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196 0 CHAPTER 3 DIFFERENTIATION RULES
APPLIED PROJECT Where Should a Pilot Start Descent?
1. Condition (i) will hold if and only if all of the following four conditions hold:
(0) P(O) == 0
((3) P' (0) == 0 (for a smooth landing)
(r) p' (£) == 0 (since the plane is cruising horizontally when it begins its descent)
(5) P(£) == h.
First of all, condition 0 implies that P (0) == d == 0, so P (x) == ax3 + bx2 + cx =? p' (x) == 3ax2 + 2bx + c. But
2bp' (0) == c == 0 by condition (3. So p' (£) == 3al:2 + 2b£ == £ (3a£ + 2b). Now by condition r, 3a£ + 2b == 0 =? a == - 3£'
Therefore, P(x) = - ~~X3 + bic", Setting P(£) = h for condition 8, we get P(£) = - ~~£3 + b£2 = h =}
2 2 1 2 3h 2h 2h 3 3h 2-"3 b£2 + bl: == h =? "3 b£ == h =? b == f2 =? a == -l}' So y == P(x) == -l}x + f2x .
2. By condition (ii), ~~ = -v for all t, so x (t) = £ - vt. Condition (iii) states that I~:; I < k. By the Chain Rule,
h dy _ dy dx __ 2h (3 2) dx 3h (2 ) dx _ 6hx2v
_ 6hxv (D < £) we ave dt - dx dt - £3 x dt + £2 x dt - £3 £2 or x_=?
d2y 6hv dx 6hv dx 12hv2 6hv2 . dt 2 == ~ (2x) dt - 7:2 dt == -~x + ~. In partIcular, when t == 0, x == £ and so
d2y 2 2 2 2y 2 ! 12hv 6hv 6hv I d I 6hv .. . .
dt 2 t~O = -~£ + ~ = -~. Thus, dt 2 t=o = ~ :<:; k. (This condition also follows from taking x = 0.)
3. We substitute k = 860 mi/h", h = 35,000 ft x ~, and v = 300 mi/h into the result ofpart (b):
6(35,000 };k) (300? <_ 860 =? I: >_ 300 6. 35,000 ~ 64.5 mile. 5280 . 860 s
4. Substituting the values of hand £ in Problem 3 into 7---------.........
2h 3h . P(x) == -l}x3 + f2x 2 gives us P(x) == ax 3 + bx",
where a ~ -4.937 x 10-5 and b ~ 4.78 X 10-3.
0......-:;;;..------- 64.52
3.5 Implicit Differentiation
1. (a) dxd
(xy + 2x + 3x 2) == dx
d (4) =? (x· y' + y. 1) + 2 + 6x == 0 =? xy' == -y - 2 - 6x =?
, -y - 2 - 6x, y + 2 y == or y == -6 - --.
x x
4 - 2x - 3x 2 4 ,4(b) xy + 2x + 3x 2 == 4 =? xy == 4 - 2x - 3x 2 =? Y == == - - 2 - 3x, so y == -2 - 3.
x x x
-y-2-6x -(4/x-2-3x)-2-6x -4/x-3x 4(c) From part (a), y' == == == ==-- - 3.2x x X x
SECTION 3.5 IMPLICIT DIFFERENTIATION D 197
2. (a) dd (4x 2 + 9y2) == dd (36) ::::} 8x + 18y . yl == 0 ::::} y' == _~ == _ 4x x x 18y 9y
(b) 4x2 + 9y2 == 36 ::::} 9y2 == 36 - 4x2 ::::} y2 == ~ (9 - x 2) ::::} Y == ± ~ V9 - x 2, so
y' == ±~ . 1.(9 _ X2)-1/2 (-2x) == =j= 2x 3 2 3 y!9 - x 2
I 4x 4x 2x (c) Frompart(a),y ==--==- ( 2 ~) ===j= .~.
9y 9 ± 3" v 9 - x 2 3 v 9 - x 2
3. (a) ~ (!+ !) = ~ (1) =? _~ - J:..- y' = 0 =? _~ y' = ~ =? y' = _ y2dx x y dx x 2 y2 y2 x 2 x 2
1 1 1 1 x-I x, (x - 1)(1) - (x)(l) -1 (b) ;; + Y== 1 ::::} Y== 1 - ;; == -x- ::::} y == x-I' so Y == (x _ 1) 2 == (x _ 1) 2 .
