3.10 Implicit Differentiation

Transcription

1 300 C HAPTER 3 DIFFERENTIATION (b) B art (a), Alternatel, ln f./g./d f 0./ f./ C g0./ g./ D f 0./g./ C f./g 0./ : f./g./.f./g.//0 ln f./g./d f./g./ : Thus, or.f./g.// 0 f./g./ D f 0./g./ C f./g 0./ ; f./g./.f./g.// 0 D f 0./g./ C f./g 0./: 85. Use the formula log b D log a for a; b > 0 to verif the formula log a b log b D.ln b/ log b D ln ln b D.ln b/. 3.0 Imlicit Differentiation Preliminar Questions. Which ifferentiation rule is use to show sin D cos? The chain rule is use to show that sin D cos.. One of (a) (c) is incorrect. Fin an correct the mistake. (a) sin. / D cos. / (b) sin. / D cos. / (c) sin. / D cos. / (a) This is correct. Note that the ifferentiation is with resect to the variable. (b) This is correct. Note that the ifferentiation is with resect to the variable. (c) This is incorrect. Because the ifferentiation is with resect to the variable,thechainruleisneeetoobtain sin. / D cos. / : 3. On an eam, Jason was aske to ifferentiate the equation Fin the errors in Jason s answer: C 0 C 3 D 0 obtain C C 3 D 7 There are two mistakes in Jason s answer. First, Jason shoul have alie the rouct rule to the secon term to./ D C : Secon, he shoul have alie the general ower rule to the thir term to obtain 3 D 3 :. Which of (a) or (b) is equal to. sin t/? (a). cos t/ t (b). cos t/ t C sin t

2 SECTION 3.0 Imlicit Differentiation 30 Using the rouct rule an the chain rule we see that so the correct answer is (b). t. sin t/ D cos t C sin t; Eercises. Show that if ou ifferentiate both sies of C 3 D 6, theresultis C 6 D 0. Then solve for = an evaluate it at the oint.; /. At.; /, D 6 D 3.. C 3 / D 6 C 6 D 0 C 6 D 0 6 D D 6 :. Show that if ou ifferentiate both sies of C C D, theresultis. C / C C D 0. Then solve for = an evaluate it at the oint.; /. Aling the rouct rule At.; /, = D 3=3 D.. C C / D C C C D 0. C / D. C / D C C : In Eercises 3 8, ifferentiate the eression with resect to,assumingthat D f./ Assuming that eens on,then Assuming that eens on,then 3 D 3 0 C 3 D 3 0 C 3 :! 3 D.3 / 3 0 D : 5.. C / 3= Assuming that eens on,then C 3= D 3 C = C 0 D 3 C 0 q C :

3 30 C HAPTER 3 DIFFERENTIATION 6. tan./ 7. C 8. e = Assuming that eens on,then.tan.// D 0 C sec./. Assuming that eens on,then Assuming that eens on,then C D. C / C / D. C /. e= D e = 0 : InEercises9 6,calculatetheerivativewithresectto C D 5 0. D 3 C Let 3 3 C D 5. Then9 0 C D 0, an 0 D 9. Let D 3 C.Then. C 3 D C Let C 3 D C. Then. / D.3 C / D C 0. 3 / D C 0 D C 3 0 C C 3 0 C 6 D C 0 0 C D 6. C 5 3 D 3 Let C 5 3 D 3. Then 0 D 6 C 3 : 3. 3 R 5 D 0 C C 5 0 C 5 3 D 0. C 5 / 0 D D 3 5 C 5 Let 3 R 5 D.Then 3 5R R 0 C R 5 3 D 0, anr 0 D 3 R 5 3R 5 3 D R 5.. C z D Let C z D. Then 3 C z 3 z 0 D 0,anz 0 D 3 =z C D

