# If the solve the problem

Question:

If $\mathrm{x}=\mathrm{a}\left(\cos \mathrm{t}+\log \tan \frac{\mathrm{t}}{2}\right), \mathrm{y}=\mathrm{a} \sin \mathrm{t}$, evaluate $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ at $\mathrm{t}=\frac{\pi}{3}$

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

(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 $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

Given: -

$x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=\sin t$

Differentiating with respect to $t$, we have

$\Rightarrow \frac{d x}{d t}=-a \sin t+a \frac{1}{\tan \frac{t}{2}} \times \sec ^{2}\left(\frac{t}{2}\right) \times \frac{1}{2}$

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

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

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

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

Now find the value of $\frac{d y}{d t}$

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

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

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

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

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{\sin t}{a \cos ^{4} t}$

$\left(\frac{d^{2} y}{d x^{2}}\right)_{t=\frac{\pi}{3}}=\frac{\sin \frac{\pi}{3}}{a \cos ^{4} \frac{\pi}{3}}=\frac{8 \sqrt{3}}{a}$