (/, 3/), (/, 3/) 0. + = 0; (0, 3), (0, 3) 4 True False Determine whether the statement is true or false. Eplain our answer.. If an equation in an efines a function = f() implicitl, then the graph of the equation an the graph of f are ientical.. The function f() = {, 0 <, 0 is efine implicitl b the equation + =. 3. The function is not efine implicitl b the equation ( + )( ) = If is efine implicitl as a function of b the equation + =, then / = /. 5 8 Use implicit ifferentiation to fin the slope of the tangent line to the curve at the specifie point, an check that our answer is consistent with the accompaning graph on the net page = 6; (, 4 5) [Lamé s special quartic] = 0; (0, 3) [trisectri] 7. ( + ) = 5( ); (3, ) [lemniscate] 8. /3 + /3 = 4; (, 3 3) [four-cuspe hpoccloi]

7 3. Implicit Differentiation 9 Figure E Figure E-7 FOCUS ON CONCEPTS Figure E Figure E-8 9. In the accompaning figure, it appears that the ellipse + + = 3 has horizontal tangent lines at the points of intersection of the ellipse an the line =. Use implicit ifferentiation to eplain wh this is the case. = = These eercises eal with the rotate ellipse C whose equation is + = Show that the line = intersects C at two points P an Q an that the tangent lines to C at P an Q are parallel. 34. Prove that if P(a,b)is a point on C, then so is Q( a, b) an that the tangent lines to C through P an through Q are parallel. 35. Fin the values of a an b for the curve + a = b if the point (, ) is on its graph an the tangent line at (, ) has the equation = At what point(s) is the tangent line to the curve 3 = perpenicular to the line + = 0? Two curves are sai to be orthogonal if their tangent lines are perpenicular at each point of intersection, an two families of curves are sai to be orthogonal trajectories of one another if each member of one famil is orthogonal to each member of the other famil. This terminolog is use in these eercises. 37. The accompaning figure shows some tpical members of the families of circles + ( c) = c (black curves) an ( k) + = k (gra curves). Show that these families are orthogonal trajectories of one another. [Hint: For the tangent lines to be perpenicular at a point of intersection, the slopes of those tangent lines must be negative reciprocals of one another.] 38. The accompaning figure shows some tpical members of the families of hperbolas = c (black curves) an = k (gra curves), where c = 0 an k = 0. Use the hint in Eercise 37 to show that these families are orthogonal trajectories of one another. 3 Figure E-9 C 30. (a) A stuent claims that the ellipse + = has a horizontal tangent line at the point (, ). Without oing an computations, eplain wh the stuent s claim must be incorrect. (b) Fin all points on the ellipse + = at which the tangent line is horizontal. 3. (a) Use the implicit plotting capabilit of a CAS to graph the equation 4 + = ( ). (b) Use implicit ifferentiation to help eplain wh the graph in part (a) has no horizontal tangent lines. (c) Solve the equation 4 + = ( ) for in terms of an eplain wh the graph in part (a) consists of two parabolas. 3. Use implicit ifferentiation to fin all points on the graph of 4 + = ( ) at which the tangent line is vertical. C C Figure E-37 Figure E-38 (a) Use the implicit plotting capabilit of a CAS to graph the curve C whose equation is = 0. (b) Use the graph in part (a) to estimate the -coorinates of a point in the first quarant that is on C an at which the tangent line to C is parallel to the -ais. (c) Fin the eact value of the -coorinate in part (b). (a) Use the implicit plotting capabilit of a CAS to graph the curve C whose equation is = 0. (b) Use the graph to guess the coorinates of a point in the first quarant that is on C an at which the tangent line to C is parallel to the line =. (cont.)

8 9 Chapter 3 / Topics in Differentiation (c) Use implicit ifferentiation to verif our conjecture in part (b). 4. Prove that for ever nonzero rational number r, the tangent line to the graph of r + r = at the point (, ) has slope. 4. Fin equations for two lines through the origin that are tangent to the ellipse = Writing Write a paragraph that compares the concept of an eplicit efinition of a function with that of an implicit efinition of a function. 44. Writing A stuent asks: Suppose implicit ifferentiation iels an unefine epression at a point. Does this mean that / is unefine at that point? Using the equation + = 0 as a basis for our iscussion, write a paragraph that answers the stuent s question. QUICK CHECK ANSWERS = = sec tan 3. DERIVATIVES OF LOGARITHMIC FUNCTIONS In this section we will obtain erivative formulas for logarithmic functions, an we will eplain wh the natural logarithm function is preferre over logarithms with other bases in calculus. DERIVATIVES OF LOGARITHMIC FUNCTIONS We will establish that f() = ln is ifferentiable for >0 b appling the erivative efinition to f(). To evaluate the resulting it, we will nee the fact that ln is continuous for >0 (Theorem.6.3), an we will nee the it ( + v 0 v)/v = e () This it can be obtaine from its (7) an (8) of Section.3 b making the substitution v = / an using the fact that v 0 + as + an v 0 as. This prouces two equal one-sie its that together impl () (see Eercise 64 of Section.3). ln( + h) ln [ln ] = h 0 h = h 0 h ln = h 0 h ln = v 0 v ( ) + h ( + h ln( + v) Let ) The quotient propert of logarithms in Theorem 0.5. v = h/ an note that v 0 if an onl if h 0. = v 0 v ln( + v) is fie in this it computation, so / can be move through the it sign. = ln( + v)/v v 0 = [ ln ln ( + v)/v] = ln e v 0 The power propert of logarithms in Theorem 0.5. is continuous on (0, + ) so we can move the it through the function smbol. = Since ln e =

9 3. Derivatives of Logarithmic Functions 93 Note that, among all possible bases, the base b = e prouces the simplest formula for the erivative of log b. This is one of the reasons wh the natural logarithm function is preferre over other logarithms in calculus. Thus, [ln ] =, > 0 () A erivative formula for the general logarithmic function log b can be obtaine from () b using Formula (6) of Section 0.5 to write [log b ]= [ ] ln ln b It follows from this that Eample = ln b [ln ] [log b ]= ln b, > 0 (3) = ln with tangent lines (a) Figure 3.. shows the graph of = ln an its tangent lines at the points =,, 3, an 5. Fin the slopes of those tangent lines. (b) Does the graph of = ln have an horizontal tangent lines? Use the erivative of ln to justif our answer. Solution (a). From (), the slopes of the tangent lines at the points =,, 3, an 5 are / =,, 3, an, respectivel, which is consistent with Figure Figure 3.. Solution (b). It oes not appear from the graph of = ln that there are an horizontal tangent lines. This is confirme b the fact that / = / is not equal to zero for an real value of. If u is a ifferentiable function of, an if u() > 0, then appling the chain rule to () an (3) prouces the following generalize erivative formulas: [ln u] = u u an [log b u]= u ln b u (4 5) Eample Fin [ln( + )]. Solution. Using (4) with u = + we obtain [ln( + )] = + [ + ] = + = + When possible, the properties of logarithms in Theorem 0.5. shoul be use to convert proucts, quotients, an eponents into sums, ifferences, an constant multiples before ifferentiating a function involving logarithms. Eample 3 [ ( )] sin ln = [ln + ln(sin ) ] + ln( + ) = + cos sin ( + ) = + cot +

10 94 Chapter 3 / Topics in Differentiation Figure 3.. shows the graph of f() = ln. This function is important because it etens the omain of the natural logarithm function in the sense that the values of ln an ln are the same for >0, but ln is efine for all nonzero values of, an ln is onl efine for positive values of. Figure 3.. = ln The erivative of ln for = 0 can be obtaine b consiering the cases >0 an <0 separatel: Case >0. In this case =, so [ln ] = [ln ] = Case <0. In this case =, so it follows from (4) that [ln ] = [ln( )]= ( ) [ ] = Since the same formula results in both cases, we have shown that [ln ] = if = 0 (6) Eample 4 From (6) an the chain rule, [ln sin ] = sin [sin ] =cos sin = cot LOGARITHMIC DIFFERENTIATION We now consier a technique calle logarithmic ifferentiation that is useful for ifferentiating functions that are compose of proucts, quotients, an powers. Eample 5 The erivative of 3 = 7 4 (7) ( + ) 4 is mess to calculate irectl. However, if we first take the natural logarithm of both sies an then use its properties, we can write ln = ln + 3 ln(7 4) 4ln( + ) Differentiating both sies with respect to iels = + 7 /

