Parametric trig deriatives

**I-Think** · November 7th 2009, 07:27 PM

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

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

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

And from here I'm stuck. Assistance appreciated.

**tonio** · November 8th 2009, 03:41 AM

Originally Posted by I-Think

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

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

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

And from here I'm stuck. Assistance appreciated.

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

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

Tonio

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