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Thread: Parametric trig deriatives

  1. #1
    Senior Member I-Think's Avatar
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    Parametric trig deriatives

    Given that$\displaystyle x=t-sint$ and $\displaystyle y=1-cost$ for $\displaystyle 0\leq{\theta}\geq2\pi$
    prove that
    $\displaystyle \frac{d^2y}{dx^2}=-\frac{1}{4}cosec^4(\frac{1}{2}t)$

    My working
    $\displaystyle x=t-sint$ $\displaystyle y=1-cost$
    $\displaystyle \frac{dx}{dt}=1-cost $ $\displaystyle \frac{dy}{dt}=sint$
    $\displaystyle \frac{d^2x}{dt^2}=sint$ $\displaystyle \frac{d^2y}{dt^2}=cost$

    $\displaystyle \frac{d^2y}{dx^2}=\frac{cost(1-cost)-sin^2t}{(1-cost)^3}=\frac{-1}{(1-cost)^2}$

    And from here I'm stuck. Assistance appreciated.
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  2. #2
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    Quote Originally Posted by I-Think View Post
    Given that$\displaystyle x=t-sint$ and $\displaystyle y=1-cost$ for $\displaystyle 0\leq{\theta}\geq2\pi$
    prove that
    $\displaystyle \frac{d^2y}{dx^2}=-\frac{1}{4}cosec^4(\frac{1}{2}t)$

    My working
    $\displaystyle x=t-sint$ $\displaystyle y=1-cost$
    $\displaystyle \frac{dx}{dt}=1-cost $ $\displaystyle \frac{dy}{dt}=sint$
    $\displaystyle \frac{d^2x}{dt^2}=sint$ $\displaystyle \frac{d^2y}{dt^2}=cost$

    $\displaystyle \frac{d^2y}{dx^2}=\frac{cost(1-cost)-sin^2t}{(1-cost)^3}=\frac{-1}{(1-cost)^2}$

    And from here I'm stuck. Assistance appreciated.

    You're almost there! As $\displaystyle \cos 2x=\cos^2x-\sin^2x$, we get that: $\displaystyle 1-\cos t=1-\left(\cos^2\frac{t}{2}-\sin^2\frac{t}{2}\right)=2\sin^2\frac{t}{2}$ , so:

    $\displaystyle \frac{-1}{(1-\cos t)^2}=\frac{-1}{4\sin^4\frac{t}{2}}=-\frac{1}{4}\csc^4\frac{t}{2}$

    Tonio
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