11 Thus, on solving for / an using (7) we obtain 3 = 7 4 ( + ) 4 3. Derivatives of Logarithmic Functions 95 [ ] + REMARK Since ln is onl efine for >0, the computations in Eample 5 are onl vali for > (verif). However, because the erivative of ln is the same as the erivative of ln, an because ln is efine for <0 as well as >0, it follows that the formula obtaine for / is vali for < as well as >. In general, whenever a erivative / is obtaine b logarithmic ifferentiation, the resulting erivative formula will be vali for all values of for which = 0. It ma be vali at those points as well, but it is not guarantee. DERIVATIVES OF REAL POWERS OF We know from Theorem.3. an Eercise 8 in Section.3 that the ifferentiation formula [r ]=r r (8) In the net section we will iscuss ifferentiating functions that have eponents which are not constant. hols for constant integer values of r. We will now use logarithmic ifferentiation to show that this formula hols if r is an real number (rational or irrational). In our computations we will assume that r is a ifferentiable function an that the familiar laws of eponents hol for real eponents. Let = r, where r is a real number. The erivative / can be obtaine b logarithmic ifferentiation as follows: ln =ln r =rln [ln ] = [r ln ] = r = r = r r = r r QUICK CHECK EXERCISES 3. (See page 96 for answers.). The equation of the tangent line to the graph of = ln at = e is.. Fin /. (a) = ln 3 (b) = ln (c) = log(/ ) 3. Use logarithmic ifferentiation to fin the erivative of + f() = 3 ln( + h) 4. = h 0 h EXERCISE SET 3. 6 Fin /.. = ln 5. = ln 3 3. = ln + 4. = ln( + ) 5. = ln 6. = ln ( ) 7. = ln 8. = ln = ln 0. = (ln ) 3. = ln. = ln 3. = ln 4. = 3 ln 5. = log (3 ) 6. = [log ( )] 3 7. = + log 8. = log + log

12 96 Chapter 3 / Topics in Differentiation 9. = ln(ln ) 0. = ln(ln(ln )). = ln(tan ). = ln(cos ) 3. = cos(ln ) 4. = sin (ln ) 5. = log(sin ) 6. = log( sin ) 7 30 Use the metho of Eample 3 to help perform the inicate ifferentiation. 7. [ln(( )3 ( + ) 4 )] 8. [ln((cos ) + 4 )] [ ] [ ] cos 9. ln 30. ln True False Determine whether the statement is true or false. Eplain our answer. 3. The slope of the tangent line to the graph of = ln at = a approaches infinit as a If + f () = 0, then the graph of = f() has a horizontal asmptote. 33. The erivative of ln is an o function. 34. We have ((ln ) ) = ((ln )) = Fin / using logarithmic ifferentiation. 35. = = = ( 8) / Fin (a) 40. Fin (a) 38. = sin cos tan3 [log e] (b) [log ]. [log (/) e] (b) [log (ln ) e] Fin the equation of the tangent line to the graph of = f() at = f() = ln ; 0 = e 4. f() = log ; 0 = f() = ln( ); 0 = e 44. f() = ln ; 0 = FOCUS ON CONCEPTS 45. (a) Fin the equation of a line through the origin that is tangent to the graph of = ln. (b) Eplain wh the -intercept of a tangent line to the curve = ln must be unit less than the -coorinate of the point of tangenc. 46. Use logarithmic ifferentiation to verif the prouct an quotient rules. Eplain what properties of ln are important for this verification. 47. Fin a formula for the area A(w) of the triangle boune b the tangent line to the graph of = ln at P(w,ln w), the horizontal line through P, an the -ais. 48. Fin a formula for the area A(w) of the triangle boune b the tangent line to the graph of = ln at P(w,ln w ), the horizontal line through P, an the -ais. 49. Verif that = ln( + e) satisfies / = e, with = when = Verif that = ln(e ) satisfies / = e, with = when = Fin a function f such that = f()satisfies / = e, with = 0 when = Fin a function f such that = f() satisfies / = e, with = ln when = Fin the it b interpreting the epression as an appropriate erivative. ln( + 3) 53. (a) (a) 0 ln(e + ) ln( 5) (b) 0 ln w (b) w w ( + h) (b) h 0 ln(cos ) 55. (a) 0 h 56. Moif the erivation of Equation () to give another proof of Equation (3). 57. Writing Review the erivation of the formula [ln ] = an then write a paragraph that iscusses all the ingreients (theorems, it properties, etc.) that are neee for this erivation. 58. Writing Write a paragraph that eplains how logarithmic ifferentiation can replace a ifficult ifferentiation computation with a simpler computation. QUICK CHECK ANSWERS 3.. = +. (a) e = (b) = (c) = ln 0 3. [ ] + 3 ( + ) 3( ) 4.

13 3.3 Derivatives of Eponential an Inverse Trigonometric Functions DERIVATIVES OF EXPONENTIAL AND INVERSE TRIGONOMETRIC FUNCTIONS See Section 0.4 for a review of one-toone functions an inverse functions. In this section we will show how the erivative of a one-to-one function can be use to obtain the erivative of its inverse function. This will provie the tools we nee to obtain erivative formulas for eponential functions from the erivative formulas for logarithmic functions an to obtain erivative formulas for inverse trigonometric functions from the erivative formulas for trigonometric functions. Our first goal in this section is to obtain a formula relating the erivative of the inverse function f to the erivative of the function f. = f () 3 (, ) Figure 3.3. Slope = /f () = Slope = f () = f() (, ) 3 Eample Suppose that f is a one-to-one ifferentiable function such that f() = an f () = 3. Then the tangent line to = f() at the point (, ) has equation 4 = 3 ( ) 4 The tangent line to = f () at the point (, ) is the reflection about the line = of the tangent line to = f() at the point (, ) (Figure 3.3.), an its equation can be obtaine b interchanging an : = 3 4 ( ) or = 4 ( ) 3 Notice that the slope of the tangent line to = f () at = is the reciprocal of the slope of the tangent line to = f() at =. That is, (f ) () = f () = 4 () 3 Since = f () for the function f in Eample, it follows that f () = f (f ()). Thus, Formula () can also be epresse as (f ) () = f (f ()) In general, if f is a ifferentiable an one-to-one function, then (f ) () = f (f ()) () provie f (f ()) = 0. Formula () can be confirme using implicit ifferentiation. The equation = f () is equivalent to = f(). Differentiating with respect to we obtain = [] = [f()]=f () so that = f () = f (f ()) Also from = f()we have / = f (), which gives the following alternative version of Formula (): = (3) / Figure 3.3. The graph of an increasing function (blue) or a ecreasing function (purple) is cut at most once b an horizontal line. INCREASING OR DECREASING FUNCTIONS ARE ONE-TO-ONE If the graph of a function f is alwas increasing or alwas ecreasing over the omain of f, then a horizontal line will cut the graph of f in at most one point (Figure 3.3.), so f

14 98 Chapter 3 / Topics in Differentiation must have an inverse function (see Section 0.4). We will prove in the net chapter that f is increasing on an interval on which f () > 0 (since the graph has positive slope) an that f is ecreasing on an interval on which f () < 0 (since the graph has negative slope). These intuitive observations, together with Formula (), suggest the following theorem, which we state without formal proof theorem Suppose that the omain of a function f is an open interval on which f () > 0 or on which f () < 0. Then f is one-to-one, f () is ifferentiable at all values of in the range of f, an the erivative of f () is given b Formula (). Eample Consier the function f() = In general, once it is establishe that f is ifferentiable, one has the option of calculating the erivative of f using Formula () or (3), or b ifferentiating implicitl, as in Eample. (a) Show that f is one-to-one on the interval (, + ). (b) Fin a formula for the erivative of f. (c) Compute (f ) (). Solution (a). Since f () = > 0 for all real values of, it follows from Theorem 3.3. that f is one-to-one on the interval (, + ). Solution (b). Let = f (). Differentiating = f() = implicitl with respect to iels [] = [5 + + ] = (5 4 + ) = (4) We cannot solve = for in terms of, so we leave the epression for / in Equation (4) in terms of. Solution (c). From Equation (4), (f ) () = = = Thus, we nee to know the value of = f () at =, which we can obtain b solving the equation f() = for. This equation is =, which, b inspection, is satisfie b = 0. Thus, (f ) () = = =0 = DERIVATIVES OF EXPONENTIAL FUNCTIONS Our net objective is to show that the general eponential function b (b > 0,b = ) is ifferentiable everwhere an to fin its erivative. To o this, we will use the fact that

15 3.3 Derivatives of Eponential an Inverse Trigonometric Functions 99 How oes the erivation of Formula (5) change if 0 <b<? In Section 0.5 we state that b = e is the onl base for which the slope of the tangent line to the curve = b at an point P on the curve is the -coorinate at P (see page 54). Verif this statement. b is the inverse of the function f() = log b. We will assume that b>. With this assumption we have ln b>0, so f () = [log b ]= ln b > 0 for all in the interval (0, + ) It now follows from Theorem 3.3. that f () = b is ifferentiable for all in the range of f() = log b. But we know from Table that the range of log b is (, + ), so we have establishe that b is ifferentiable everwhere. To obtain a erivative formula for b we rewrite = b as = log b an ifferentiate implicitl using Formula (5) of Section 3. to obtain = ln b Solving for / an replacing b b we have = ln b = b ln b Thus, we have shown that [b ]=b ln b (5) In the special case where b = e we have ln e =, so that (5) becomes [e ]=e (6) Moreover, if u is a ifferentiable function of, then it follows from (5) an (6) that [bu ]=b u ln b u an [eu ]=e u u (7 8) It is important to istinguish between ifferentiating an eponential function b (variable eponent an constant base) an a power function b (variable base an constant eponent). For eample, compare the erivative [ ]= to the erivative of in Eample 3. Eample 3 The following computations use Formulas (7) an (8). [ ]= ln [e ]=e [ ] = e ]=e [e3 3 [3 ]=3 e 3 [ecos ]=e cos [cos ] = (sin )ecos Functions of the form f() = u v in which u an v are nonconstant functions of are neither eponential functions nor power functions. Functions of this form can be ifferentiate using logarithmic ifferentiation. Eample 4 Use logarithmic ifferentiation to fin [( + ) sin ]. Solution. Setting = ( + ) sin we have ln = ln[( + ) sin ]=(sin )ln( + )