4 SECTION 3.0 Imlicit Differentiation 303 Then 6. Let C s D C s Let. C s/ = D C s.then C D : 0 0 C D 0 0 D Multiling b s C s an then solving for s 0 gives 7. =3 C 3= D 8. = C =3 D 9. C D C Let =3 C 3= D.Then 0 D 0 D. /. / :. C s/ = C s 0 D s s 0 : s C s 0 D s C s s 0 C s s s 0 C s 0 C s D s C s s s C C s s 0 D s C C s s 0 D s C C s s C C s : 3 5=3 0 C 3 = D 0 or 0 D 9 = 5=3 : Let = C =3 D.Then = C 3 =3 0 D 5 0,an 0 D = 3 =3 C 5 : Let C D C. Then 0. sin.t/ D t 0 0 D C or 0 D C. C / D : In what follows, t 0 D t. Aling the chain rule an the rouct rule, we get: sin.t/ D t cos.t/.t 0 C t/ D t 0 cos.t/t 0 C t cos.t/ D t 0 cos.t/t 0 t 0 D t cos.t/ t 0. cos.t/ / D t cos.t/ t 0 D t cos.t/ cos.t/ :

5 30 C HAPTER 3 DIFFERENTIATION. sin. C / D C cos. tan. / D. C / 3 3. e D C 3. e D sin. / Let sin. C / D C cos.then Let tan D. C / 3.Then. C 0 / cos. C / D 0 sin cos. C / C 0 cos. C / D 0 sin.cos. C / C sin / 0 D cos. C / 0 D cos. C / cos. C / C sin : sec. /. 0 C / D 3. C /. C 0 / sec. / 0 C sec. / D 3. C / C 3. C / 0 sec. / 3. C / 0 D 3. C / sec. / 0 D 3. C / sec sec 3. C / : Let e D C 3.Then 0 e C e D 0 C C 3 0,whence 0 D e C 3 e : Let e D sin. /.Thene 0 C D cos. / 0,whence 5. ln C ln D Let ln C ln D.Then 0 D e cos. / e : 6. ln. C / D C Let ln. C / D C. Then C 0 D 0 or 0 D C D C : C 0 C D or 0 D C : 7. Show that C D an D efine the same curve (ecet that.0; 0/ is not a solution of the first equation) an that imlicit ifferentiation iels 0 D an 0 D. Elainwhtheseformulasroucethesamevaluesforthe erivative. Multil the first equation b an then isolate the term to obtain Imlicit ifferentiation alie to the first equation iels C D ) D : C 0 D 0 or 0 D : From the first equation, we fin D ;uonsubstitutingthiseressionintothereviouserivative,wefin 0 D D ; which is the erivative of the secon equation. 8. Use the metho of Eamle to comute ˇ ˇP at P D.; / on the curve 3 C 3 0 C D 5.

6 SECTION 3.0 Imlicit Differentiation 305 Imlicit ifferentiation iels 3 C 3 0 C 3 3 C C 0 D 0 or 0 D C 3 C : Thus, at P D.; /, ˇ D 0 ˇP 3./././ 3./ 3./ 3./ C 3././ C D 3 65 : InEercises9an30,fin= at the given oint. 9.. C / 6. C 3/ D 3,.; / B the scaling an shifting rule,. C /. C 3/ 0 D 0: If D an D, then.3/./ 0 D 0: so that 0 D 6, or 0 D : 30. sin.3/ D C, ; Taking the erivative of both sies of sin.3/ D C iels sin.3/ cos.3/.3 0 / D C 0 : If D an D, we get Using we fin sin 6 sin 3 3 cos 0 D C 0 : 3 D 6 an cos 3 D!! 0 D C D C 0 0 D : InEercises3 38,finanequationofthetangentlineatthegivenoint. 3. C D 5,.; / Taking the erivative of both sies of C D 5 iels Substituting D ; D,wefin 0 C C C 0 D 0: 0 C C C 8 0 D 0 or 0 D : Hence, the equation of the tangent line at.; / is D. / or D C. 3. =3 C =3 D,.; / Taking the erivative of both sies of =3 C =3 D iels 3 =3 C 3 =3 0 D 0: Substituting D, D iels 3 C 3 0 D 0, so that C 0 D 0,or 0 D. Hence,theequationofthetangentlineat.; / is D. /,or D.

