Question

If $x=a(t-\sin t)$ and $y=a(1+\cos t)$, then the value of $y_2$ at $t=\frac{\pi }{2}$ is

• A. $a$
• B. $a^2$
• C. $\frac{1}{a}$
• D. $\frac{1}{a^2}$

Answer 1

Answer: (C)

$\frac{1}{a}$

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

$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\left(\frac{cost(cost-1)+sint\left(sint\right)}{(cost-1{)}^2}\right))\frac{1}{a(1-cost)}$

$=(cost-1{)}^2$

$\frac{d^{2}y}{d{x}^2}{|}_{t=\frac{\pi }{2}}=\frac{1}{a}$