A) \[-\frac{1}{{{t}^{2}}}\]
B) \[-\frac{1}{2a{{t}^{3}}}\]
C) \[\frac{1}{{{t}^{2}}}\]
D) \[-\frac{a}{2{{t}^{3}}}\]
E) zero
Correct Answer: B
Solution :
\[x=a{{t}^{2}},y=2at\] On differentiating w.r.t. t respectively \[\Rightarrow \] \[\frac{dx}{dt}=2at,\frac{dy}{dt}=2a\] \[\therefore \] \[\frac{dy}{dx}=\frac{2a}{2at}=\frac{1}{t},\frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{1}{{{t}^{2}}}\frac{dt}{dx}\] \[=-\frac{1}{{{t}^{2}}}.\frac{1}{2at}=-\frac{1}{2a{{t}^{3}}}\]
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