Question

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

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

Answer 1

Answer: (D)

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

We have

$x=a{t}^2$ and $y=2at$

$\frac{x}{y}=\frac{t}{2}\therefore y=2a\times \frac{2x}{y}⟹y^2=4ax$

Now differenciating wrt $x$

$2y\frac{dy}{dx}=4a\dots \dots \left(1\right)$

Let $\frac{dy}{dx}=y^{\prime }$ and $\frac{d^{2}y}{d{x}^2}=y^{\prime \prime }$

Differentiating (1) again wrt $x$, we get

$2y{y}^{\prime \prime }+2(y^{\prime }{)}^2=0$

$\Rightarrow y^{\prime \prime }=-\frac{(y^{\prime }{)}^2}{y}=-\frac{(2a/y{)}^2}{y}=-\frac{4a^2}{y^3}$

$\Rightarrow y^{\prime \prime }=-\frac{4a^2}{(2at{)}^3}=-\frac{1}{2a{t}^3}$