16 00 Chapter 3 / Topics in Differentiation Differentiating both sies with respect to iels = [(sin )ln( + )] = (sin ) + () + (cos )ln( + ) Thus, [ ] sin = + + (cos )ln( + ) = ( + ) sin [ sin + + (cos )ln( + ) ] DERIVATIVES OF THE INVERSE TRIGONOMETRIC FUNCTIONS To obtain formulas for the erivatives of the inverse trigonometric functions, we will nee to use some of the ientities given in Formulas () to (7) of Section 0.4. Rather than memorize those ientities, we recommen that ou review the triangle technique that we use to obtain them. To begin, consier the function sin. Ifweletf() = sin ( π/ π/), then it follows from Formula () that f () = sin will be ifferentiable at an point where cos(sin ) = 0. This is equivalent to the conition sin = π an sin = π Observe that sin is onl ifferentiable on the interval (, ), even though its omain is [, ]. This is because the graph of = sin has horizontal tangent lines at the points (π/, ) an ( π/, ), so the graph of = sin has vertical tangent lines at =±. sin cos(sin ) = Figure so it follows that sin is ifferentiable on the interval (, ). A erivative formula for sin on (, ) can be obtaine b using Formula () or (3) or b ifferentiating implicitl. We will use the latter metho. Rewriting the equation = sin as = sin an ifferentiating implicitl with respect to, we obtain [] = [sin ] = cos = cos = cos(sin ) At this point we have succeee in obtaining the erivative; however, this erivative formula can be simplifie using the ientit inicate in Figure This iels = Thus, we have shown that [sin ]= ( <<) More generall, if u is a ifferentiable function of, then the chain rule prouces the following generalize version of this formula: u [sin u]= ( <u<) u The metho use to erive this formula can be use to obtain generalize erivative formulas for the remaining inverse trigonometric functions. The following is a complete list of these

17 3.3 Derivatives of Eponential an Inverse Trigonometric Functions 0 The appearance of u in (3) an (4) will be eplaine in Eercise 58. formulas, each of which is vali on the natural omain of the function that multiplies u/. u [sin u]= u [tan u]= u + u [sec u]= u u u u [cos u]= u [cot u]= u + u [csc u]= u u u (9 0) ( ) (3 4) Eample 5 Fin / if (a) = sin ( 3 ) (b) = sec (e ) Solution (a). From (9) = (3 ) (3 ) = 3 6 Solution (b). From (3) = e (e ) (e ) = e QUICK CHECK EXERCISES 3.3 (See page 03 for answers.). Suppose that a one-to-one function f has tangent line = at the point (, 8). Evaluate (f ) (8).. In each case, from the given erivative, etermine whether the function f is invertible. (a) f () = + (b) f () = (c) f () = sin () f () = π + tan 3. Evaluate the erivative. (a) [e ] (b) [7 ] (c) [cos(e + )] () [e3 ] 4. Let f() = e 3 +. Use f () to verif that f is one-to-one. EXERCISE SET 3.3 Graphing Utilit FOCUS ON CONCEPTS. Let f() = (a) Show that f is one-to-one an confirm that f() = 3. (b) Fin (f ) (3).. Let f() = 3 + e. (a) Show that f is one-to-one an confirm that f(0) =. (b) Fin (f ) (). 3 4 Fin (f ) () using Formula (), an check our answer b ifferentiating f irectl. 3. f() = /( + 3) 4. f() = ln( + ) 5 6 Determine whether the function f is one-to-one b eamining the sign of f (). 5. (a) f() = (b) f() = (c) f() = + sin () f() = ( ) 6. (a) f() = (b) f() = (c) f() = + () f() = log b, 0 <b<

18 0 Chapter 3 / Topics in Differentiation 7 0 Fin the erivative of f b using Formula (3), an check our result b ifferentiating implicitl. 7. f() = f() = /, > 0 9. f() = f() = 5 sin, π 4 << π 4 FOCUS ON CONCEPTS. Figure is a proof b picture that the reflection of a point P(a,b)about the line = is the point Q(b, a). Establish this result rigorousl b completing each part. (a) Prove that if P is not on the line =, then P an Q are istinct, an the line PQ is perpenicular to the line =. (b) Prove that if P is not on the line =, the mipoint of segment PQ is on the line =. (c) Carefull eplain what it means geometricall to reflect P about the line =. () Use the results of parts (a) (c) to prove that Q is the reflection of P about the line =.. Prove that the reflection about the line = of a line with slope m, m = 0, is a line with slope /m. [Hint: Appl the result of the previous eercise to a pair of points on the line of slope m an to a corresponing pair of points on the reflection of this line about the line =.] 3. Suppose that f an g are increasing functions. Determine which of the functions f() + g(), f()g(), an f(g()) must also be increasing. 4. Suppose that f an g are one-to-one functions. Determine which of the functions f() + g(), f()g(), an f(g()) must also be one-to-one. 5 6 Fin /. 5. = e 7 6. = e 5 7. = 3 e 8. = e / 9. = e e 0. = sin(e ) e + e. = e tan. = e ln 3. = e ( e3 ) 4. = ep( ) 5. = ln( e ) 6. = ln(cos e ) 7 30 Fin f () b Formula (7) an then b logarithmic ifferentiation. 7. f() = 8. f() = 3 9. f() = π sin 30. f() = π tan 3 35 Fin / using the metho of logarithmic ifferentiation. 3. = ( 3 ) ln 3. = sin 33. = (ln ) tan 34. = ( + 3) ln 35. = (ln ) ln 36. (a) Eplain wh Formula (5) cannot be use to fin (/)[ ]. (b) Fin this erivative b logarithmic ifferentiation Fin /. 37. = sin (3) 38. = cos ( = sin (/) 40. = cos (cos ) 4. = tan ( 3 ) 4. = sec ( 5 ) 43. = (tan ) 44. = tan 45. = e sec 46. = ln(cos ) 47. = sin + cos 48. = (sin ) = sec + csc 50. = csc (e ) 5. = cot ( ) 5. = cot True False Determine whether the statement is true or false. Eplain our answer. 53. If a function = f() satisfies / =, then = e. 54. If = f() is a function such that / is a rational function, then f() is also a rational function. 55. (log b ) = ln b 56. We can conclue from the erivatives of sin an cos that sin + cos is constant. 57. (a) Use Formula () to prove that [cot ] = =0 (b) Use part (a) above, part (a) of Eercise 48 in Section 0.4, an the chain rule to show that [cot ]= + for <<+. (c) Conclue from part (b) that [cot u]= u + u for <u< (a) Use part (c) of Eercise 48 in Section 0.4 an the chain rule to show that [csc ]= for <. (b) Conclue from part (a) that [csc u]= u u u for < u. ) (cont.)

19 3.3 Derivatives of Eponential an Inverse Trigonometric Functions 03 (c) Use Equation () in Section 0.4 an parts (b) an (c) of Eercise 48 in that section to show that if then, sec + csc = π/. Conclue from part (a) that [sec ]= () Conclue from part (c) that [sec u]= u u u Fin / b implicit ifferentiation tan = e 60. sin () = cos ( ) 6. (a) Show that f() = is not one-to-one on (, + ). (b) Fin the largest value of k such that f is one-to-one on the interval ( k, k). 6. (a) Show that the function f() = 4 3 is not one-toone on (, + ). (b) Fin the smallest value of k such that f is one-to-one on the interval [k, + ). 63. Let f() = , 0. (a) Show that f is one-to-one. (b) Let g() = f () an efine F() = f(g()). Fin an equation for the tangent line to = F()at = Let f() = ep(4 ), >0. (a) Show that f is one-to-one. (b) Let g() = f () an efine F() = f([g()] ). Fin F ( ). 65. Show that for an constants A an k, the function = Ae kt satisfies the equation /t = k. 66. Show that for an constants A an B, the function = Ae + Be 4 satisfies the equation + 8 = Show that (a) = e satisfies the equation = ( ) (b) = e / satisfies the equation = ( ). 68. Show that the rate of change of = 00e 0. with respect to is proportional to. 69. Show that 60 ( = satisfies = r ) 5 + 7e t t K for some constants r an K, an etermine the values of these constants. 70. Suppose that the population of ogen-epenent bacteria in a pon is moele b the equation 60 P(t) = 5 + 7e t where P(t) is the population (in billions) t as after an initial observation at time t = 0. (a) Use a graphing utilit to graph the function P(t). (b) In wors, eplain what happens to the population over time. Check our conclusion b fining t + P(t). (c) In wors, what happens to the rate of population growth over time? Check our conclusion b graphing P (t) Fin the it b interpreting the epression as an appropriate erivative. e h 73. h w h [ ( sin )] 3 + π 3 sec w π w 7. 0 ep( ) 74. h 0 tan ( + h) π/4 h 77. Writing Let G enote the graph of an invertible function f an consier G as a fie set of points in the plane. Suppose we relabel the coorinate aes so that the -ais becomes the -ais an vice versa. Carefull eplain wh now the same set of points G becomes the graph of f (with the coorinate aes in a nonstanar position). Use this result to eplain Formula (). 78. Writing Suppose that f has an inverse function. Carefull eplain the connection between Formula () an implicit ifferentiation of the equation = f(). QUICK CHECK ANSWERS (a) es (b) no (c) no () es 3. (a) e (b) 7 ln 7 (c) e sin(e + ) () 3e f () = e 3 + (3 + ) >0for all

