300 CHAPTER 3 DIFFERENTIATION
(b) By part (a),
d f .x/ g .x/ f .x/g.x/ f .x/g .x/ ln f .x/g.x/ 0 0 0 C 0 : dx D f .x/ C g.x/ D f .x/g.x/ Alternately,
d .f .x/g.x// ln f .x/g.x/ 0 : dx D f .x/g.x/ Thus, .f .x/g.x// f .x/g.x/ f .x/g .x/ 0 0 C 0 ; f .x/g.x/ D f .x/g.x/ or
.f .x/g.x// f .x/g.x/ f .x/g .x/: 0 D 0 C 0 log a x 85. Use the formula log b x for a;b > 0 to verify the formula D log a b d 1 log x dx b D .ln b/x
d d ln x 1 SOLUTION log x . dx b D dx ln b D .ln b/x
3.10 Implicit Differentiation
Preliminary Questions d dy 1. Which differentiation rule is used to show sin y cos y ? dx D dx d dy SOLUTION The chain rule is used to show that sin y cos y . dx D dx 2. One of (a)–(c) is incorrect. Find and correct the mistake. d d d (a) sin .y 2/ 2y cos .y 2/ (b) sin .x 2/ 2x cos .x 2/ (c) sin .y 2/ 2y cos .y 2/ dy D dx D dx D
SOLUTION (a) This is correct. Note that the differentiation is with respect to the variable y. (b) This is correct. Note that the differentiation is with respect to the variable x. (c) This is incorrect. Because the differentiation is with respect to the variable x, the chain rule is needed to obtain d dy sin .y 2/ 2y cos .y 2/ : dx D dx 3. On an exam, Jason was asked to differentiate the equation
x2 2xy y3 7 C C D Find the errors in Jason’s answer: 2x 2xy 3y 2 0 C 0 C D SOLUTION There are two mistakes in Jason’s answer. First, Jason should have applied the product rule to the second term to obtain d dy .2xy/ 2x 2y: dx D dx C Second, he should have applied the general power rule to the third term to obtain d dy y3 3y2 : dx D dx d 4. Which of (a) or (b) is equal to .x sin t/ ? dx dt dt (a) .x cos t/ (b) .x cos t/ sin t dx dx C SECT ION 3.10 Implicit Differentiation 301
SOLUTION Using the product rule and the chain rule we see that
d dt .x sin t/ x cos t sin t; dx D dx C so the correct answer is (b) .
Exercises 2 3 2 dy 1. Show that if you differentiate both sides of x 2y 6, the result is 2x 6y dx 0. Then solve for dy=dx and evaluate it at the point .2; 1/ . C D C D
SOLUTION
d d .x2 2y 3/ 6 dx C D dx dy 2x 6y 2 0 C dx D dy 2x 6y 2 0 C dx D dy 6y2 2x dx D dy 2x : dx D 6y 2
At .2; 1/ , dy 4 2 . dx D 6 D 3 dy 2. Show that if you differentiate both sides of xy 4x 2y 1, the result is .x 2/ dx y 4 0. Then solve for dy=dx and evaluate it at the point .1; 1/ . C C D C C C D
SOLUTION Applying the product rule
d d .xy 4x 2y/ 1 dx C C D dx dy dy x y 4 2 0 dx C C C dx D dy .x 2/ .y 4/ C dx D C dy y 4 C : dx D x 2 C At .1; 1/ , dy=dx 3=3 1. D D In Exercises 3–8, differentiate the expression with respect to x, assuming that y f .x/ . D 3. x2y3
SOLUTION Assuming that y depends on x, then
d x2y3 x2 3y2y y3 2x 3x 2y2y 2xy 3: dx D 0 C D 0 C x3 4. y2
SOLUTION Assuming that y depends on x, then
3 2 2 3 2 3 d x y .3x / x 2yy 0 3x 2x y0 2 4 2 3 : dx y ! D y D y y 5. .x 2 y2/3=2 C SOLUTION Assuming that y depends on x, then
d 3=2 3 1=2 x2 y2 x2 y2 2x 2yy 3 x yy x2 y2: dx C D 2 C C 0 D C 0 C " " q 302 CHAPTER 3 DIFFERENTIATION
6. tan .xy/ d SOLUTION Assuming that y depends on x, then .tan .xy// xy y sec 2.xy/ . dx D 0 C y " 7. y 1 C d y .y 1/y 0 yy 0 y0 SOLUTION Assuming that y depends on x, then C . dx y 1 D .y 1/2 D .y 1/2 C C C 8. ey=x
SOLUTION Assuming that y depends on x, then
d xy y ey=x ey=x 0 : dx D x2
In Exercises 9–26, calculate the derivative with respect to x.
