Question

If $x=a{t}^2,y=2at,$ at, 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. $-\frac{a}{2t^3}$
• E. zero

Answer 1

Answer: (B)

$x=a{t}^2,y=2at$ On differentiating w.r.t. t respectively $\Rightarrow$ $\frac{dx}{dt}=2at,\frac{dy}{dt}=2a$ $\therefore$ $\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t},\frac{d^{2}y}{d{x}^2}=-\frac{1}{t^2}\frac{dt}{dx}$ $=-\frac{1}{t^2}.\frac{1}{2at}=-\frac{1}{2a{t}^3}$