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Question
If x = at2, y = 2 at, then \[\frac{d^2 y}{d x^2} =\]
Options
\[- \frac{1}{t^2}\]
\[\frac{1}{2 \ at^3}\]
\[- \frac{1}{t^3}\]
\[- \frac{1}{ 2 \ at^3}\]
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Solution
(d) \[- \frac{1}{2 \ a t^3}\]
\[x = a t^2 \text { and y } = 2\text { at }\]
\[\text { Differentiating w . r . t . t, we get }\]
\[\frac{d x}{d t} = 2\text { at and } \frac{d y}{d t} = 2a\]
\[ \therefore \frac{d y}{d x} = \frac{2a}{2at} = \frac{1}{t}\]
\[\text{ Differentiating again w . r . t . t, we get }\]
\[\frac{d^2 y}{d x^2} = \frac{- 1}{t^2}\frac{d t}{d x} = \frac{- 1}{2 \ a t^3}\]
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