7 306 C HAPTER 3 DIFFERENTIATION 33. C sin D C,.; 0/ Taking the erivative of both sies of C sin D C iels Substituting D ; D 0,wefin C cos 0 D C 0 : C 0 D 0 or 0 D : Hence, the equation of the tangent line is 0 D. / or D C. 3. sin. / D cos C, ; Taking the erivative of both sies of sin. / D cos C iels Substituting D ; D, we fin Hence, the equation of the tangent line is 35. = C = D,.; / cos. /. 0 / D cos C sin C 0 :. 0 / D 0 0 or 0 D C : D : C Taking the erivative of both sies of = C = D iels = 3= 0 D 0 C : Substituting D ; D,wefin 0 D 0 C or 0 D 8 5 : Hence, the equation of the tangent line is D 5. / or D 5 C e C e D,.; 0/ Taking the erivative of both sies of e C e D iels Substituting D ; D 0,wefin Hence, the equation of the tangent line is 37. e D,.; / e 0 C e C e C e 0 D 0: 0 C C 0 C e 0 D 0 or 0 D C e : D. /: C e taking the erivative of both sies of e D iels e. 0 / D 0 : Substituting D ; D,we fin e 0. 0 / D or 0 D 3 : Hence, the equation of the tangent line is D 3. / or D 3 C 3.

8 SECTION 3.0 Imlicit Differentiation e 6 D,.; / Taking the erivative of both sies of e 6 D iels e 6 C 0 e 6 C 0 D 0: Substituting D ; D,wefin 3e 0 C 0 e 0 C 0 D 0 or 0 D 63 0 : Hence, the equation of the tangent line is D / or D 63 0 C Fin the oints on the grah of D 3 3 C (Figure ) where the tangent line is horizontal. (a) First show that 0 D 3 3, where 0 D =. (b) Do not solve for 0.Rather,set 0 D 0 an solve for.thisielstwovaluesof where the sloe ma be zero. (c) Show that the ositive value of oes not correson to a oint on the grah. () The negative value corresons to the two oints on the grah where the tangent line is horizontal. Fin their coorinates. FIGURE Grah of D 3 3 C. (a) Aling imlicit ifferentiation to D 3 3 C, wehave D 3 3: (b) Setting 0 D 0 we have 0 D 3 3,so D or D. (c) If we return to the equation D 3 3 C an substitute D, weobtaintheequation D, whichhasnoreal solutions. () Substituting D into D 3 3 C iels D. / 3 3. / C D C 3 C D 3; so D 3 or 3. The tangent is horizontal at the oints. ; 3/ an. ; 3/. 0. Show, b ifferentiating the equation, that if the tangent line at a oint.; / on the curve C 8 D is horizontal, then D. Thensubstitute D in C 8 D to show that the tangent line is horizontal at the oints ; an ;. Taking the erivative on both sies of the equation C 8 D iels 0 C C 8 0 D 0 or 0 D. / C 8 : Thus, if the tangent line to the given curve is horizontal, it must be that D 0, or D. Substituting D into C 8 D then iels C 8 D or C 8 D. C /. / D 0: Hence, the given curve has a horizontal tangent line when D an when D. Thecorresoningointsonthecurveare thus ; an ;.. Fin all oints on the grah of 3 C C 3 D where the tangent line is horizontal (Figure ). FIGURE Grah of 3 C C 3 D.