2 (c) y' = _ y2 = _ [x/(x - 1)] 2 = _ x 1
2 2 2(x x x x 1)2 (x - 1)2
2 1 ,. '2 r. : 4. (a) :x (cosx+JY) = d~(5) =? -sinx+h-1/ · y' = O ::::} r:. . y == SIn x ::::} y == y y SIn x
2yy
(b) COS x + JY == 5 ::::} JY == 5 - cos x ::::} y == (5 - cos x) 2, so y' == 2(5 - cos x)' (sin x) == 2 sin x (5 - cos x) .
(c) From part (a), y' == 2 JYsinx == 2 )(5 - COSX)2 == 2(5 - cos x) sinx [since 5 - cosx > 0].
1 y'::::} -+--==0 ::::}
V; 2JY
y' 1 , 2JY::::} y == --2JY
-V; V;
7. ~(X2+xy_y2)==~(4) ::::} 2x+x·y'+y·1-2yy'==0 ::::}dx dx
, , - 2x - y 2x + yxy' - 2yy' == -2x - y ::::} (x - 2y) y == -2x - y ::::} y == ==-
x - 2y 2y - x
-6x2 - 2xy + y3::::} (x 2 _ 3x y2) y' == -6x2 - 2xy + y3 ::::} y' == - _
- x 2 - 3x y2
x 4 + x 4 y' + 4x4 + 4x3y == 3y2 - y2 yl + 6xy y' - 2y2 y' ::::} x 4 y' + 3y2 y' - 6xy y' == 3y2 - 5x4 - 4x3y ::::}
3y2 - 5x4 - 4x3y (x 4 + 3y2 - 6xy) y' == 3y2 - 5x4
- 4x3y ::::} y' == 4 2 6 x + 3y - xy
198 D CHAPTER 3 DIFFERENTIATION RULES
__
5y4 + 3X2y2 _ ex 2
2xy (e x 2 y2)
11. d~ (X2y2 + xsiny) = d~ (4) =? x 2. 2yy' + y2 . 2x + xcosy· y' + siny' 1 = 0 =?
2 .y' = - 2xy - sIn y2x 2y y' + x cos y . y' = -2xy2 - sin y =? (2x 2y + Xcos y)y' = -2x y2 - sin y =?
2x 2y + Xcosy
13.
14.
15.
1 - y2 COS(xy2) = 2xy COS(xy2) y'
d~ (4 cos x siny) = d~ (1) =?
y' (4 cos x cos y) = 4 sin x sin y
d~ [y sin(x2)] = d~ [x sin(y2)]
sin(x2) - 2xy COS(y2)
~(ex/y) = ~(x - y) =? eX/ Y . ~ (~) = 1 - y' =?
dx dx dx y
yy. 1 - x· y' 1 xex /eX/ Y . = 1 - y' eX
/ Y . - - -- • y' = 1 - y'y2 Y y2
y _ eX / Y
x/ y) Y y y(y _ eX / Y )'( xe y - e
X
/y 1--- =~-- =? y' y2 - xex/yy2 Y
1 ' --- + y = 2x 2yy' + 2x y2 =? 1 + y' = 4x 2y Jx + yy' + 4xy2JX + y =? 2Jx+y 2Jx+y
y' - 4x2y J x + y y' = 4x y2 J x + y - 1 =? y' ( 1 - 4x2y Jx + y ) = 4x y2J x + y - 1 =?
4xy2~-1y'
1- 4x 2y Jx + y
, 1 - y2 COS(xy2) =? Y = -------:....--..,;,..
2xy cos (x y 2)
4 [cos x· cosy· y' + siny' (~sinx)] = 0 =?
