If x = `a cos^(3) theta, y = a sin ^(3) theta then sqrt(1+((dy)/(dx))^(2))`=?

Options

A. `sec theta`
B. `3 a sin theta cos theta`
C. `3 a tan theta`
D. `a sin theta cos theta`

Answer

To solve the problem, we need to find the expression \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \) given the parametric equations \( x = a \cos^3 \theta \) and \( y = a \sin^3 \theta \).

### Step-by-Step Solution:

1. **Differentiate \( x \) with respect to \( \theta \)**:
\[ x = a \cos^3 \theta \]
Using the chain rule, we differentiate:
\[ \frac{dx}{d\theta} = a \cdot 3 \cos^2 \theta \cdot (-\sin \theta) = -3a \cos^2 \theta \sin \theta \]

2. **Differentiate \( y \) with respect to \( \theta \)**:
\[ y = a \sin^3 \theta \]
Similarly, we differentiate:
\[ \frac{dy}{d\theta} = a \cdot 3 \sin^2 \theta \cdot \cos \theta = 3a \sin^2 \theta \cos \theta \]

3. **Find \( \frac{dy}{dx} \)**:
Using the chain rule, we can express \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{3a \sin^2 \theta \cos \theta}{-3a \cos^2 \theta \sin \theta} \]
Simplifying this expression:
\[ \frac{dy}{dx} = \frac{3 \sin^2 \theta \cos \theta}{-3 \cos^2 \theta \sin \theta} = -\frac{\sin \theta}{\cos \theta} = -\tan \theta \]

4. **Calculate \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \)**:
Now we substitute \( \frac{dy}{dx} \) into the expression:
\[ \sqrt{1 + \left(-\tan \theta\right)^2} = \sqrt{1 + \tan^2 \theta} \]
Using the identity \( 1 + \tan^2 \theta = \sec^2 \theta \):
\[ \sqrt{1 + \tan^2 \theta} = \sqrt{\sec^2 \theta} = \sec \theta \]

### Final Answer:
Thus, the value of \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \) is:
\[ \sec \theta \]