20 04 Chapter 3 / Topics in Differentiation 3.4 RELATED RATES In this section we will stu relate rates problems. In such problems one tries to fin the rate at which some quantit is changing b relating the quantit to other quantities whose rates of change are known. DIFFERENTIATING EQUATIONS TO RELATE RATES Figure 3.4. shows a liqui raining through a conical filter. As the liqui rains, its volume V, height h, an raius r are functions of the elapse time t, an at each instant these variables are relate b the equation V = π 3 r h If we were intereste in fining the rate of change of the volume V with respect to the time t, we coul begin b ifferentiating both sies of this equation with respect to t to obtain V = π [ r h ( t 3 t + h r r )] = π ( r h ) t 3 t + rhr t Thus, to fin V /t at a specific time t from this equation we woul nee to have values for r, h, h/t, an r/t at that time. This is calle a relate rates problem because the goal is to fin an unknown rate of change b relating it to other variables whose values an whose rates of change at time t are known or can be foun in some wa. Let us begin with a simple eample. r V h Figure 3.4. Eample Suppose that an are ifferentiable functions of t an are relate b the equation = 3. Fin /t at time t = if = an /t = 4 at time t =. Solution. Using the chain rule to ifferentiate both sies of the equation = 3 with respect to t iels = t t [3 ]=3 t Thus, the value of /t at time t = is t = 3() t= t = 4 = 48 t=

21 3.4 Relate Rates 05 Eample Assume that oil spille from a rupture tanker spreas in a circular pattern whose raius increases at a constant rate of ft/s. How fast is the area of the spill increasing when the raius of the spill is 60 ft? Solution. Let t = number of secons elapse from the time of the spill r = raius of the spill in feet after t secons A = area of the spill in square feet after t secons (Figure 3.4.). We know the rate at which the raius is increasing, an we want to fin the rate at which the area is increasing at the instant when r = 60; that is, we want to fin A r given that t t = ft /s r=60 Arni Katz/Phototake Oil spill from a rupture tanker. This suggests that we look for an equation relating A an r that we can ifferentiate with respect to t to prouce a relationship between A/t an r/t. But A is the area of a circle of raius r, so A = πr () Oil spill Differentiating both sies of () with respect to t iels A t = πr r t Thus, when r = 60 the area of the spill is increasing at the rate of A t = π(60)() = 40π ft /s 754 ft /s r=60 () r With some minor variations, the metho use in Eample can be use to solve a variet of relate rates problems. We can break the metho own into five steps. Figure 3.4. WARNING We have italicize the wor After in Step 5 because it is a common error to substitute numerical values before performing the ifferentiation. For instance, in Eample ha we substitute the known value of r = 60 in () before ifferentiating, we woul have obtaine A/t = 0, which is obviousl incorrect. A Strateg for Solving Relate Rates Problems Step. Assign letters to all quantities that var with time an an others that seem relevant to the problem. Give a efinition for each letter. Step. Ientif the rates of change that are known an the rate of change that is to be foun. Interpret each rate as a erivative. Step 3. Fin an equation that relates the variables whose rates of change were ientifie in Step. To o this, it will often be helpful to raw an appropriatel labele figure that illustrates the relationship. Step 4. Differentiate both sies of the equation obtaine in Step 3 with respect to time to prouce a relationship between the known rates of change an the unknown rate of change. Step 5. After completing Step 4, substitute all known values for the rates of change an the variables, an then solve for the unknown rate of change.

22 06 Chapter 3 / Topics in Differentiation 3r 90 ft Figure The quantit n Home t =0 is negative because is ecreasing with respect to t. st Eample 3 A baseball iamon is a square whose sies are 90 ft long (Figure 3.4.3). Suppose that a plaer running from secon base to thir base has a spee of 30 ft/s atthe instant when he is 0 ft from thir base. At what rate is the plaer s istance from home plate changing at that instant? Solution. We are given a constant spee with which the plaer is approaching thir base, an we want to fin the rate of change of the istance between the plaer an home plate at a particular instant. Thus, let t = number of secons since the plaer left secon base = istance in feet from the plaer to thir base = istance in feet from the plaer to home plate (Figure 3.4.4). Thus, we want to fin given that t =0 t = 30 ft/s =0 As suggeste b Figure 3.4.4, an equation relating the variables an can be obtaine using the Theorem of Pthagoras: + 90 = (3) Differentiating both sies of this equation with respect to t iels n from which we obtain t t = t = t (4) 3r st When = 0, it follows from (3) that = = 8500 = Figure Home so that (4) iels t = 0 = ( 30) = 6.5 ft/s The negative sign in the answer tells us that is ecreasing, which makes sense phsicall from Figure Rocket Eample 4 In Figure we have shown a camera mounte at a point 3000 ft from the base of a rocket launching pa. If the rocket is rising verticall at 880 ft/s when it is 4000 ft above the launching pa, how fast must the camera elevation angle change at that instant to keep the camera aime at the rocket? Camera Figure Elevation angle 3000 ft Launching pa Solution. Let t = number of secons elapse from the time of launch φ = camera elevation angle in raians after t secons h = height of the rocket in feet after t secons (Figure 3.4.6). At each instant the rate at which the camera elevation angle must change

23 3.4 Relate Rates 07 Rocket h f 3000 ft Camera Figure f 3000 Figure cm is φ/t, an the rate at which the rocket is rising is h/t. We want to fin φ h t given that h=4000 t = 880 ft/s h=4000 From Figure we see that tan φ = h 3000 Differentiating both sies of (5) with respect to t iels (sec φ) φ t = h 3000 t When h = 4000, it follows that (sec φ) h=4000 = = 5 3 (see Figure 3.4.7), so that from (6) ( ) 5 φ 3 t φ t = 3000 h=4000 = h= = = ra /s 6.05 eg/s Eample 5 Suppose that liqui is to be cleare of seiment b allowing it to rain through a conical filter that is 6 cm high an has a raius of 4 cm at the top (Figure 3.4.8). Suppose also that the liqui is force out of the cone at a constant rate of cm 3 /min. (5) (6) Filter Funnel to hol filter Figure r 6 cm (a) Do ou think that the epth of the liqui will ecrease at a constant rate? Give a verbal argument that justifies our conclusion. (b) Fin a formula that epresses the rate at which the epth of the liqui is changing in terms of the epth, an use that formula to etermine whether our conclusion in part (a) is correct. (c) At what rate is the epth of the liqui changing at the instant when the liqui in the cone is 8 cm eep? Solution (a). For the volume of liqui to ecrease b a fie amount, it requires a greater ecrease in epth when the cone is close to empt than when it is almost full (Figure 3.4.9). This suggests that for the volume to ecrease at a constant rate, the epth must ecrease at an increasing rate. The same volume has raine, but the change in height is greater near the bottom than near the top. Figure Solution (b). Let t = time elapse from the initial observation (min) V = volume of liqui in the cone at time t (cm 3 ) = epth of the liqui in the cone at time t (cm) r = raius of the liqui surface at time t (cm) (Figure 3.4.8). At each instant the rate at which the volume of liqui is changing is V /t, an the rate at which the epth is changing is /t. We want to epress /t in terms of given that V /t has a constant value of V /t =. (We must use a minus sign here because V ecreases as t increases.) From the formula for the volume of a cone, the volume V, the raius r, an the epth are relate b V = 3 πr (7)

24 08 Chapter 3 / Topics in Differentiation If we ifferentiate both sies of (7) with respect to t, the right sie will involve the quantit r/t. Since we have no irect information about r/t, it is esirable to einate r from (7) before ifferentiating. This can be one using similar triangles. From Figure we see that r = 4 6 or r = 4 Substituting this epression in (7) gives V = π 48 3 (8) Differentiating both sies of (8) with respect to t we obtain V = π ( 3 ) t 48 t or = 6 V = 6 ( ) = 3 (9) t π t π π which epresses /t in terms of. The minus sign tells us that is ecreasing with time, an t = 3 π tells us how fast is ecreasing. From this formula we see that /t increases as ecreases, which confirms our conjecture in part (a) that the epth of the liqui ecreases more quickl as the liqui rains through the filter. Solution (c). The rate at which the epth is changing when the epth is 8 cm can be obtaine from (9) with = 8: t = 3 =8 π(8 ) = π 0.6 cm /min QUICK CHECK EXERCISES 3.4. If A = an t. If A = an A t = 3, fin A t = 3, fin t. =0. =0 (See page for answers.) 3. A 0-foot laer stans on a horizontal floor an leans against a vertical wall. Use to enote the istance along the floor from the wall to the foot of the laer, an use to enote the istance along the wall from the floor to the top of the laer. If the foot of the laer is ragge awa from the wall, fin an equation that relates rates of change of an with respect to time. 4. Suppose that a block of ice in the shape of a right circular cliner melts so that it retains its clinrical shape. Fin an equation that relates the rates of change of the volume (V ), height (h), an raius (r) of the block of ice. EXERCISE SET Both an enote functions of t that are relate b the given equation. Use this equation an the given erivative information to fin the specifie erivative.. Equation: = (a) Given that /t =, fin /t when =. (b) Given that /t =, fin /t when = 0.. Equation: + 4 = 3. (a) Given that /t =, fin /t when =. (b) Given that /t = 4, fin /t when = Equation: =. (a) Given that ( /t = 3, fin /t when (, ) =, 3 ). (cont.)

