Question

If \( x^3=\sin\theta,\ y^3=\cos\theta \), then \( x\dfrac{dy}{dx} \) is

• A. \( \dfrac{y^5-1}{y^5} \)
• B. \( \dfrac{y^6-1}{y^5} \)
• C. \( \dfrac{y^6-1}{y^6} \)
• D. \( \dfrac{y^3-1}{y^3} \)
• E. \( \dfrac{y^2-1}{y^2} \)

Hint:

In parametric differentiation, first find \( \dfrac{dx}{d\theta} \) and \( \dfrac{dy}{d\theta} \), then use \( \dfrac{dy}{dx}=\dfrac{dy/d\theta}{dx/d\theta} \). Also use identities like \( \sin^2\theta+\cos^2\theta=1 \) to simplify the final answer.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We are given parametric equations for \( x \) and \( y \) in terms of the parameter \( \theta \). To find \( \frac{dy}{dx} \), we use parametric differentiation. After finding the derivative, we need to express the result in terms of \( y \) as suggested by the options.
Step 2: Key Formula or Approach:
1. Use the formula for parametric differentiation: \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \).
2. Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \) by differentiating the given equations.
3. Compute the ratio to find \( \frac{dy}{dx} \).
4. Multiply the result by \( x \).
5. Use the fundamental trigonometric identity \( \sin^2\theta + \cos^2\theta = 1 \) to relate \( x \) and \( y \) and substitute to get the final answer in terms of \( y \).
Step 3: Detailed Explanation:
Differentiate the given equations with respect to \( \theta \):
For \( x^3 = \sin\theta \):
\[ 3x^2 \frac{dx}{d\theta} = \cos\theta \implies \frac{dx}{d\theta} = \frac{\cos\theta}{3x^2} \] For \( y^3 = \cos\theta \):
\[ 3y^2 \frac{dy}{d\theta} = -\sin\theta \implies \frac{dy}{d\theta} = \frac{-\sin\theta}{3y^2} \] Now, find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{-\sin\theta / 3y^2}{\cos\theta / 3x^2} = \frac{-\sin\theta}{3y^2} \cdot \frac{3x^2}{\cos\theta} = \frac{-x^2 \sin\theta}{y^2 \cos\theta} \] Substitute \( \sin\theta = x^3 \) and \( \cos\theta = y^3 \):
\[ \frac{dy}{dx} = \frac{-x^2 (x^3)}{y^2 (y^3)} = \frac{-x^5}{y^5} \] The question asks for \( x\frac{dy}{dx} \):
\[ x\frac{dy}{dx} = x \left( \frac{-x^5}{y^5} \right) = \frac{-x^6}{y^5} \] Now we need to express \( x^6 \) in terms of \( y \). We use the identity \( \sin^2\theta + \cos^2\theta = 1 \).
\[ (x^3)^2 + (y^3)^2 = 1 \] \[ x^6 + y^6 = 1 \implies x^6 = 1 - y^6 \] Substitute this into our expression for \( x\frac{dy}{dx} \):
\[ x\frac{dy}{dx} = \frac{-(1-y^6)}{y^5} = \frac{y^6-1}{y^5} \] Step 4: Final Answer:
The expression \( x\frac{dy}{dx} \) is equal to \( \frac{y^6-1}{y^5} \). This corresponds to option (B).