If $a
eq 0, x=a(t+\sin t)$ and $y=a(1-\cos t)$, then $\frac{d^2 y}{d x^2}$ at $t=\frac{2 \pi}{3}$ is

Question

If $a 
eq 0, x=a(t+\sin t)$ and $y=a(1-\cos t)$, then $\frac{d^2 y}{d x^2}$ at $t=\frac{2 \pi}{3}$ is

Aicharya · Accepted Answer

Given,

Differentiate Eq. (i) w.r.t. $t$, we get $ \frac{d x}{d t}=a(1+\cos t) $ Differentiate Eq. (ii) w.r.t. $t$, we get $ \begin{aligned} & \frac{d y}{d x}=a \sin t \\ & \therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{a \sin t}{a(1+\cos t)}=\frac{\sin t}{(1+\cos t)} \\ & \text { Now, } \quad \frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d t}\left(\frac{d y}{d x}\right) \times \frac{d t}{d x} \\ & =\frac{d}{d t}\left(\frac{\sin t}{1+\cos t}\right) \times \frac{1}{a(1+\cos t)} \\ & =\frac{(1+\cos t) \cos t-\sin t(-\sin t)}{(1+\cos t)^2} \times \frac{1}{a(1+\cos t)} \\ & \frac{d^2 y}{d x^2}=\frac{(\cos t+1)}{a(1+\cos t)^3}=\frac{1}{a(1+\cos t)^2} \\ & \text { At } t=2 \pi / 3 \\ & \frac{d^2 y}{d x^2}=\frac{1}{a\left(\frac{1}{2}\right)^2}=\frac{4}{a} \\ & \end{aligned} $

If $a \neq 0, x=a(t+\sin t)$ and $y=a(1-\cos t)$, then $\frac{d^2 y}{d x^2}$ at $t=\frac{2 \pi}{3}$ is

Solution