25 3.4 Relate Rates 09 (b) Given that ( /t = 8, fin /t when (, ) = 3, ) Equation: + = + 4. (a) Given that /t = 5, fin /t when (, ) = (3, ). (b) Given that /t = 6, fin /t when (, ) = ( +, + 3 ). FOCUS ON CONCEPTS 5. Let A be the area of a square whose sies have length, an assume that varies with the time t. (a) Draw a picture of the square with the labels A an place appropriatel. (b) Write an equation that relates A an. (c) Use the equation in part (b) to fin an equation that relates A/t an /t. () At a certain instant the sies are 3 ft long an increasing at a rate of ft/min. How fast is the area increasing at that instant? 6. In parts (a) (), let A be the area of a circle of raius r, an assume that r increases with the time t. (a) Draw a picture of the circle with the labels A an r place appropriatel. (b) Write an equation that relates A an r. (c) Use the equation in part (b) to fin an equation that relates A/t an r/t. () At a certain instant the raius is 5 cm an increasing at the rate of cm/s. How fast is the area increasing at that instant? 7. Let V be the volume of a cliner having height h an raius r, an assume that h an r var with time. (a) How are V /t, h/t, an r/t relate? (b) At a certain instant, the height is 6 in an increasing atin/s, while the raius is 0 in an ecreasing atin/s. How fast is the volume changing at that instant? Is the volume increasing or ecreasing at that instant? 8. Let l be the length of a iagonal of a rectangle whose sies have lengths an, an assume that an var with time. (a) How are l/t, /t, an /t relate? (b) If increases at a constant rate of ft /s an ecreases at a constant rate of 4 ft /s, how fast is the size of the iagonal changing when = 3ftan = 4 ft? Is the iagonal increasing or ecreasing at that instant? 9. Let θ (in raians) be an acute angle in a right triangle, an let an, respectivel, be the lengths of the sies ajacent to an opposite θ. Suppose also that an var with time. (a) How are θ/t, /t, an /t relate? (b) At a certain instant, = units an is increasing at unit/s, while = units an is ecreasing at 4 unit/s. How fast is θ changing at that instant? Is θ increasing or ecreasing at that instant? 0. Suppose that z = 3, where both an are changing with time. At a certain instant when = an =, is ecreasing at the rate of units/s, an is increasing at the rate of 3 units/s. How fast is z changing at this instant? Is z increasing or ecreasing?. The minute han of a certain clock is 4 in long. Starting from the moment when the han is pointing straight up, how fast is the area of the sector that is swept out b the han increasing at an instant uring the net revolution of the han?. A stone roppe into a still pon sens out a circular ripple whose raius increases at a constant rate of 3 ft/s. How rapil is the area enclose b the ripple increasing at the en of 0 s? 3. Oil spille from a rupture tanker spreas in a circle whose area increases at a constant rate of 6 mi /h. How fast is the raius of the spill increasing when the area is 9 mi? 4. A spherical balloon is inflate so that its volume is increasing at the rate of 3 ft 3 /min. How fast is the iameter of the balloon increasing when the raius is ft? 5. A spherical balloon is to be eflate so that its raius ecreases at a constant rate of 5 cm/min. At what rate must air be remove when the raius is 9 cm? 6. A 7 ft laer is leaning against a wall. If the bottom of the laer is pulle along the groun awa from the wall at a constant rate of 5 ft/s, how fast will the top of the laer be moving own the wall when it is 8 ft above the groun? 7. A 3 ft laer is leaning against a wall. If the top of the laer slips own the wall at a rate of ft/s, how fast will the foot be moving awa from the wall when the top is 5 ft above the groun? 8. A 0 ft plank is leaning against a wall. If at a certain instant the bottom of the plank is ft from the wall an is being pushe towar the wall at the rate of 6 in/s, how fast is the acute angle that the plank makes with the groun increasing? 9. A softball iamon is a square whose sies are 60 ft long. Suppose that a plaer running from first to secon base has a spee of 5 ft/s at the instant when she is 0 ft from secon base. At what rate is the plaer s istance from home plate changing at that instant? 0. A rocket, rising verticall, is tracke b a raar station that is on the groun 5 mi from the launchpa. How fast is the rocket rising when it is 4 mi high an its istance from the raar station is increasing at a rate of 000 mi/h?. For the camera an rocket shown in Figure 3.4.5, at what rate is the camera-to-rocket istance changing when the rocket is 4000 ft up an rising verticall at 880 ft/s?

26 0 Chapter 3 / Topics in Differentiation. For the camera an rocket shown in Figure 3.4.5, at what rate is the rocket rising when the elevation angle is π/4 raians an increasing at a rate of 0. ra/s? 3. A satellite is in an elliptical orbit aroun the Earth. Its istance r (in miles) from the center of the Earth is given b 4995 r = + 0. cos θ where θ is the angle measure from the point on the orbit nearest the Earth s surface (see the accompaning figure). (a) Fin the altitue of the satellite at perigee (the point nearest the surface of the Earth) an at apogee (the point farthest from the surface of the Earth). Use 3960 mi as the raius of the Earth. (b) At the instant when θ is 0, the angle θ is increasing at the rate of.7 /min. Fin the altitue of the satellite an the rate at which the altitue is changing at this instant. Epress the rate in units of mi/min. constant rate of 5 ft/min, at what rate is san pouring from the chute when the pile is 0 ft high? 8. Wheat is poure through a chute at the rate of 0 ft 3 /min an falls in a conical pile whose bottom raius is alwas half the altitue. How fast will the circumference of the base be increasing when the pile is 8 ft high? 9. An aircraft is cbing at a 30 angle to the horizontal. How fast is the aircraft gaining altitue if its spee is 500 mi/h? 30. A boat is pulle into a ock b means of a rope attache to a pulle on the ock (see the accompaning figure). The rope is attache to the bow of the boat at a point 0 ft below the pulle. If the rope is pulle through the pulle at a rate of 0 ft/min, at what rate will the boat be approaching the ock when 5 ft of rope is out? Pulle Boat Dock Figure E-30 Apogee r u Perigee Figure E-3 4. An aircraft is fling horizontall at a constant height of 4000 ft above a fie observation point (see the accompaning figure). At a certain instant the angle of elevation θ is 30 an ecreasing, an the spee of the aircraft is 300 mi/h. (a) How fast is θ ecreasing at this instant? Epress the result in units of eg/s. (b) How fast is the istance between the aircraft an the observation point changing at this instant? Epress the result in units of ft/s. Use mi = 580 ft. u 4000 ft Figure E-4 5. A conical water tank with verte own has a raius of 0 ft at the top an is 4 ft high. If water flows into the tank at a rate of 0 ft 3 /min, how fast is the epth of the water increasing when the water is 6 ft eep? 6. Grain pouring from a chute at the rate of 8 ft 3 /min forms a conical pile whose height is alwas twice its raius. How fast is the height of the pile increasing at the instant when the pile is 6 ft high? 7. San pouring from a chute forms a conical pile whose height is alwas equal to the iameter. If the height increases at a 3. For the boat in Eercise 30, how fast must the rope be pulle if we want the boat to approach the ock at a rate of ft/min at the instant when 5 ft of rope is out? 3. A man 6 ft tall is walking at the rate of 3 ft/s towar a streetlight 8 ft high (see the accompaning figure). (a) At what rate is his shaow length changing? (b) How fast is the tip of his shaow moving? Figure E A beacon that makes one revolution ever 0 s is locate on a ship anchore 4 kilometers from a straight shoreline. How fast is the beam moving along the shoreline when it makes an angle of 45 with the shore? 34. An aircraft is fling at a constant altitue with a constant spee of 600 mi/h. An antiaircraft missile is fire on a straight line perpenicular to the flight path of the aircraft so that it will hit the aircraft at a point P (see the accompaning figure). At the instant the aircraft is mi from the impact point P the missile is 4 mi from P an fling at 00 mi/h. At that instant, how rapil is the istance between missile an aircraft ecreasing? P Figure E-34

