Question

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

• A. $\frac{4\sqrt{3}}{a}$
• B. $\frac{\sqrt{3}}{a}$
• C. $\frac{8\sqrt{3}}{a}$
• D. $\frac{8\sqrt{3}}{5a}$

Answer 1

The correct option is C $\frac{8\sqrt{3}}{a}$

$x=a(cost+logtan\frac{t}{2})$

$1=a(-sint+\frac{se{c}^2\frac{t}{2}}{2tan\frac{t}{2}})\frac{dt}{dx}$

$\frac{dt}{dx}=\frac{1}{a(-sint+\frac{se{c}^2\frac{t}{2}}{2tan\frac{t}{2}})}$

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

$\frac{dy}{dx}=\frac{cost}{(-sint+\frac{se{c}^2\frac{t}{2}}{2tan\frac{t}{2}})}$

$\frac{dy}{dx}=\frac{cost}{(-sint+\frac{1}{2sin\frac{t}{2}cos\frac{t}{2}})}$

$\frac{dy}{dx}=\frac{cost}{(-sint+\frac{1}{sint})}$

$\frac{dy}{dx}=\frac{costsint}{1-si{n}^{2}t}$

$\frac{dy}{dx}=tant$

$\frac{d^{2}y}{d{x}^2}=se{c}^{2}t\frac{dt}{dx}$

At $t=\frac{\pi }{3}$

$\frac{d^{2}y}{d{x}^2}=\frac{4}{a(-sint+\frac{se{c}^2\frac{t}{2}}{2tan\frac{t}{2}})}=\frac{8\sqrt{3}}{a}$