Section 3.5 Implicit Differentiation

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Section 3.5 Implicit Differentiation

44. If x²+ xy + y³= 1, find the value of y⁰⁰⁰ at the point where x = 1.

Solution:

216 ¤ CHAPTER 3 DIFFERENTIATION RULES

37.sin  + cos  = 1 ⇒ cos  · ⁰− sin  = 0 ⇒ ⁰= sin  cos  ⇒

⁰⁰=cos  cos  − sin (− sin ) ⁰

(cos )² =cos  cos  + sin  sin (sin  cos ) cos²

=cos² cos  + sin² sin 

cos² cos  =cos² cos  + sin² sin  cos³

Using sin  + cos  = 1, the expression for ⁰⁰can be simpliﬁed to ⁰⁰= (cos² + sin ) cos³

38.³− ³ = 7 ⇒ 3²− 3²⁰ = 0 ⇒ ⁰ = ²

² ⇒

⁰⁰=²(2) − ²(2 ⁰)

(²)² = 2[ − (²²)]

⁴ = 2( − ³²)

³ =2(³− ³)

³² =2(−7)

⁵ = −14

⁵

39.If  = 0 in  + ^= , then we get 0 + ^= , so  = 1 and the point where  = 0 is (0 1). Differentiating implicitly with respect to  gives us ⁰+  · 1 + ^⁰= 0. Substituting 0 for  and 1 for  gives us

0 + 1 + ⁰= 0 ⇒ ⁰= −1 ⇒ ⁰= −1. Differentiating ⁰+  + ^⁰= 0implicitly with respect to  gives us ⁰⁰+ ⁰· 1 + ⁰+ ^⁰⁰+ ⁰· ^⁰= 0. Now substitute 0 for , 1 for , and −1 for ⁰.

0 +



−1



 +



−1





+ ⁰⁰+



−1



 ()



−1





= 0 ⇒ −2

+ ⁰⁰+1

 = 0 ⇒ ⁰⁰= 1

 ⇒ ⁰⁰= 1

². 40.If  = 1 in ²+  + ³= 1, then we get 1 +  + ³= 1 ⇒ ³+  = 0 ⇒ (²+ 1) ⇒  = 0, so the point

where  = 1 is (1 0). Differentiating implicitly with respect to  gives us 2 + ⁰+  · 1 + 3²· ⁰= 0. Substituting 1 for

and 0 for  gives us 2 + ⁰+ 0 + 0 = 0 ⇒ ⁰= −2. Differentiating 2 + ⁰+  + 3²⁰= 0implicitly with respect to  gives us 2 + ⁰⁰+ ⁰· 1 + ⁰+ 3(²⁰⁰+ ⁰· 2⁰) = 0. Now substitute 1 for , 0 for , and −2 for ⁰.

2 + ⁰⁰+ (−2) + (−2) + 3(0 + 0) = 0 ⇒ ⁰⁰= 2. Differentiating 2 + ⁰⁰+ 2⁰+ 3²⁰⁰+ 6(⁰)²= 0implicitly with respect to  gives us ⁰⁰⁰+ ⁰⁰· 1 + 2⁰⁰+ 3(²⁰⁰⁰+ ⁰⁰· 2⁰) + 6[ · 2⁰⁰⁰+ (⁰)²⁰] = 0. Now substitute 1 for , 0for , −2 for ⁰, and 2 for ⁰⁰. ⁰⁰⁰+ 2 + 4 + 3(0 + 0) + 6[0 + (−8)] = 0 ⇒ ⁰⁰⁰= −2 − 4 + 48 = 42.

41. (a) There are eight points with horizontal tangents: four at  ≈ 157735 and four at  ≈ 042265.

(b) ⁰= 3²− 6 + 2

2(2³− 3²−  + 1) ⇒ ⁰= −1 at (0 1) and ⁰=¹₃ at (0 2).

Equations of the tangent lines are  = − + 1 and  =¹3 + 2.

(c) ⁰= 0 ⇒ 3²− 6 + 2 = 0 ⇒  = 1 ±¹3

√3

(d) By multiplying the right side of the equation by  − 3, we obtain the ﬁrst graph. By modifying the equation in other ways, we can generate the other graphs.

