IMPLICIT DIFFERENTIATION

CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES

3.6 IMPLICIT DIFFERENTIATION

Step 2: 2x 2y ^dy_dx ^dy_dx^dy_dxœ 1 2y Step 3: ^dy_dx(2x2y1)œ " 2y Step 4: ^dy_dxœ _2x¹_2y^2y₁

22. x^$xyy^$œ1 Ê 3x^# y x 3y 0 ^dy_dx ^{# dy}_dxœ Ê a3y^#x yb^dy_dx œ 3x ^# Ê ^dy_dx œ^y_3y#^3x_x^#

23. x (x^# y)^#œx^#y :^#

Step 1: x^#’2(xy) 1Š ^dy_dx‹“ (x y) (2x)^# œ2x2y ^dy_dx Step 2: 2x (x ^# y) 2y 2x^dy_dx ^dy_dxœ 2x (x^# y) 2x(xy)^# Step 3: ^dy_dxc2x (x^# y) 2ydœ2x 1c x(x y) (x y)^#d

Step 4: ^dy_dx œ ^{2x 1c}_{2x (x}^x(x# ^y)_y) ^(x_2y^y)#d œ^{x 1c} _y^x(x _{x (x}#^y)_y)^{(x y)#d} œ^{x 1a x} _{x y}#^xy _x^x$ _y^2xyyb

# # #

œ ^x2x_{x y}#^{$3x y}_x$^# _y^xy#

24. (3xy7)^#œ6y Ê 2(3xy7)†Š3x 3y^dy_dx ‹œ6 ^dy_dx Ê 2(3xy7)(3x) 6 ^dy_dx ^dy_dx œ 6y(3xy7) Ê ^dy_dx[6x(3xy 7) 6]œ 6y(3xy7) Ê ^dy_dx œ _x(3xy^y(3xy₇₎⁷⁾₁ œ ₁^3xy_{3x y}^##^7y_7x

25. y^# œ^x_x ^"₁ Ê 2y _dx^dy œ ^(x _(x¹⁾₁₎^(x# ¹⁾ œ _(x²₁₎# Ê _dx^dy œ_y(x^"₁₎#

26. x^# œ^x_x^y_y Ê x^$x y^# œ Êx y 3x^#2xyx y^{# w} œ 1 y ^w Ê ax^#1 yb ^wœ 1 3x^#2xy Ê y^wœ ¹_x^3x#^#₁^2xy

27. xœtan y Ê 1œasec y ^# b^dy_dx Ê ^dy_dx œ_{sec y}^"# œcos y^#

28. xyœcot a bxy xÊ ^dy_dx œ y csc (xy) x^# Š ^dy_dxy x‹ Ê ^dy_dxx csc (xy)^{# dy}_dxœ y csc (xy)^# y xÊ ^dy_dx x csc (xy)^# ‘œ y csc (xy) ^# " Ê‘ ^dy_dxœ _x^{y csc (xy)}_{" csc (xy)}^# _# ^"_‘^‘ œ ^y_x

29. xtan (xy)œ ! Ê 1 csec (xy)^# d Šyx ^dy_dx‹œ0 Ê x sec (xy) ^{# dy}_dx œ 1 y sec (xy) ^# Ê ^dy_dx œ ^" _{x sec (xy)}^{y sec (xy)}# ^#

œ _{x sec (xy)}#¹ œ^y_x ^{cos (xy)}_x œ_x^{y cos (xy)}_x ^y

# #

30. xsin yœxy Ê 1 (cos y) ^dy_dx œ y x ^dy_dx Ê (cos yx) ^dy_dx œ Êy 1 ^dy_dx œ_{cos y}^y1_x

31. y sinŠ ‹^"_y œ 1 xy Ê y cos’ Š ‹^"_y †( 1) _y^"#†_dx^dy“sin Š ‹^"_y †_dx^dy œ x _dx^dy Êy

^dy_{dx y}cos _y sin _y x y ^dy_dx ^{y y}

cos sin x y sin cos xy

’^" Š ‹^" Š ‹^" “œ Ê œ œ

^{" " " " "}

#

y Š ‹y Š ‹y Š ‹y Š ‹y

32. y cos^# Š ‹^"_y œ2x2y Ê y^#’sinŠ ‹^"_y †( 1) _y^"#†_dx^dy“cosŠ ‹^"_y †2y _dx^dy œ 2 2 _dx^dy Ê