4sinx siny=? y' 4 = tanx tany
cosx cosy
=? y cos(x2) . 2x + sin(x2) . y' = X COS(y2) . 2y y' + sin(y2) . 1 =?
sin(y2) - 2xy cos(x2)
SECTION 3.5 IMPLICIT DIFFERENTIATION D 199
2y17. v;i; == 1 + x ::::} ~(xy)-1/2(xy' + y. 1) == 0 + x 2y' + y. 2x::::} x~ y' + y~ == x 2y' + 2xy ::::} 2 yxy 2 y xy
2v;i;)'( X 2 y ,(X-2x 4xyv;i;-y , 4xyv;i;-yy -- - x 2) == xy - -- ::::} y == ::::} y == ---'---
2 v;i; 2 v;i; 2 v;i; 2 v;i; x - 2x 2 v;i;
2) 2(x18. tan(x - y) == ~ ::::} (1 + x 2) tan(x - y) == y ::::} (1 + x sec - y) . (1 - y') + tan(x - y) . 2x == y' ::::}l+x
2) 2(x 2) 2(x(1 + x sec - y) - (1 + x sec - y) . y' + 2x tan(x - y) == y' ::::}
2) 2(x 2) 2(x(1 + x sec - y) + 2xtan(x - y) == [1 + (1 + x sec - y)J . y' ::::}
2 2(x , (1 + x ) sec - y) + 2xtan(x - y) y == 2 ) sec2(x1 + (1 + x - y)
19. d~ (eY cos x) = d~ [1+ sin(xy)] =? eY (- sin x) + cos x . e" . y' = cos(xy) . (xy' + y . 1) =?
- eY sin x + eY cos x . y' == x cos (xy) . y' + yeas (xy) ::::} eY cos x . y' - x cos (xy) . y' == eY sin x + yeas (xy) ::::}
, eY sin x + yeas (xy)[eY cosx - xcos(xy)] y' == eY sinx + ycos(xy) ::::} y == -----~....;.
eY cos x - x cos(xy)
20. sin x + cos y == sin x cos y ::::} cos x - sin y . y' == sin x (- sin y . y') + cos y cos x ::::}
, cos x (cos y - 1)(sin x sin y - sin y) y' == cos x cos y - cos x ::::} y ==. (si 1)SIn y sInx
2[f(x)]3} 221. dx d
{f(x) + x == dx d
(10) ::::} f'(x) + x . 3[f(x)]2 . f'(x) + [f(x)]3 . 2x == O. Ifx == 1, we have
f' (1) + 12. 3 [f (1)] 2 . i' (1) + [f (1)] 3 . 2(1) == 0 ::::} f' (1) + 1 . 3 . 22
. f' (1) + 23. 2 == 0 ::::}
f'(I) + 12f'(I) == -16 ::::} 13f'(I) == -16 ::::} f'(I) == -~.
22. d~ [g(x) + xsing(x)] = d~ (x 2 ) =? g'(x) + xcosg(x) . g'(x) + sing(x) ·1 = 2x. Ifx = 0, we have
g' (0) + 0 + sin 9 (0) == 2(0) ::::} g' (0) + sin 0 == 0 ::::} g' (0) + 0 == 0 ::::} g' (0) == O.
, dx -2x4y + x 3 - 6x y2 X - - - ------
- dy - 4X3y2 - 3x2y + 2y3
d d 224. dy (y sec x) = dy (xtany) =? y. secx tanx· x' + sec x ·1 = x· sec y + tany· x' =?
y sec x tan x . x' - tan y . x' == x sec" y - sec x ::::} (y sec x tan x - tan y) x' == x sec" y - sec x ::::}
dx x sec 2 y ~ sec xx'
dy y sec x tan x - tan y
200 D CHAPTER 3 DIFFERENTIATION RULES
25. x 2 + xy + y2 == 3 ~ 2x + X y' + y . 1 + 2yy' == 0 ~ X y' + 2y y' == - 2x - y ~ y' (x + 2y) == - 2x - y ~
, -2x - y , -2 - 1 -3 . . . y == x + 2y· When x == 1 and y == 1, we have y == 1 + 2 . 1 == 3 == -1, so an equation of the tangent hne IS
y - 1 == -l(x - 1) or y == -x + 2.
2+2xy_ y2+ X==226. x ~ 2x+2(xy'+y·1)-2yy'+1==0 ~ 2xy'-2yy'==-2x-2y-1 ~
-2x - 2y-1y'(2x-2y)==-2x-2y-1 ~ y'== . Whenx==landy==2,wehave
2x - 2y
-2 - 4 - 1 -7 7 7 7 3y' == == - == -, so an equation of the tangent line is y - 2 == "2 (x - 1) or y == "2 x - "2.
2 - 4 -2 2
27. x 2 + y2 == (2x2 + 2y2 - X)2 ~ 2x + 2yy' == 2(2x2 + 2y2 - x)(4x + 4yy' - 1). When x == 0 and y == ~,we have
o+ y' == 2(~ ) (2y' - 1) ~ y' == 2y' - 1 ~ y' == 1, so an equation of the tangent line is y - ~ == 1(x - 0)
or y == x + ~.
