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Math Help - Are my book answers wrong?

  1. #1
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    Are my book answers wrong?

    I've been doing some questions that I thought I was doing right but then I check the book answer and i'm wrong..:

    1. Find the slope of the function y=2cosxsin2x at x=pi/2.

    So first I differentiated:

    y' = (-2sinx)(sin2x) + (2cosx)(cos2x)(2)
    = -2sinxsin2x + 4cosxcos2x

    Then I determined that sin(pi/2) = 1 and that cos pi/2 = 0 because at 90 degrees, y=1, r=1, x=0.. and sin(pi/2) = y/r = 1/1 = 1.. and cos(pi/2) = x/r = 0/1 = 0 So I subbed these values into my derived equation:

    y'(pi/2) = -2(1)(2)(1) + 4(0)(0)(2)
    = -4

    My answer is the slope is -4, however the book answer says 0. Did I go wrong?


    2. Find the equation of the line that is tangent to the curve y=2sinx+4cosx at x=pi/4.

    So this is what I got when I differentiated y=2cosx+4cosx:

    y'=2cosx+4(-sinx)
    =2cosx-4sinx

    At pi/4: cosx=1/square root of 2, sinx = 1/square root of 2

    y'(pi/4) = 2(1/square root of 2) - 4(1/square root of 2)
    =0

    To get y value:
    y(pi/4) = 2(1/square root of 2) + 4(1/square root of 2)
    =square root of 2 + square root of 2

    y-y1=m(x-x1)

    y-(square root of 2+ square root of 2)=0(x-(pi/2))
    y=square root of 2 + square root of 2

    The teacher said the answer is: y=-(square root of 2)x + (square root of 2 / 4) pi + 3(square root of 2)

    Where did I go wrong?

    3. Find the equation of the line that is tangent to the curve y=2cosx-(1/2)sinx at x=3pi/2.

    y'= -2sinx-(1/2)cosx

    At 3pi/2: sinx = y/r = 1/1 =1, cosx = x/r = 0/1 = 0

    y'(3pi/2) = -2(1) - 1/2(0)
    =-2

    Finding y value:
    y(3pi/2) = 2(0) - (1/2)(1)
    = -1/2

    y-y1=m(x-x1)
    y-(-1/2)=-2(x - 3pi/2)
    y=-2x + 3pi + 1/2

    Book answer is: 2x + 1/2 -3pi

    Why is their 3pi negative, and there 2x is positive?

    4. Find the slope of the line that is tangent to the curve y=2sinxcosx at x=pi/4.

    y'= -2sinxcosx

    At pi/4: sinx= 1/square root of 2, cosx= 1/square root of 2

    y'(pi/4) = -2(1/square root of 2)(1/square root of 2)
    =-1

    Book answer: 0

    Where did I go wrong?

    5. Find the equation of the line that is tangent to the curve y=2cos^3x at x=pi/3.

    y'= -6sin^2x

    At x=pi/3, sinx = (square root of 3) / 2

    y'(pi/3) = -6[(square root of 3)/2]^2

    I'm not really sure.. how would I work that?

    Book answer:



    THANKS!
    Last edited by kmjt; April 19th 2010 at 09:46 AM.
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  2. #2
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    I'll just do the first one real quick

    d/dx cos x sin 2x use product rule

    d/dx uv = v du/dx + u dv/dx

    u = cos x, v = sin 2x
    du/dx = -sin x

    Use chain rule for d/dx sin 2x
    dy/dx = dy/du du/dx
    u = 2x
    d/dx sin 2x = 2cos 2x

    d/dx cos x sin 2x = sin 2x * -sin x + cos x * 2 cos 2x

    f'(x) = 2cos 2x cos x - sin 2x sin x
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  3. #3
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    Quote Originally Posted by kmjt View Post

    2. Find the equation of the line that is tangent to the curve y=2cosx+4cosx at x=pi/4.
    y= 2\cos x +4\cos x= 6\cos x \implies \frac{dy}{dx}= -6\sin x

    The tangent then becomes

    y-f\left(\frac{\pi}{4}\right) = f'\left(\frac{\pi}{4}\right) \left(x - \frac{\pi}{4}\right)
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  4. #4
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    for number three the slope is 2 (not -2)

    sin 3pi/2 = -1 not 1
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  5. #5
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    Quote Originally Posted by kmjt View Post
    4. Find the slope of the line that is tangent to the curve y=2sinxcosx at x=pi/4.

    y'= -2sinxcosx

    At pi/4: sinx= 1/square root of 2, cosx= 1/square root of 2

    y'(pi/4) = -2(1/square root of 2)(1/square root of 2)
    =-1

    Book answer: 0

    Where did I go wrong?
    d/dx 2 sin x cos x = 2 d/dx sin x cos x

    Product rule d/dx uv = v du/dx + u dv/dx

    u = cos x, v = sin x
    du/dx = -sin x, dv/dx = cos x

    so

    2 d/dx sin x cos x = sin x * -sin x + cos x * cos x = 2(cos^2 x - sin^2 x)
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  6. #6
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    Where you have gone wrong

    Quote Originally Posted by kmjt View Post
    I've been doing some questions that I thought I was doing right but then I check the book answer and i'm wrong..:

    1. Find the slope of the function y=2cosxsin2x at x=pi/2.

    So first I differentiated:

    y' = (-2sinx)(sin2x) + (2cosx)(cos2x)(2)
    = -2sinxsin2x + 4cosxcos2x

    Then I determined that sin(pi/2) = 1 and that cos pi/2 = 0 because at 90 degrees, y=1, r=1, x=0.. and sin(pi/2) = y/r = 1/1 = 1.. and cos(pi/2) = x/r = 0/1 = 0 So I subbed these values into my derived equation:

    y'(pi/2) = -2(1)(2)(1) + 4(0)(0)(2)
    = -4

    My answer is the slope is -4, however the book answer says 0. Did I go wrong?