^dy_dx sin _y 2y cos _y 2 2 ^dy_dx ²

sin 2y cos

’ Š ‹^" Š ‹^" “

#

œ Ê œ

Š ‹^"_y Š ‹^"_y

33. )^"Î#r^"Î#œ1 Ê ^"_#)^"Î#"_#r^"Î#†_d^dr œ0 Ê _d^dr _#^"_r œ _#^" Ê _d^dr œ ^{2 r} œ ^r

) )’ _È “ _È) ) 2_È^È) _È^È)

34. r2È)œ ³_#)^#Î$4₃)^$Î% Ê ^dr_d))^"Î#œ)^"Î$)^"Î% Ê _d^dr₎ œ)^"Î#)^"Î$)^"Î%

35. sin (r )) œ ^"_# Ê [cos (r )] r) ˆ ) _d^dr₎‰œ0 Ê _d^dr₎[ cos (r )]) ) œ r cos (r ) ) Ê _d^dr₎ œ₎^{r cos (r )}_{cos (r ))}⁾ œ ^r₎, cos (r )) Á0

36. cos rcot )œr ) Ê ( sin r) _d^dr₎csc ^#)œ r )_d^dr₎ Ê _d^dr₎[ sin r )]œ r csc ^#) Ê _d^dr₎ œ ^r_{sin r}^{csc #}₎⁾

37. x^#y^#œ1 Ê 2x2yy^wœ0 Ê 2yy^wœ 2x Ê ^dy_dx œy^wœ _y^x; now to find ^{d y}_dx^##, _dx^d a by^w œ _dx^d Š_y^x‹

y since y y

Ê ^wwœ ^{y( 1) xy}_y# ^w œ ^{y x}_y# ^wœ Ê_y^{x d y}_dx^## œ ^wwœ ^y_y^#$^x# œ ^"y _y$ ^y œ^"_y$

# #

Š ^xy‹ a b

38. x^#Î$y^#Î$œ1 Ê ²₃x^"Î$2₃y^"Î$ 0 y_dx^dyœ Ê _dx^dy2₃ ^"Î$‘œ ²₃x^"Î$ Ê y^wœ _dx^dyœ ^x_y^"Î$_"Î$ œ ˆ ‰_x^{y "Î$}; Differentiating again, y^wwœ^{x"Î$ "y#Î$}_x^y#Î$^{wy"Î$ "x#Î$} œ^{x y} _x#Î$ ^{y x}

"Î$ " #Î$ "Î$ "Î$ " #Î$

"Î$

†ˆ ‰ ˆ ‰ †ˆ ‰Œ ˆ ‰

3 3 3 3

y x

x y y x

Ê ^{d y}_dx^## œ₃" #Î$ "Î$₃" "Î$ %Î$œ _3x^y"Î$%Î$ _3y"Î$ #Î$"_x

39. y^# œx^#2x Ê 2yy^wœ2x Ê2 y^wœ ^2x_2y² œ ^x_y¹; then y^wwœ^{y (x}_y# ^1)yw œ ^{y (x} _y¹⁾#

Š^{x 1}y‹

Ê ^{d y}_dx^## œywwœ^{y# (x}_y$ ^1)#

40. y^#2xœ 1 2y Ê 2y y† ^w œ 2 2y ^w Ê y (2y^w 2)œ2 Ê y^wœ _y^"₁ œ(y1)^"; then y^wwœ (y 1)^#†y^w (yœ 1)^#(y1)^" Ê ^{d y}_dx^## œy^wwœ _(y^"₁₎$

41. 2Èyœ Êx y y^{"Î# w}y œ 1 y ^w Ê y y^wˆ ^"Î#1‰œ1 Ê ^dy_dxœy^wœ _y"Î#^"₁ œ _y^y₁; we can

È È

differentiate the equation y y^wˆ ^"Î#1‰œ1 again to find y : y^{ww w}ˆ^"_#y^{$Î# w}y‰ˆy^"Î#1 y‰ ^wwœ0

y 1 y y y y

Ê ˆ ^"Î# ‰ ^wwœ ^"_#c d^{w# $Î#} Ê ^{d y}_dx œ ^wwœ œ ^" œ_# ^"

y

y 1 2y y 1 1 y

#

#

"

#

# $Î#

"Î# $Î# "Î# $ $

Œ

a b a b ˆ È ‰

"