(²− 1)( − 2)

= ( − 1)( − 2)( − 3)

58. Find the value of the number a such that the families of curves y = (x + c)⁻¹ and y = a(x + k)^1/3 are orthogonal trajectories.

Solution:

²

² + ²

² = ²

² −²

² ⇒ ²²+ ²²

²² =²²− ²²

²² ⇒ ²(²+ ²)

²² = ²(²− ²)

²² (4). Since

²− ² = ²+ ², we have ²− ²= ²+ ². Thus, equation (4) becomes ²

²² = ²

²² ⇒ ²

² =²²

²², and substituting for²

² in equation (3) gives us 12= −²²

²² ·²²

²² = −1. Hence, the ellipse and hyperbola are orthogonal trajectories.

70. = ( + )⁻¹ ⇒ ⁰= −( + )⁻²and  = ( + )^13 ⇒ ⁰= ¹₃( + )^−23, so the curves are othogonal if the product of the slopes is −1, that is, −1

( + )² · 

3( + )^23 = −1 ⇒  = 3( + )²( + )^23 ⇒

 = 3

1



2 



2

[since ²= ( + )⁻²and ²= ²( + )^23] ⇒  = 3

1

²



⇒ ³= 3 ⇒  =√³3.

71. (a)



 +²

²



( − ) =  ⇒   −   +²

 −³

² =  ⇒



(  −   + ²⁻¹− ³⁻²) = 

( ) ⇒

 ⁰+  · 1 −  − ²⁻²· ⁰+ 2³⁻³· ⁰= 0 ⇒ ⁰( − ²⁻²+ 2³⁻³) =  −  ⇒

⁰=  − 

 − ²⁻²+ 2³⁻³ or

 = ³( −  )

 ³− ² + 2³

(b) Using the last expression for  from part (a), we get



 = (10L)³[(1mole)(004267 Lmole) − 10 L]



(25 atm)(10L)³− (1 mole)²(3592L²- atm mole²)(10L)

+ 2(1mole)³(3592L²- atm mole²)(004267L mole)





= −995733 L⁴

2464386541L³- atm ≈ −404L atm

72. (a) ²+  + ²+ 1 = 0 ⇒ 2 + ⁰+  · 1 + 2⁰+ 0 = 0 ⇒ ⁰( + 2) = −2 −  ⇒ ⁰= −2 − 

 + 2

(b) Plotting the curve in part (a) gives us an empty graph, that is, there are no points that satisfy the equation. If there were any points that satisﬁed the equation, then  and  would have opposite signs; otherwise, all the terms are positive and their sum can not equal 0. ²+  + ²+ 1 = 0 ⇒ ²+ 2 + ²−  + 1 = 0 ⇒ ( + )²=  − 1. The left side of the last equation is nonnegative, but the right side is at most −1, so that proves there are no points that satisfy the equation.

Another solution: ²+  + ²+ 1 = ¹₂²+  +¹₂²+¹₂²+¹₂²+ 1 = ¹₂(²+ 2 + ²) +¹₂(²+ ²) + 1

= ¹₂( + )²+¹₂(²+ ²) + 1 ≥ 1

Another solution: Regarding ²+  + ²+ 1 = 0as a quadratic in , the discriminant is ²− 4(²+ 1) = −3²− 4.

This is negative, so there are no real solutions.

(c) The expression for ⁰in part (a) is meaningless; that is, since the equation in part (a) has no solution, it does not implicitly deﬁne a function  of , and therefore it is meaningless to consider ⁰.

65. Use implicit differentiation to find dy/dx for the equation x

y = y²+ 1 y 6= 0 and for the equivalent equation

x = y³+ y y 6= 0

Show that although the expressions you get for dy/dx look different, they agree for all points that satisfy the given equation.

