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Math Help - questions about differentiation of trigo

  1. #1
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    questions about differentiation of trigo

    1.
    Find the equation of the tangent and normal to the curve y= 1 + sin 2x at the point x= \frac{\pi}{12}.


    2.
    The volume of a cone V cm^3 is given by the formula V= kr^3 cot \theta , where k is constant, r is the radius of the cone and \theta is a semi vertical angle. Given that r is fixed and \theta is increasing at a rate of 0.2 radian per second, calculate the rate of change of volume with respect to time when \theta = \frac{\pi}{6}.


    ================
    i think i know how to do, but the problem is my answer is not the same as the book's answer. Can someone please tell me where i went wrong ??

    1.
    \frac{dy}{dx} = -2 cos 2x
    When x= \frac{\pi}{12}, \frac{dy}{dx} = -\sqrt3
    When x= \frac{\pi}{12}, y= \frac{3}{2}

    equation of tangent :
    y - \frac{3}{2} = - \sqrt3 (x - \frac{\pi}{12} )
    2y -3 = -2 \sqrt3 x + \frac{2 \sqrt3 \pi}{12}
    2y= -2 \sqrt3 x + 3 + \frac{\sqrt3 \pi}{6}

    the book's answer is 2y= 2 \sqrt3 x +3 - \frac{\sqrt3 \pi}{6}
    so my answer is different from the book's answer in terms of the positive sign and negative sign .....

    gradient of normal= \frac{1}{\sqrt3}

    equation of normal:
    y- \frac{3}{2} = \frac{1}{\sqrt3} (x - \frac{\pi}{12})
    2y-3 = \frac{2}{\sqrt3} x - \frac{2\pi}{12 \sqrt3}
    2y= \frac{2}{\sqrt3} x + 3 - \frac{\pi}{6 \sqrt3}
    2 \sqrt3 y - 2x = 3 \sqrt3 - \frac{\pi}{6}

    the book's answer is 2 \sqrt3 y + 2x = 3 \sqrt3 + \frac{\pi}{6}

    ======================

    2.
    V = kr^3 cot \theta
    = kr^3 (tan \theta)^{-1}

    \frac{dv}{d \theta}
    =  - kr^3 (tan \theta)^{-2} sec^2 \theta
    =  - kr^3 sec^4 \theta

    \frac{d \theta}{dt} = 0.2

    Find \frac{dV}{dt} when \theta = \frac{\pi}{6} .

    \frac{dV}{dt} = \frac{d \theta}{dt} * \frac{dV}{d \theta}
    = 0.2 * ( - kr^3 sec^4 \frac{\pi}{6})
    = ( - \frac{kr^3}{5})(\frac{1}{tan \frac{\pi}{6}}^4
    = ( - \frac{kr^3}{5})(\frac{3}{\sqrt3} )^4
    =( - \frac{kr^3}{5})( \frac{81}{9})
    =  - \frac{9kr^3}{5}

    the book's answer is - \frac{4kr^3}{5}.....

    Can someone please tell me where i went wrong ?? Thanks.
    Last edited by wintersoltice; March 11th 2009 at 06:13 AM.
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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    1.
    Find the equation of the tangent and normal to the curve y= 1 + sin 2x at the point x= \frac{\pi}{12}.


    2.
    The volume of a cone V cm^3 is given by the formula V= kr^3 cot \theta , where k is constant, r is the radius of the cone and \theta is a semi vertical angle. Given that r is fixed and \theta is increasing at a rate of 0.2 radian per second, calculate the rate of change of volume with respect to time when \theta = \frac{\pi}{6}.


    ================
    i think i know how to do, but the problem is my answer is not the same as the book's answer. Can someone please tell me where i went wrong ??

    1.
    \frac{dy}{dx} = -2 cos 2x
    When x= \frac{\pi}{12}, \frac{dy}{dx} = -\sqrt3
    When x= \frac{\pi}{12}, y= \frac{3}{2}

    equation of tangent :
    y - \frac{3}{2} = - \sqrt3 (x - \frac{\pi}{12} )
    2y -3 = -2 \sqrt3 x + \frac{2 \sqrt3 \pi}{12}
    2y= -2 \sqrt3 x + 3 + \frac{\sqrt3 \pi}{6}

    the book's answer is 2y= 2 \sqrt3 x +3 - \frac{\sqrt3 \pi}{6}
    so my answer is different from the book's answer in terms of the positive sign and negative sign .....

    gradient of normal= \frac{1}{\sqrt3}

    equation of normal:
    y- \frac{3}{2} = \frac{1}{\sqrt3} (x - \frac{\pi}{12})
    2y-3 = \frac{2}{\sqrt3} x - \frac{2\pi}{12 \sqrt3}
    2y= \frac{2}{\sqrt3} x + 3 - \frac{\pi}{6 \sqrt3}
    2 \sqrt3 y - 2x = 3 \sqrt3 - \frac{\pi}{6}

    the book's answer is 2 \sqrt3 y + 2x = 3 \sqrt3 + \frac{\pi}{6}
    y= 1 + sin 2x
    \therefore \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \cos 2x

    If you are differentiating \cos x then you get -\sin x (Negative) but if you are differentiating \sin x then you get \cos x (Positive).

    Quote Originally Posted by wintersoltice View Post
    ======================

    2.
    V = kr^3 cot \theta
    = kr^3 (tan \theta)^-1

    \frac{dv}{d \theta}
    =  - kr^3 (tan \theta)^(-2) sec^2 \theta
    =  - kr^3 sec^4 \theta

    \frac{d \theta}{dt} = 0.2

    Find \frac{dV}{dt} when \theta = \frac{\pi}{6} .

    \frac{dV}{dt} = \frac{d \theta}{dt} * \frac{dV}{d \theta}
    = 0.2 * ( - kr^3 sec^4 \frac{\pi}{6})
    = ( - \frac{kr^3}{5})(\frac{1}{tan \frac{\pi}{6}}^4
    = ( - \frac{kr^3}{5})(\frac{3}{\sqrt3} )^4
    =( - \frac{kr^3}{5})( \frac{81}{9})
    =  - \frac{9kr^3}{5}

    the book's answer is - \frac{4kr^3}{5}.....

    Can someone please tell me where i went wrong ?? Thanks.
    You correctly wrote that \cot x = \frac{1}{\tan x} but further down you changed your \frac{1}{\tan ^2 x} to \sec ^2 x which is incorrect.

    \frac{\mathrm{d}v}{\mathrm{d} \theta} = - kr^3 (\tan ^{-2} \theta)\sec ^2 \theta = -\frac{kr^3}{\tan ^2\theta \cos ^2 \theta}
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