"Î#

y 1

42. xyy^#œ1 Ê xy^w y 2yy^wœ0 Ê xy^w2yy^wœ Êy y (x^w 2y)œ Êy y^wœ _{(x 2y)}^y ; ^{d y}_dx^## œy^ww œ ^{(x 2y)y}_(x^w_2y)^y(1# ^{2y )w} œ ^{(x 2y)} _(x2y)^{y 1}# ² œ ^y(x_(x^2y)_2y)^y(x#^2y)2y

’_(x^y_2y)“ ’ Š_(x^y_2y)‹“ _(x^"_2y)c #d

œ ^2y(x_(x^2y)_2y)$^2y# œ ^2y_(x^#_2y)^2xy$ œ ^2y(x_(x2y)^y)$

43. x^$y^$œ16 Ê 3x^#3y y^{# w} œ0 Ê 3y y^{# w}œ 3x ^# Ê y^wœ ^x_y^##; we differentiate y y^{# w} œ x to find y :^{# ww} y y^{# ww}y 2y y^wc † ^wdœ 2x Ê y y^{# ww}œ 2x 2y yc d^{w #} Ê y^wwœ ^{2x 2y}_y^{Šxy ‹} œ ^2x_y ^2xy

# %

# $

#

# #

œ ^2xy_y^$&^2x% Ê ^{d y}_dx^##¹ œ ³²₃₂³²œ 2

(2 2)ß

44. xyy^#œ1 Ê xy^w y 2yy^wœ0 Ê y (x^w 2y)œ Êy y^wœ _{(x 2y)}^y Ê y^wwœ ^(x2y)a_(x^ywb_2y)^{( y) 1}# ^{a 2ywb}; since y^wk_(0ß1)œ ^"_# we obtain y^wwk_(0ß1)œ^{( 2) ˆ ‰"#}₄⁽¹⁾⁽⁰⁾ œ ^"₄

45. y^#x^#œy^%2x at (#ß ") and (#ß 1) Ê 2y ^dy_dx2xœ4y ^{$ dy}_dx Ê2 2y ^dy_dx4y ^{$ dy}_dxœ 2 2x

2y 4y 2 2x 1 and 1

Ê ^dy_dxa ^$bœ Ê ^dy_dx œ_# ^x_y^"$ _y Ê ^dy_dx¹ œ ^dy_dx¹ œ

( 2 1) ß ( 2 ß1)

46. xa ^#y^#b^#œ(xy) at(^# "ß !) and ("ß 1) Ê 2 xa ^#y^#b Š2x2y ^dy_dx‹œ2(xy) 1Š ^dy_dx‹

2y x y (x y) 2x x y (x y) 1

Ê ^dy_dxc a ^{# #}b dœ a ^{# #}b Ê ^dy_dx œ _{2y x}^{2x x}_a^a#^#_y^y#_b^#b _(x^(x _y)^y) Ê ^dy_dx¹ œ

(1 0)ß

and ^dy_dx¹ 1

(1ß1)œ

47. x^#xyy^#œ1 Ê 2x y xy^w2yy^wœ0 Ê (x2y)y^wœ Ê2x y y^wœ ^2x_2y^y_x;

(a) the slope of the tangent line mœy^wk_{(2 3)ß} œ ⁷₄ Ê the tangent line is y œ3 ⁷₄(x2) Ê yœ⁷₄x^"_# (b) the normal line is y œ 3 ⁴₇(x2) Ê yœ ⁴₇x²⁹₇

48. x^#y^#œ25 Ê 2x2yy^wœ0 Ê y^wœ ^x_y;

(a) the slope of the tangent line mœy^wk(3 4)œ ¹ œ Ê the tangent line is y œ4 (x3)

(3 4)

ß ß

x 3 3

y 4 4

y x

Ê œ³₄ ²⁵₄

(b) the normal line is y œ 4 ⁴₃(x3) Ê yœ ⁴₃x

49. x y^{# #}œ9 Ê 2xy^#2x yy^{# w}œ0 Ê x yy^{# w}œ xy ^# Ê y^wœ ^y_x;

(a) the slope of the tangent line mœy^wk_{( 1 3)}œ ¸ œ3 Ê the tangent line is y œ3 3(x1)

( 1 3)

ß ß

y x

y 3x 6

Ê œ

(b) the normal line is y œ 3 ^"₃(x1) Ê yœ ^"₃x₃⁸

50. y^#2x4y " œ ! Ê 2yy^w 2 4y^wœ0 Ê 2(y2)y^wœ2 Ê y^wœ _y#^" ;

(a) the slope of the tangent line mœy^wk( 2 1) ß œ Ê1 the tangent line is y œ 1 1(x2) Ê yœ x 1 (b) the normal line is y œ1 1(x2) Ê yœ x 3

