Question

If \[ x = a\cos^3\theta, \qquad y = a\sin^3\theta, \] then find \[ \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \, ? \] 

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

Hint:

For parametric forms, first compute \(\frac{dy}{dx}\), square it if needed, and then simplify using standard identities like: \[ 1+\tan^2\theta=\sec^2\theta \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We first calculate the derivative \(dy/dx\) using parametric differentiation. Then we evaluate the required radical expression.
Step 2: Key Formula or Approach:
1. \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\).
2. Identity: \(1 + \tan^2 \theta = \sec^2 \theta\).
Step 3: Detailed Explanation:
Calculate derivatives with respect to \(\theta\):
\(\frac{dx}{d\theta} = 3a\cos^2\theta (-\sin\theta) = -3a\cos^2\theta\sin\theta\).
\(\frac{dy}{d\theta} = 3a\sin^2\theta (\cos\theta) = 3a\sin^2\theta\cos\theta\).
Find \(dy/dx\):
\[ \frac{dy}{dx} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta \] Substitute into the expression:
\[ \sqrt{1 + (-\tan\theta)^2} = \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta \] Step 4: Final Answer:
The value is \(\sec \theta\).