Solution:

SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 237 64.²+ 4² = 36 ⇒ 2 + 8⁰= 0 ⇒ ⁰= − 

4. Let ( ) be a point on ²+ 4² = 36whose tangent line passes through (12 3). The tangent line is then  − 3 = −

4( − 12), so  − 3 = −

4( − 12). Multiplying both sides by 4

gives 4²− 12 = −²+ 12, so 4²+ ²= 12( + ). But 4²+ ²= 36, so 36 = 12( + ) ⇒  +  = 3 ⇒

 = 3 − . Substituting 3 −  for  into ²+ 4²= 36gives ²+ 4(3 − )²= 36 ⇔ ²+ 36 − 24 + 4²= 36 ⇔ 5²− 24 = 0 ⇔ (5 − 24) = 0, so  = 0 or  =²⁴5. If  = 0,  = 3 − 0 = 3, and if  =²⁴5,  = 3 −²⁴5 = −⁹5. So the two points on the ellipse are (0 3) and24

5 −⁹5

. Using

 − 3 = −

4( − 12) with ( ) = (0 3) gives us the tangent line

 − 3 = 0 or  = 3. With ( ) =24 5 −⁹5

, we have

 − 3 = −4(^245−95)( − 12) ⇔  − 3 =²3( − 12) ⇔  =²3 − 5.

A graph of the ellipse and the tangent lines confirms our results.

65.For

 = ²+ 1,  6= 0, we have 









= 

(²+ 1) ⇒  · 1 −  · ⁰

² = 2 ⁰ ⇒  −  ⁰= 2³⁰ ⇒ 2³⁰+  ⁰=  ⇒ ⁰(2³+ ) =  ⇒ ⁰= 

2³+ . For  = ³+ ,  6= 0, we have 

() = 

(³+ ) ⇒ 1 = 3²⁰+ ⁰ ⇒ 1 = ⁰(3²+ 1) ⇒

⁰= 1 3²+ 1.

From part (a), ⁰ = 

2³+ . Since  6= 0, we substitute ³+ for  to get



2³+ = 

2³+ (³+ )= 

3³+ = 

(3²+ 1) = 1

3²+ 1, which agrees with part (b).

66. (a)  = () and ⁰⁰+ ⁰+  = 0 ⇒ ⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + ⁰(0) + 0 = 0, so ⁰(0) = 0.

(b) Differentiating ⁰⁰+ ⁰+  = 0implicitly, we get ⁰⁰⁰+ ⁰⁰· 1 + ⁰⁰+ ⁰+  · 1 = 0 ⇒

⁰⁰⁰+ 2⁰⁰+ ⁰+  = 0, so ⁰⁰⁰() + 2⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + 2⁰⁰(0) + 0 + 1 [(0) = 1 is given] = 0 ⇒ 2⁰⁰(0) = −1 ⇒ ⁰⁰(0) = −¹2. 67.²+ 4² = 5 ⇒ 2 + 4(2⁰) = 0 ⇒ ⁰= −

4. Now let  be the height of the lamp, and let ( ) be the point of tangency of the line passing through the points (3 ) and (−5 0). This line has slope ( − 0)[3 − (−5)] = ¹8. But the slope of the tangent line through the point ( ) can be expressed as ⁰= −

4, or as  − 0

 − (−5)= 

 + 5[since the line passes through (−5 0) and ( )], so −

4 = 

 + 5 ⇔ 4²= −²− 5 ⇔ ²+ 4² = −5. But ²+ 4²= 5

1

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66. The Bessel function of order 0, y = J (x), satisfies the differential equation xy⁰⁰+ y⁰+ xy = 0 for all values of x and its value at 0 is J (0) = 1.

(a) Find J⁰(0).

(b) Use implicit differentiation to find J⁰⁰(0).

Solution:

77. (a) If  = ⁻¹(), then () = . Differentiating implicitly with respect to  and remembering that  is a function of , we get ⁰()

= 1, so

= 1

⁰() ⇒ 

⁻¹0

() = 1

⁰(⁻¹()). (b) (4) = 5 ⇒ ⁻¹(5) = 4. By part (a),

⁻¹₀

(5) = 1

⁰(⁻¹(5)) = 1

⁰(4)= 1₂

3

 =³₂.

78. (a) Assume   . Since ^is an increasing function, ^ ^, and hence,  + ^  + ^; that is, ()  ()

So () =  + ^is an increasing function and therefore one-to-one.

