Question

If \(x = 4t^3+3,y = 3t^4+4\) and \(\frac{\frac{d^2x}{dy^2}}{(\frac{dx}{dy})^n}\) is constant then the value of \(n\) is

• A. \(1\)
• B. \(2\)
• C. \(6\)
• D. \(5\)

Hint:

Find dx/dy as a function of t, then differentiate again with respect to y.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Both \(x\) and \(y\) depend on \(t\), so we find \(\dfrac{dx}{dy}\) first and then differentiate it with respect to \(y\) using the chain rule.

Step 2: First derivative
\[ \frac{dx}{dt}=12t^2,\quad\frac{dy}{dt}=12t^3\Rightarrow\frac{dx}{dy}=\frac1t \]

Step 3: Second derivative
\[ \frac{d^2x}{dy^2}=\frac{d}{dt}\left(\frac1t\right)\cdot\frac{dt}{dy}=-\frac1{t^2}\cdot\frac1{12t^3}=-\frac1{12t^5} \]

Step 4: Form the ratio
\[ \frac{d^2x/dy^2}{(dx/dy)^n}=\frac{-1/(12t^5)}{t^{-n}}=-\frac{t^{n-5}}{12} \]

Step 5: Constant condition
The ratio does not depend on \(t\) only when \(n-5=0\), so \(n=5\). Option (D).

Final Answer:
The ratio equals -t^(n-5)/12, which is constant only for n = 5, option (D). \[ \boxed{5} \]