A) \[-\frac{1}{{{t}^{2}}}\]
B) \[\frac{1}{2a{{t}^{3}}}\]
C) \[-\frac{1}{{{t}^{3}}}\]
D) \[-\frac{1}{2a{{t}^{3}}}\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2a}{2at}\] Þ \[\frac{dy}{dx}=\frac{1}{t}=\frac{2a}{y}\] Þ \[y\frac{dy}{dx}=2a\]Þ \[y\frac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \frac{dy}{dx} \right)}^{2}}=0\] \[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{-{{(dy/dx)}^{2}}}{y}=-\frac{1}{2a{{t}^{3}}}\].
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