1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of...

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1 1.1 Homework Solutions 1. f ( x ) = x 2 + 3x 4 f ( x ) = 2 x + 3 3. y = x 2 3 dy dx = 2 3 x 5 3 5. v(r ) = 4 3 πr 3 v(r ) = 4 πr 2 7. y = x 2 + 4 x + 3 x = x 3 2 + 4 x 1 2 + 3x 1 2 dy dx = 3 2 x 1 2 + 2 x 1 2 3 2 x 3 2 = 3 x 2 + 2 x 3 2 x 3 9. z = A y 10 + Be y = Ay 10 + Be y z = 10 Ay 11 + Be y 11. If f ( x ) = 3x 5 5x 3 + 3 , find f '( x ). f x ( ) = 45 x 14 15 x 2 You can see that when f is decreasing the graph of f is below the x - axis, i.e. f is negative - if f is increasing the graph of f is above the x - axis. 13. y = e x = e x 1 2 dy dx = e x 1 2 1 2 x 1 2 = e x 1 2 2 x 1 2 = e x 2 x y = e x

Transcript of 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of...

Page 1: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.1 Homework Solutions 1. f (x)= x

2 +3x− 4 → f ′(x)= 2x+3

3. y = x−

23 → dy

dx = − 23x−

53

5.

v(r)= 4

3πr3 → v′(r)= 4πr2

7. y = x

2 + 4x+3x

= x3

2 + 4x1

2 +3x−1

2

dydx =

32x

12 + 2x

−12 − 3

2x−3

2 = 3 x2

+ 2x− 3

2 x3

9. z = Ay10 + Be

y = Ay−10 + Bey

′z = −10Ay−11 + Bey 11. If f (x) = 3x5 − 5x3 + 3 , find f '(x). ′f x( ) = 45x14 −15x2 You can see that when f is decreasing the graph of ′f is below the x - axis, i.e. ′f is negative - if f is increasing the graph of ′f is above the x - axis.

13. y = e x = ex12 ⇒ dy

dx = ex12 ⋅12 x

−12 = ex12

2x12= e x

2 x y = e x

Page 2: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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15. Given the following table of values, find the indicated derivatives.

x f (x) ′f (x)2 1 78 5 −3

a. g x( ) = f x( )⎡⎣ ⎤⎦

3

′g x( ) = 3 f x( )⎡⎣ ⎤⎦2⋅ ′f x( )⇒ ′g 2( ) = 3 f 2( )⎡⎣ ⎤⎦

2⋅ ′f 2( ) = 3⋅12 ⋅7 = 21

b. h x( ) = f x3( ) ′h x( ) = ′f x3( )⋅3x2 ⇒ ′h 2( ) = ′f 23( )⋅3⋅22 =12 ′f 8( ) =12 ⋅−3= −36

17. If

f (x) = x3 + 2x( )37

, find f '(x) .

⇒ ′f x( ) = 37 x3 + 2x( )36 ⋅ 3x2 + 2( ) = 37 3x2 + 2( )⋅ x3 + 2x( )36

19. f x( ) = 4 − 49 x

2 ; find f ' 5( )

f x( ) = 4 − 49 x2 = 4 − 49 x

2⎛⎝⎜

⎞⎠⎟

12⇒ ′f x( ) = 12 4 − 49 x

2⎛⎝⎜

⎞⎠⎟

−12⋅ − 89 x⎛⎝⎜

⎞⎠⎟= −4x

9 4 − 49 x2

21. y = x3 − 2x7 ; find dy

dx

y = x3 − 2x7 = x3 − 2x( )17 ⇒ ′f x( ) = 17 x3 − 2x( )−67 ⋅ 3x2 − 2( ) = 3x2 − 2

7 x3 − 2x( )67

Page 3: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.2 Homework Solutions 1. sin 4y x=

′y = 4cos4x 3. y = a3 + cos3 x

′y = 3cos2 x ⋅ −sin x( ) = −3cos2 xsin x 5. f (t) = 1+ tant3

′f t( ) = 13 1+ tant( )−23 ⋅sec2 t = sec2 t1+ tant( )23

7.

