Question

If x = 2at, y = at², where a is a constant, then find \[\frac{d^2 y}{d x^2} \text { at }x = \frac{1}{2}\] ?

Answer 1

Here,

\[x = 2\text { at and y } = a t^2 \]
\[\text { Differentiating w . r . t . t, we get }\]
\[\frac{d x}{d t} = 2\text { a and} \frac{d y}{d t} = 2\text { at } \]
\[ \therefore \frac{d y}{d x} = \frac{2at}{2a} = t\]
\[\text { Differentiating again w . r . t . t, we get} \]
\[\frac{d^2 y}{d x^2} = 1 \times \frac{dt}{dx} = \frac{1}{2a}\]
\[\text { Now,} \left[ \frac{d^2 y}{d x^2} \right]_{x = \frac{1}{2}} = \frac{1}{2a}\]