9. 3y 3 x2 5 C D 2x SOLUTION Let 3y 3 x2 5. Then 9y 2y 2x 0, and y . C D 0 C D 0 D 9y 2 10. y4 2y 4x 3 x D C SOLUTION Let y4 2y 4x 3 x. Then D C d d .y4 2y/ .4x 3 x/ dx D dx C 4y 3y 2y 12x 2 1 0 0 D C y .4y 3 2/ 12x 2 1 0 D C 12x 2 1 y C 0 D 4y 3 2
11. x2y 2x 3y x y C D C SOLUTION Let x2y 2x 3y x y. Then C D C x2y 2xy 2x 3y 6x 2y 1 y 0 C C 0 C D C 0 x2y 2x 3y y 1 2xy 6x 2y 0 C 0 0 D 1 2xy 6x 2y y : 0 D x2 2x3 1 C 12. xy 2 x2y5 x3 3 C D SOLUTION Let xy 2 x2y5 x3 3. Then C D 2xyy y2 5x 2y4y 2xy 5 3x 2 0 0 C C 0 C D .2xy 5x 2y4/y 3x 2 y2 2xy 5 C 0 D 3x 2 y2 2xy 5 y 0 D 2xy 5x2y4 C 13. x3R5 1 D 3x 2R5 3R SOLUTION Let x3R5 1. Then x3 5R 4R R5 3x 2 0, and R . D 0 C D 0 D 5x 3R4 D 5x 14. x4 z4 1 C D SOLUTION Let x4 z4 1. Then 4x 3 4z 3z 0, and z x3=z 3. C D C 0 D 0 D y x 15. 2y x C y D SECT ION 3.10 Implicit Differentiation 303
SOLUTION Let y x 2y: x C y D Then xy y y xy 0 0 2y x2 C y2 D 0 1 x y 1 2 y x y2 0 D x2 y y2 x2 2xy 2 y2 x2 y xy 2 0 D x2y y.y 2 x2/ y : 0 D x.y 2 x2 2xy 2/
1 1 16. px s C D x C s 1=2 SOLUTION Let .x s/ x 1 s 1. Then C D C 1 .x s/ 1=2 1 s x 2 s 2s : 2 C C 0 D 0 " Multiplying by 2x 2s2px s and then solving for s gives C 0 x2s2 1 s 2s 2px s 2x2s px s C 0 D C 0 C x2s2s 2x2s "px s 2s 2px s x2s2 0 C 0 C D C x2 s2 2px s s s2 x2 2px s C C 0 D C C s2 x2 2px s s C C : 0 D x2 s2 2px s " C C 17. y 2=3 x3=2 1 " C D SOLUTION Let y 2=3 x3=2 1. Then C D 2 3 9 y 5=3 y x1=2 0 or y x1=2 y5=3 : 3 0 C 2 D 0 D 4 18. x1=2 y2=3 4y C D 1 2 SOLUTION Let x1=2 y2=3 y 4. Then x 1=2 y 1=3 y 4y 5y , and C D 2 C 3 0 D 0 1 1=2 2 x y0 : D 2 y 1=3 4y 5 3 C 1 19. y x2 x C y D C 1 SOLUTION Let y x2 x. Then C y D C 1 2x 1 .2x 1/y 2 y y 2x 1 or y C C : 0 y2 0 D C 0 D 1 y 2 D y2 1
20. sin .xt/ t D dt SOLUTION In what follows, t . Applying the chain rule and the product rule, we get: 0 D dx d d sin .xt/ t dx D dx cos .xt/.xt t/ t 0 C D 0 x cos .xt/t t cos .xt/ t 0 C D 0 x cos .xt/t t t cos .xt/ 0 0 D t .x cos .xt/ 1/ t cos .xt/ 0 D t cos .xt/ t : 0 D x cos .xt/ 1
304 CHAPTER 3 DIFFERENTIATION
21. sin .x y/ x cos y C D C SOLUTION Let sin .x y/ x cos y. Then C D C .1 y / cos .x y/ 1 y sin y C 0 C D 0 cos .x y/ y cos .x y/ 1 y sin y C C 0 C D 0 .cos .x y/ sin y/ y 1 cos .x y/ C C 0 D C 1 cos .x y/ y C : 0 D cos .x y/ sin y C C 22. tan .x 2y/ .x y/ 3 D C SOLUTION Let tan x2y .x y/ 3. Then D C " sec 2.x 2y/ .x 2y 2xy/ 3.x y/ 2.1 y / 0 C D C C 0 x2 sec 2.x 2y/y 2xy sec 2.x 2y/ 3.x y/ 2 3.x y/ 2y 0 C D C C C 0 x2 sec 2.x 2y/ 3.x y/ 2 y 3.x y/ 2 2xy sec 2.x 2y/ C 0 D C 3.x y/ 2 2xy sec 2 x2y y C : 0 D x2 sec 2 x2y 3.x y/ 2 C" 23. xe y 2xy y3 " D C SOLUTION Let xe y 2xy y3. Then xy ey ey 2xy 2y 3y 2y , whence D C 0 C D 0 C C 0 ey 2y y : 0 D 2x 3y2 xe y C 24. exy sin .y 2/ D SOLUTION Let exy sin .y 2/. Then exy xy y 2y cos .y 2/y , whence D 0 C D 0 " ye xy y : 0 D 2y cos .y 2/ xe xy
25. ln x ln y x y C D SOLUTION Let ln x ln y x y. Then C D 1 1 y0 1 x xy y 1 y0 or y0 : x C y D D 1 1 D xy x C y C 26. ln .x 2 y2/ x 4 C D C SOLUTION Let ln .x 2 y2/ x 4. Then C D C 2x 2yy x2 y2 2x C 0 1 or y C : x2 y2 D 0 D 2y C 1 2 27. Show that x yx 1 and y x x define the same curve (except that .0; 0/ is not a solution of the first equation) and C D D 1 that implicit differentiation yields y0 yx x and y0 1 2x . Explain why these formulas produce the same values for the derivative. D D
SOLUTION Multiply the first equation by x and then isolate the y term to obtain
x2 y x y x x2: C D ) D Implicit differentiation applied to the first equation yields
1 yx 2 x 1y 0 or y yx 1 x: C 0 D 0 D From the first equation, we find yx 1 1 x; upon substituting this expression into the previous derivative, we find D y 1 x x 1 2x; 0 D D which is the derivative of the second equation. 28. Use the method of Example 4 to compute dy at P .2; 1/ on the curve y2x3 y3x4 10x y 5. dx P D C C D ˇ ˇ SECT ION 3.10 Implicit Differentiation 305
SOLUTION Implicit differentiation yields
10 3x 2y2 4x 3y3 3x 2y2 2x 3yy 4x 3y3 3x 4y2y 10 y 0 or y : C 0 C C 0 C 0 D 0 D 2x 3y 3x4y2 1 C C Thus, at P .2; 1/ , D dy 10 3.2/ 2.1/ 2 4.2/ 3.1/ 3 34 : dx P D 2.2/ 3.1/ 3.2/ 4.1/ 2 1 D 65 ˇ C C ˇ ˇ In Exercises 29 and 30, find dy=dx at the given point.