y = cos a3+x3( )

′y = −sin a3 + x3( )⋅3x2 = −3x2 sin a3 + x3( ) 9.

f (x) = cos ln x( )

′f x( ) = −sin ln x( )⋅1x = −sin ln x( )

x

11.

f (x) = log10 2+ sin x( ) ′f x( ) = 1

2+ sin x ⋅1ln10 ⋅cosx =

cosxln10 2+ sin x( )

13.

y = sin−1 ex( )

′y = 11− e2x

⋅ex = ex

1− e2x

15.

y = sin−1 2x +1( )

′y = 11− 2x+1( )2

⋅2 = 21− 4x2 + 4x+1( )

= 2 −4x2 − 4x( )−12 = −1

−x2 − x( )12

Page 4: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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17. y = csc−1 x2 +1( ) ′y = −1

x2 +1( ) x2 +1( )2 −1⋅2x = −2x

x2 +1( ) x2 +1( )2 −1

Page 5: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.3 Homework Solutions 1. Find the equation of the tangent line to f (x) = x5 − 5x +1 at x = −2 and use it to get an approximate value of f (-1.9).

f −2( ) = −21 ′f x( ) = 5x4 − 5⇒ ′f −2( ) = 75 y+ 21= 75 x+ 2( )y = −21+ 75 x+ 2( )⇒ y x=−1.9 = −13.5

f −1.9( ) ≈ −13.5 3. Find an equation of the line tangent to the curve y = x

4 + 2ex at the point

(0, 2) .

′y = 4x3 + 2ex ⇒ ′y x=0 = 2 Tangent Line: y− 2 = 2 x− 0( )

5. Find the equation of the tangent line to cosy x x= + at the point

0,1( ) .

′y =1− sin x⇒ ′y x=0 =1 Tangent Line: y−1=1 x− 0( )

7. Find the equation of the line tangent to y = 2

π x + cos 4x( ) when x = π

2.

′y = 2π − 4sin 4x( )⇒ ′yx=π 2

= 2π − 4sin 2π( ) = 2π

yx=π 2

= 2π ⋅π2 + cos 2π( ) = 2

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9. Use Euler’s Method with 4 equal step sizes to find an approximation for

f 1.4( ) , given that

f x( ) = ln 2x −1( ) and

f 1( ) = 0 .

f 1.4( )≈ 0.635

11. Given the differential equation,

dydx

= y8

6− y( ) where y = f t( ) and

f 0( ) = 8 ,

use Euler’s Method with two steps of equal size to approximate f 1( ) .

Page 7: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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13. Let y = f x( ) be the solution to the differential equation dy x y

dx= + with the

initial condition f 1( ) = 2 . What is the approximation for

f 2( ) if Euler’s method is

used, starting at x = 1 with a step size of 0.5? (A) 3 (B) 5 (C) 6 (D) 10 (E) 12

Page 8: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.4 Homework Solutions 1. 3 cosy t t=

′y = t 3 ⋅ −sint( )+ cost ⋅3t2 = t2 3cost − t sint( )

3. y = tan x −1

sec x

′y =secx ⋅sec2 x− tan x−1( )⋅secx tan x

sec2 x =sec2 x− tan x−1( )tan x

secx = sec2 x− tan2 x+ tan x

secx Recall 1= sec2 x− tan2 x , so substitute into numerator:

′y = 1+ tan xsecx

5. y = xe−x2 ′y = x ⋅e−x2 ⋅ −2x( )+ e−x2 = e−x2 1− 2x2⎡⎣ ⎤⎦ 7. y = excosx ′y = excosx x ⋅−sin x+ cosx⎡⎣ ⎤⎦ = e

xcosx cosx− xsin x⎡⎣ ⎤⎦

9. y = xsin 1

x

′y = x ⋅cos 1x ⋅−1x2

+ sin 1x = − 1x cos1x + sin

1x

11. f (x) = x ln x

′f x( ) = x ⋅12 ln x( )−12 ⋅1x + ln x( )12 = 1+ 2ln x2 ln x( )12

Page 9: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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13. Find the equation of the line tangent to y = x2e−x at the point 1, 1

e( ) .