1 y' , VY ~+ vy=O =}- y = - ~. When x = -3J3
(_3J3)2/3 and y == 1, we have y
, == -
1 1/3 3 ~ ~' so an equation of the tangent line is
(-3 J3) -3J3
y - 1 == ~ (x+ 3 J3) or y == ~x + 4.
29. 2(x2 + y2)2 == 25(x2 - y2) ~ 4(x2 + y2) (2x + 2y y') == 25(2x - 2y y') ~
4(x + yy')(x2 + y2) == 25(x - yy') ~ 4yy'(x2 + y2) + 25yy' == 25x _ 4x(x2 + y2) ~
, 25x - 4x(x2 + y2) , 75 -120 45 9
Y == (2 2)· When x == 3 and y == 1, we have y == 25 + 40 == -65" == -13'25y + 4y x + y
so an equation of the tangent line is y - 1 == - ft(x - 3) or y == - ftx + ~.
When x == 0 and y == -2, we have -32y' + 16y' == 0 ~ -16y' == 0 ~ y' == 0, so an equation of the tangent line is
" y + 2 == 0(x - 0) or y == - 2.
10x3 - X 5 (b)
Y
. () , 10(1)3 - 1 9 d .So at the point 1, 2 we have y == 2 == "2' an an equation
of the tangent line is y - 2 == ~ (x - 1) or y == ~ x - %. -2
-2 r----+--+---f,L------I 2
SECTION 3.5 IMPLICIT DIFFERENTIATION D 201
, 3x2 + 6x32. (a) y2 = x 3 + 3x2 =? 2y y' = 3x2 + 3(2x) =? y =
2y So at the point (1, -2) we have
y' = 3(1;~ ~2~(1) = -~, and an equation of the tangent line is y + 2 = -~(x - 1) or y = -~x + ~.
(b) The curve has a horizontal tangent where y' = 0 {:} (c)
3x2 + 6x = 0 {:} 3x(x + 2) = 0 {:} x = 0 or x = -2.
But note that at x = 0, y = 0 also, so the derivative does not exist. -
At x = -2, y2 = (_2)3 + 3( _2)2 = -8 + 12 = 4, so y = ±2.
So the two points at which the curve has a horizontal tangent are
(-2, -2) and (-2,2).
33. 9x2 + y2 = 9 =? 18x + 2yy' = 0 =? 2yy' = -18x =? y' = -9x/y =?
y)) 2 y" = -9 (y . 1 - x . y') = -9 (y - x(-9X/ = -9 . y2 + 9x = -9 . ~ [since x andy must satisfy the original
y2 y2 y3 y3
equation, 9x2 + y2 = 9]. Thus, y" = -81/y3.
1 y' r: 34 /" + r::. 1 ~ + 0 ~ y' = - 'Y.....J!.. ~ .yx yy= ;- r:: -T-T -T
2yx 2yy V;;
V;; [_1]y' _ yY [_1]2yY 2 V;;
y" x
1 since x and y must satisfy the original equation, V;; + yY = 1.
2xV;;
3 1---+---~---1 2
(1, -2)
-3
2x
2 2x2y( _X2/y2)y2(2x) - x . 2y y' 2x y2 2x y" =
5 '(y2)2 y4 Y
since x and y must satisfy the original equation, x 3 + y3 = 1.
37. (a) There are eight points with horizontal tangents: four at x ~ 1.57735 and 4
four at x ~ 0.42265.
, 3x2 - 6x + 2 (b) y = 2(2y 3 _ 3
y 2 _ Y + 1) '* y' = -1 at (0,1) and y' = ~ at (0, 2).
Equations of the tangent lines are y = -x + 1 and y = ~x + 2.