29. .x 2/ 2 6.2y 3/ 2 3, .1; 1/ C C D SOLUTION By the scaling and shifting rule,
2.x 2/ 24.2y 3/y 0: C C 0 D If x 1 and y 1, then D D 2.3/ 24.1/y 0: 0 D so that 24y 6, or y 1 : 0 D 0 D 4 2 30. sin 2.3y/ x y, ; D C 4 4 SOLUTION Taking the derivative of both sides of sin 2.3y/ x y yields D C 2 sin .3y/ cos .3y/.3y / 1 y : 0 D C 0 If x 2 and y , we get D 4 D 4 3 3 6 sin cos y 1 y : 4 4 0 D C 0 Using
3 p2 3 p2 sin and cos 4 D 2 4 D 2 we find
p2 p2 6 y0 1 y0 2 ! 2 ! D C 3y 1 y 0 D C 0 1 y : 0 D 4
In Exercises 31–38, find an equation of the tangent line at the given point.
31. xy x2y2 5, .2; 1/ C D SOLUTION Taking the derivative of both sides of xy x2y2 5 yields C D xy y 2xy 2 2x 2yy 0: 0 C C C 0 D Substituting x 2; y 1, we find D D 1 2y 1 4 8y 0 or y : 0 C C C 0 D 0 D 2 Hence, the equation of the tangent line at .2; 1/ is y 1 1 .x 2/ or y 1 x 2. D 2 D 2 C 32. x2=3 y2=3 2, .1; 1/ C D SOLUTION Taking the derivative of both sides of x2=3 y2=3 2 yields C D 2 2 x 1=3 y 1=3 y 0: 3 C 3 0 D 2 2 Substituting x 1, y 1 yields 3 3 y0 0, so that 1 y0 0, or y0 1. Hence, the equation of the tangent line at .1; 1/ is y 1 .x D1/ , or Dy 2 x. C D C D D D D 306 CHAPTER 3 DIFFERENTIATION
33. x2 sin y xy 2 1, .1; 0/ C D C SOLUTION Taking the derivative of both sides of x2 sin y xy 2 1 yields C D C 2x cos yy y2 2xyy : C 0 D C 0 Substituting x 1; y 0, we find D D 2 y 0 or y 2: C 0 D 0 D Hence, the equation of the tangent line is y 0 2.x 1/ or y 2x 2. D D C 34. sin .x y/ x cos y , ; D C 4 4 4 SOLUTION Taking the" derivative of" both sides of sin .x y/ x cos y yields D C 4 " cos .x y/.1 y / cos y x sin y y : 0 D C 4 C 4 0 " " Substituting x ; y , we find D 4 D 4 4 1.1 y / 0 y or y : 0 D 4 0 0 D 4 C Hence, the equation of the tangent line is 4 y x : 4 D 4 4 C 35. 2x 1=2 4y 1=2 xy , .1; 4/ C D SOLUTION Taking the derivative of both sides of 2x 1=2 4y 1=2 xy yields C D x 1=2 2y 3=2 y xy y: 0 D 0 C Substituting x 1; y 4, we find D D 1 12 1 2 y y 4 or y : 8 0 D 0 C 0 D 5 Hence, the equation of the tangent line is y 4 12 .x 1/ or y 12 x 32 . D 5 D 5 C 5 36. x2ey ye x 4, .2; 0/ C D SOLUTION Taking the derivative of both sides of x2ey ye x 4 yields C D x2eyy 2xe y ye x exy 0: 0 C C C 0 D Substituting x 2; y 0, we find D D 4 4y 4 0 e2y 0 or y : 0 C C C 0 D 0 D 4 e2 C Hence, the equation of the tangent line is 4 y .x 2/: D 4 e2 C x2 37. e2x y , .2; 4/ D y 2 2x y x SOLUTION taking the derivative of both sides of e yields D y
2xy x2y e2x y.2 y / 0 : 0 D y2 Substituting x 2; y 4, we find D D 16 4y 4 e0.2 y / 0 or y : 0 D 16 0 D 3 Hence, the equation of the tangent line is y 4 4 .x 2/ or y 4 x 4 . D 3 D 3 C 3 SECT ION 3.10 Implicit Differentiation 307
2 x2 16 1 38. y e xy 2, .4; 2/ D 2 SOLUTION Taking the derivative of both sides of y2ex 16 xy 1 2 yields D 2 2 2xy 2ex 16 2yy ex 16 xy 2y y 1 0: C 0 C 0 D Substituting x 4; y 2, we find D D 1 63 32e 0 4y e0 y 0 or y : C 0 C 0 2 D 0 D 10 Hence, the equation of the tangent line is y 2 63 .x 4/ or y 63 x 136 . D 10 D 10 C 5 39. Find the points on the graph of y2 x3 3x 1 (Figure 1) where the tangent line is horizontal. 2 D C (a) First show that 2yy 0 3x 3, where y0 dy=dx . (b) Do not solve for y . Rather,D set y 0 andD solve for x. This yields two values of x where the slope may be zero. 0 0 D (c) Show that the positive value of x does not correspond to a point on the graph. (d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates.
