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: Write the given relations clearly.
We are given \[ x^3=\sin\theta \quad \text{and} \quad y^3=\cos\theta \] We need to find \[ x\frac{dy}{dx} \] Since both \( x \) and \( y \) are given in terms of the parameter \( \theta \), this is a parametric differentiation problem.

Step 2: Differentiate \( x^3=\sin\theta \) with respect to \( \theta \).
Differentiating both sides with respect to \( \theta \), we get \[ 3x^2\frac{dx}{d\theta}=\cos\theta \] Now using \[ \cos\theta=y^3 \] we obtain \[ 3x^2\frac{dx}{d\theta}=y^3 \] Hence, \[ \frac{dx}{d\theta}=\frac{y^3}{3x^2} \]

Step 3: Differentiate \( y^3=\cos\theta \) with respect to \( \theta \).
Differentiating both sides with respect to \( \theta \), we get \[ 3y^2\frac{dy}{d\theta}=-\sin\theta \] Now using \[ \sin\theta=x^3 \] we obtain \[ 3y^2\frac{dy}{d\theta}=-x^3 \] Therefore, \[ \frac{dy}{d\theta}=\frac{-x^3}{3y^2} \]

Step 4: Use the parametric derivative formula.
We know that \[ \frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] Substituting the values found above: \[ \frac{dy}{dx} = \frac{-x^3/(3y^2)}{y^3/(3x^2)} \] \[ = \frac{-x^3}{3y^2}\cdot\frac{3x^2}{y^3} \] \[ = -\frac{x^5}{y^5} \]

Step 5: Multiply by \( x \).
We need \[ x\frac{dy}{dx} \] So, \[ x\frac{dy}{dx}=x\left(-\frac{x^5}{y^5}\right) =-\frac{x^6}{y^5} \]

Step 6: Express \( x^6 \) in terms of \( y \).
From the given relations, \[ x^3=\sin\theta,\qquad y^3=\cos\theta \] So, \[ x^6=\sin^2\theta,\qquad y^6=\cos^2\theta \] Using the identity \[ \sin^2\theta+\cos^2\theta=1 \] we get \[ x^6+y^6=1 \] Thus, \[ x^6=1-y^6 \] Substitute into the expression: \[ x\frac{dy}{dx} = -\frac{1-y^6}{y^5} = \frac{y^6-1}{y^5} \]

Step 7: Final conclusion.
Therefore, \[ \boxed{x\frac{dy}{dx}=\frac{y^6-1}{y^5}} \] Hence, the correct option is \[ \boxed{(2)\ \dfrac{y^6-1}{y^5}} \]