′y = x2 ⋅e−x ⋅ −1( )+ e−x ⋅2x m = ′y x=1 = −1e +2e =

1e

y− 1e =1e x−1( )

15. h(t) = 1+ x2

1− x2

⎝⎜⎞

⎠⎟

17

, find h '(t)

′h t( ) =17 1+ t2

1− t2⎛

⎝⎜⎞

⎠⎟

16

⋅1− t2( )⋅2t − 1+ t2( )⋅−2t

1− t2( )2=17 1+ t2( )16 2t − 2t 3 + 2t + 2t 3⎡⎣ ⎤⎦

1− t2( )18

′h t( ) = 68t 1+ t22( )16

1− t 2( )18

17. f (x) = xsin 2x( ) + tan4 x7( )⎡

⎣⎤⎦

5, find f ' x( ) .

′f x( ) = 5 xsin 2x( )+ tan4 x7( )⎡⎣⎢

⎤⎦⎥4⋅ xcos2x ⋅2+ sin2x+ 4 tan3 x7( )⋅sec2 x7( )⋅7x6⎡⎣⎢

⎤⎦⎥

= 5 xsin 2x( )+ tan4 x7( )⎡⎣⎢

⎤⎦⎥4⋅ 2xcos2x+ sin2x+ 28x6 tan3 x7( )sec2 x7( )⎡⎣⎢

⎤⎦⎥

19. Find the equation of the line tangent to y = ex sin 4x( ) + 2 when x = 0

y x=0 = e0sin0 + 2 = 3⇒ 0,3( )

′y = exsin4x x ⋅cos4x ⋅4 + sin4x⎡⎣ ⎤⎦

′y x=0 = e0 0 ⋅cos0 ⋅4 + sin0⎡⎣ ⎤⎦ = 0

y− 3= 0 x− 3( )⇒ y = 3

20. Find the equation of the lines tangent and normal to y = xsin 1

x⎛

⎝⎜⎞

⎠⎟ when

21.

H(x) = 1+ x2( ) tan−1(x)

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′H x( ) = 1+ x2( )⋅ 11+ x2 + tan

−1 x ⋅2x

23. If f (x) = ex − x2 arctan x , find

f ' x( ) .

′f x( ) = ex − x2 ⋅ 1

1+ x2 + arctan x ⋅2x⎡⎣⎢

⎤⎦⎥

= ex − x21+ x2 − 2xarctan x

25. y = sec−1 xx

′y =x ⋅ 1x x2 −1

− sec−1 x

x2=

1x2 −1

− sec−1 x

x2

Page 11: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.5 Homework Solutions 1. f (x) = x5 + 6x2 − 7x

′f x( ) = 5x4 +12x− 7′′f x( ) = 20x3 +12

3. y = x3 +1( )23

′y = 23 x3 +1( )−13 ⋅3x2 = 2x2

x3 +1( )13

′′y =x3 +1( )13 ⋅4x− 2x2 ⋅13 x3 +1( )−23 ⋅3x2

x3 +1( )23=x3 +1( )−23 x3 +1( )⋅4x− 2x4⎡

⎣⎢⎤⎦⎥

x3 +1( )23

= 4x4 + 4x− 2x4

x3 +1( )4 3=2x x3 + 2( )x3 +1( )4 3

5. g(t) = t3e5t

′g t( ) = t 3 ⋅e5t ⋅5+ e5t ⋅3t2 = t2e5t 5t + 3⎡⎣ ⎤⎦

′′g t( ) = t2e5t ⋅5+ t2 5t + 3( )⋅e5t ⋅5+ e5t ⋅ 5t + 3( )⋅2t= te5t 5t + 5t 5t + 3( )+ 2 5t + 3( )⎡

