Question

If \( x = \frac{t^2}{1+t^5} \), \( y = \frac{2t^3}{1+t^5} \) and \( t \neq -1 \) is a parameter then \( \frac{dy}{dx} = \)

• A. \( \frac{2(3+2t^5)}{(2-3t^5)} \)
• B. \( \frac{2t(3-2t^5)}{(2-3t^5)} \)
• C. \( \frac{2t(3-2t^5)}{(2+3t^5)} \)
• D. \( \frac{2(3+2t^5)}{(2+3t^5)} \)

Hint:

When differentiating parametric equations where both denominators are the same, the squared denominator terms will usually cancel out in the final division step. You can sometimes ignore the denominator squared term during intermediate calculation to save time, focusing only on the numerator part \( v u' - u v' \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

This is a problem of parametric differentiation. We are given \( x \) and \( y \) as functions of a parameter \( t \). We need to find \( \frac{dy}{dx} \) using the formula \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \).
Step 2: Key Formula or Approach:

Quotient Rule: \( \frac{d}{dt}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2} \).
Step 3: Detailed Explanation:

Given \( x = \frac{t^2}{1+t^5} \). Differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{(1+t^5)(2t) - (t^2)(5t^4)}{(1+t^5)^2} \] \[ \frac{dx}{dt} = \frac{2t + 2t^6 - 5t^6}{(1+t^5)^2} = \frac{2t - 3t^6}{(1+t^5)^2} = \frac{t(2-3t^5)}{(1+t^5)^2} \] Given \( y = \frac{2t^3}{1+t^5} \). Differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt} = \frac{(1+t^5)(6t^2) - (2t^3)(5t^4)}{(1+t^5)^2} \] \[ \frac{dy}{dt} = \frac{6t^2 + 6t^7 - 10t^7}{(1+t^5)^2} = \frac{6t^2 - 4t^7}{(1+t^5)^2} = \frac{2t^2(3-2t^5)}{(1+t^5)^2} \] Now find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{2t^2(3-2t^5)}{(1+t^5)^2}}{\frac{t(2-3t^5)}{(1+t^5)^2}} \] The term \( (1+t^5)^2 \) cancels out. \[ \frac{dy}{dx} = \frac{2t^2(3-2t^5)}{t(2-3t^5)} \] Cancel one \( t \) from numerator and denominator: \[ \frac{dy}{dx} = \frac{2t(3-2t^5)}{2-3t^5} \]
Step 4: Final Answer:

The derivative is \( \frac{2t(3-2t^5)}{2-3t^5} \).