9 308 C HAPTER 3 DIFFERENTIATION Differentiating the equation 3 C C 3 D imlicitl iels 6 C 8 0 C 3 0 C 3 D 0; so 0 6 C 3 D 8 C 3 : Setting 0 D 0 leas to 6 C 3 D 0, or D.Substituting D into the equation 3 C C 3 D iels 3 C. / C 3. / D ; or 3 D. Thus, D 78=3, an the coorinates of the two oints on the grah of 3 C C 3 D where the tangent line is horizontal are 78 3 ;! 78 an ;! 78 : 3. Show that no oint on the grah of 3 C D has a horizontal tangent line. Let the imlicit curve 3 C D be given. Then C 0 D 0; so 0 D 3 3 : Setting 0 D 0 leas to D 3. Substituting D 3 into the equation of the imlicit curve gives 3 3 C 3 D ; or 5 9 D, which has no real solutions. Accoringl, there are no oints on the imlicit curve where the tangent line has sloe zero. 3. Figure shows the grah of C D 3 C. Fin= at the two oints on the grah with -coorinate 0 an fin an equation of the tangent line at.; /. Consier the equation C D 3 C.Then 3 0 C 0 C D 3, an 0 D 3 C 3 : Substituting D 0 into C D 3 C gives D, whichhastworealsolutions, D =.When D =, we have 0 D = C D :35: 3= 8 When D =,wehave 0 D = 3= D :083: 8 At the oint.; /,wehave 0 D 5. At this oint the tangent line is D 5. / or D 5 C 5.. Folium of Descartes The curve 3 C 3 D 3 (Figure 3) was first iscusse in 638 b the French hilosoher-mathematician René Descartes, who calle it the folium (meaning leaf ). Descartes s scientific colleague Gilles e Roberval calle it the jasmine flower. Both men believe incorrectl that the leaf shae in the first quarant was reeate in each quarant, giving the aearance of etals of a flower. Fin an equation of the tangent line at the oint 3 ; 3. FIGURE 3 Folium of Descartes: 3 C 3 D 3.

10 SECTION 3.0 Imlicit Differentiation 309 Let 3 C 3 D 3. Then3 C 3 0 D 3 0 C 3,an 0 D. At the oint 3 ; 3, we have The tangent line at P is thus 3 D 5 0 D or D 5 C 5. D D 5 : 5. Fin a oint on the folium 3 C 3 D 3 other than the origin at which the tangent line is horizontal. Using imlicit ifferentiation, we fin 3 C 3 D.3/ 3 C 3 0 D 3. 0 C / Setting 0 D 0 in this equation iels 3 D 3 or D.Ifwesubstitutethiseressionintotheoriginalequation 3 C 3 D 3, weobtain: 3 C 6 D 3. / D 3 3 or 3. 3 / D 0: One solution of this equation is D 0 an the other is D =3.Thus,thetwoointsonthefolium 3 C 3 D 3 at which the tangent line is horizontal are.0; 0/ an. =3 ; =3 /. 6. Plot 3 C 3 D 3 C b for several values of b an escribe how the grah changes as b! 0.Thencomute = at the oint.b =3 ;0/.Howoesthisvaluechangeasb!?Doourlotsconfirmthisconclusion? Consier the first row of figures below. When b < 0,thegrahof 3 C 3 D 3 C b consists of two ieces. As b! 0, thetwoiecesmoveclosertointersectingattheorigin.fromtheseconrowoffigures,weseethatthegrahof 3 C 3 D 3 C b when b>0consists of a single iece that has a loo in the first quarant. As b! 0C, theloocomes closer to inching off at the origin. b = 0..5 b = b = b = 0. b = 0.0 b = Differentiating the equation 3 C 3 D 3 C b with resect to iels 3 C 3 0 D 3 0 C 3,so At.b =3 ;0/,wehave 0 D 0 D : 0 0 D D 3 b: Consequentl, as b!, 0!at the oint on the grah where D 0. Thisconclusionissuortebthefiguresshown below, which correson to b D, b D 0, anb D 00.

11 30 C HAPTER 3 DIFFERENTIATION b = 0 b = 0 b = Fin the -coorinates of the oints where the tangent line is horizontal on the trient curve D 3 5 C, so name b Isaac Newton in his treatise on curves ublishe in 70 (Figure ). Hint: 3 5 C D. /. / FIGURE Trient curve: D 3 5 C. Take the erivative of the equation of a trient curve: D 3 5 C to obtain 0 C D 3 0 C : Setting 0 D 0 gives D 3 0 C. Substitutingthisintotheequationofthetrient,wehave or Collecting like terms an setting to zero, we have Hence, D ;. D.3 0 C / D 3 5 C C D 3 5 C 0 D 3 5 C D. /. /: 8. Fin an equation of the tangent line at each of the four oints on the curve. C / D. C / where D. This curve (Figure 5) is an eamle of a limaçon of Pascal, nameafterthefatherofthefrenchhilosoherblaisepascal,who first escribe it in FIGURE 5 Limaçon:. C / D. C /. Plugging D into the equation for the limaçon an solving for,wefinthattheointsonthecurvewhere D are:.; /,.; /,.; 7/,.; 7/. Using imlicit ifferentiation, we obtain We lug in D an get. C /. C 0 / D. C 0 /:. C /. C 0 / D. C 0 /

