If the solve the problem
Question:

If $x=2 \cos t-\cos 2 t, y=2 \sin t-\sin 2 t$, find $\frac{d^{2} y}{d x^{2}}$ at $t=\frac{\pi}{2}$

Solution:

Formula:

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

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

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

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

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

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

Given: –

$x=2 \cos t-\cos 2 t$

$y=2 \sin t-\sin 2 t$

differentiating w.r.t t

$\frac{\mathrm{dy}}{\mathrm{dx}}=2(-\sin \mathrm{t})-2(-\sin 2 \mathrm{t})$

$\Rightarrow \frac{d y}{d t}=2 \cos t-2 \cos 2 t$

Dividing both

$\frac{d y}{d x}=\frac{2(\cos t-\cos 2 t)}{2(\sin 2 t-\sin t)}$

Differentiating w.r.t t

$\Rightarrow \frac{d \frac{d y}{d x}}{d t}=\frac{(\sin 2 t-\sin t)(-\sin t+2 \sin 2 t)-(\cos t-\cos 2 t)(2 \cos 2 t-\cos t)}{(\sin 2 t-\sin t)^{2}}$

Dividing

$\frac{d^{2} y}{d x^{2}}=\frac{(\sin 2 t-\sin t)(2 \sin t-\sin t)-(\cos t-\cos 2 t)(2 \cos 2 t-\cos t)}{2(\sin 2 t-\sin t)^{3}}$

Putting $t=\frac{\pi}{2}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1+2}{-2}=-\frac{3}{2}$

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