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If $x=2 \cos t-\cos 2 t, \quad y=2 \sin t-\sin 2 t$, then
the value of $\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}$ is
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the value of $\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}$ is
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Verified Answer
The correct answer is:
$-3 / 2$
$\frac{d x}{d t}=-2 \sin t+2 \sin 2 t$
$$
\begin{aligned}
\frac{d y}{d t} &=2 \cos t-2 \cos 2 t \\
\therefore \quad \frac{d y}{d x} &=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \\
&=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t} \\
&=\frac{2 \sin \frac{3 t}{2} \cdot \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \cdot \sin \frac{t}{2}} \\
&=\tan \frac{3 t}{2}
\end{aligned}
$$
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\sec ^{2} \frac{3 t}{2} \cdot \frac{3}{2} \cdot \frac{d t}{d x} \\ &=\frac{3}{2} \sec ^{2} \frac{3 t}{2} \frac{1}{(2 \sin 2 t-2 \sin t)} \\ \Rightarrow \quad &\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}=-3 / 2 \end{aligned}$
$$
\begin{aligned}
\frac{d y}{d t} &=2 \cos t-2 \cos 2 t \\
\therefore \quad \frac{d y}{d x} &=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \\
&=\frac{\cos t-\cos 2 t}{\sin 2 t-\sin t} \\
&=\frac{2 \sin \frac{3 t}{2} \cdot \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \cdot \sin \frac{t}{2}} \\
&=\tan \frac{3 t}{2}
\end{aligned}
$$
$\begin{aligned} \frac{d^{2} y}{d x^{2}} &=\sec ^{2} \frac{3 t}{2} \cdot \frac{3}{2} \cdot \frac{d t}{d x} \\ &=\frac{3}{2} \sec ^{2} \frac{3 t}{2} \frac{1}{(2 \sin 2 t-2 \sin t)} \\ \Rightarrow \quad &\left.\frac{d^{2} y}{d x^{2}}\right|_{t=\pi / 2}=-3 / 2 \end{aligned}$
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