If the solve the problem

Question:

If $x=\cos t+\log \tan \frac{t}{2}, y=\sin t$, then find the value of $\frac{d^{2} y}{d t^{2}}$ and $\frac{d^{2} y}{d x^{2}}$ at $t=\frac{\pi}{4}$.

Solution:

Formula: -

(i) $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{y}_{1}$ and $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\mathrm{y}_{2}$

(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) $\frac{\mathrm{d}}{\mathrm{dx}} \log \mathrm{x}=\frac{1}{\mathrm{x}}$

(v) $\frac{d}{d x} \tan x=\sec ^{2} x$

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

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

(vi) parameteric forms $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

Given: -

$x=\cos t+\log \tan \frac{t}{2}, y=\sin t$

Differentiating with respect to $t$, we have

$\frac{\mathrm{dx}}{\mathrm{dt}}=-\sin \mathrm{t}+\frac{1}{\tan \frac{\mathrm{t}}{2}} \times \sec ^{2}\left(\frac{\mathrm{t}}{2}\right) \times \frac{1}{2}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin t+\frac{1}{\frac{\sin \left(\frac{\mathrm{t}}{2}\right)}{\cos \left(\frac{\mathrm{t}}{2}\right)}} \times \frac{1}{\cos ^{2} \frac{\mathrm{t}}{2}} \times \frac{1}{2}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin t+\frac{1}{2 \sin \left(\frac{\mathrm{t}}{2}\right) \cos \left(\frac{\mathrm{t}}{2}\right)}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=-\sin \mathrm{t}+\frac{1}{\sin \mathrm{t}}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{1-\sin ^{2} \mathrm{t}}{\sin \mathrm{t}}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\frac{\cos ^{2} \mathrm{t}}{\sin \mathrm{t}}$

$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=$ cost. cott

Now find the value of $\frac{\mathrm{dy}}{\mathrm{dt}}$

$\frac{\mathrm{dy}}{\mathrm{dt}}=\cos t$

Now

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

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=$ cost $\times \frac{1}{\text { cost. cott }}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\operatorname{tant}$

We have

$\frac{d y}{d t}=\cos t$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}=-\sin t$

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

$\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dt}^{2}}\right)_{\mathrm{t}=\frac{\pi}{4}}=-\sin \left(\frac{\pi}{4}\right)=-\frac{1}{\sqrt{2}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}(\tan \mathrm{t})}{\operatorname{cost} \cdot \operatorname{cott}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{2} \mathrm{t}}{\operatorname{cost} \cdot \operatorname{cott}}$

$\Rightarrow \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\sec ^{2} \mathrm{t}}{\cos ^{2} \mathrm{t}} \cdot \sin \mathrm{t}$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\sec ^{4} t \times \sin t$

Now putting $t=\frac{\pi}{4}$

$\left(\frac{d^{2} y}{d x^{2}}\right)_{t=\frac{\pi}{4}}=\sec ^{4} \frac{\pi}{4} \cdot \sin \left(\frac{\pi}{4}\right)=2$

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