27 3.4 Relate Rates 35. Solve Eercise 34 uner the assumption that the angle between the flight paths is 0 instea of the assumption that the paths are perpenicular. [Hint: Use the law of cosines.] 36. A police helicopter is fling ue north at 00 mi/h an at a constant altitue of mi. Below, a car is traveling west on a highwa at 75 mi/h. At the moment the helicopter crosses over the highwa the car is mi east of the helicopter. (a) How fast is the istance between the car an helicopter changing at the moment the helicopter crosses the highwa? (b) Is the istance between the car an helicopter increasing or ecreasing at that moment? 37. A particle is moving along the curve whose equation is 3 + = 8 5 Assume that the -coorinate is increasing at the rate of 6 units/s when the particle is at the point (, ). (a) At what rate is the -coorinate of the point changing at that instant? (b) Is the particle rising or falling at that instant? 38. A point P is moving along the curve whose equation is = When P is at (, 5), is increasing at the rate of units/s. How fast is changing? 39. A point P is moving along the line whose equation is =. How fast is the istance between P an the point (3, 0) changing at the instant when P is at (3, 6) if is ecreasing at the rate of units/s at that instant? 40. A point P is moving along the curve whose equation is =. Suppose that is increasing at the rate of 4 units/s when = 3. (a) How fast is the istance between P an the point (, 0) changing at this instant? (b) How fast is the angle of inclination of the line segment from P to (, 0) changing at this instant? 4. A particle is moving along the curve = /( + ). Fin all values of at which the rate of change of with respect to time is three times that of. [Assume that /t is never zero.] 4. A particle is moving along the curve = 44. Fin all points (, ) at which the rates of change of an with respect to time are equal. [Assume that /t an /t are never both zero at the same point.] 43. The thin lens equation in phsics is s + S = f where s is the object istance from the lens, S is the image istance from the lens, an f is the focal length of the lens. Suppose that a certain lens has a focal length of 6 cm an that an object is moving towar the lens at the rate of cm/s. How fast is the image istance changing at the instant when the object is 0 cm from the lens? Is the image moving awa from the lens or towar the lens? 44. Water is store in a cone-shape reservoir (verte own). Assuming the water evaporates at a rate proportional to the surface area epose to the air, show that the epth of the water will ecrease at a constant rate that oes not epen on the imensions of the reservoir. 45. A meteor enters the Earth s atmosphere an burns up at a rate that, at each instant, is proportional to its surface area. Assuming that the meteor is alwas spherical, show that the raius ecreases at a constant rate. 46. On a certain clock the minute han is 4 in long an the hour han is 3 in long. How fast is the istance between the tips of the hans changing at 9 o clock? 47. Coffee is poure at a uniform rate of 0 cm 3 /s into a cup whose insie is shape like a truncate cone (see the accompaning figure). If the upper an lower raii of the cup are 4 cm an cm an the height of the cup is 6 cm, how fast will the coffee level be rising when the coffee is halfwa up? [Hint: Eten the cup ownwar to form a cone.] Figure E-47 QUICK CHECK ANSWERS t + = 0 4. t V t = πrh r t + πr h t

28 Chapter 3 / Topics in Differentiation 3.5 LOCAL LINEAR APPROXIMATION; DIFFERENTIALS In this section we will show how erivatives can be use to approimate nonlinear functions b linear functions. Also, up to now we have been interpreting / as a single entit representing the erivative. In this section we will efine the quantities an themselves, thereb allowing us to interpret / as an actual ratio. Recall from Section. that if a function f is ifferentiable at 0, then a sufficientl magnifie portion of the graph of f centere at the point P( 0,f( 0 )) takes on the appearance of a straight line segment. Figure 3.5. illustrates this at several points on the graph of = +. For this reason, a function that is ifferentiable at 0 is sometimes sai to be locall linear at 0. The line that best approimates the graph of f in the vicinit of P( 0,f( 0 )) is the tangent line to the graph of f at 0, given b the equation = f( 0 ) + f ( 0 )( 0 ) [see Formula (3) of Section.]. Thus, for values of near 0 we can approimate values of f() b f() f( 0 ) + f ( 0 )( 0 ) () This is calle the local linear approimation of f at 0. This formula can also be epresse in terms of the increment = 0 as Magnifing portions of the graph of = + f( 0 + ) f( 0 ) + f ( 0 ) () Figure 3.5. Eample (a) Fin the local linear approimation of f() = at 0 =. (b) Use the local linear approimation obtaine in part (a) to approimate., an compare our approimation to the result prouce irectl b a calculating utilit. Solution (a). Since f () = /( ), it follows from () that the local linear approimation of at a point 0 is 0 + ( 0 ) 0 Thus, the local linear approimation at 0 = is + ( ) (3) (, ) Figure 3.5. = + ( ) = f() = 3 4 The graphs of = an the local linear approimation = + ( ) are shown in Figure Solution (b). Appling (3) with =. iels. + (. ) =.05 Since the tangent line = + ( ) in Figure 3.5. lies above the graph of f() =, we woul epect this approimation to be slightl too large. This epectation is confirme b the calculator approimation

29 3.5 Local Linear Approimation; Differentials 3 Eamples an illustrate important ieas an are not meant to suggest that ou shoul use local linear approimations for computations that our calculating utilit can perform. The main application of local linear approimation is in moeling problems where it is useful to replace complicate functions b simpler ones Figure = = sin Eample (a) Fin the local linear approimation of f() = sin at 0 = 0. (b) Use the local linear approimation obtaine in part (a) to approimate sin, an compare our approimation to the result prouce irectl b our calculating evice. Solution (a). Since f () = cos, it follows from () that the local linear approimation of sin at a point 0 is sin sin 0 + (cos 0 )( 0 ) Thus, the local linear approimation at 0 = 0is sin sin 0 + (cos 0)( 0) which simplifies to sin (4) Solution (b). The variable in (4) is in raian measure, so we must first convert to raians before we can appl this approimation. Since ( π ) = = π raian it follows from (4) that sin Comparing the two graphs in Figure 3.5.3, we woul epect this approimation to be slightl larger than the eact value. The calculator approimation sin shows that this is inee the case. E E() = sin Figure ERROR IN LOCAL LINEAR APPROXIMATIONS As a general rule, the accurac of the local linear approimation to f()at 0 will eteriorate as gets progressivel farther from 0. To illustrate this for the approimation sin in Eample, let us graph the function E() = sin which is the absolute value of the error in the approimation (Figure 3.5.4). In Figure 3.5.4, the graph shows how the absolute error in the local linear approimation of sin increases as moves progressivel farther from 0 in either the positive or negative irection. The graph also tells us that for values of between the two vertical lines, the absolute error oes not ecee 0.0. Thus, for eample, we coul use the local linear approimation sin for all values of in the interval 0.35 <<0.35 (raians) with confience that the approimation is within ±0.0 of the eact value. DIFFERENTIALS Newton an Leibniz each use a ifferent notation when the publishe their iscoveries of calculus, thereb creating a notational ivie between Britain an the European continent that laste for more than 50 ears. The Leibniz notation / eventuall prevaile because it suggests correct formulas in a natural wa, the chain rule = u u being a goo eample. Up to now we have interprete / as a single entit representing the erivative of with respect to ; the smbols an, which are calle ifferentials, have ha no meanings attache to them. Our net goal is to efine these smbols in such a wa that / can be treate as an actual ratio. To o this, assume that f is ifferentiable at a point, efine to be an inepenent variable that can have an real value, an efine b the formula = f () (5)

30 4 Chapter 3 / Topics in Differentiation = f() Slope = f () Rise = Run = + Figure If = 0, then we can ivie both sies of (5) b to obtain = f () (6) Thus, we have achieve our goal of efining an so their ratio is f (). Formula (5) is sai to epress (6) in ifferential form. To interpret (5) geometricall, note that f () is the slope of the tangent line to the graph of f at. The ifferentials an can be viewe as a corresponing rise an run of this tangent line (Figure 3.5.5). 6 5 = Eample 3 Epress the erivative with respect to of = in ifferential form, an iscuss the relationship between an at =. Solution. The erivative of with respect to is / =, which can be epresse in ifferential form as = When = this becomes = This tells us that if we travel along the tangent line to the curve = at =, then a change of units in prouces a change of units in. Thus, for eample, a run of = units prouces a rise of = 4 units along the tangent line (Figure 3.5.6). Figure = f () Δ It is important to unerstan the istinction between the increment an the ifferential. To see the ifference, let us assign the inepenent variables an the same value, so =. Then represents the change in that occurs when we start at an travel along the curve = f() until we have move (= ) units in the -irection, while represents the change in that occurs if we start at an travel along the tangent line until we have move (= ) units in the -irection (Figure 3.5.7). Figure = Figure Δ = Δ ( + ) = 0.75 Δ 0.65 Eample 4 Let =. Fin an at = 4 with = = 3. Then make a sketch of =, showing an in the picture. Solution. With f() = we obtain = f( + ) f() = + = If =, then =, so = = 4 (3) = 3 4 = 0.75 Figure shows the curve = together with an. LOCAL LINEAR APPROXIMATION FROM THE DIFFERENTIAL POINT OF VIEW Although an are generall ifferent, the ifferential will nonetheless be a goo approimation of provie = is close to 0. To see this, recall from Section. that f () = 0 It follows that if is close to 0, then we will have f () / or, equivalentl, f () If we agree to let =, then we can rewrite this as f () = (7)

