Question

If $ x=a{{t}^{2}},y=2at, $ at, then $ \frac{{{d}^{2}}y}{d{{x}^{2}}} $ is equal to:

• A. $ -\frac{1}{{{t}^{2}}} $
• B. $ -\frac{1}{2a{{t}^{3}}} $
• C. $ \frac{1}{{{t}^{2}}} $
• D. $ -\frac{a}{2{{t}^{3}}} $
• E. zero

Answer 1

Answer: (B)

$ 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}}} $