(b) ⁻¹(1) =  ⇔ () = 1, so we need to ﬁnd  such that () = 1. By inspection, we see that (0) = 0 + ⁰= 1, so

 = 0, and hence, ⁻¹(1) = 0.

(c) (⁻¹)⁰(1) = 1

⁰(⁻¹(1)) = 1

⁰(0) [by part (b)]. Now () =  + ^ ⇒ ⁰() = 1 + ^, so ⁰(0) = 1 + ⁰= 2.

Thus, (⁻¹)⁰(1) = ¹₂.

79. (a)  = () and ⁰⁰+ ⁰+  = 0 ⇒ ⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + ⁰(0) + 0 = 0, so ⁰(0) = 0.

⁰⁰⁰+ 2⁰⁰+ ⁰+  = 0, so ⁰⁰⁰() + 2⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + 2⁰⁰(0) + 0 + 1 [(0) = 1 is given] = 0 ⇒ 2⁰⁰(0) = −1 ⇒ ⁰⁰(0) = −¹2. 80.²+ 4²= 5 ⇒ 2 + 4(2⁰) = 0 ⇒ ⁰= −

4, or as  − 0

 − (−5)= 

4 = 

 + 5 ⇔ 4²= −²− 5 ⇔ ²+ 4²= −5. But ²+ 4²= 5 [since ( ) is on the ellipse], so 5 = −5 ⇔  = −1. Then 4²= −²− 5 = −1 − 5(−1) = 4 ⇒  = 1, since the point is on the top half of the ellipse. So 

8 = 

 + 5 = 1

−1 + 5= 1

4 ⇒  = 2. So the lamp is located 2 units above the

-axis.

LABORATORY PROJECT Families of Implicit Curves

1. (a) There appear to be nine points of intersection. The “inner four” near the origin are about (±02 −09) and (±03 −11).

The “outer ﬁve” are about (20 −89), (−28 −88), (−75 −77), (−78 −47), and (−80 15).

67. The figure shows a lamp located three units to the right of the y-axis and a shadow created by the elliptical region x²+ 4y²≤ 5. If the point (−5, 0) is on the edge of the shadow, how far above the x-axis is the lamp located?

Laboratory Project Families of Implicit Curves 217

(b) Illustrate part (a) by graphing the ellipse and the normal line.

75. Find all points on the curve x²y²1xy − 2 where the slope of the tangent line is 21.

76. Find equations of both the tangent lines to the ellipse x²14y²− 36 that pass through the point s12, 3d.

77. (a) Suppose f is a one-to-one differentiable function and its inverse function f²¹ is also differentiable. Use implicit differentiation to show that

s f²¹d9sxd − 1 f 9s f²¹sxdd provided that the denominator is not 0.

(b) If fs4d − 5 and f 9s4d −²₃, find s f²¹d9s5d.

78. (a) Show that fsxd − x 1 e^x is one-to-one.

(b) What is the value of f ²¹s1d?

(c) Use the formula from Exercise 77(a) to find s f²¹d9s1d.

79. The Bessel function of order 0, y − Jsxd, satisfies the differential equation xy99 1 y9 1 xy − 0 for all values of x and its value at 0 is Js0d − 1.

(a) Find J9s0d.

(b) Use implicit differentiation to find J99s0d.

80. The figure shows a lamp located three units to the right of the y-axis and a shadow created by the elliptical region x²14y²<5. If the point s25, 0d is on the edge of the shadow, how far above the x-axis is the lamp located?

? x y

0 3 _5

≈+4¥=5

; 69. Show that the ellipse x²ya²1y²yb²− 1 and the hyperbola

x²yA²2y²yB²− 1 are orthogonal trajectories if A²,a² and a²2b²− A²1B² (so the ellipse and hyperbola have the same foci).

70. Find the value of the number a such that the families of curves y −sx 1 cd²¹ and y − asx 1 kd^1y3 are orthogonal trajectories.

71. (a) The van der Waals equation for n moles of a gas is

S

^{P 1} ⁿ^V²^a²

D

sV 2 nbd − nRT

where P is the pressure, V is the volume, and T is the temperature of the gas. The constant R is the universal gas constant and a and b are positive constants that are characteristic of a particular gas. If T remains constant, use implicit differentiation to find dVydP.

