Question

If \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \), then \( 1 + \left( \frac{dy}{dx} \right)^2 \) is:

• A. \( \tan \theta \)
• B. \( \tan^2 \theta \)
• C. \( 1 \)
• D. \( \sec^2 \theta \)
• E. \( \sec \theta \)

Hint:

The parametric equations $x = a \cos^3 \theta, y = a \sin^3 \theta$ represent an Astroid. For such curves, the derivative often simplifies to a simple power or ratio of $\tan \theta$.

Answer 1

The Correct Option is D

Concept: This is a parametric differentiation problem. We first find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \), then compute \( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \). Finally, we substitute this into the given expression and simplify using trigonometric identities.

Step 1: Differentiate \( x \) and \( y \) with respect to \( \theta \).
Given \( x = a \cos^3 \theta \): \[ \frac{dx}{d\theta} = a \cdot 3\cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \] Given \( y = a \sin^3 \theta \): \[ \frac{dy}{d\theta} = a \cdot 3\sin^2 \theta \cdot (\cos \theta) = 3a \sin^2 \theta \cos \theta \]

Step 2: Find \( \frac{dy}{dx} \).
\[ \frac{dy}{dx} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \] Canceling common terms: \[ \frac{dy}{dx} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

Step 3: Evaluate the expression \( 1 + \left( \frac{dy}{dx} \right)^2 \).
\[ 1 + (-\tan \theta)^2 = 1 + \tan^2 \theta \] Using the fundamental trigonometric identity \( 1 + \tan^2 \theta = \sec^2 \theta \): \[ = \sec^2 \theta \]