If the solve the problem

Question:

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t)$, then find the value of $\frac{d^{2} y}{d x^{2}}$ at $t=\frac{\pi}{4}$

Solution:

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 $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Given: -

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

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

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=$ atcost $+$ asint

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\mathrm{asin} \mathrm{t}+\mathrm{atcos} \mathrm{t}+\mathrm{asin} \mathrm{t}=\mathrm{atcost}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=-\mathrm{atsint}+\mathrm{acos} \mathrm{t}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{\frac{d x}{d t} \frac{d^{2} y}{d t^{2}}-\frac{d y}{d t} \frac{d^{2} x}{d t^{2}}}{\left(\frac{d x}{d t}\right)^{3}}$

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

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{1}{(\text { atcost })^{3}}$

Putting $t=\frac{\pi}{4}$

$\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{t}=\frac{\pi}{4}}=\frac{1}{\mathrm{a} \cos ^{3} \frac{\pi}{4} \cdot \mathrm{a} \frac{\pi}{4}}=\frac{8 \sqrt{2}}{\pi \mathrm{a}}$

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