Question

The equations \( x = a(\theta + \sin \theta) \) and \( y = a(1 - \cos \theta) \) represent the equation of a curve. If \( \theta \) changes at a constant rate \( k \), then the rate of change of the slope of the tangent to the curve at \( \theta = \frac{\pi}{3} \) is

• A. 2k
• B. \( \frac{k}{3} \)
• C. \( \frac{2k}{\sqrt{3}} \)
• D. \( \frac{2k}{3} \)

Hint:

When differentiating a quotient, use the quotient rule: \( \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2} \).

Answer 1

The Correct Option is D

Step 1: Find the expressions for \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \).
The given equations are:
\( x = a(\theta + \sin \theta) \)
\( y = a(1 - \cos \theta) \)
Now, differentiate both equations with respect to \( \theta \):
\[ \frac{dx}{d\theta} = a(1 + \cos \theta) \] \[ \frac{dy}{d\theta} = a\sin \theta \]

Step 2: Find the slope of the tangent.
The slope of the tangent is given by:
\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta} \]

Step 3: Differentiate the slope with respect to \( \theta \).
Now, differentiate \( \frac{dy}{dx} \) with respect to \( \theta \):
\[ \frac{d}{d\theta} \left( \frac{\sin \theta}{1 + \cos \theta} \right) \]
Using the quotient rule:
\[ \frac{d}{d\theta} \left( \frac{\sin \theta}{1 + \cos \theta} \right) = \frac{(1 + \cos \theta) \cos \theta - \sin \theta(-\sin \theta)}{(1 + \cos \theta)^2} \]
Simplifying:
\[ \frac{d}{d\theta} \left( \frac{\sin \theta}{1 + \cos \theta} \right) = \frac{(1 + \cos \theta) \cos \theta + \sin^2 \theta}{(1 + \cos \theta)^2} \]
Thus, the rate of change of the slope of the tangent is:
\[ \frac{d}{d\theta} \left( \frac{\sin \theta}{1 + \cos \theta} \right) = \frac{2k}{3} \quad \text{(at \( \theta = \frac{\pi}{3} \))} \]

Step 4: Final Answer.
The rate of change of the slope of the tangent at \( \theta = \frac{\pi}{3} \) is \( \frac{2k}{3} \), which corresponds to option (D).