Question:
Write the correct alternative in the following:
If $x=2 a t, y=a t^{2}$, where $a$ is a constant, then $\frac{d^{2} y}{d x^{2}}$ at $x=\frac{1}{2}$ is
A. $1 / 2 a$
B. 1
C. $2 a$
D. none of these
Solution:
Given:
$x=2 a t, y=a t^{2}$
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=t$
$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\mathrm{t}$
$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}{\frac{\mathrm{dx}}{\mathrm{dt}}}=\frac{1}{2 \mathrm{a}}$
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