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

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

  • 1

    1.1 Homework Solutions 1. f (x)= x

    2 +3x 4 f (x)= 2x+3

    3. y = x

    23 dydx =

    23x

    53

    5.

    v(r)= 4

    3r3 v(r)= 4r2

    7. y = x

    2 + 4x+3x

    = x3

    2 + 4x1

    2 +3x1

    2

    dydx =

    32x

    12 + 2x

    12 3

    2x3

    2 = 3 x2

    + 2x 3

    2 x3

    9. z = Ay10 + Bey = Ay10 + Bey

    z = 10Ay11 + Bey 11. If f (x) = 3x

    5 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 dydx = e

    x12 12 x

    12 = ex12

    2x12= e

    x

    2 x y = e x

  • 2

    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( ) = 312 7 = 21

    b. h x( ) = f x3( ) h x( ) = f x3( )3x2 h 2( ) = f 23( )322 =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( ) = 3x

    2 2

    7 x3 2x( )67

  • 3

    1.2 Homework Solutions 1. sin 4y x=

    y = 4cos4x 3. y = a

    3 + cos3 x y = 3cos2 x sin x( ) = 3cos2 xsin x

    5. f (t) = 1+ tant

    3

    f t( ) = 13 1+ tant( )23 sec2 t = sec

    2 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( ) = 12+ sin x

    1ln10 cosx =

    cosxln10 2+ sin x( )

    13.

    y = sin1 ex( )

    y = 11 e2x

    ex = ex

    1 e2x

    15.

    y = sin1 2x +1( )

    y = 11 2x+1( )2

    2 = 21 4x2 + 4x+1( )

    = 2 4x2 4x( )12 = 1x2 x( )12

  • 4

    17. y = csc1 x2 +1( ) y = 1

    x2 +1( ) x2 +1( )2 12x = 2x

    x2 +1( ) x2 +1( )2 1

  • 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 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: y1=1 x 0( )

    7. Find the equation of the line tangent to y = 2 x + cos 4x( ) when x =

    2

    .

    y = 2 4sin 4x( ) y x= 2 =2 4sin 2( ) =

    2

    yx= 2

    = 2 2 + cos 2( ) = 2

  • 6

    9. Use Eulers 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 Eulers Method with two steps of equal size to approximate

    f 1( ) .

  • 7

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

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

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

  • 8

    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 x1( )secx tan x

    sec2 x =sec2 x tan x1( )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 ex

    2 2x( )+ ex2 = ex2 1 2x2 7. y = e

    xcosx y = excosx x sin x+ cosx = e

    xcosx cosx xsin x

    9. y = xsin 1

    x

    y = x cos 1x 1x2

    + sin 1x = 1x cos

    1x + 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

  • 9

    13. Find the equation of the line tangent to y = x2ex at the point

    1, 1e( ) .

    y = x2 ex 1( )+ ex 2x m = y x=1 = 1e +2e =

    1e

    y 1e =1e x1( )

    15. h(t) = 1+ x

    2

    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+ t 22( )161 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 = e

    x sin 4x( ) + 2 when x = 0 y x=0 = e

    0sin0 + 2 = 3 0,3( ) y = exsin4x x cos4x 4 + sin4x y x=0 = e

    0 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( ) tan1(x)

  • 10

    H x( ) = 1+ x2( ) 11+ x2 + tan1 x 2x

    23. If f (x) = e

    x x2 arctan x , find f ' x( ) .

    f x( ) = ex x2 11+ x2 + arctan x 2x

    = ex x2

    1+ x2 2xarctan x

    25. y = sec1 xx

    y =x 1x x2 1

    sec1 x

    x2=

    1x2 1

    sec1 x

    x2

  • 11

    1.5 Homework Solutions 1. f (x) = x

    5 + 6x2 7x

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

    3. y = x3 +1( )23

    y = 23 x3 +1( )13 3x2 = 2x

    2

    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) = t

    3e5t

    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

  • 12

    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 20x

    3 + 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

  • 13

    17. h x( ) = x2 + 5 , find h x( ) h x( ) = x2 + 5 = x2 + 5( )12 h x( ) = x2 + 5 = 12 x

    2 + 5( )12 2x( ) = xx2 + 5( )12

    h x( ) =x2 + 5( )12 1( ) x x

    x2 + 5( )12x2 + 5( )1

    =x2 + 5( )1 x2x2 + 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

  • 14

    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( )= x

    2

    8 x2( )12+ 8 x2( )12

    =x2 + 8 x2( )18 x2( )12

    = 8 2x2

    8 x2( )12

  • 15

    d2ydx2

    =

    8 x2( )12 4x( ) 8 2x2( ) x28 x2( )12

    8 x2( )1

    =8 x2( )1 4x( ) x2( ) 8 2x2( )

    8 x2( )32= 2x

    4 + 4x3 + 8x2 32x

    8 x2( )32

    27. y = xex

    dydx = xe

    x 1( )+ ex 1( ) = ex x+1( )d2ydx2

    = ex 1( )+ x+1( )ex 1( ) = ex x 2( )

    29. y = x

    x2 9

    dydx =

    x2 9( ) 1( ) x( ) 2x( )x2 9( )2

    = x2 9

    x2 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

  • 16

    1.6 Homework Solutions 1. xy + 2x + 3x

    2 = 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 3x2

    x2

    dydx =

    6x y 2x =

    6x 4x 2 3x

    2

    x = 64x2

    + 2x + 32x