51. 6x^#3xy2y^#17y œ6 0 12xÊ 3y3xy^w4yy^w17y^wœ0 Ê y (3x^w 4y17)œ 12x3y

y ;

Ê ^wœ _3x^12x_4y^3y₁₇

(a) the slope of the tangent line mœy^wk_{( 1 0)}œ ^" ¹ œ Ê the tangent line is y œ0 (x1)

( 1 0)

ß ß

2x 3y

3x 4y 17 7 7

6 6

y x

Ê œ⁶₇ ⁶₇

(b) the normal line is y œ 0 ⁷₆(x1) Ê yœ ⁷₆x⁷₆

52. x^# 3xy 2y^# 5 2x 3xy^w 3y 4yy^w 0 y^w 4y 3x 3y 2x y^w ;

È œ Ê È È œ Ê Š È ‹œÈ Ê œ ^È_4y^3y_È^2x_3x

(a) the slope of the tangent line mœy^wk_ŠÈ _‹œ ¹ œ0 Ê the tangent line is yœ2

ŠÈ ‹

3 2ß 3 2ß

È È 3y 2x

4y 3x

(b) the normal line is xœÈ3

53. 2xy1 sin yœ2 1 Ê 2xy^w2y1(cos y)y^wœ0 Ê y (2x^w 1 cos y)œ 2y Ê y^wœ_2x1^2y _{cos y}; (a) the slope of the tangent line mœy^wkˆ1 ‰œ ¹ˆ ‰œ _# Ê the tangent line is

ß¹² 1ß¹₂ 2y

2x 1 cos y 1

y œ ¹_# ¹_#(x1) Ê yœ ¹_#x1

(b) the normal line is y œ¹_{# 1}²(x1) Ê yœ ₁²x ₁^{2 1}_#

54. x sin 2yœy cos 2x Ê x(cos 2y)2y^wsin 2yœ 2y sin 2xy cos 2x ^w Ê y (2x cos 2y^w cos 2x)

sin 2y 2y sin 2x y ;

œ Ê ^wœ _{cos 2x}^{sin 2y2y sin 2x}_{2x cos 2y}

(a) the slope of the tangent line mœy^wkˆ^{1 14 2}ß ‰œ ¹ˆ^{1 1}_{4 2}ß ‰œ œ2 Ê the tangent line is

sin 2y 2y sin 2x

cos 2x 2x cos 2y 1₁

#

y œ¹_# 2 xˆ ¹₄‰ Ê yœ2x

(b) the normal line is y œ ¹_{# #}^"ˆx¹₄‰ Ê yœ _#^"x⁵₈¹

55. yœ2 sin ( x1 y) Ê y^wœ2 [cos ( x1 y)]†a1y ^wb Ê y [1^w 2 cos ( x1 y)]œ2 cos ( x1 1 y)

y ;

Ê ^wœ 2 cos ( x _# y) 1 cos ( x y)

1 1

1

(a) the slope of the tangent line mœy^wk(1 0)œ ¹ œ2 Ê the tangent line is

(1 0)

ß ß

2 cos ( x y) 1 2 cos ( x y)

1 1

1 1

y œ0 2 (x1 1) Ê yœ2 x1 21

(b) the normal line is y œ 0 _#^"₁(x1) Ê yœ ₂^x_{1 #}^"₁

56. x cos y^{# #} sin yœ0 Ê x (2 cos y)( sin y)y^{# w}2x cos y^# y cos y^w œ0 Ê y^wc2x cos y sin y^# cos yd

2x cos y y ;

œ ^# Ê ^wœ 2x cos y sin y^{2x cos y}cos y

#

#

(a) the slope of the tangent line mœy^wk_{(0 )}œ ¹ œ0 Ê the tangent line is yœ

ß1 (0 )ß1

2x cos y 2x cos y sin y cos y

#

# 1

(b) the normal line is xœ0

57. Solving x^#xyy^#œ7 and yœ0 Ê x^#œ7 Ê xœ „È7 Ê Š È7ß !‹ and ŠÈ7ß !‹ are the points where the curve crosses the x-axis. Now x^#xyy^#œ7 Ê 2x y xy^w2yy^wœ0 Ê (x2y)y^wœ 2x y

y m the slope at 7 is m 2 and the slope at 7 is

Ê ^wœ ^2x_x2y^y Ê œ ^2x_x2y^y Ê ŠÈ ß !‹ œ ^2È₇⁷ œ ŠÈ ß !‹

È

mœ ²_È^È₇⁷ œ 2. Since the slope is 2 in each case, the corresponding tangents must be parallel.

