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

Concept:
For parametric curves, \[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} \] ip

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

Step 2: Find \(\frac{dy}{dx}\).
\[ \frac{dy}{dx} = \frac{3a\sin^2\theta\cos\theta}{-3a\cos^2\theta\sin\theta} = -\frac{\sin\theta}{\cos\theta} = -\tan\theta \] So, \[ \left(\frac{dy}{dx}\right)^2=\tan^2\theta \] ip

Step 3: Evaluate the required expression.
\[ \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \sqrt{1+\tan^2\theta} = \sqrt{\sec^2\theta} = \sec\theta \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \sec\theta} \]