(c) y' = 0 =? 3x2 - 6x + 2 = 0 =? x = 1 ± ~ V3 -3
-21----i'----i1~~--+---+-__l
202 D CHAPTER 3 DIFFERENTIATION RULES
(d) By multiplying the right side of the equation by x - 3, we obtain the first
graph. By modifying the equation in other ways, we can generate the other
graphs. -
4
2 1---+--+---#--...:lIr--+---l----l 5
-3
y(y2 _ l)(y - 2)
4.5 == x(x - l)(x - 2)(x - 3)
- 3 I---l---+--+--+--+----,if------+--+----l 6- 2 I----l----::ool'-+-f"'!~--+---+----l 6
-3
y(y2 _ 4)(y - 2) y(y + 1)(y2 - l)(y - 2) (y + 1)(y2 - l)(y - 2)
== x(x- l)(x- 2) == x(x- l)(x - 2) == (x - l)(x - 2)
4 4
-3.5 -3
- 3 I----+---+--~-l,....---+-----l
3 I-+---I---'k::---:if---;----I-----I 4 -
-
-3 -4
y(y2 + l)(y - 2) y(y + 1)(y2 - 2)
== x(x2 - l)(x - 2) == x(x - 1)(x2 - 2)
38. (a) (b) There are 9 points with horizontal tangents: 3 at x == 0, 3 at x == ~,
and 3 at x == 1. The three horizontal tangents along the top of the
wagon are hard to find, but by limiting the y-range of the graph - 3 I---+--+---¥--~-+----I-----I 4
(to [1.6,1.7]' for example) they are distinguishable.
-3
x(y + 1)(y2 - l)(y - 2)
== y(x - l)(x- 2)
4 1----I--+-I-+-----'"k----:>oI-r-+---+----l 4
-4
39. From Exercise 29, a tangent to the lemniscate will be horizontal if y' == 0 =} 25x - 4x (x 2 + y2) == 0 =}
x[25 - 4(x 2 + y2)] == 0 =} x2+ y2 == ¥ (1). (Note that when x is 0, y is also 0, and there is no horizontal tangent
at the origin.) Substituting ¥ for x2 + y2 in the equation of the lemniscate, 2(x2 + y2)2 == 25 (x 2 - y2), we get
2 2x - y2 = ¥- (2). Solving (1) and (2), we have x = Hand y2 = ~, so the four points are (±sf! , ±~ ).
SECTION 3.5 IMPLICIT DIFFERENTIATION D 203
an equation of the tangent line at (xo, Yo) is
2xo-b '" Yo, Yoy Y6 xox x6 . . . y - yo = -2- (x - xo). Multiplying both sides by b2 gives -b - -b = - -2 + 2' SInce (xo, Yo) lies on the ellipse, 2 2a yo a a
xox yoY X6 Y6 we have --;;} + ----,;2 = a 2 + b2 = I,
=? an equation of the tangent line at (xo, Yo) is
2b x o ... Yo. YoY Y6 xox X6 . .Y - yo = -2- (x - xo). Multiplying both sides by b2 grves -b - -b = -2 - 2' SInce (xo, Yo) Iies on the hyperbola, 2 2a yo a a
xox YoY X6 Y6 we have --;;} - ----,;2 = a2 - b2 = 1.
1 y' /Y42. vf;; + /Y = Vc =? -- + -- = 0 =? y' = - IX =? an equation of the tangent line at (xo, Yo)
2vf;; 2/Y v z
is y - yo = - y);; (x - xo). Now x = 0 =? Y = yo - vJij (-xo) = yo + ~ y);;, so the y-intercept is ~ yXO
. rr: c: y);; ( yo ~ Yo + y Xo V yo. And y = 0 =? -Yo = - ~ x - xo) =? x - Xo = c- =? yXO vYo
x = Xo + ~ y);;, so the x-intercept is Xo + ~ y);;. The sum of the intercepts is
(yO + ~ y);;) + (xo + ~ y);;) = xo + 2 ~ y);; + Yo = (~ + y);;) 2 = (vc) 2 = c.
43. If the circle has radius r, its equation is x 2 + y2 = r2 =? 2x + 2yy' = 0 =? y' = - ~, so the slope of the tangent line y
at P (xo, Yo) is - Xo . The negative reciprocal of that slope is ~/ = Yo, which is the slope of 0 P, so the tangent line at yo -xo yo Xo
P is perpendicular to the radius 0 P.
46. y = vtan- 1 x = (tan- 1 X)1/2 =?
dy' = -21 (tan- 1 X)-1/2 . -(tan- 1 x) =
dx
1 1 _2Vtan 1 x'1+x2
1
47.
y'
y = sin- 1 (2x + 1) =?