y
2
x −2 −1 1 2
−2
FIGURE 1 Graph of y2 x3 3x 1. D C SOLUTION (a) Applying implicit differentiation to y2 x3 3x 1, we have D C dy 2y 3x2 3: dx D 2 (b) Setting y0 0 we have 0 3x 3, so x 1 or x 1. (c) If we returnD to the equationD y2 x3 3x D 1 andD substitute x 1, we obtain the equation y2 1, which has no real solutions. D C D D (d) Substituting x 1 into y2 x3 3x 1 yields D D C y2 . 1/ 3 3. 1/ 1 1 3 1 3; D C D C C D so y p3 or p3. The tangent is horizontal at the points . 1; p3/ and . 1; p3/ . D 40. Show, by differentiating the equation, that if the tangent line at a point .x; y/ on the curve x2y 2x 8y 2 is horizontal, then xy 1. Then substitute y x 1 in x2y 2x 8y 2 to show that the tangent line is horizontal C at theD points 2; 1 and D D C D 2 4; 1 . 4 SOLUTION Taking the derivative on both sides of the equation x2y 2x 8y 2 yields C D 2.1 xy/ x2y 2xy 2 8y 0 or y : 0 C C 0 D 0 D x2 8 C 1 Thus, if the tangent line to the given curve is horizontal, it must be that 1 xy 0, or xy 1. Substituting y x into x2y 2x 8y 2 then yields D D D C D 8 x 2x 2 or x2 2x 8 .x 4/.x 2/ 0: C x D C D C D Hence, the given curve has a horizontal tangent line when x 2 and when x 4. The corresponding points on the curve are thus 2; 1 and 4; 1 . D D 2 4 41. Find all points on the graph of 3x 2 4y 2 3xy 24 where the tangent line is horizontal (Figure 2). C C D y
x
FIGURE 2 Graph of 3x 2 4y 2 3xy 24 . C C D 308 CHAPTER 3 DIFFERENTIATION
SOLUTION Differentiating the equation 3x 2 4y 2 3xy 24 implicitly yields C C D 6x 8yy 3xy 3y 0; C 0 C 0 C D so 6x 3y y C : 0 D 8y 3x C Setting y 0 leads to 6x 3y 0, or y 2x . Substituting y 2x into the equation 3x 2 4y 2 3xy 24 yields 0 D C D D D C C D 3x 2 4. 2x/ 2 3x. 2x/ 24; C C D or 13x 2 24 . Thus, x 2p78=13 , and the coordinates of the two points on the graph of 3x 2 4y 2 3xy 24 where the tangent lineD is horizontalD are ˙ C C D
2p78 4p78 2p78 4p78 ; and ; : 13 13 ! 13 13 ! 42. Show that no point on the graph of x2 3xy y2 1 has a horizontal tangent line. C D SOLUTION Let the implicit curve x2 3xy y2 1 be given. Then C D 2x 3xy 3y 2yy 0; 0 C 0 D so 2x 3y y : 0 D 3x 2y
Setting y 0 leads to y 2 x. Substituting y 2 x into the equation of the implicit curve gives 0 D D 3 D 3 2 2 2 x2 3x x x 1; 3 C 3 D 5 2 or 9 x 1, which has no real solutions. Accordingly, there are no points on the implicit curve where the tangent line has slope zero. D 43. Figure 1 shows the graph of y4 xy x3 x 2. Find dy=dx at the two points on the graph with x-coordinate 0 and find an equation of the tangent line at .1;C 1/ . D C SOLUTION Consider the equation y4 xy x3 x 2. Then 4y 3y xy y 3x 2 1, and C D C 0 C 0 C D 3x 2 y 1 y : 0 D x 4y3 C Substituting x 0 into y4 xy x3 x 2 gives y4 2, which has two real solutions, y 21=4 . When y 21=4 , we have D C D C D D ˙ D 4 21=4 1 p2 p2 y0 C :3254: D 4 23=4 D 8
When y 21=4 , we have D 4 21=4 1 p2 p2 y0 :02813: D 4 23=4 D 8
At the point .1; 1/ , we have y 1 . At this point the tangent line is y 1 1 .x 1/ or y 1 x 4 . 0 D 5 D 5 D 5 C 5 44. Folium of Descartes The curve x3 y3 3xy (Figure 3) was first discussed in 1638 by the French philosopher-mathematician Ren´eDescartes, who called it the foliumC (meaningD “leaf”). Descartes’s scientific colleague Gilles de Roberval called it the jasmine flower. Both men believed incorrectly that the leaf shape in the first quadrant was repeated in each quadrant, giving the appearance 2 4 of petals of a flower. Find an equation of the tangent line at the point 3 ; 3 . y 2
x −2 2
−2 FIGURE 3 Folium of Descartes: x3 y3 3xy . C D SECT ION 3.10 Implicit Differentiation 309
2 x y 2 4 SOLUTION Let x3 y3 3xy . Then 3x 2 3y 2y 3xy 3y , and y . At the point ; , we have C D C 0 D 0 C 0 D x y2 3 3 4 4 8 4 y 9 3 9 : 0 D 2 16 D 10 D 5 3 9 9 The tangent line at P is thus y 4 4 x 2 or y 4 x 4 . 3 D 5 3 D 5 C 5 45. Find a point on the folium x3 y3 3xy other than the origin at which the tangent line is horizontal. C D SOLUTION Using implicit differentiation, we find
d d x3 y3 .3xy/ dx C D dx 3x 2 3y 2y 3.xy y/ C 0 D 0 C 2 2 3 3 Setting y0 0 in this equation yields 3x 3y or y x . If we substitute this expression into the original equation x y 3xy , we obtain:D D D C D