(b) Find the rate of change of volume with respect to pressure of 1 mole of carbon dioxide at a volume of V − 10 L and a pressure of P − 2.5 atm. Use a − 3.592 L²-atmymole² and b − 0.04267 Lymole.

72. (a) Use implicit differentiation to find y9 if x²1xy 1 y²11 − 0

(b) Plot the curve in part (a). What do you see? Prove that what you see is correct.

(c) In view of part (b), what can you say about the expression for y9 that you found in part (a)?

73. The equation x²2xy 1 y²− 3 represents a “rotated ellipse,” that is, an ellipse whose axes are not parallel to the coordinate axes. Find the points at which this ellipse crosses the x-axis and show that the tangent lines at these points are parallel.

74. (a) Where does the normal line to the ellipse

x²2xy 1 y²− 3 at the point s21, 1d intersect the ellipse a second time?

CAS

laboratory Project

^CAS

Families oF imPlicit curves

In this project you will explore the changing shapes of implicitly defined curves as you vary the constants in a family, and determine which features are common to all members of the family.

1. Consider the family of curves

y²22x²sx 1 8d − cfsy 1 1d²sy 1 9d 2 x²g

(a) By graphing the curves with c − 0 and c − 2, determine how many points of inter- section there are. (You might have to zoom in to find all of them.)

(b) Now add the curves with c − 5 and c − 10 to your graphs in part (a). What do you notice? What about other values of c?

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Solution:

77. (a) If  = ⁻¹(), then () = . Differentiating implicitly with respect to  and remembering that  is a function of , we get ⁰()

= 1, so

= 1

⁰() ⇒ 

⁻¹0

() = 1

⁰(⁻¹()). (b) (4) = 5 ⇒ ⁻¹(5) = 4. By part (a),

⁻¹0(5) = 1

⁰(⁻¹(5)) = 1

⁰(4)= 12 3

 =³₂.

78. (a) Assume   . Since ^is an increasing function, ^ ^, and hence,  + ^  + ^; that is, ()  ()

So () =  + ^is an increasing function and therefore one-to-one.

(b) ⁻¹(1) =  ⇔ () = 1, so we need to ﬁnd  such that () = 1. By inspection, we see that (0) = 0 + ⁰= 1, so

 = 0, and hence, ⁻¹(1) = 0.

(c) (⁻¹)⁰(1) = 1

⁰(⁻¹(1)) = 1

⁰(0) [by part (b)]. Now () =  + ^ ⇒ ⁰() = 1 + ^, so ⁰(0) = 1 + ⁰= 2.

Thus, (⁻¹)⁰(1) = ¹₂.

79. (a)  = () and ⁰⁰+ ⁰+  = 0 ⇒ ⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + ⁰(0) + 0 = 0, so ⁰(0) = 0.

⁰⁰⁰+ 2⁰⁰+ ⁰+  = 0, so ⁰⁰⁰() + 2⁰⁰() + ⁰() + () = 0. If  = 0, we have 0 + 2⁰⁰(0) + 0 + 1 [(0) = 1 is given] = 0 ⇒ 2⁰⁰(0) = −1 ⇒ ⁰⁰(0) = −¹2. 80.²+ 4²= 5 ⇒ 2 + 4(2⁰) = 0 ⇒ ⁰= −

4, or as  − 0

 − (−5)= 

4 = 

 + 5 ⇔ 4²= −²− 5 ⇔ ²+ 4²= −5. But ²+ 4²= 5 [since ( ) is on the ellipse], so 5 = −5 ⇔  = −1. Then 4²= −²− 5 = −1 − 5(−1) = 4 ⇒  = 1, since the point is on the top half of the ellipse. So 

8 = 

 + 5 = 1

−1 + 5= 1

4 ⇒  = 2. So the lamp is located 2 units above the

-axis.

LABORATORY PROJECT Families of Implicit Curves

1. (a) There appear to be nine points of intersection. The “inner four” near the origin are about (±02 −09) and (±03 −11).

The “outer ﬁve” are about (20 −89), (−28 −88), (−75 −77), (−78 −47), and (−80 15).

2