⎣⎤⎦

= te5t 25t2 + 30t + 6⎡⎣ ⎤⎦

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7. y = sin 3x

′y = 3sin2 xcosx

′′y = 3 sin2 x ⋅−sin x+ cosx ⋅2sin xcosx⎡⎣ ⎤⎦ = 3sin x 2cos2 x− sin2 x⎡⎣ ⎤⎦

9. d2 dx2

5x4 + 9x3 − 4x2 + x − 8⎡⎣ ⎤⎦

= ddx −20x3 + 27x2 − 8x+1⎡⎣ ⎤⎦

= −60x2 + 54x− 8

11. y = cosx2 , find ′′y

dydx = −sin x

2 2x( ) = −2xsin x2

d2ydx2

= −2x cosx2 2x( )⎡⎣ ⎤⎦ + sin x2 −2( ) = −2 2x2 cosx2 + sin x2⎡⎣ ⎤⎦

13. y = sec3x , find d2ydx2

dydx = sec3x tan3x 3( ) = 3sec3x tan3xd2ydx2

= 3sec3x sec2 3x 3( )( ) + 3tan3x 3sec3x tan3x( ) = 3sec3x 3sec2 3x+ tan2 3x( )

15. f x( ) = ln x2 + 3( ) , find

′′f x( )

f ′ x( ) = 2xx2 + 3

′′f x( ) =x2 + 3( ) 2( )− 2x( ) 2x( )

x2 + 3( )2=

2x2 + 6( )− 4x2( )x2 + 3( )2

=−2 x2 − 3( ) x2 + 3( )2

Page 13: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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17. h x( ) = x2 + 5 , find ′′h x( )

h x( ) = x2 + 5 = x2 + 5( )12 h′ x( ) = x2 + 5 = 12 x2 + 5( )−12 2x( ) = x

x2 + 5( )12

′′h x( ) =x2 + 5( )12 1( )− x ⋅ x

x2 + 5( )12

x2 + 5( )1

=x2 + 5( )1 − x2

x2 + 5( )32

= 5

x2 + 5( )32

19. y = x2 − 3

x2 −10, find d

2ydx2

dydx =

x2 −10( ) 2x( )− x2 − 3( ) 2x( )x2 −10( )2

=2x( ) x2 −10( )− x2 − 3( )⎡

⎣⎢⎤⎦⎥

x2 −10( )2= −14xx2 −10( )2

d2ydx2 =

x2 −10( )2 −14( )− −14x( )2 x2 −10( )1 2x( )x2 −10( )4

=−14 x2 −10( ) x2 −10( )− 4x2( )⎡

⎣⎢⎤⎦⎥

x2 −10( )4

=−14 −3x2 −10( )

x2 −10( )3=

14 3x2 +10( )x2 −10( )3

21. y = x3 + x2 − 7x −15

y′ = 3x2 + 2x− 7y″ = 6x+ 2

Page 14: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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23. y = −4xx2 + 4

dydx =

x2 + 4( ) −4( )− −4x( ) 2x( )x2 + 4( )2

=−4x2 −16( )− −8x2( )

x2 + 4( )2= 4x

2 −16x2 + 4( )2

d2ydx2

=x2 + 4( )2 8x( )− 4x2 −16( )2 x2 + 4( )1 2x( )

x2 + 4( )4

=x2 + 4( ) 8x( )− 4x2 −16( )2 2x( )

x2 + 4( )3

=8x3 + 32x( )− 16x3 − 64( )

x2 + 4( )3

=−8x3 + 96x( )x2 + 4( )3

=−8x x2 −12( )x2 + 4( )3

25. y = x 8− x2 = x 8− x2( )12 dydx = x ⋅

12 ⋅ 8− x

2( )−12 −2x( )+ 8− x2( )12 1( )

= −x2

8− x2( )12+ 8− x2( )12

=−x2 + 8− x2( )18− x2( )12

= 8− 2x2

8− x2( )12

Page 15: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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d2ydx2

=

8− x2( )12 −4x( )− 8− 2x2( ) −x2

8− x2( )128− x2( )1

=8− x2( )1 −4x( )− −x2( ) 8− 2x2( )

