If the solve the problem


If $x=a(\cos 2 t+2 t \sin 2 t)$ and $y=a(\sin 2 t-2 t \cos 2 t)$, then find $\frac{d^{2} y}{d x^{2}}$


Formula: -

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$

(iii) $\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=-\cos \mathrm{x}$

(iv) $\frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}^{\mathrm{n}}=\mathrm{n} \mathrm{x}^{\mathrm{n}-1}$

(v) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d} \text { (wou) }}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

(vi) parameteric forms $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given: -

$x=a(\cos 2 t+2 t \sin 2 t)$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-2 \mathrm{a} \sin 2 \mathrm{t}+2 \mathrm{asin} 2 \mathrm{t}+4 \mathrm{atco} 2 \mathrm{t}=4 \mathrm{atcos} 2 \mathrm{t}$

and $y=a(\sin 2 t-2 t \cos 2 t)$

$\Rightarrow \frac{d y}{d t}=2 a \cos 2 t-2 a \cos 2 t+4 a t \sin 2 t=4 a t \sin 2 t$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{dx}} \times \frac{\mathrm{dy}}{\mathrm{dt}}=\frac{\sin 2 \mathrm{t}}{\cos 2 \mathrm{t}}=\tan 2 \mathrm{t}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}(\tan 2 \mathrm{t})}{\mathrm{dx}}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\sec ^{2} 2 t \frac{d(2 t)}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=2 \sec ^{2} 2 t \frac{d(t)}{d x}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{\sec ^{3} 2 t}{2 a}$

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