# Get Answers to all your Questions

### Answers (1)

Answer:

$\frac{-3}{2}$

Hint:

You must know the derivative of cos and sin function

\begin{aligned} &x=2 \cos t-\cos 2 t \\ \end{aligned}

Given:

\begin{aligned} &y=2 \sin t-\sin 2 t , \text { find } \frac{d^{2} y}{d x^{2}} \text { at } t=\frac{\pi}{2} \end{aligned}

Solution:

\begin{aligned} &x=2 \cos t-\cos 2 t \\ &y=2 \sin t-\sin 2 t \\ &\frac{d x}{d t}=-2 \sin t+2 \sin 2 t \\ &\frac{d y}{d t}=2 \cos t-2 \cos 2 t \end{aligned}

\begin{aligned} &\text { Now, }\\ &\frac{d y}{d x}=\frac{2 \cos t-2 \cos 2 t}{-2 \sin t+2 \sin 2 t} \Rightarrow \frac{\cos t-\cos 2 t}{\sin 2 t-\sin t}\\ &=\frac{2 \sin \frac{3 t}{2} \sin \frac{t}{2}}{2 \cos \frac{3 t}{2} \sin \frac{t}{2}} \Rightarrow \tan \frac{3 t}{2} \end{aligned}

\begin{aligned} &\text { Therefore, }\\ &\frac{d^{2} y}{d x^{2}}=\sec ^{2} \frac{3 t}{2} \times \frac{3}{2} \times \frac{d t}{d x}\\ &=3 \sec ^{2} \frac{3 t}{2} \cdot \frac{1}{2 \sin 2 t-2 \sin t}\\ &\left.\frac{d^{2} y}{d x^{2}}\right]_{t=\frac{\pi}{2}}=\frac{-3}{2} \end{aligned}

View full answer

## Crack CUET with india's "Best Teachers"

• HD Video Lectures
• Unlimited Mock Tests
• Faculty Support