x3 x6 3x.x 2/ 3x 3 or x3.x 3 2/ 0: C D D D One solution of this equation is x 0 and the other is x 21=3 . Thus, the two points on the folium x3 y3 3xy at which the tangent line is horizontal are .0; 0/ Dand .2 1=3 ; 2 2=3 /. D C D
46. Plot x3 y3 3xy b for several values of b and describe how the graph changes as b 0. Then compute C D C ! dy=dx at the point .b 1=3 ; 0/ . How does this value change as b ? Do your plots confirm this conclusion? ! 1 SOLUTION Consider the first row of figures below. When b < 0 , the graph of x3 y3 3xy b consists of two pieces. As b 0 , the two pieces move closer to intersecting at the origin. From the secondC rowD of figures,C we see that the graph of x3 !y3 3xy b when b > 0 consists of a single piece that has a “loop” in the first quadrant. As b 0 , the loop comes closerC to “pinchingD C off” at the origin. ! C
b = −0.1 b = −0.01 b = −0.001 y y y 1.5 1.5 1.5
1 1 1
0.5 0.5 0.5
x x x −0.5 0.5 1 1.5 −0.5 0.5 1 1.5 −0.5 0.5 1 1.5 −0.5 −0.5 −0.5
b = 0.1 b = 0.01 b = 0.001 y y y 1.5 1.5 1.5
1 1 1
0.5 0.5 0.5
x x x −0.5 0.5 1 1.5 −0.5 0.5 1 1.5 −0.5 0.5 1 1.5 −0.5 −0.5 −0.5
Differentiating the equation x3 y3 3xy b with respect to x yields 3x 2 3y 2y 3xy 3y , so C D C C 0 D 0 C y x2 y : 0 D y2 x
At .b 1=3 ; 0/ , we have
2 0 x 3 y x pb: 0 D 02 x D D
Consequently, as b , y0 at the point on the graph where y 0. This conclusion is supported by the figures shown below, which correspond! 1 to b !1, 1 b 10 , and b 100 . D D D D 310 CHAPTER 3 DIFFERENTIATION
b = 100 b = 01 b = 10 y y y 4 4 4
2 2 2
x x x −4 −2 2 4 −4 −2 2 4 −4 −2 2 4 −2 −2 −2
−4 −4 −4
47. Find the x-coordinates of the points where the tangent line is horizontal on the trident curve xy x3 5x 2 2x 1, so named by Isaac Newton in his treatise on curves published in 1710 (Figure 4). D C Hint: 2x 3 5x 2 1 .2x 1/.x 2 2x 1/ . C D y
20
4 x −2 2 6 8
−20
FIGURE 4 Trident curve: xy x3 5x 2 2x 1. D C
SOLUTION Take the derivative of the equation of a trident curve:
xy x3 5x 2 2x 1 D C to obtain
xy y 3x 2 10x 2: 0 C D C Setting y 0 gives y 3x 2 10x 2. Substituting this into the equation of the trident, we have 0 D D C xy x.3x 2 10x 2/ x3 5x 2 2x 1 D C D C or
3x 3 10x 2 2x x3 5x 2 2x 1 C D C Collecting like terms and setting to zero, we have
0 2x 3 5x 2 1 .2x 1/.x 2 2x 1/: D C D Hence, x 1 ; 1 p2. D 2 ˙ 48. Find an equation of the tangent line at each of the four points on the curve .x 2 y2 4x/ 2 2.x 2 y2/ where x 1. This curve (Figure 5) is an example of a limac¸on of Pascal , named after the father ofC the French philosopherD C Blaise Pascal,D who first described it in 1650.
y
3
x 1 3 5
−3 FIGURE 5 Limac¸on: .x 2 y2 4x/ 2 2.x 2 y2/. C D C
SOLUTION Plugging x 1 into the equation for the limac¸on and solving for y, we find that the points on the curve where x 1 D D are: .1; 1/ , .1; 1/ , .1; p7/ , .1; p7/ . Using implicit differentiation, we obtain
2.x 2 y2 4x/.2x 2yy 4/ 2.2x 2yy /: C C 0 D C 0 We plug in x 1 and get D 2.1 y2 4/.2 2yy 4/ 2.2 2yy / C C 0 D C 0 SECT ION 3.10 Implicit Differentiation 311 or
.2y 2 6/.2yy 2/ 4 4yy : 0 D C 0
After collecting like terms and solving for y0, we have 2 y2 y C : 0 D y3 4y
1 At the point .1; 1/ the slope of the tangent is 3 and the tangent line is 1 1 2 y 1 .x 1/ or y x : D 3 D 3 C 3 At the point .1; 1/ the slope of the tangent is 1 and the tangent line is 3 1 1 2 y 1 .x 1/ or y x : C D 3 D 3 3 At the point .1; p7/ the slope of the tangent is 5=3 p7 and the tangent line is 5 5 5 y p7 .x 1/ or y x p7 : D 3p7 D 3p7 C 3p7
At the point .1; p7/ the slope of the tangent is 5=3 p7 and the tangent line is
5 5 5 y p7 .x 1/ or y x p7: C D 3p7 D 3p7 C 3p7 49. Find the derivative at the points where x 1 on the folium .x 2 y2/2 25 xy2. See Figure 6. D C D 4 y 2
x 1
−2 25 FIGURE 6 Folium curve: .x 2 y2/2 xy2 C D 4
25 SOLUTION First, find the points .1; y/ on the curve. Setting x 1 in the equation .x 2 y2/2 xy2 yields D C D 4 25 .1 y2/2 y2 C D 4 25 y4 2y2 1 y2 C C D 4 4y4 8y 2 4 25y 2 C C D 4y 4 17y 2 4 0 C D .4y 2 1/.y 2 4/ 0 D 1 y2 or y2 4 D 4 D Hence y 1 or y 2. Taking d of both sides of the original equation yields D ˙ 2 D ˙ dx 25 25 2.x 2 y2/.2x 2yy / y2 xyy C C 0 D 4 C 2 0 25 25 4.x 2 y2/x 4.x 2 y2/yy y2 xyy C C C 0 D 4 C 2 0 25 25 .4.x 2 y2/ x/yy y2 4.x 2 y2/x C 2 0 D 4 C 25 2 2 2 4 y 4.x y /x y0 C D y.4.x 2 y2/ 25 x/ C 2 312 CHAPTER 3 DIFFERENTIATION
