A) \[\frac{-1}{16a}\]
B) \[\frac{1}{16a}\]
C) \[\frac{1}{16}\]
D) \[\frac{1}{a}\]
Correct Answer: A
Solution :
We have, \[x=a{{t}^{2}}\]and \[y=2at\] \[\Rightarrow \] \[\frac{dx}{dt}=2at\]and \[\frac{dy}{dt}=2a\] \[\therefore \] \[\frac{dy}{dx}=\frac{\left( \frac{dy}{dx} \right)}{\left( \frac{dx}{dt} \right)}=\frac{1}{t}\] Now, \[\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{1}{{{t}^{2}}}.\frac{dt}{dx}=\frac{1}{{{t}^{2}}}.\frac{1}{2at}=-\frac{1}{2a{{t}^{3}}}\] At \[t=2,\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-1}{2a{{(2)}^{3}}}=-\frac{1}{16a}\]
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