8− x2( )32= −2x4 + 4x3 + 8x2 − 32x

8− x2( )32

27. y = xe−x

dydx = xe

−x −1( )+ e−x 1( ) = e−x x+1( )d2ydx2

= e−x 1( )+ x+1( )e−x −1( ) = e−x x− 2( )

29. y = x

x2 − 9

dydx =

x2 − 9( ) 1( )− x( ) 2x( )x2 − 9( )2

= −x2 − 9x2 − 9( )2

d2ydx2

=x2 − 9( )2 −2x( )− −x2 − 9( )2 x2 − 9( )1 2x( )

x2 − 9( )4

=x2 − 9( ) −2x( )− −x2 − 9( )2 2x( )

x2 − 9( )3

=−2x3 +18x( )− −4x3 − 36x( )

x2 − 9( )3

=2x3 + 54x( )x2 − 9( )3

=2x x2 + 27( )x2 − 9( )3

Page 16: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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1.6 Homework Solutions 1. xy + 2x + 3x2 = 4 Implicit Explicit

x dydx + y ⋅1+ 2+ 6x = 0 xy = 4 − 2x− 3x2

dydx =

−6x− y− 2x y = 4x − 2− 3x = −4x

−2 − 3= − 4x2

− 3

dydx =

−4 − 3x2x2

dydx =

−6x− y− 2x =

−6x− 4x − 2− 3x

⎛⎝⎜

⎞⎠⎟− 2

x = −6− 4x2

+ 2x + 3−2x = −3− 4

x2= −4 − 3x2

x2

3. x + y = 4 Implicit Explicit 12 x

−12 + 12 y−12 dy

dx = 0 y = 4 − x⇒ y =16− 8 x + x

dydx =

− 12 x−12

12 y

−12= − y

12

x12

dydx = −4x

−12 +1= − 4x+1= −4 + x

x

dydx = − y

12

x12= − 4 − x

x= −4 + x

x

5. x

3 +10x2 y + 7 y2 = 60

3x2 +10 x2 dydx + y ⋅2x⎛⎝⎜

⎞⎠⎟+14y dydx = 0

3x2 +10x2 dydx + 20xy+14ydydx = 0

dydx =

−3x2 − 20xy10x2 +14y

Page 17: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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7. 4cos x( )sin y( ) =1

4 cosx ⋅cosy dydx + sin y ⋅−sin x⎡

⎣⎢

⎦⎥ = 0

4cosxcosy dydx = 4sin xsin y

dydx = tan x tan y

9. tan x - y( ) = y

1+ x2

sec2 x− y( ) 1− dydx⎛⎝⎜

⎞⎠⎟=1+ x2( )− y ⋅2x1+ x2( )2

sec2 x− y( )− sec2 x− y( )dydx⎡

⎣⎢

⎦⎥ ⋅ 1+ x2( )2 = dydx + x2

dydx − 2xy

sec2 x− y( ) 1+ x2( )2 − sec2 x− y( ) 1+ x2( )2 dydx =dydx + x

2 dydx − 2xy

sec2 x− y( ) 1+ x2( )2 + 2xy = dydx 1+ x2 + sec2 x− y( ) 1+ x2( )2⎡⎣⎢

⎤⎦⎥

dydx =

sec2 x− y( ) 1+ x2( )2 + 2xy1+ x2 + sec2 x− y( ) 1+ x2( )2

11. 9x2 + 4y2 + 36x −8y + 4 = 0 at

0, − 2( )

18x+ 8y dydx + 36− 8dydx = 0

−16 dydx + 36− 8dydx = 0

dydx =

−36−24 =

32

y+ 2 = 32 x

Page 18: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

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13. Find the equation of the lines tangent and normal to

y − 4

π 2 x2 = 2eysin x + y3 − 3 through the point

π2

,0⎛

⎝⎜⎞

⎠⎟.

dydx −

8π 2 x = 2e

ysinx ycosx+ sin x dydx⎡

⎣⎢

⎦⎥+ 3y2

dydx

dydx −

8π 2

π2

⎛⎝⎜

⎞⎠⎟= 2e0 0+ sinπ2

dydx

⎣⎢

⎦⎥+ 0

dydx −

4π = 2 dydx⇒

dydx = −

Tangent line: y− 0 = −4π x−π 2( ) , Normal line: y− 0 =π 4 x−π 2( )

15. Find the equation of the line tangent to x

2 + 3xy + y2 =11, through the point (1,2).

