#### Question

If *x* = at^{2}, y = 2 at, then \[\frac{d^2 y}{d x^2} =\]

(a) \[- \frac{1}{t^2}\]

(b) \[\frac{1}{2 \ at^3}\]

(c) \[- \frac{1}{t^3}\]

(d) \[- \frac{1}{ 2 \ at^3}\]

#### Solution

(d) \[- \frac{1}{2 \ a t^3}\]

\[x = a t^2 \text { and y } = 2\text { at }\]

\[\text { Differentiating w . r . t . t, we get }\]

\[\frac{d x}{d t} = 2\text { at and } \frac{d y}{d t} = 2a\]

\[ \therefore \frac{d y}{d x} = \frac{2a}{2at} = \frac{1}{t}\]

\[\text{ Differentiating again w . r . t . t, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{- 1}{t^2}\frac{d t}{d x} = \frac{- 1}{2 \ a t^3}\]

Is there an error in this question or solution?

Solution for question: If X = At2, Y = 2 At, Then D 2 Y D X 2 = (A) − 1 T 2 concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Commerce), CBSE (Arts), PUC Karnataka Science, CBSE (Science)