At .1; 2/ , x2 y2 5, and C D 25 2 4 2 4.5/.1/ 1 y0 : D 2.4.5/ 25 .1// D 3 2 At .1; 2/ , x2 y2 5 as well, and C D 25 2 4 . 2/ 4.5/.1/ 1 y0 : D 2.4.5/ 25 .1// D 3 2 At .1; 1 /, x2 y2 5 , and 2 C D 4 2 25 1 4 5 .1/ 4 2 4 11 y0 : D 1 4 5 25 .1/ D 12 2 4 2 At .1; 1 /, x2 y2 5 , and 2 C D 4 2 25 1 4 5 .1/ 4 2 4 11 y0 : D 1 4 5 25 .1/ D 12 2 4 2 The folium and its tangent lines are plotted below:
y
2
1
x 0.5 1 1.5 2 −1
−2
50. Plot .x2 y2/2 12.x 2 y2/ 2 for 4 x 4, 4 y 4 using a computer algebra system. How many horizontal tangent linesC does theD curve appear toC have? Find the points where these occur. SOLUTION A plot of the curve .x 2 y2/2 12.x 2 y2/ 2 is shown below. From this plot, it appears that the curve has a horizontal tangent line at six differentC locations.D C
y
1
x −3 −2 −1 1 2 3
−1
Differentiating the equation .x2 y2/2 12.x 2 y2/ 2 with respect to x yields C D C 2.x 2 y2/.2x 2yy / 12.2x 2yy /; C C 0 D 0 so x.6 x2 y2/ y : 0 D y.x 2 y2 6/ C C Thus, horizontal tangent lines occur when x 0 and when x2 y2 6. Substituting x 0 into the equation for the curve leaves D C D D y4 12y 2 2 0, from which it follows that y2 p38 6 or y p38 6. Substituting x2 y2 6 into the equation forC the curve leavesD x2 y2 17 . From here, it followsD that D ˙ C D D 6 p p159 p57 x and y : D ˙ 6 D ˙ 6 The six points at which horizontal tangent lines occur are therefore
0; p38 6 ; 0; p38 6
q q SECT ION 3.10 Implicit Differentiation 313
p159 p57 p159 p57 p159 p57 p159 p57 ; ; ; ; ; ; ; 6 6 ! 6 6 ! 6 6 ! 6 6 !
Exercises 51–53: If the derivative dx=dy (instead of dy=dx 0) exists at a point and dx=dy 0, then the tangent line at that point is vertical. D D
51. Calculate dx=dy for the equation y4 1 y2 x2 and find the points on the graph where the tangent line is vertical. C D C SOLUTION Let y4 1 y2 x2. Differentiating this equation with respect to y yields C D C dx 4y 3 2y 2x ; D C dy so dx 4y3 2y y.2y 2 1/ : dy D 2x D x
dx p2 Thus, 0 when y 0 and when y . Substituting y 0 into the equation y4 1 y2 x2 gives 1 x2, so dy D D D ˙ 2 D C D C D p2 p3 x 1. Substituting y , gives x2 3=4 , so x . Thus, there are six points on the graph of y4 1 y2 x2 D ˙ D ˙ 2 D D ˙ 2 C D C where the tangent line is vertical:
p3 p2 p3 p2 p3 p2 p3 p2 .1; 0/; . 1; 0/; ; ; ; ; ; ; ; : 2 2 ! 2 2 ! 2 2 ! 2 2 ! 52. Show that the tangent lines at x 1 p2 to the conchoid with equation .x 1/ 2.x 2 y2/ 2x 2 are vertical (Figure 7). D ˙ C D y
2 1 x 1 2 −1 −2
FIGURE 7 Conchoid: .x 1/ 2.x 2 y2/ 2x 2. C D
SOLUTION Consider the equation of a conchoid:
.x 1/ 2 x2 y2 2x 2: C D Taking the derivative of both sides of this equation gives
dx dx dx .x 1/ 2 2x 2y x2 y2 2 .x 1/ 4x ; dy C C C dy D dy so that
dx .x 1/ 2 y : dy D 2x .1 x/ x2 y2 x .x 1/ 2 C C Setting dx=dy 0 yields x 1 or y 0. We can’t have x " 1, lest 0 2 in the conchoid’s equation. Plugging y 0 into the D D D D D D equation gives .x 1/ 2 x2 2x 2 or x2 .x 1/ 2 2 0, which implies x 0 (a double root) or x 1 p2. [Plugging D D D D ˙ x 0 into the conchoid’s equation gives y2 0 or y 0: At .x; y/ .0; 0/ the expression for dx=dy is undefined ( 0=0 ). Via an alternativeD parametric analysis, the slopes ofD the tangentD lines at the originD turn out to be p3.] Accordingly, the tangent lines to the conchoid are vertical at .x; y/ .1 p2; 0/ . ˙ D ˙ 53. Use a computer algebra system to plot y2 x3 4x for 4 x 4, 4 y 4. Show that if dx=dy 0, then y 0. Conclude that the tangent line is vertical at theD points where the curve intersects the x-axis. Does your plot confirmD this conclusion?D 314 CHAPTER 3 DIFFERENTIATION
SOLUTION A plot of the curve y2 x3 4x is shown below. D y
2
1
x −2 −1 1 2 3
−1
−2
Differentiating the equation y2 x3 4x with respect to y yields D dx dx 2y 3x 2 4 ; D dy dy or dx 2y : dy D 3x 2 4
From here, it follows that dx 0 when y 0, so the tangent line to this curve is vertical at the points where the curve intersects dy D D the x-axis. This conclusion is confirmed by the plot of the curve shown above. 54. Show that for all points P on the graph in Figure 8, the segments OP and PR have equal length.
