If x=2-2 and y=2-2, [0,2], then d^2ydx^2 at = is:

Question

If $x=2\sin	heta-\sin2	heta$ and $y=2\cos	heta-\cos2	heta$, $	heta\in[0,2\pi]$, then $\dfrac{d^2y}{dx^2}$ at $	heta=\pi$ is:

Accepted Answer

$\dfrac{dx}{d\theta}=2\cos\theta-2\cos2\theta$, $\dfrac{dy}{d\theta}=-2\sin\theta+2\sin2\theta$.

$$\frac{dy}{dx}=\frac{\sin2\theta-\sin\theta}{\cos\theta-\cos2\theta}=\cot\frac{3\theta}{2}$$

$$\frac{d^2y}{dx^2}=-\frac32\csc^2\frac{3\theta}{2}\cdot\frac{1}{2(\cos\theta-\cos2\theta)}$$

At $\theta=\pi$: $\csc^2\left(\frac{3\pi}{2}\right)=1$, $\cos\pi-\cos2\pi=-1-1=-2$, so $\dfrac{d^2y}{dx^2}=-\dfrac{3/2}{2(-2)}=\dfrac38$.

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