2x+ 3 x dydx + y⎡

⎣⎢

⎦⎥+ 2y

dydx = 0

2+ 3 dydx + 2

⎣⎢

⎦⎥+ 4

dydx = 0⇒ 2+ 3dydx + 6+ 4

dydx = 0⇒ 7 dydx = −8⇒

dydx = −

8 7

y− 2 = −8 7 x−1( )

17. Find d2ydx2

if 4x2 + 9y2 = 36

8x+18y dydx = 0⇒dydx =

−8x18y =

−4x9y

d2ydx2

=9y ⋅−4 − −4x( )⋅9 dydx

81y2 =−36y+ 36x ⋅ −4x9y

81y2 =−36y−144x

2

9y81y2

= −324y2 −144x2729y3 = −36y2 −16x2

81y2

Page 19: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  19  

18. y = 2x +1( )4 x3 − 3( )5

19.

z = y3 − 3( )e 2 y+1( )

lnz = ln y3 − 3( )+ 2y+11zdzdy =

3y2y3 − 3+ 2

dzdy = z

3y2y3 − 3+ 2

⎣⎢

⎦⎥

dzdy = y3 − 3( )e2y+1 3y2

y3 − 3+ 2⎡

⎣⎢

⎦⎥

20.

y = sin2 x tan4 x

x2 +5( )2

ln y = 2lnsin x+ 4ln tan x− 2ln x2 + 5( )1ydydx = 2 ⋅

1sin x ⋅cosx+ 4

1tan x ⋅sec

2 x− 4xx2 + 5

dydx = y 2cot x+

4sin xcosx −

4xx2 + 5

⎡⎣⎢

⎤⎦⎥

dydx =

sin2 x+ tan4 xx2 + 5( )2

2cot x+ 4sin xcosx −

4xx2 + 5

⎡⎣⎢

⎤⎦⎥

Page 20: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  20  

21.

g t( ) = t ln t( )

′g t( ) = t ⋅1t + lnt =1+ lnt 22.

y = lnx x( )

ln y = x ln ln x( )1ydydx = x ⋅

1ln x ⋅

1x + ln ln x( )⋅1

dydx = y

1ln x + ln ln x( )⎡

⎣⎢⎤⎦⎥

dydx = ln

x x 1ln x + ln ln x( )⎡

⎣⎢⎤⎦⎥

23.

p v( ) = vev

ln p = ev lnv1pdpdv = e

v ⋅1v + lnv ⋅ev

dpdv = p

evv + ev lnv⎡

⎣⎢

⎦⎥

dpdv = v

ev evv + ev lnv⎡

⎣⎢

⎦⎥

Page 21: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  21  

24. Use logarithmic differentiation to prove the product rule.

y = uv⇒ ln y = lnu − lnv

1ydydx =

1u ⋅dudx −

1v ⋅dvdx⇒

dydx = y

1u ⋅dudx −

1v ⋅dvdx

⎣⎢

⎦⎥

⇒ dydx =

v dudx −udvdx

v2

25. Find dtdu if tu = ut .

ddt t

u = ut( )⇒ utu−1 dudt = ut ⋅lnu ⋅1⇒ du

dt =ut lnuutu−1 ⇒ du

dt =ut−1 lnutu−1

26. For the function,

f x( ) = xln x , find an equation for a tangent line at x = e, and

use that to approximate the value for f (2.7). Find the percent difference between this and the actual value of f (2.7). ln y = ln x ⋅ln x = ln x( )21ydydx = 2ln x ⋅

1x⇒

dydx = y 2ln x ⋅

1x

⎡⎣⎢

⎤⎦⎥⇒ dy

dx = xlnx 2ln x ⋅1x

⎡⎣⎢

⎤⎦⎥

dydx x=e

= elne 2lne ⋅1e⎡⎣⎢

⎤⎦⎥= 2

Tangent Line: y− e= 2 x− e( ) f 2.7( ) ≈ e+ 2 2.7− e( ) = 2.682 f 2.7( ) = 2.7( )ln 2.7( ) = 2.682

% Difference = 0

Page 22: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  22  

1.7 Homework Solutions 1. If V is the volume of a cube with edge length, s, and the cube expands as

time passes, find dVdt in terms of

dsdt .