y Tangent line
P
x O R
FIGURE 8 Graph of x2 y2 a2. D
SOLUTION Because of the symmetry of the graph, we may restrict attention to any point P in the first quadrant. Suppose P has 2 2 2 2 2 coordinates .p; p a /. Taking the derivative of both sides of the equation x y a yields 2x 2yy 0 0, or y0 x=y . It follows that the slope of the line tangent to the graph at P has slope D D D p p p2 a2 and the slope of the normal line is p
p2 a2 : p p Thus, the equation of the normal line is
p2 a2 y p2 a2 .x p/; D p q p and the coordinates of the point R are .2p; 0/ . Finally, the length of the line segment OP is
p2 p2 a2 2p 2 a2; C D q q while the length of the segment PR is
.2p p/2 p2 a2 2p 2 a2: C D q q SECT ION 3.10 Implicit Differentiation 315
In Exercises 55–58, use implicit differentiation to calculate higher derivatives.
55. Consider the equation y3 3 x2 1. 2 D (a) Show that y x=y 2 and differentiate again to show that 0 D y2 2xyy y 0 00 D y4
(b) Express y00 in terms of x and y using part (a).
SOLUTION (a) Let y3 3 x2 1. Then 3y 2y 3x 0, and y x=y 2. Therefore, 2 D 0 D 0 D y2 1 x 2yy y2 2xyy y 0 0 : 00 D y4 D y4
(b) Substituting the expression for y0 into the result for y00 gives
y2 2xy x=y 2 y3 2x2 y00 4 5 : D y D y 56. Use the method of the previous exercise to show that y y 3 on the circle x2 y2 1. 00 D C D x SOLUTION Let x2 y2 1. Then 2x 2yy 0, and y . Thus C D C 0 D 0 D y
x y 1 xy y x y y2 x2 1 y 0 C y 3: 00 D y2 D y2 D y3 D y3 D 57. Calculate y at the point .1; 1/ on the curve xy 2 y 2 0 by the following steps: 00 C D (a) Find y0 by implicit differentiation and calculate y0 at the point .1; 1/ . (b) Differentiate the expression for y0 found in (a). Then compute y00 at .1; 1/ by substituting x 1, y 1, and the value of y0 found in (a). D D
SOLUTION Let xy 2 y 2 0. C D y2 1 (a) Then x 2yy y2 1 y 0, and y . At .x; y/ .1; 1/ , we have y . 0 C C 0 D 0 D 2xy 1 D 0 D 3 C (b) Therefore,
2 .3/ 2 .1/ 2 2 .2xy 1/ 2yy 0 y 2xy 0 2y 3 3 C 6 2 6 10 y00 C C C D .2xy 1/2 D 32 D 27 D 27 C given that .x; y/ .1; 1/ and y 1 . D 0 D 3 58. Use the method of the previous exercise to compute y at the point .1; 1/ on the curve x3 y3 3x y 2. 00 C D C 3.1 x2/ SOLUTION Let x3 y3 3x y 2. Then 3x 2 3y 2y 3 y , and y . At .x; y/ .1; 1/ , we find C D C C 0 D C 0 0 D 3y 2 1 D
3.1 1/ y 0: 0 D 3.1/ 1 D
Similarly,
2 2 3y 1 . 6x/ 3 3x 6yy 0 y00 2 3 D 3y2 1 D when .x; y/ .1; 1/ and y 0. D 0 D In Exercises 59–61, x and y are functions of a variable t and use implicit differentiation to relate dy=dt and dx=dt . dy y dx 59. Differentiate xy 1 with respect to t and derive the relation . D dt D x dt dy dx dy y dx SOLUTION Let xy 1. Then x y 0, and . D dt C dt D dt D x dt 316 CHAPTER 3 DIFFERENTIATION
60. Differentiate x3 3xy 2 1 with respect to t and express dy=dt in terms of dx=dt , as in Exercise 59. C D SOLUTION Let x3 3xy 2 1. Then C D dx dy dx 3x 2 6xy 3y2 0; dt C dt C dt D and dy x2 y2 dx C : dt D 2xy dt 61. Calculate dy=dt in terms of dx=dt . (a) x3 y3 1 (b) y4 2xy x2 0 D C C D SOLUTION (a) Taking the derivative of both sides of the equation x3 y3 1 with respect to t yields D dx dy dy x2 dx 3x 2 3y2 0 or : dt dt D dt D y2 dt
(b) Taking the derivative of both sides of the equation y4 2xy x2 0 with respect to t yields C C D dy dy dx dx 4y 3 2x 2y 2x 0; dt C dt C dt C dt D or dy x y dx C : dt D 2y 3 x dt C 62. The volume V and pressure P of gas in a piston (which vary in time t) satisfy P V 3=2 C , where C is a constant. Prove that D dP =dt 3 P dV =dt D 2 V
The ratio of the derivatives is negative. Could you have predicted this from the relation P V 3=2 C ? D SOLUTION Let P V 3=2 C , where C is a constant. Then D 3 dV dP dP=dt 3 P P V 1=2 V 3=2 0; so : 2 dt C dt D dV=dt D 2 V
If P is increasing (respectively, decreasing), then V .C=P / 2=3 is decreasing (respectively, increasing). Hence the ratio of the derivatives ( = or = ) is negative. D C C Further Insights and Challenges
63. Show that if P lies on the intersection of the two curves x2 y2 c and xy d (c; d constants), then the tangents to the curves at P are perpendicular. D D
SOLUTION Let C1 be the curve described by x2 y2 c, and let C 2 be the curve described by xy d. Suppose that 2 D 2 2 2 D P .x 0; y 0/ lies on the intersection of the two curves x y c and xy d. Since x y c, the chain rule gives us D 2x x D Dx0 D 2x 2yy 0 0, so that y0 2y y . The slope to the tangent line to C1 is y0 . On the curve C 2 , since xy d, the product D D D y y0 D rule yields that xy 0 y 0, so that y0 x . Therefore the slope to the tangent line to C 2 is x0 . The two slopes are negative reciprocals of one another,C D hence the tangentsD to the two curves are perpendicular. 64. The lemniscate curve .x 2 y2/2 4.x 2 y2/ was discovered by Jacob Bernoulli in 1694, who noted that it is “shaped like a figure 8, or a knot, or the bowC of a ribbon.”D Find the coordinates of the four points at which the tangent line is horizontal (Figure 9).
