Question

If $x = \frac{3t}{1+t^3}$ and $y = \frac{3t^2}{1+t^3}$, then $\frac{dy}{dx}$ at $t=1$ equals

• A. $-6$
• B. $-1$
• C. $1$
• D. $6$
• E. $4$

Hint:

In parametric problems, always simplify before substituting values.

Answer 1

The Correct Option is B

Step 1: Use parametric differentiation. \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \]

Step 2: Differentiate $x$. \[ x = \frac{3t}{1+t^3} \] Using quotient rule: \[ \frac{dx}{dt} = \frac{3(1+t^3) - 3t(3t^2)}{(1+t^3)^2} \] \[ = \frac{3 + 3t^3 - 9t^3}{(1+t^3)^2} = \frac{3 - 6t^3}{(1+t^3)^2} \]

Step 3: Differentiate $y$. \[ y = \frac{3t^2}{1+t^3} \] \[ \frac{dy}{dt} = \frac{6t(1+t^3) - 3t^2(3t^2)}{(1+t^3)^2} \] \[ = \frac{6t + 6t^4 - 9t^4}{(1+t^3)^2} = \frac{6t - 3t^4}{(1+t^3)^2} \]

Step 4: Compute $\frac{dy{dx}$.} \[ \frac{dy}{dx} = \frac{6t - 3t^4}{3 - 6t^3} \]

Step 5: Substitute $t=1$. \[ \frac{dy}{dx} = \frac{6 - 3}{3 - 6} = \frac{3}{-3} = -1 \] \[ \boxed{-1} \]