31 3.5 Local Linear Approimation; Differentials 5 In wors, this states that for values of near zero the ifferential closel approimates the increment (Figure 3.5.7). But this is to be epecte since the graph of the tangent line at is the local linear approimation of the graph of f. Michael Newman/PhotoEit Real-worl measurements inevitabl have small errors. Note that measurement error is positive if the measure value is greater than the eact value an is negative if it is less than the eact value. The sign of the propagate error conves similar information. Eplain wh an error estimate of at most ± 3 inch is reasonable for a ruler that is calibrate in siteenths of an inch. ERROR PROPAGATION In real-worl applications, small errors in measure quantities will invariabl occur. These measurement errors are of importance in scientific research all scientific measurements come with measurement errors inclue. For eample, our height might be measure as 70 ± 0.5 cm, meaning that our eact height lies somewhere between 69.5 an 70.5 cm. Researchers often must use these ineactl measure quantities to compute other quantities, thereb propagating the errors from the measure quantities to the compute quantities. This phenomenon is calle error propagation. Researchers must be able to estimate errors in the compute quantities. Our goal is to show how to estimate these errors using local linear approimation an ifferentials. For this purpose, suppose 0 is the eact value of the quantit being measure 0 = f( 0 ) is the eact value of the quantit being compute is the measure value of 0 = f() is the compute value of We efine (= ) = 0 to be the measurement error of = f() f( 0 ) to be the propagate error of It follows from (7) with 0 replacing that the propagate error can be approimate b = f ( 0 ) (8) Unfortunatel, there is a practical ifficult in appling this formula since the value of 0 is unknown. (Keep in min that onl the measure value is known to the researcher.) This being the case, it is stanar practice in research to use the measure value in place of 0 in (8) an use the approimation = f () (9) for the propagate error. Eample 5 Suppose that the sie of a square is measure with a ruler to be 0 inches with a measurement error of at most ± in. Estimate the error in the compute area of the 3 square. Solution. Let enote the eact length of a sie an the eact area so that =.It follows from (9) with f() = that if is the measurement error, then the propagate error can be approimate as = Substituting the measure value = 0 into this equation iels = 0 (0) But to sa that the measurement error is at most ± means that Multipling these inequalities through b 0 an appling (0) iels 0 ( 3) ( 0 ) or equivalentl Thus, the propagate error in the area is estimate to be within ± 5 8 in.

32 6 Chapter 3 / Topics in Differentiation If the true value of a quantit is q an a measurement or calculation prouces an error q, then q/q is calle the relative error in the measurement or calculation; when epresse as a percentage, q/q is calle the percentage error. As a practical matter, the true value q is usuall unknown, so that the measure or calculate value of q is use instea; an the relative error is approimate b q/q. Eample 6 The raius of a sphere is measure with a percentage error within ±0.04%. Estimate the percentage error in the calculate volume of the sphere. Formula () tells us that, as a rule of thumb, the percentage error in the compute volume of a sphere is approimatel 3 times the percentage error in the measure value of its raius. As a rule of thumb, how is the percentage error in the compute area of a square relate to the percentage error in the measure value of a sie? Solution. The volume V of a sphere is V = 4 3 πr3,so V r = 4πr from which it follows that V = 4πr r. Thus, the relative error in V is approimatel V V = 4πr r = 3 r 4 3 πr3 r We are given that the relative error in the measure value of r is ±0.04%, which means that r r Multipling these inequalities through b 3 an appling () iels () 3( ) V V 3(0.0004) or equivalentl 0.00 V V 0.00 Thus, we estimate the percentage error in the calculate value of V to be within ±0.%. MORE NOTATION; DIFFERENTIAL FORMULAS The smbol f is another common notation for the ifferential of a function = f(). For eample, if f() = sin, then we can write f = cos. We can also view the smbol as an operator that acts on a function to prouce the corresponing ifferential. For eample, [ ]=,[sin ] =cos, an so on. All of the general rules of ifferentiation then have corresponing ifferential versions: erivative formula ifferential formula [c] = 0 [c] = 0 f [cf] = c f [cf ] = c f g [ f + g] = f + g [ f + g] = + g f [ fg] = f g + g f [ fg] = f + g f g g f f f g f f g = g g = g g For eample, [ sin ] =( cos + sin ) = (cos )+ ()sin = [sin ]+(sin )[ ] illustrates the ifferential version of the prouct rule.

33 3.5 Local Linear Approimation; Differentials 7 QUICK CHECK EXERCISES 3.5 (See page 9 for answers.). The local linear approimation of f at 0 uses the line to the graph of = f() at = 0 to approimate values of for values of near.. Fin an equation for the local linear approimation to = 5 at 0 =. 3. Let = 5. Fin an at = with = = The intensit of light from a light source is a function I = f() of the istance from the light source. Suppose that a small gemstone is measure to be 0 m from a light source, f(0) = 0.W/m, an f (0) = 0.04 W/m 3.If the istance = 0 m was obtaine with a measurement error within ±0.05 m, estimate the percentage error in the calculate intensit of the light on the gemstone. EXERCISE SET 3.5 Graphing Utilit. (a) Use Formula () to obtain the local linear approimation of 3 at 0 =. (b) Use Formula () to rewrite the approimation obtaine in part (a) in terms of. (c) Use the result obtaine in part (a) to approimate (.0) 3, an confirm that the formula obtaine in part (b) prouces the same result.. (a) Use Formula () to obtain the local linear approimation of / at 0 =. (b) Use Formula () to rewrite the approimation obtaine in part (a) in terms of. (c) Use the result obtaine in part (a) to approimate /.05, an confirm that the formula obtaine in part (b) prouces the same result. FOCUS ON CONCEPTS 3. (a) Fin the local linear approimation of the function f() = + at 0 = 0, an use it to approimate 0.9 an.. (b) Graph f an its tangent line at 0 together, an use the graphs to illustrate the relationship between the eact values an the approimations of 0.9 an.. 4. Astuent claims that whenever a local linear approimation is use to approimate the square root of a number, the approimation is too large. (a) Write a few sentences that make the stuent s claim precise, an justif this claim geometricall. (b) Verif the stuent s claim algebraicall using approimation (). 5 0 Confirm that the state formula is the local linear approimation at 0 = ( + ) tan e + 0. ln( + ) 6 Confirm that the state formula is the local linear approimation of f at 0 =, where =.. f() = 4 ; ( + ) f() = ; f() = + ; f() = (4 + ) 3 ; (5 + ) tan ; tan ( + ) π 4 + ( ( 6. sin ); sin + ) π Confirm that the formula is the local linear approimation at 0 = 0, an use a graphing utilit to estimate an interval of -values on which the error is at most ± tan 0. ( + ) 0 5. (a) Use the local linear approimation of sin at 0 = 0 obtaine in Eample to approimate sin, an compare the approimation to the result prouce irectl b our calculating evice. (b) How woul ou choose 0 to approimate sin 44? (c) Approimate sin 44 ; compare the approimation to the result prouce irectl b our calculating evice.. (a) Use the local linear approimation of tan at 0 = 0to approimate tan, an compare the approimation to the result prouce irectl b our calculating evice. (b) How woul ou choose 0 to approimate tan 6? (c) Approimate tan 6 ; compare the approimation to the result prouce irectl b our calculating evice. 3 3 Use an appropriate local linear approimation to estimate the value of the given quantit. 3. (3.0) 4 4. (.97)

34 8 Chapter 3 / Topics in Differentiation sin tan cos 3 3. ln(.0) 33. tan (0.99) FOCUS ON CONCEPTS 34. The approimation ( + ) k + k is commonl use b engineers for quick calculations. (a) Derive this result, an use it to make a rough estimate of (.00) 37. (b) Compare our estimate to that prouce irectl b our calculating evice. (c) If k is a positive integer, how is the approimation ( + ) k + k relate to the epansion of ( + ) k using the binomial theorem? 35. Use the approimation ( + ) k + k, along with some mental arithmetic to show that an / Referring to the accompaning figure, suppose that the angle of elevation of the top of the builing, as measure from a point 500 ft from its base, is foun to be θ = 6. Use an appropriate local linear approimation, along with some mental arithmetic to show that the builing is about 5 ft high. u h 500 ft Figure E (a) Let =. Fin an at = with = =. (b) Sketch the graph of =, showing an in the picture. 38. (a) Let = 3. Fin an at = with = =. (b) Sketch the graph of = 3, showing an in the picture Fin formulas for an. 39. = = = + 4. = sin Fin the ifferential. 43. (a) = (b) = cos 44. (a) = / (b) = 5 tan 45. (a) = (b) = ( + ) (a) = 3 (b) = True False Determine whether the statement is true or false. Eplain our answer. 47. A ifferential is efine to be a ver small change in. 48. The error in approimation () is the same as the error in approimation (7). 49. Alocal linear approimation to a function can never be ienticall equal to the function. 50. A local linear approimation to a nonconstant function can never be constant Use the ifferential to approimate when changes as inicate. 5. = 3 ; from = to = = + 8; from = to = = ; from = to = = 8 + ; from = 3to = The sie of a square is measure to be 0 ft, with a possible error of ±0. ft. (a) Use ifferentials to estimate the error in the calculate area. (b) Estimate the percentage errors in the sie an the area. 56. The sie of a cube is measure to be 5 cm, with a possible error of ± cm. (a) Use ifferentials to estimate the error in the calculate volume. (b) Estimate the percentage errors in the sie an volume. 57. The hpotenuse of a right triangle is known to be 0 in eactl, an one of the acute angles is measure to be 30, with a possible error of ±. (a) Use ifferentials to estimate the errors in the sies opposite an ajacent to the measure angle. (b) Estimate the percentage errors in the sies. 58. One sie of a right triangle is known to be 5 cm eactl. The angle opposite to this sie is measure to be 60, with a possible error of ±0.5. (a) Use ifferentials to estimate the errors in the ajacent sie an the hpotenuse. (b) Estimate the percentage errors in the ajacent sie an hpotenuse. 59. The electrical resistance R of a certain wire is given b R = k/r, where k is a constant an r is the raius of the wire. Assuming that the raius r has a possible error of ±5%, use ifferentials to estimate the percentage error in R. (Assume k is eact.) 60. A -foot laer leaning against a wall makes an angle θ with the floor. If the top of the laer is h feet up the wall, epress h in terms of θ an then use h to estimate the change in h if θ changes from 60 to The area of a right triangle with a hpotenuse of H is calculate using the formula A = 4 H sin θ, where θ is one of the acute angles. Use ifferentials to approimate the error in calculating A if H = 4 cm (eactl) an θ is measure to be 30, with a possible error of ±5.