1 d V 1 (2x + 1)2 . dx (2x + 1) = V1 _
1 2 (4x2 + 4x + 1) ·2= V-4x2 4x
1
v-x2 - X
204 D CHAPTER 3 DIFFERENTIATION RULES
1 ~ 1 1 1 (2 ) -1/2 1 x sec-1
x48. g(x) = VX2=1 sec- x ::::} g/(x) = VX2 -1· + sec- x· - x -1 (2x) = - + -- Xvx2 - 1 2 X vx2 - 1
/ ~ -1 1 2 1/2 x arccos x49. G(x) = Vl- x2arccos x ::::} G(x)=v1-x2. +arccosx·-(l-x)- (-2x)=-1---
VI - x 2 2 VI - x 2
50. y = tan- 1 (x - vx2+ 1) ::::}
y/ = 1 (1 _ x ) = 1 (VX2+1 -x)1 + (x - Vx2 + 1 ) 2 V x2 + 1 1 + x2 - 2x Vx2 + 1 + x2 + 1 Vx2 + 1
vx2+ 1- x vx2+ 1- x vx2+ 1- x 2 2)2(1 + x - Xvx2 + 1) vx2 + 1 2 [vx2 + 1 (1 + x - x(x2 + l)J
1
51. h(t) = cot-1(t ) + cot-1(1/ t ) ::::}
2/ lId 1 1 t (1 ) 1 1
h (t) = -1 + t2 - 1 + (1/t)2 . dt t = -1 + t2 - t2 + l' -(i = -1 + t2 + t2 + 1 = O.
Note that this makes sense because h(t) = ~ for t > 0 and h(t) = 3; for t < O.
52. F(()) = arcsin vsin () = arcsin(sin ())1/2 ::::}
F /(()) __ 1 d (. ())1/2 1 1 (. ())-1/2 () cos(). - SIn = . - SIn . COS = -;=:=:====:::::;----r====::
JI- (VSiner de vI-sine 2 2vl-sinevsinB
54. y = arctan JI- x = arctan (1 -X)1/2 ::::}l+x l+x
/= 1 .~(1_X)1/2 = 1 .~(1-X)-1/2.(1+X)(-1)-(1-X)(1) y (f1=X)2 dx l+x 1+ I-x 2 l+x (1~x)2
1+ V~ l+x
1 1(1+X)1/2 -2 l+x 1 (1+X)1/2 -2 1 + x 1 - x . 2 1 - x . (1 + X)2 -2-'2' (1-X)1/2' (1+x)2--+-l+x l+x
-1 -1 2(1 - X)1/2(1 + X)1/2 2 Vl- x 2
SECTION 3.5 IMPLICIT DIFFERENTIATION 0 205
~ , ~ 1 . 1 (2) - 1/2 x arcsin x55. j (x ) = v l - x 2arcsin x =} j(x) =v1 - x2 . ~ +arcsm x· - I- x (-2x ) =I - ~
v I - x 2 2 1 - x 2
2
»: - -.........----1-J \ f'
Note that l' = 0 where the graph of f has a horizontal tangent. Also note
- 1 that l' is negative when f is decreasing and l' is positive when f is
increas ing.
- 5
56. f (x ) = arcta n( x2 - x)
2 Note that l' = 0 where the graph of j has a horizontal tangent. Also note
f that .f' is negative when f is decreasing and .f' is positive when j is
- 4 increasing. 4 I--~=----,\;:-h''---==-t
- 2
157. Let y = cos - x. Then cos y = x and 0 :S y :S 7[" =} - sin y ~~ = 1 =}
dy 1 1 1 [Note that sin y 2: 0 for 0 :S y :S 7[" .]
dx sin y \,/1 - cos? y Vl - x 2 '
158. (a) Let y = sec - x . Then sec y = x and y E (0, ~ ] U (7[" , 3; ]. Differentiate with respect to x : sec y t an y (~~) = 1 =}
dy 1 1 1JX2=1' Note that t an2 y = sec 2 y - 1 =} t an y = Jsec2 y - 1 dx sec y t an y secy Jsec2 y - 1 X x 2 - 1
since t an y > 0 when 0 < Y < ~ or 7[" < Y < 3271" .
