Q98 / 160 · AP EAMCET 2015 · Mathematics
If $x=a t^2$ and $y=2 a t$, then $\frac{d^2 y}{d x^2}$ at $t=\frac{1}{2}$ is
- A$-\frac{2}{a}$
- B$\frac{4}{a}$
- C$\frac{8}{a}$
- Correct answerD$-\frac{4}{a}$
Solution
We have, $x=a t^2$
$\Rightarrow \quad \frac{d x}{d t}=2 a t \quad \Rightarrow \quad \frac{d t}{d x}=\frac{1}{2 a t}$
and $y=2 a t \Rightarrow \frac{d y}{d t}=2 a$
$\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t}=\frac{1}{t}$
Differentiating w.r.t $x$, we get
$\begin{aligned}
& \text { Now, } \frac{d^2 y}{d x^2}=\frac{-1}{t^2} \cdot \frac{d t}{d x}=\frac{-1}{t^2} \cdot \frac{1}{(2 a t)} \quad\left[\because \frac{d t}{d x}=\frac{1}{2 a t}\right] \\
& =\frac{-1}{2 a t^3}
\end{aligned}$
At $\quad t=\frac{1}{2}\left(\frac{d^2 y}{d x^2}\right)_{t=\frac{1}{2}}=\frac{-1}{2 a \times \frac{1}{8}}=\frac{-4}{a}$
Hence, the correct answer is option: 'D'.