Question

If $x=a{t}^2$ and $y=2at$, then $\frac{d^{2}y}{d{x}^2}$ at $t=2$, is

• A. $-\frac{1}{16a}$
• B. $\frac{1}{16a}$
• C. $\frac{1}{16}$
• D. $\frac{1}{a}$

Answer 1

Answer: (A)

$-\frac{1}{16a}$

We have, $x=a{t}^2$ and $y=2at$
$\frac{dx}{dt}=2at$ and $\frac{dy}{dt}=2a$
$\therefore \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{1}{t}$
Now, $\frac{d^{2}y}{d{x}^2}=-\frac{1}{t^2}\cdot \frac{dt}{dx}=-\frac{1}{t^2}\cdot \frac{1}{2at}=-\frac{1}{2a{t}^3}$
At $t=2,\frac{d^{2}y}{d{x}^2}=-\frac{1}{2a(2{)}^3}=-\frac{1}{16a}$

Hence, option A is correct.