58. x^#xyy^#œ7 Ê 2x y x 2y 0 ^dy_dx ^dy_dxœ Ê (x2y) ^dy_dxœ Ê2x y ^dy_dx œ _x^2x_2y^yand _dy^dxœ ^x_2x^2y_y; (a) Solving ^dy_dx œ0 Ê œ2x y 0 Ê yœ 2x and substitution into the original equation gives

x^# x( 2x) ( 2x)^#œ7 Ê 3x^#œ7 Ê xœ „É⁷₃ and yœ …2É⁷₃ when the tangents are parallel to the x-axis.

(b) Solving ^dx_dy œ0 Ê x 2yœ0 Ê yœ ^x_# and substitution gives x^#xˆ^x_#‰ ˆ ^x_#‰^#œ7 Ê ^3x₄^# œ7 x 2 and y when the tangents are parallel to the y-axis.

Ê œ „ É⁷₃ œ …É⁷₃

59. y^% œy^#x ^# Ê 4y y^{$ w}œ2yy^w2x Ê 2 2ya ^$y yb ^wœ 2x Ê y^wœ _y^x_2y$; the slope of the tangent line at

is 1; the slope of the tangent line at

Š^È₄³ß^È_#³‹ _y^x_2y¹ œ œ œ_{# 3} œ Š^È₄³ß_#‹

" "

$ "

#

"

Œ^{È È}4³ß2³ # È

È È

3

4 4

3 6 3

8

3 4

is ^x ¹ È3

y 2y 4 2

2 3

^$ _Œ^{È3 "}_#

4 2

ß1 œ ^È342 œ œ

8

È

60. y (2^# x)œx ^$ Ê 2yy (2^w x) y ( 1)^# œ3x ^# Ê y^wœ _2y(2^y#3x_x)^# ; the slope of the tangent line is mœ_2y(2^y#3x_x)^#¹ œ œ⁴_# 2 Ê the tangent line is y œ1 2(x1) Ê yœ2x1; the normal line is

(1 1)ß

y œ 1 ^"_#(x1) Ê yœ ^"_#x_#³

61. y^%4y^#œx^%9x ^# Ê 4y y^{$ w}8yy^wœ4x^$18x Ê y 4y^wa ^$8ybœ4x^$18x Ê y^wœ^4x_4y^$$^18x_8y œ^2x_2y^$$^9x_4y

m; ( 3 2): m ; ( ): m ; (3 ): m ; (3 ): m œ ^{x 2x}_{y 2y}^a_a ^##⁹₄^b_b œ ß œ ^{( 3)(18}_{2(8 4)}⁹⁾ œ ²⁷₈ $ß # œ²⁷₈ ß # œ ²⁷₈ ß # œ ²⁷₈

62. x^$y^$9xyœ0 Ê 3x^#3y y^{# w}9xy^w9yœ0 Ê y 3y^wa ^#9xbœ9y3x ^# Ê y^wœ ^9y_3y#^3x_9x^# œ ^3y_y#_3x^x#

(a) y^wk_{(4 2)ß} œ⁵₄ and y^wk_{(2 4)ß} œ⁴₅;

(b) y^wœ0 Ê ^3y_y#_3x^x# œ0 Ê 3yx^#œ0 Ê yœ^x₃ Ê x^{$ x}₃ ^$9x ^x₃ œ0 Ê x^'54x^$œ0

# # #

Š ‹ Š ‹

x x 54 0 x 0 or x 54 3 2 there is a horizontal tangent at x 3 2 . To find the Ê ^$a ^$ bœ Ê œ œ ^$È œ ^$È Ê œ ^$È

corresponding y-value, we will use part (c).

(c) ^dx_dyœ0 Ê _3y^y#3x_x# œ0 Ê y 3xœ0 Ê yœ „ 3x ; yœ 3x Ê x 3x 9x 3xœ0

# $ $

È È ŠÈ ‹ È

x 6 3 x 0 x x 6 3 0 x 0 or x 6 3 x 0 or x 108 3 4 .

Ê ^$ È ^$Î#œ Ê ^$Î#Š ^$Î# È ‹œ Ê ^$Î#œ ^$Î#œ È Ê œ œ^$È œ ^$È Since the equation x^$y^$9xyœ0 is symmetric in x and y, the graph is symmetric about the line yœx.