12 SECTION 3.0 Imlicit Differentiation 3 or After collecting like terms an solving for 0,wehave. 6/. 0 / D C 0 : 0 D C 3 : At the oint.; / the sloe of the tangent is 3 an the tangent line is D 3. / or D 3 C 3 : At theoint.; / the sloe of the tangent is 3 an the tangent line is C D 3. / or D 3 3 : At theoint.; 7/ the sloe of the tangent is 5=3 7 an the tangent line is 7 D / or D C : At theoint.; 7/ the sloe of the tangent is 5=3 7 an the tangent line is C 7 D / or D C : 9. Fin the erivative at the oints where D on the folium. C / D 5.SeeFigure6. FIGURE 6 Folium curve:. C / D 5 First, fin the oints.; / on the curve. Setting D in the equation. C / D 5 iels. C / D 5 C C D 5 C 8 C D 5 7 C D 0. /. / D 0 D or D Hence D or D. Taking ofbothsiesoftheoriginalequationiels. C /. C 0 / D 5 C 5 0. C / C. C / 0 D 5 C C / 5 /0 D 5. C / 0 5 D. C /.. C / 5 /

13 3 C HAPTER 3 DIFFERENTIATION At.; /, C D 5, an 0 D 5.5/./..5/ 5.// D 3 : At.; /, C D 5 as well, an 0 D 5. /.5/./..5/ 5.// D 3 : At.; /, C D 5, an 0 D 5 5./ D 5 5./ : At.; /, C D 5, an 0 D 5 5./ D 5 5./ : The folium an its tangent lines are lotte below: Plot. C / D. / C for, using a comuter algebra sstem. How man horizontal tangent lines oes the curve aear to have? Fin the oints where these occur. Alotofthecurve. C / D. / C is shown below. From this lot, it aears that the curve has a horizontal tangent line at si ifferent locations. 3 3 Differentiating the equation. C / D. / C with resect to iels so. C /. C 0 / D. 0 /; 0 D.6 /. C C 6/ : Thus, horizontal tangent lines occur when D 0 an when C D 6.Substituting D 0 into the equation for the curve leaves C D 0, fromwhichitfollowsthat D 38 6 or D Substituting C D 6 into the equation for the curve leaves D 7 6. From here, it follows that D an D 6 6 : The si oints at which horizontal tangent lines occur are therefore q q ; 6 ; 0; 6

14 ! ; ; 6 6 SECTION 3.0 Imlicit Differentiation 33!!! ; ; ; ; ; Eercises 5 53: If the erivative = (instea of = D 0) eistsataointan= D 0, thenthetangentlineatthat oint is vertical. 5. Calculate = for the equation C D C an fin the oints on the grah where the tangent line is vertical. Let C D C.Differentiatingthisequationwithresectto iels so 3 D C ; D 3 D. / : Thus, D 0 when D 0 an when D. Substituting D 0 into the equation C D C gives D,so 3 D. Substituting D, gives D 3=, so D. Thus, there are si oints on the grah of C D C where the tangent line is vertical:!!!! ; 0/;. ; 0/; ; ; ; ; ; ; ; : 5. Show that the tangent lines at D totheconchoi with equation. /. C / D are vertical (Figure 7). FIGURE 7 Conchoi:. /. C / D. Consier the equation of a conchoi:. / C D : Taking the erivative of both sies of this equation gives. / C C C. / D ; sothat D. / C. / C. / : Setting = D 0 iels D or D 0. Wecan thave D, lest0 D in the conchoi s equation. Plugging D 0 into the equation gives. / D or. / D 0, whichimlies D 0 (a ouble root) or D. [Plugging D 0 into the conchoi s equation gives D 0 or D 0: At.; / D.0; 0/ the eression for = is unefine (0=0). Via an alternative arametric analsis, the sloes of the tangent lines at the origin turn out to be 3.] Accoringl,thetangentlinesto the conchoi are vertical at.; / D. ; 0/. 53. Use a comuter algebra sstem to lot D 3 for,. Showthatif= D 0, then D 0. Concluethatthetangentlineisverticalattheointswherethecurveintersectsthe-ais. Does our lot confirm this conclusion?