]dy 2 2 2(b) y = sec - 1 x =} sec y = x =} sec y tan y ~~ = 1 =} - - - - . Now tan y = sec y - 1 = x - 1, dx secy tan y
so tany = ± v x 2 - 1. Fory E [0, ~) , x 2: 1, so sec y = x = Ix l and tany 2: 0 =}
dy = ~= ~. FOr Y E ( ~ , 7["], X :S - l , So l x l = -x and tan Y = -Vx2 - 1 =}2dx x x - 1 Ixl x 2 - 1 .
dy 1 1 1 1
dx sec y tany x (-JX2=1) (- x) v x 2 - 1
59. x 2 + y2 = 1'2 is a circle with center 0 and ax + by = 0 is a line through 0 [assume a y
and b are not both zero ]. x 2 + y2 = 1'2 =} 2x + 2yy' = 0 =} y' = -x/y, so the
slope of the tangent line at Po (xo ,Yo) is - xo/yo. The slope of the line OPo is yo/ xo ,
which is the negative reciprocal of - x o/ yo. Hence, the curves are orthogonal, and the
families of curves are orthogonal trajectories of each other.
x
206 D CHAPTER 3 DIFFERENTIATION RULES
60. The circles x 2 + y2 = ax and x 2 + y 2 = by intersect at the origin where the tangents are vertica l and horizontal [assume a
and b are both nonzero]. If (x o, Yo) is the other point of intersection, then X6 + Y5 = axo (1) and X6 + Y5 = byo (2).
ya - 2x
Now x 2 + y 2 = ax =} 2x + 2y y' = a =} y' = -- and x 2 + y 2 = by2y
, 2x ( )2x + 2yy' = by' =} y = -b--. Thus, the curves are orthogonal at Xo, yo ¢} - 2y
a - 2x o b - 2yo---- ¢} 2a x o - 4X6 = 4Y5 - 2byo ¢} axo + byo = 2(x 6 + Y5 ),
2yo 2xo
which is true by ( I) and (2).
x
x
261 . Y = cx =} y' = 2cx and x 2 + 2y 2 = k [assume k > 0] =} 2x + 4yy' = 0 =}
, x x 2yy' = - x __1_, so the curves are orthogonal if
=} y = - 2(y ) 2cx
c =1= O. If c = 0, then the horizontal line y = cx2 = 0 intersects x 2 + 2y 2 = k orthogonally
at ( ±Vk, 0) ,since the ellipse x 2 + 2y 2 = k has vertica l tangents at those two points.
362. Y = a x =} y ' = 3a x 2 and x 2 + 3 y2 = b [assume b > 0] =} 2x + 6yy' = 0 =}
, , x x l . 3y y = - x =} y = - 3(y ) = - 3(a x 3 ) = - 3ax 2' so the curves are orthogonal If
3a =1= O. If a = 0, then the horizontal line y = ax = 0 intesects x 2 + 3y 2 = b orthogonally
at (±V'b,0) ,since the ellipse x 2 + 3y 2 = b has vertical tangents at those two points.
63. To find the points at which the ellipse x 2 - x y + y2 = 3 crosses the x-axis, let y = 0 and solve for x .
y = 0 =} x 2 - x (O) + 02 = 3 ¢} x = ±V3. So the graph of the ellipse crosses the x-axis at the points (±V3,0) .
' (2 ) , y - 2x Using implicit differentiation to find y ' , we get 2x - x y ' - y + 2y y' = 0 =} y y - x = y - 2x ¢} y = -2--. y - x
So y' at (V3,0) is 0 - 2 ~ = 2 and y' at (-V3 ,0) is 0 + 2 ~ = 2. Thus, the tangent lines at these points are parallel. 2(0) - 3 2(0) + 3
64. (a) We use implicit differentiation to find y ' = y - 2x as in Exercise 63. The slope (b) 2y - x
of the tangent line at (- 1, 1) is m = ~ ) 2((1 )) = ~ = 1, so the slope of the 2 1 - - 1 3
normal line is -~ = - 1, and its equation is y - 1 = - 1(x + 1) ¢}7n
y = -x. Substitutin g -x for y in the equation of the ellipse, we get
2 x - x(-x) + ( _X)2 = 3 =} 3x2 = 3 ¢} x = ±1. So the normal line
must intersect the ellipse again at x = 1, and since the equation of the line is
y = - x , the other point of intersection must be (1, - 1).
