Question

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

• A. $-\frac{1}{t^2}$
• B. $-\frac{1}{2a{t}^3}$
• C. $\frac{1}{t^2}$
• D. $-\frac{a}{2t^2}$

Answer 1

The correct option is B $-\frac{1}{2a{t}^3}$

$x=a{t}^2\Rightarrow \frac{dx}{dt}=2at$.....(1)

And $y=2at\Rightarrow \frac{dy}{dt}=2a$

Thus $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{1}{t}$

$\therefore \frac{d^{2}y}{d{x}^2}=-\frac{1}{t^2}\cdot \frac{dt}{dx}$, use chain rule

$=-\frac{1}{t^2}\cdot \frac{1}{dx/dt}=-\frac{1}{t^2}\cdot \frac{1}{2at}=-\frac{1}{2a{t}^3}$