That is, if (a b) is a point on the folium, then so is (b a). Moreover, if yß ß ^wk(a b)ß œm, then y^wk(b a)ß œ m^" . Thus, if the folium has a horizontal tangent at (a b), it has a vertical tangent at (b a) so one might expectß ß that with a horizontal tangent at xœ ^$È54 and a vertical tangent at xœ3^$È4, the points of tangency are

54 3 4 and 3 4 54 , respectively. One can check that these points do satisfy the equation Š^$È ß ^$È ‹ Š ^$È ß^$È ‹

x^$y^$9xyœ0.

63. x^#2tx2t^#œ4 Ê 2x 2x^dx_dt 2t 4t^dx_dt œ0 Ê (2x2t) 2x^dx_dt œ 4t Ê ^dx_dt œ ^{2x 4t}_{2x 2t} œ ^{x 2t}_{x t} ; 2y^$3t^#œ4 Ê 6y ^{# dy}_dt œ6t 0 Ê ^dy_dt œ_6y^6t# œ_y^t#; thus ^dy_dxœ ^dy/dt_dx/dtœ # œ _{y (x 2t)}#^{t(x t)} ; tœ2

Š ‹ ˆ ‰

t y x 2t

x t

x 2(2)x 2(2) 4 x 4x 4 0 (x 2) 0 x 2; t 2 2y 3(2) 4

Ê ^{# #}œ Ê ^# œ Ê ^#œ Ê œ œ Ê ^{$ #}œ

2y 16 y 8 y 2; therefore 0

Ê ^$œ Ê ^$œ Ê œ ^dy_dx¹ œ _{(2) (2}²⁽²²⁾₂₍₂₎₎ œ

t 2œ ^#

64. xœÉ5Èt Ê ^dx_dt œ ˆ5Èt‰ ˆ t ‰œ ; y(t1)œÈt yÊ (t 1)_dt œ t

4 t 5 t

" " " dy "

# # #

"Î# "Î# "Î#

ÈÉ È

Ê at1 b^dy_dt œ_#È^"_t Êy ^dy_dt œ _at1b^y œ _#t^{" #}_Èt^y₂^È_È^t_t; thus ^dy_dxœ œ œ †

"

# #

" #

"

È È ÈÈ

ÈÉ È

t t t 2 t

y t

4 t 5 t

dy dt dx dt

" #

# "

"

y t t t

4 t 5 t

È È a b

ÈÉ È

; t 4 x 5 4 3; t 4 y(3) 4 2

œ ^{# " #ˆ yÈ}_" ^t‰_t^{É& Èt} œ Ê œÉ È œÈ œ Ê œ È œ

therefore, ^dy_dx¹ ¹⁴₃

t 4œ œ ^{2Š" 2 2a bÈ}_" ⁴₄^{‹É& È4} œ

65. x2x^$Î#œ Êt^# t 3x^dx_dt ^"Î# 2t^dx_dt œ Ê1 ˆ13x^"Î#‰ 2t^dx_dt œ Ê1 ^dx_dt œ _{1 3x}^{2t 1}"Î#; yÈt 1 2tÈyœ4

t 1 y (t 1) 2 y 2t y 0 t 1 2 y 0

Ê ^dy_dt È ˆ ‰^"_# ^"Î# È ˆ^"_# ^"Î#‰ ^dy_dt œ Ê ^dy_dt È _2È^y_t1 È Š_È^t_y‹^dy_dt œ

Ê ŠÈt 1 ^t_y‹^dy_dt œ ^y 2Èy Ê ^dy_dt œ œ ; thus

2 t 1

È È

Š È ‹

ŠÈ ‹

È È

È È

c b y 2 t 1

t y È

È

2 y

t 1

y y 4y t 1 2 y (t 1) 2t t 1

; t 0 x 2x 0 x 1 2x 0 x 0; t 0

dy dy/dt

dxœ dx/dtœ ^Œ œ Ê œ Ê œ Ê œ œ

Š ‹

c c b

b b b

b b "Î#

y y 4y t 1 2 y (t 1) 2t t 1

2t 1 1 3x

È È

È È

$Î# ˆ "Î#‰

y 0 1 2(0) y 4 y 4; therefore 6

Ê È È œ Ê œ ^dy_dx¹ œ œ

t 0œ

Œ

Œ

"Î#

4 4 4(4) 0 1 2 4(0 1) 2(0) 0 1

2(0) 1 1 3(0)

