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Math Help - the equation of normal lline/tangent line

  1. #1
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    the equation of normal lline/tangent line

    the question is

    find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3

    how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ?????
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  2. #2
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    Quote Originally Posted by haebinpark View Post
    the question is

    find the equation of the normal line and tangent line to the graph of y = cos^2 x at x = Pi/3

    how do i do this question without knowing the equation (something like 2x^2 + 2y^2 = 9xy that i can diffrentiate or derivate) ?????
    You can differentiate \cos^2(x) directly using the chain rule:

    \frac{d}{dx} \cos^2(x) = -2 \cos(x) \sin(x)


    If you prefer to use an identity: \cos^2(x) = \frac{\cos(2x)}{2}+\frac{1}{2}

    \frac{d}{dx} \left[\frac{1}{2}\cos(2x)\right] = -\sin(2x)

    Note that \frac{d}{dx}\, 0.5 = 0

    These answers are equal as \sin(2x) = 2\sin(x) \cos(x)


    To find your gradient find f' \left(\frac{\pi}{3}\right)
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  3. #3
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    i found slope of tangent

    f'(pi/3) = -2(cos pi/3)(sin pi/3)
    = -2(1/2)(sqrt 3 /2)
    = - sqrt(3) / 2

    i also found the point (pi/3 , 1/4)

    then what do i do next?
    Last edited by haebinpark; February 17th 2010 at 10:57 AM.
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  4. #4
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    here's what i tried
    if i did something wrong, could you please point me out?
    because i still am not sure if i did it right or wrong... the answer doesn't seem to right...


    f(x)=y=cos^2 x
    f(pi/3)= (cos pi/3)^2 = (1/2)^2 = 1/4
    point: (pi/3 , 1/4) (x0,y0)


    f'(x)= -2(cosx)(sinx)
    f'(pi/3) = -2(1/2)(sqrt3/2) = -sqrt3 / 2 <-- slope of tangent


    y=m(x-x0)+y0
    y= -sqrt3/2 (x-pi/3)+1/4
    = -(sqrt3/2 x)+[(2sqrt3 pi + 3)/12] <-- eq. of tangent


    slope of normal line = -1/-sqrt3 / 2 = 2/sqrt3 = 2sqrt3/3
    y=m(x-x0)+y0
    y= 2sqrt3/3 (x-pi/3) + 1/4
    = 2sqrt3/3x - [(8sqrt3 + 9)/36] <--- eq. of normal line
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  5. #5
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    Quote Originally Posted by haebinpark View Post
    here's what i tried
    if i did something wrong, could you please point me out?
    because i still am not sure if i did it right or wrong... the answer doesn't seem to right...

    slope of normal line = -1/-sqrt3 / 2 = 2/sqrt3 = 2sqrt3/3
    y=m(x-x0)+y0
    y= 2sqrt3/3 (x-pi/3) + 1/4
    = 2sqrt3/3x - [(8sqrt3 + 9)/36] <--- eq. of normal line
    You made an error in the last step of this series of equations (what happened to the factor of pi?)
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  6. #6
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    = 2sqrt3/3x - [(8sqrt3 pi + 9)/36]

    is this what you're taking about?
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  7. #7
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    Quote Originally Posted by haebinpark View Post
    = 2sqrt3/3x - [(8sqrt3 pi + 9)/36]

    is this what you're taking about?
    Yes. Although the correct equation is actually

    y = \frac{2\sqrt{3}}{3}x + \frac{9 - 8\pi\sqrt{3}}{36}
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