Question

If $x=a{t}^2$ and $y=2at$, what is $\frac{d^{2}y}{d{x}^2}$ at $t=\frac{1}{2}$?

• A. $-\frac{4}{a}$
• B. $\frac{a}{4}$
• C. $\frac{2}{a}$
• D. $-2a$

Answer 1

The correct option is A $-\frac{4}{a}$

$x=a{t}^2$

So, $\frac{dx}{dt}=2at$

$y=2at$

So, $\frac{dy}{dt}=2a$

Now,

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

$=\frac{2a}{2at}$

$\frac{dy}{dx}=\frac{1}{t}$ ............................ (1)

Differentiate equation (1) again w.r.t. $x$ ,

$\frac{d^{2}y}{d{x}^2}=-\frac{1}{t^2}\times \frac{dt}{dx}$

$=-\frac{1}{t^2}\times \frac{1}{2at}$

$=-\frac{1}{2a{t}^3}$

At $t=\frac{1}{2}$ ,

$\frac{d^{2}y}{d{x}^2}$$=-\frac{1}{2a{\left(\frac{1}{2}\right)}^3}$

$=-\frac{4}{a}$