    2. Find the equation of the line that is tangent to the curve y=2cosx+4cosx at x=pi/4.

    So this is what I got when I differentiated y=2cosx+4cosx:

    y'=2cosx+4(-sinx)
    =2cosx-4sinx

    At pi/4: cosx=1/square root of 2, sinx = 1/square root of 2

    y'(pi/4) = 2(1/square root of 2) - 4(1/square root of 2)
    =0

    To get y value:
    y(pi/4) = 2(1/square root of 2) + 4(1/square root of 2)
    =square root of 2 + square root of 2

    y-y1=m(x-x1)

    y-(square root of 2+ square root of 2)=0(x-(pi/2))
    y=square root of 2 + square root of 2

    The teacher said the answer is: y=-(square root of 2)x + (square root of 2 / 4) pi + 3(square root of 2)

    Where did I go wrong?

    3. Find the equation of the line that is tangent to the curve y=2cosx-(1/2)sinx at x=3pi/2.

    y'= -2sinx-(1/2)cosx

    At 3pi/2: sinx = y/r = 1/1 =1, cosx = x/r = 0/1 = 0

    y'(3pi/2) = -2(1) - 1/2(0)
    =-2

    Finding y value:
    y(3pi/2) = 2(0) - (1/2)(1)
    = -1/2

    y-y1=m(x-x1)
    y-(-1/2)=-2(x - 3pi/2)
    y=-2x + 3pi + 1/2

    Book answer is: 2x + 1/2 -3pi

    Why is their 3pi negative, and there 2x is positive?

    4. Find the slope of the line that is tangent to the curve y=2sinxcosx at x=pi/4.

    y'= -2sinxcosx

    At pi/4: sinx= 1/square root of 2, cosx= 1/square root of 2

    y'(pi/4) = -2(1/square root of 2)(1/square root of 2)
    =-1

    Book answer: 0

    Where did I go wrong?

    5. Find the equation of the line that is tangent to the curve y=2cos^3x at x=pi/3.

    y'= -6sin^2x

    At x=pi/3, sinx = (square root of 3) / 2

    y'(pi/3) = -6[(square root of 3)/2]^2

    I'm not really sure.. how would I work that?

    Book answer:



    THANKS!
    1. SIN2X=0 WHEN X=PI/2
    2. aRE YOU SURE YOU QUOTED THE QUESTION CORRECTLY?
    Y=2COSX+4COSX WOULD SIMPLY BE Y=6COSX
    3. SIN(3PI/2)=-1 SO Y=1/2 AND Y'=2
    4. yOU MUST USE THE PRODUCT RULE WHEN DIFFERENTIATING HERE GIVING Y'=2COS^2X-2SIN^2X ALTERNATIVELY WE CAN SAY Y=SIN2X SO Y'=2COS2X
    5. HERE YOU NEED THE CHAIN (FUNCTION OF A FUNCTION)( RULE GIVING Y'=-6COS^2XSINX
    Hope this helps
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  7. #7
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    Quote Originally Posted by kmjt View Post
    I've been doing some questions that I thought I was doing right but then I check the book answer and i'm wrong..:

    1. Find the slope of the function y=2cosxsin2x at x=pi/2.

    So first I differentiated:

    y' = (-2sinx)(sin2x) + (2cosx)(cos2x)(2)
    = -2sinxsin2x + 4cosxcos2x

    Then I determined that sin(pi/2) = 1 and that cos pi/2 = 0 because at 90 degrees, y=1, r=1, x=0.. and sin(pi/2) = y/r = 1/1 = 1.. and cos(pi/2) = x/r = 0/1 = 0 So I subbed these values into my derived equation:

    y'(pi/2) = -2(1)(2)(1) + 4(0)(0)(2)
    = -4

    My answer is the slope is -4, however the book answer says 0. Did I go wrong?
    sin(2x)\ne 2 sin(x) and cos(2x)\ne 2cos(x)!

    With x= \pi/2 2x= \pi and sin(2x)= 0, not 2. Similarly, cos(2x)= -1, not 0.
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  8. #8
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    For number one.. my derivative looks like -2sinxsin2x +4cosxcos2x, isn't sin(pi/2)=1? At 90 degrees, y=1, r=1.. and if sin is y/r = 1/1 = 1. How are you getting 0?

    For 2.. I made a mistake in the question I'm actually asking
    2. Find the equation of the line that is tangent to the curve y=2sinx+4cosx at x=pi/4.

    I've completed 3.

    I've completed 4.

    I think I completed 5.. is this correct what i've shown my work?:



    Did I do that right.. when you multiply [(3)(square root of 3)]/4 by pi/3, the 3s in numerator and denominator cancel each other out like that?
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  9. #9
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    for 1 the correct derivative is f'(x) = 2cos 2x cos x - sin 2x sin x. See my first comment for steps

    2 is a straight foward derivative 2 cos x - 4 sin x

    2 cos pi/4 - 4 sin pi/4 = (y - y(pi/4))/(x - pi/4)

    2 cos pi/4 - 4 sin pi/4 = (y - 2 sin (pi/4) - 4 cos (pi/4))/(x - pi/4)

    y = (2 cos pi/4 - 4 sin pi/4)(x - pi/4)/(2 sin (pi/4) + 4 cos (pi/4))
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