35 3.6 L Hôpital s Rule; Ineterminate Forms 9 6. The sie of a square is measure with a possible percentage error of ±%. Use ifferentials to estimate the percentage error in the area. 63. The sie of a cube is measure with a possible percentage error of ±%. Use ifferentials to estimate the percentage error in the volume. 64. The volume of a sphere is to be compute from a measure value of its raius. Estimate the maimum permissible percentage error in the measurement if the percentage error in the volume must be kept within ±3%. (V = 4 3 πr3 is the volume of a sphere of raius r.) 65. The area of a circle is to be compute from a measure value of its iameter. Estimate the maimum permissible percentage error in the measurement if the percentage error in the area must be kept within ±%. 66. A steel cube with -inch sies is coate with 0.0 inch of copper. Use ifferentials to estimate the volume of copper in the coating. [Hint: Let V be the change in the volume of the cube.] 67. A metal ro 5 cm long an 5 cm in iameter is to be covere (ecept for the ens) with insulation that is 0. cm thick. Use ifferentials to estimate the volume of insulation. [Hint: Let V be the change in volume of the ro.] 68. The time require for one complete oscillation of a penulum is calle its perio. If L is the length of the penulum an the oscillation is small, then the perio is given b P = π L/g, where g is the constant acceleration ue to gravit. Use ifferentials to show that the percentage error in P is approimatel half the percentage error in L. 69. If the temperature T of a metal ro of length L is change b an amount T, then the length will change b the amount L = αl T, where α is calle the coefficient of linear epansion. For moerate changes in temperature α is taken as constant. (a) Suppose that a ro 40 cm long at 0 C is foun to be cm long when the temperature is raise to 30 C. Fin α. (b) If an aluminum pole is 80 cm long at 5 C, how long is the pole if the temperature is raise to 40 C? [Take α = / C.] 70. If the temperature T of a soli or liqui of volume V is change b an amount T, then the volume will change b the amount V = βv T, where β is calle the coefficient of volume epansion. For moerate changes in temperature β is taken as constant. Suppose that a tank truck loas 4000 gallons of ethl alcohol at a temperature of 35 C an elivers its loa sometime later at a temperature of 5 C. Using β = / C for ethl alcohol, fin the number of gallons elivere. 7. Writing Eplain wh the local linear approimation of a function value is equivalent to the use of a ifferential to approimate a change in the function. 7. Writing The local linear approimation sin is known as the small angle approimation an has both practical an theoretical applications. Do some research on some of these applications, an write a short report on the results of our investigations. QUICK CHECK ANSWERS 3.5. tangent; f(); 0. = + ( 4)( ) or = = 0.4, = within ±% 3.6 L HÔPITAL S RULE; INDETERMINATE FORMS In this section we will iscuss a general metho for using erivatives to fin its. This metho will enable us to establish its with certaint that earlier in the tet we were onl able to conjecture using numerical or graphical evience. The metho that we will iscuss in this section is an etremel powerful tool that is use internall b man computer programs to calculate its of various tpes. INDETERMINATE FORMS OF TYPE 0/0 Recall that a it of the form f() () a g() in which f() 0 an g() 0 as a is calle an ineterminate form of tpe 0/0. Some eamples encountere earlier in the tet are =, sin 0 =, 0 cos = 0

36 0 Chapter 3 / Topics in Differentiation The first it was obtaine algebraicall b factoring the numerator an canceling the common factor of, an the secon two its were obtaine using geometric methos. However, there are man ineterminate forms for which neither algebraic nor geometric methos will prouce the it, so we nee to evelop a more general metho. To motivate such a metho, suppose that () is an ineterminate form of tpe 0/0 in which f an g are continuous at = a an g (a) = 0. Since f an g can be closel approimate b their local linear approimations near a, it is reasonable to epect that f() a g() = f(a) + f (a)( a) () a g(a) + g (a)( a) Since we are assuming that f an g are continuous at = a, we have f () = f (a) an g () = g (a) a a an since the ifferentiabilit of f an g at = a implies the continuit of f an g at = a, we have f(a) = f() = 0 an g(a) = g() = 0 a a Thus, we can rewrite () as f() a g() = f (a)( a) a g (a)( a) = f (a) a g (a) = f () (3) a g () This result, calle L Hôpital s rule, converts the given ineterminate form into a it involving erivatives that is often easier to evaluate. Although we motivate (3) b assuming that f an g have continuous erivatives at = a an that g (a) = 0, the result is true uner less stringent conitions an is also vali for one-sie its an its at + an. The proof of the following precise statement of L Hôpital s rule is omitte theorem (L Hôpital s Rule for Form 0/0) Suppose that f an g are ifferentiable functions on an open interval containing = a, ecept possibl at = a, an that f() = 0 an g() = 0 a a If [f ()/g ()] eists, or if this it is + or, then a WARNING Note that in L Hôpital s rule the numerator an enominator are ifferentiate iniviuall. This is not the same as ifferentiating f()/g(). f() a g() = f () a g () Moreover, this statement is also true in the case of a it as a, a +,, or as +. In the eamples that follow we will appl L Hôpital s rule using the following three-step process: Appling L Hôpital s Rule Step. Check that the it of f()/g() is an ineterminate form of tpe 0/0. Step. Differentiate f an g separatel. Step 3. Fin the it of f ()/g (). If this it is finite, +, or, then it is equal to the it of f()/g().

37 3.6 L Hôpital s Rule; Ineterminate Forms Eample Fin the it 4 using L Hôpital s rule, an check the result b factoring. The it in Eample can be interprete as the it form of a certain erivative. Use that erivative to evaluate the it. Solution. The numerator an enominator have a it of 0, so the it is an ineterminate form of tpe 0/0. Appling L Hôpital s rule iels 4 = This agrees with the computation [ 4] = [ ] = 4 4 = ( )( + ) = ( + ) = 4 Eample In each part confirm that the it is an ineterminate form of tpe 0/0, an evaluate it using L Hôpital s rule. (a) () sin 0 (b) sin π/ cos (c) e 0 tan cos (e) (f ) /3 sin(/) WARNING Appling L Hôpital s rule to its that are not ineterminate forms can prouce incorrect results. For eample, the computation + 6 [ + 6] 0 + = 0 [ + ] = 0 = is not vali, since the it is not an ineterminate form. The correct result is = = 3 Solution (a). The numerator an enominator have a it of 0, so the it is an ineterminate form of tpe 0/0. Appling L Hôpital s rule iels sin 0 = 0 [sin ] [] = 0 cos Observe that this result agrees with that obtaine b substitution in Eample 4(b) of Section.6. Solution (b). The numerator an enominator have a it of 0, so the it is an ineterminate form of tpe 0/0. Appling L Hôpital s rule iels sin π/ cos = π/ [ sin ] = [cos ] π/ = cos sin = 0 = 0 Guillaume FrançoisAntoine e L Hôpital (66 704) French mathematician. L Hôpital, born to parents of the French high nobilit, hel the title of Marquis e Sainte- Mesme Comte Autrement. He showe mathematical talent quite earl an at age 5 solve a ifficult problem about cclois pose b Pascal. As a oung man he serve briefl as a cavalr officer, but resigne because of nearsighteness. In his own time he gaine fame as the author of the first tetbook ever publishe on ifferential calculus, L Analse es Infiniment Petits pour l Intelligence es Lignes Courbes (696). L Hôpital s rule appeare for the first time in that book. Actuall, L Hôpital s rule an most of the material in the calculus tet were ue to John Bernoulli, who was L Hôpital s teacher. L Hôpital roppe his plans for a book on integral calculus when Leibniz informe him that he intene to write such a tet. L Hôpital was apparentl generous an personable, an his man contacts with major mathematicians provie the vehicle for isseminating major iscoveries in calculus throughout Europe.