Parametric trig deriatives

• Nov 7th 2009, 07:27 PM
I-Think
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.
• Nov 8th 2009, 03:41 AM
tonio
Quote:

Originally Posted by I-Think
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