V =πs3 → dV

dt = 3π 2 dsdt

2. A particle is moving along the curve y = 1+ x3 . As it reaches the point (2,3), the y-coordinate is increasing at a rate of 4 cm/sec. How fast is the x-coordinate changing at that moment? dydt =

12 1+ x

3( )−12 ⋅ 3x2( )⋅dxdt ; dydt = 4

cm s

⇒ 4 = 12 1+ 23( )−12 ⋅3⋅22 ⋅dxdt

dxdt = 2

cm s

3. A plane flying horizontally at an altitude of 1 mile and a speed of 500 mph flies directly over a radar station. Find the rate at which the horizontal distance is increasing when it is 2 miles from the station. Find the rate at which the distance between the station and the plane is increasing when the plane is 2 miles from the station.

x2 +1= z2 ⇒ 2x dxdt = 2z

dzdt ; dxdt = 500

mih ; z = 2⇒ x = 3

⇒ 2 3 ⋅500 = 2 ⋅2 dzdt

Light

Plane

1 mi

z

x

Page 23: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  23  

⇒ dzdt = 250 3mih

4. If a snowball melts so that its surface area is decreasing at a rate of 1 cm2/min, find the rate at which the diameter is decreasing when it has a diameter of 10 cm.

S = 4πr2 ⇒ dSdt = 8πr

drdt ; dSdt = −1cm2

min ; D =10cm⇒ r = 5cm

⇒−1= 8π ⋅5 drdt ⇒drdt = −

140π

cmmin

D = 2r⇒ dDdt = 2 drdt ⇒

dDdt = − 1

20πcmmin

5. A street light is mounted at the top of a 15 foot tall pole. A 6 foot tall man walks away from the pole at a speed of 5 ft/sec in a straight line. How fast is the tip of his shadow moving away from the pole when the man is 40 feet from the pole?

15x+ y =

6y⇒ x+ y = 52 y⇒ x = 32 y ; dxdt = 5

fts ; x = 40 ft

⇒ dxdt =

32dydt ⇒ 5 = 32

dydt ⇒

dydt =

103ftmin

⇒d x+ y( )dt = dxdt +

dydt = 5+

103 = 253

ftmin

Dude

Light

6ft.15ft.

yx

Page 24: 1.1 Homework Solutions - Kevin Quattrin, EdD · ! 5! 1.3 Homework Solutions 1. Find the equation of the tangent line to f(x)=x5−5x+1 at x=−2 and use it to get an approximate value

  24  

6. Two cars start moving away from the same point. One travels south at 60 mph, and the other travels west at 25 mph. At what rate is the distance between the cars increasing two hours later?

x2 + y2 = z2 ; dydt = 60

mih ; dxdt = 25

mih

⇒ 2x dxdt + 2ydydt = 2z

dzdt

⇒ 2 ⋅50 ⋅25+ 2 ⋅120 ⋅60 = 2 ⋅130 ⋅dzdt

dzdt = 65

mih

7. The altitude of a triangle is increasing at a rate of 1 cm/min, while the area of the triangle is increasing at a rate of 2 cm2/min. At what rate is the base of the triangle increasing when the altitude is 10 cm and the area is 100 cm2?

A = 12bh ; dhdt =1

cmmin ; dAdt = 2

cm2

min ; h =10cm ; A =100cm2

⇒ dAdt =

12 b dhdt + h

dbdt

⎣⎢

⎦⎥⇒ 2 = 12 20 ⋅1+10 ⋅

dbdt

⎣⎢

⎦⎥⇒

dbdt = −1.6

cmmin

Car

Car

zy

x

h

b