È È

È È

66. x sin t2xœ Êt ^dx_dt sin tx cos t2 ^dx_dt œ1 Ê (sin t2) ^dx_dt œ 1 x cos t Ê ^dx_dt œ ¹_{sin t 2}^{x cos t} ; t sin t œ2t y Ê sin tt cos t œ2 ^dy_dt ; thus ^dy_dx œ ^{sin t}_ˆ1 ^{t cos t}x cos t_‰²; tœ Ê x sin 2xœ

sin tc 2

b 1 1 1

x ; therefore 4

Ê œ¹_# ^dy_dx¹ œ ^{sin 1 1 cos 12} œ ₂⁴¹₁⁸ œ

tœ1 –^{1 #sin 2cos} —

Š ‹1 1 1

67. (a) if f(x)œ ³_#x^#Î$3, then f (x)^w œx^"Î$ and f (x)^ww œ ₃^"x^%Î$ so the claim f (x)^ww œx^"Î$ is false (b) if f(x)œ ₁₀⁹ x^&Î$7, then f (x)^w œ³_#x^#Î$ and f (x)^ww œx^"Î$ is true

(c) f (x)^ww œx^"Î$ Ê f (x)^www œ ^"₃x^%Î$ is true (d) if f (x)^w œ_#³x^#Î$6, then f (x)^ww œx^"Î$ is true

68. 2x^#3y^#œ5 Ê 4x6yy^wœ0 Ê y^wœ ^2x_3y Ê y^wk(1 1) œ ^2x_3y¹ œ ²₃ and y^wk(1 1)œ ^2x_3y¹ œ²₃;

(1 1) (1 1)

ß ß ß ß

also, y^#œx ^$ Ê 2yy^wœ3x ^# Ê y^wœ^3x_2y^# Ê y^wk_{(1 1)}œ ^3x_2y^#¹ œ _#³ and y^wk_{(1 1)}œ ^3x_2y^#¹ œ _#³. Therefore

(1 1) (1 1)

ß ß ß ß

the tangents to the curves are perpendicular at (1 1) and (1ß ß 1) (i.e., the curves are orthogonal at these two points of intersection).

69. x^#2xy3y^# œ0 Ê 2x2xy^w2y6yy^wœ0 Ê y (2x^w 6y)œ 2x 2y Ê y^wœ _3y^xy_x Ê the slope of the tangent line mœy^wk(1 1)œ ¹ œ1 Ê the equation of the normal line at (1 1) is yß œ 1 1(x1)

(1 1)

ß ß

x y 3y x

y x 2. To find where the normal line intersects the curve we substitute into its equation:

Ê œ

x^#2x(2 x) 3(2x)^#œ0 Ê x^#4x2x^#3 4a 4xx^#bœ0 Ê 4x^#16x12œ0

x 4x 3 0 (x 3)(x 1) 0 x 3 and y x 2 1. Therefore, the normal to the curve Ê ^# œ Ê œ Ê œ œ œ

at (1 1) intersects the curve at the point (3ß ß 1). Note that it also intersects the curve at (1 1).ß

70. xy2x œy 0 Ê x ^dy_dx y 2 ^dy_dx œ0 Ê ^dy_dxœ ^y₁²_x; the slope of the line 2x œy 0 is 2. In order to be parallel, the normal lines must also have slope of 2. Since a normal is perpendicular to a tangent, the slope of the tangent is . Therefore, ^"_# ^y₁²_x œ^"_# Ê 2y œ Ê4 1 x xœ 3 2y. Substituting in the original equation, y( 3 2y) 2( 3 2y) œy 0 Ê y^#4y œ3 0 Ê yœ 3 or yœ 1. If yœ 3, then xœ3 and y œ 3 2(x3) Ê yœ 2x 3. If yœ 1, then xœ 1 and y œ 1 2(x1) Ê yœ 2x 3.

71. y^# œx Ê ^dy_dxœ _#^"_y. If a normal is drawn from (a 0) to (x y ) on the curve its slope satisfies ß _"ß _" ^y_x^"_"⁰_a œ 2y_"

y 2y (x a) or a x . Since x 0 on the curve, we must have that a . By symmetry, the

Ê " œ " " œ "" "   "

# #

two points on the parabola are xˆ _"ßÈx and x_"‰ ˆ _"ß Èx_"‰. For the normal to be perpendicular,

1 1 x (a x ) x x x x and y .