265. X2y 2 + x y = 2 =} x . 2yy' + y 2 . 2x + x · y' + y . 1 = 0 ¢} y'(2x 2y + x) = -2x y2 - Y ¢}
x y x y2 + yy' = _ 2 2 + y . So _ 2 = - 1 ¢} 2x y 2 + y = 2x 2y + X ¢} y (2xy + 1) = x (2xy + 1) ¢} 2x 2y + x 2x 2y + x
y
x
2
- 21-+ - - :----1--1 2
SECTION 3.5 IMPLICIT DIFFERENTIATION 0 207
y(2xy + 1) - x(2xy + 1) == 0 q (2xy + l)(y - x) == 0 q xy == -~ or y == x. But xy == -~ =}
X2y2 + xy == %- ~ #- 2, so we must have x == y. Then X2y2 + xy == 2 =} x 4 + x 2 == 2 q x 4 + x 2 - 2 == 0 q
(x 2 + 2)(x2 - 1) == O. So x 2 == -2, which is impossible, or x 2 == 1 q x == ±1. Since x == y, the points on the curve
where the tangent line has a slope of -1 are (-1, -1) and (1, 1).
66. x 2 + 4y2 = 36 '* 2x + 8yy' = 0 '* y' = - :y' Let (a, b) be a point on x 2 + 4y2 = 36 whose tangent line passes
through (12,3). The tangent line is then y - 3 = - :b (x - 12), so b - 3 = - :b (a - 12). Multiplying both sides by 4b
2 2gives 4b2 - 12b == _a2 + 12a, so 4b2 + a == 12(a + b). But 4b2 + a == 36, so 36 == 12(a + b) =} a + b == 3 =}
b == 3 - a. Substituting 3 - a for b into a2 + 4b2 == 36 gives a2 + 4(3 - a)2 == 36 q a2 + 36 - 24a + 4a2 == 36 q
5a2 - 24a == 0 q a(5a - 24) == 0, so a == 0 or a == ¥. If a == 0, b == 3 - 0 == 3, and if a == ¥, b == 3 - ¥ == -to So the two points on the ellipse are (0, 3) and (¥, - ~ ). Using
y=3 y
(0,3)
2
° 2
y=~X-5
(12,3)
y - 3 = - :b (x - 12) with (a, b) = (0,3) gives us the tangent line
xy - 3 == 0 or y == 3. With (a, b) == (¥, -~), we have (li _2)
5' 5 y - 3 == - 4(~tJ5) (x - 12) q Y - 3 == ~(x - 12) q Y == ~x - 5.
A graph of the ellipse and the tangent lines confirms our results.
67. (a) If y == r:' (x), then f(y) == X. Differentiating implicitly with respect to x and remembering that y is a function of x,
I dy dy 1 ( -1)1 1 we get f (y) dx == 1, so dx == fl(y) f (x) == fl(f-l(X))'
-1 ( -1)1 1 1 /(2) 3(b) f(4) == 5 =} f (5) == 4. By part (a), f (5) == fl(f-l(5)) == fl(4) == 1 3" 2'
68. (a) f (x) == 2x + cos x =} fl (x) == 2 - sin x > 0 for all X. Thus, f is increasing for all x and is therefore one-to-one.
(b) Since f is one-to-one, f-l(l) == k q f(k) == 1. By inspection, we see that f(O) == 2(0) + cosO == 1,
so k == t : (1) == O.
(c) (j-l)' (1) = 1'(f~l (1)) = l'~O) = 2 - ~in 0 = ~
69. x 2 + 4y2 = 5 '* 2x + 4(2yy') = 0 '* y' = - :y' Now let h be the height of the lamp, and let (a, b) be the point of
tangency of the line passing through the points (3, h) and (-5,0). This line has slope (h - 0)/[3 - (-5)] == ~h. But the
ab' slope of the tangent line through the point (a, b) can be expressed as yl == - 4 or as b (0 ) = --'!-5 [since the line
a - -5 a +
2passes through (-5,0) and (a, b)], so -4ab == _b_ q 4b2 == _a - 5a q a2 + 4b2 == -5a. But a2 + 4b2 == 5
a+5
[since (a, b) is on the ellipse], so 5 == -5a q a == -1. Then 4b2 == _a2 - 5a == -1- 5(-1) == 4 =} b == 1, since the
point is on the top half of the ellipse. So !!.. == _b_ == _1_ == -41
=} h == 2. So the lamp is located 2 units above the 8 a+5 -1+5
x-axis.