15 3 C HAPTER 3 DIFFERENTIATION Alotofthecurve D 3 is shown below. 3 Differentiating the equation D 3 with resect to iels or D 3 ; D 3 : From here, it follows that D 0 when D 0, sothetangentlinetothiscurveisverticalattheointswherethecurveintersects the -ais. This conclusion is confirme b the lot of the curve shown above. 5. Show that for all oints P on the grah in Figure 8, the segments OP an PR have equal length. Tangent line P O R FIGURE 8 Grah of D a. Because of the smmetr of the grah, we ma restrict attention to an oint P in the first quarant. Suose P has coorinates.; a /. Taking the erivative of both sies of the equation D a iels 0 D 0,or 0 D =. It follows that the sloe of the line tangent to the grah at P has sloe a an the sloe of the normal line is Thus, the equation of the normal line is a : q a D a. /; an the coorinates of the oint R are.; 0/.Finall,thelengthofthelinesegmentOP is q q C a D a ; while the length of the segment PR is q q. / C a D a :

16 SECTION 3.0 Imlicit Differentiation 35 InEercises55 58,useimlicitifferentiationtocalculatehighererivatives. 55. Consier the equation 3 3 D. (a) Show that 0 D = an ifferentiate again to show that (b) Eress 00 in terms of an using art (a). 00 D 0 (a) Let 3 3 D. Then D 0, an 0 D =.Therefore, (b) Substituting the eression for 0 into the result for 00 gives 00 D 0 D 0 : 00 D = D 3 5 : 56. Use the metho of the revious eercise to show that 00 D 3 on the circle C D. Let C D. Then C 0 D 0,an 0 D. Thus 00 0 D D D C 3 D 3 D 3 : 57. Calculate 00 at the oint.; / on the curve C D 0 b the following stes: (a) Fin 0 b imlicit ifferentiation an calculate 0 at the oint.; /. (b) Differentiate the eression for 0 foun in (a). Then comute 00 at.; / b substituting D, D, anthevalueof 0 foun in (a). Let C D 0. (a) Then 0 C C 0 D 0,an 0 D C. At.; / D.; /,wehave0 D 3. (b) Therefore, 00 D. C / 0 0 C.3/ 3./ 3 C 6 C 6. C / D 3 D D given that.; / D.; / an 0 D Use the metho of the revious eercise to comute 00 at the oint.; / on the curve 3 C 3 D 3 C. Let 3 C 3 D 3 C. Then3 C 3 0 D 3 C 0,an 0 D 3. / 3. At.; / D.; /,wefin Similarl, 0 D 3. / 3./ D 0: / D 3 D 3 when.; / D.; / an 0 D 0. In Eercises 59 6, an are functions of a variable t an use imlicit ifferentiation to relate =t an =t. 59. Differentiate D with resect to t an erive the relation t D Let D. Then t C D 0, an t t D t. t.

