If $x=a(\cos t+t\sin t)$ and $y=a(\sin t-t\cos t)$, then $\frac{d^{2}y}{d{x}^2}$ at $t=\pi /4$ is:

Solution

1. Given $x=a(\cos t+t\sin t)$ and $y=a(\sin t-t\cos t)$, we find the derivatives: $\frac{dx}{dt}=a(t\cos t)$ and $\frac{dy}{dt}=a(t\sin t)$.
2. The first derivative is $$\begin{array}{rl}\frac{dy}{dx}& =\frac{dy/dt}{dx/dt}\\ & =\frac{at\sin t}{at\cos t}\\ & =\tan t\end{array}$$.
3. Differentiating with respect to $x$, we get $$\begin{array}{rl}\frac{d^{2}y}{d{x}^2}& =\frac{d}{dt}(\tan t)\cdot \frac{dt}{dx}\\ & ={\sec }^{2}t\cdot \frac{1}{at\cos t}\\ & =\frac{1}{at{\cos }^{3}t}\end{array}$$.
4. At $t=\pi /4$, we have $\cos (\pi /4)=1/\sqrt{2}$, so $$\begin{array}{rl}\frac{d^{2}y}{d{x}^2}& =\frac{1}{a(\pi /4)(1/\sqrt{2}{)}^3}\\ & =\frac{1}{a(\pi /4)(1/2\sqrt{2})}\\ & =\frac{8\sqrt{2}}{a\pi }\end{array}$$.
5. Hence the answer is (A).