Š_x^Èx"_a‹ Š_a^Èx_x^"‹ _(a ^x_{x )} ˆ ‰ ₄

" " "

"

œ Ê # œ Ê "œ " # Ê "œ " #" " # Ê "œ " "œ „#"

Therefore, ˆ^"₄ß „^"_#‰ and aœ₄³.

72. Ex. 6b.) yœx^"Î# has no derivative at xœ0 because the slope of the graph becomes vertical at xœ0.

Ex. 7a.) yœa1x^{# "Î%}b has a derivative only on ("ß ") because the function is defined only on ["ß "] and the slope of the tangent becomes vertical at both xœ 1 and xœ1.

73. xy^$x y^# œ6 Ê x 3y Š ^{# dy}_dx‹y^$x 2xy^{# dy}_dx œ0 3xyÊ ^dy_dxa ^#x^#bœ y^$ 2xy Ê ^dy_dxœ _3xy^y$#^2xy_x#

; also, xy x y 6 x 3y y x y 2x 0 y 2xy 3xy x

œ _3xy^y$#^2xy_x# œ Ê _dy^dx _dy^dx œ Ê _dy^dx œ

$ # a #b $ # Š ‹ a $ b # #

; thus appears to equal . The two different treatments view the graphs as functions Ê ^dx_dyœ ^3xy_y$^#_2xy^{x# dx}_dy

"

dy dx

symmetric across the line yœx, so their slopes are reciprocals of one another at the corresponding points (a b) and (b a).ß ß

74. x^$y^#œsin y^# Ê 3x^#2y ^dy_dx œ(2 sin y)(cos y) ^dy_dx Ê ^dy_dx(2y2 sin y cos y)œ 3x^# Ê ^dy_dx œ _2y2 sin y cos y 3x^#

; also, x y sin y 3x 2y 2 sin y cos y ; thus

œ 2 sin y cos y^3x 2y œ Ê ^dxdy œ Ê ^dxdy œ 3x ^dxdy

2 sin y cos y 2y

#

$ # # # #

appears to equal dy^" . The two different treatments view the graphs as functions symmetric across the line

dx

yœx so their slopes are reciprocals of one another at the corresponding points (a b) and (b a).ß ß 75. x^%4y^#œ1:

(a) y^# œ^{1 x}₄^% Ê yœ „^"_#È1x^%

1Ê ^dy_dxœ „₄ x 4x œ ^x ;

1 x

" % "Î# $ „

a b a b _{a % "Î#}^$_b

differentiating implicitly, we find, 4x^$8y ^dy_dx œ0 Ê ^dy_dxœ _8y^4x œ ^4x œ ^x .

8 1 x 1 x

„

„

$ $ $

"

# % % "Î#

Š È ‹ a b

(b)

76. (x2)^#y^#œ4:

(a) yœ „È4 (x 2)^#

4 (x 2) ( 2(x 2))

Ê ^dy_dxœ „^"_#a ^{# "Î#}b

; differentiating implicitly, œ ^{„ (x 2)}

4 (x 2)

c ^{# "Î#}d

2(x 2) 2y ^dy_dxœ0 Ê ^dy_dxœ ^2(x_2y²⁾ œ ^(x_y ²⁾ œ _{„ c4} ^(x_(x ²⁾_2)#d"Î# œ _{c4 #}^„ _(x^{(x 2)}_)#d"Î#.

(b)

77-84. Example CAS commands:

: Maple

q1 := x^3-x*y+y^3 = 7;

pt := [x=2,y=1];

p1 := implicitplot( q1, x=-3..3, y=-3..3 ):

p1;

eval( q1, pt );

q2 := implicitdiff( q1, y, x );

m := eval( q2, pt );

tan_line := y = 1 + m*(x-2);

p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ):

p3 := pointplot( eval([x,y],pt), color=blue ):

display( [p1,p2,p3], ="Section 3.6 #77(c)" );

: (functions and x0 may vary):

Mathematica

Note use of double equal sign (logic statement) in definition of eqn and tanline.

<<Graphics`ImplicitPlot`

Clear[x, y]

{x0, y0}={1, 1/4};

eqn=x + Tan[y/x]==2;

ImplicitPlot[eqn,{ x, x03, x03},{y, y03, y03}]

eqn/.{xÄx0, yÄy0}

eqn/.{ yÄy[x]}

D[%, x]

Solve[%, y'[x]]

slope=y'[x]/.First[%]

m=slope/.{xÄx0, y[x]Äy0}

tanline=y==y0m (xx0)

ImplicitPlot[{eqn, tanline}, {x, x03, x03},{y, y03, y0 + 3}]