17 36 C HAPTER 3 DIFFERENTIATION 60. Differentiate 3 C 3 D with resect to t an eress =t in terms of =t, asineercise59. an Let 3 C 3 D. Then 3 t C 6 t C 3 t t D C t : D 0; 6. Calculate =t in terms of =t. (a) 3 3 D (b) C C D 0 (a) Taking the erivative of both sies of the equation 3 3 D with resect to t iels 3 t 3 t D 0 or t D t : (b) Taking the erivative of both sies of the equation C C D 0 with resect to t iels or 3 t C t C t C t t D C 3 C t : D 0; 6. The volume V an ressure P of gas in a iston (which var in time t) satisfpv 3= D C,whereC is a constant. Prove that P =t V =t D 3 P V The ratio of the erivatives is negative. Coul ou have reicte this from the relation PV 3= D C? Let PV 3= D C,whereC is a constant. Then P 3 = V V t 3= P C V t D 0; so P =t V=t D 3 P V : If P is increasing (resectivel, ecreasing), then V D.C=P / =3 is ecreasing (resectivel, increasing). Hence the ratio of the erivatives (C= or =C) isnegative. Further Insights an Challenges 63. Show that if P lies on the intersection of the two curves D c an D (c; constants), then the tangents to the curves at P are erenicular. Let C be the curve escribe b D c, anletc be the curve escribe b D. Suosethat P D. 0 ; 0 / lies on the intersection of the two curves D c an D. Since D c, thechainrulegivesus 0 D 0, sothat 0 D D. The sloe to the tangent line to C is 0 0. On the curve C,since D, therouct rule iels that 0 C D 0, sothat 0 D. Therefore the sloe to the tangent line to Cis 0 0. The two sloes are negative recirocals of one another, hence the tangents to the two curves are erenicular. 6. The lemniscate curve. C / D. / was iscovere b Jacob Bernoulli in 69, who note that it is shae like afigure8,oraknot,orthebowofaribbon. Finthecoorinatesofthefourointsatwhichthetangentlineishorizontal(Figure 9). FIGURE 9 Lemniscate curve:. C / D. /.

18 SECTION 3.0 Imlicit Differentiation 37 Consier the equation of a lemniscate curve: C D.Takingtheerivativeofbothsiesofthis equation, we have C C 0 D 0 : Therefore, 0 D 8 C 8 C C C D C C : If 0 D 0, theneither D 0 or C D. If D 0 in the lemniscate curve, then D or C D 0. If is real, then D 0. Theformulafor 0 in (a) is not efine at the origin (0=0). An alternative arametric analsis shows that the sloes of the tangent lines to the curve at the origin are. If C D or D,thenluggingthisintothelemniscateequationgives D which iels q 3 D D 6. Thus D q D. Accoringl, the four oints at which the tangent lines to the lemniscate curve are horizontal are 6 ;, 6 ;, 6 ;, an 6 ;. 65. Divie the curve in Figure 0 into five branches, each of which is the grah of a function. Sketch the branches. FIGURE 0 Grah of 5 D C C. The branches are: Uer branch: Lower art of lower left curve: 3 Uer art of lower left curve: 3 Uer art of lower right curve:

19 38 C HAPTER 3 DIFFERENTIATION 3 Lower art of lower right curve: 3 3. Relate Rates Preliminar Questions. Assign variables an restate the following roblem in terms of known an unknown erivatives (but o not solve it): How fast is the volume of a cube increasing if its sie increases at a rate of 0:5 cm/s? Let s an V enote the length of the sie an the corresoning volume of a cube, resectivel. Determine V t s D 0:5 cm/s. t. What is the relation between V=t an r=t if V D 3 r 3? Aling the general ower rule, we fin V t D r r t. Therefore, the ratio is r. In Questions 3 an, water ours into a clinrical glass of raius cm. LetV an h enote the volume an water level resectivel, at time t. 3. Restate this question in terms of V =t an h=t:howfastisthewaterlevelrisingifwateroursinatarateof cm 3 /min? Determine h t if V t D cm 3 /min.. Restate this question in terms of V =t an h=t: Atwhatrateiswaterouringinifthewaterlevelrisesatarateof cm/min? Determine V t if h t D cm/min. Eercises InEercisesan,consierarectangularbathtubwhosebaseis8ft. if. How fast is the water level rising if water is filling the tub at a rate of 0:7 ft 3 /min? Let h be the height of the water in the tub an V be the volume of the water. Then V D 8h an V t h D V D.0:7/ 0:039 ft=min: t 8 t 8. At what rate is water ouring into the tub if the water level rises at a rate of 0:8 ft/min? Let h be the height of the water in the tub an V its volume. Then V D 8h an D 8 h t. Thus V D 8 h D 8.0:8/D : ft 3 =min: t t 3. The raius of a circular oil slick eans at a rate of m/min. (a) How fast is the area of the oil slick increasing when the raius is 5 m? (b) If the raius is 0 at time t D 0,howfastistheareaincreasingafter3min?