y
1
x −1 1
−1 FIGURE 9 Lemniscate curve: .x 2 y2/2 4.x 2 y2/. C D SECT ION 3.10 Implicit Differentiation 317
2 SOLUTION Consider the equation of a lemniscate curve: x2 y2 4 x2 y2 . Taking the derivative of both sides of this equation, we have C D 2 x2 y2 2x 2yy 4 2x 2yy : C C 0 D 0 Therefore,
8x 4x x2 y2 x2 y2 2 x y C C : 0 D 8y 4y x2 y2 D x2 y2 2 y C C C C If y 0, then either x 0 or x2 y2 2. 0 D D C D If x 0 in the lemniscate curve, then y4 4y 2 or y2 y2 4 0. If y is real, then y 0. The formula for y in (a) is 0 not definedD at the origin ( 0=0 ). An alternativeD parametric analysisC showsD that the slopes of theD tangent lines to the curve at the origin are 1. ˙ If x2 y2 2 or y2 2 x2, then plugging this into the lemniscate equation gives 4 4 2x 2 2 which yields C D D D 3 p6 1 p2 x . Thus y . Accordingly, the four points at which the tangent lines to the lemniscate D ˙ 2 D ˙ 2 D ˙ 2 D ˙ 2 curve areq horizontal are p6 ; pq2 , p6 ; p2 , p6 ; p2 , and p6 ; p2 . 2 2 2 2 2 2 2 2 65. Divide the curve in Figure 10 into five branches, each of which is the graph of a function. Sketch the branches.
y
2
x −4−2 2 4
−2
FIGURE 10 Graph of y5 y x2y x 1. D C C
SOLUTION The branches are: Upper branch: y
2
x −4 −2 2 4 −2
Lower part of lower left curve: y
1
x −4 −3 −2 −1 −1
−2
Upper part of lower left curve: y
1
x −4 −3 −2 −1 −1
−2
Upper part of lower right curve: 318 CHAPTER 3 DIFFERENTIATION
y
1 1 2 3 4 x
−1
−2
Lower part of lower right curve: y
1 1 2 3 4 x
−1
−2
3.11 Related Rates
Preliminary Questions 1. Assign variables and restate the following problem in terms of known and unknown derivatives (but do not solve it): How fast is the volume of a cube increasing if its side increases at a rate of 0:5 cm/s? dV SOLUTION Let s and V denote the length of the side and the corresponding volume of a cube, respectively. Determine dt if ds 0:5 cm/s. dt D 2. What is the relation between dV =dt and dr=dt if V 4 r3? D 3 dV dr SOLUTION Applying the general power rule, we find 4r 2 . Therefore, the ratio is 4r 2. dt D dt In Questions 3 and 4, water pours into a cylindrical glass of radius 4 cm. Let V and h denote the volume and water level respectively, at time t.
3. Restate this question in terms of dV =dt and dh=dt : How fast is the water level rising if water pours in at a rate of 2 cm 3/min? dh dV SOLUTION Determine if 2 cm3/min. dt dt D 4. Restate this question in terms of dV =dt and dh=dt : At what rate is water pouring in if the water level rises at a rate of 1 cm/min? dV dh SOLUTION Determine if 1 cm/min. dt dt D Exercises In Exercises 1 and 2, consider a rectangular bathtub whose base is 18 ft 2.
1. How fast is the water level rising if water is filling the tub at a rate of 0:7 ft 3/min? dV dh SOLUTION Let h be the height of the water in the tub and V be the volume of the water. Then V 18h and 18 . Thus D dt D dt dh 1 dV 1 .0:7/ 0:039 ft =min : dt D 18 dt D 18 2. At what rate is water pouring into the tub if the water level rises at a rate of 0:8 ft/min?
SOLUTION Let h be the height of the water in the tub and V its volume. Then V 18h and D dV dh 18 18 .0:8/ 14:4 ft 3=min : dt D dt D D 3. The radius of a circular oil slick expands at a rate of 2 m/min. (a) How fast is the area of the oil slick increasing when the radius is 25 m? (b) If the radius is 0 at time t 0, how fast is the area increasing after 3 min? D