Let the equation of a curve be `x=a(theta+sin theta),y=a(1-cos theta)`. If `theta` changes at a constant rate k then the rate of change of the slope of the tangent to the curve at `theta=pi/3` is (a) `(2k)/sqrt3` (b) `k/sqrt3` (c) k (d) none of these

Options

A. `2k//sqrt3`
B. `k//sqrt3`
C. k
D. none of these

Explanation

To solve the problem, we need to find the rate of change of the slope of the tangent to the curve defined by the parametric equations \( x = a(\theta + \sin \theta) \) and \( y = a(1 - \cos \theta) \) at \( \theta = \frac{\pi}{3} \), given that \( \theta \) changes at a constant rate \( k \).

### Step 1: Differentiate \( x \) and \( y \) with respect to \( t \)

The equations are:
\[ x = a(\theta + \sin \theta) \]
\[ y = a(1 - \cos \theta) \]

Differentiating both with respect to \( t \):
\[ \frac{dx}{dt} = a\left(\frac{d\theta}{dt} + \cos \theta \cdot \frac{d\theta}{dt}\right) = a\frac{d\theta}{dt}(1 + \cos \theta) \]
\[ \frac{dy}{dt} = a\left(0 + \sin \theta \cdot \frac{d\theta}{dt}\right) = a \sin \theta \cdot \frac{d\theta}{dt} \]

### Step 2: Find \( \frac{dy}{dx} \)

Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin \theta \cdot \frac{d\theta}{dt}}{a(1 + \cos \theta) \cdot \frac{d\theta}{dt}} = \frac{\sin \theta}{1 + \cos \theta} \]

### Step 3: Differentiate \( \frac{dy}{dx} \) with respect to \( t \)

Let \( m = \frac{dy}{dx} = \frac{\sin \theta}{1 + \cos \theta} \).

Now differentiate \( m \) with respect to \( t \):
\[ \frac{dm}{dt} = \frac{(1 + \cos \theta) \cos \theta \cdot \frac{d\theta}{dt} - \sin \theta \cdot (-\sin \theta) \cdot \frac{d\theta}{dt}}{(1 + \cos \theta)^2} \]

This simplifies to:
\[ \frac{dm}{dt} = \frac{(1 + \cos \theta) \cos \theta + \sin^2 \theta}{(1 + \cos \theta)^2} \cdot \frac{d\theta}{dt} \]

### Step 4: Substitute \( \theta = \frac{\pi}{3} \)

Now substitute \( \theta = \frac{\pi}{3} \) into the expression:
- \( \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} \)
- \( \cos \frac{\pi}{3} = \frac{1}{2} \)

Thus,
\[ \frac{dm}{dt} = \frac{(1 + \frac{1}{2}) \cdot \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2}{(1 + \frac{1}{2})^2} \cdot k \]
Calculating the components:
\[ = \frac{\left(\frac{3}{2}\right) \cdot \frac{1}{2} + \frac{3}{4}}{\left(\frac{3}{2}\right)^2} \cdot k \]
\[ = \frac{\frac{3}{4} + \frac{3}{4}}{\frac{9}{4}} \cdot k = \frac{\frac{6}{4}}{\frac{9}{4}} \cdot k = \frac{6}{9} \cdot k = \frac{2}{3} k \]

### Final Answer

The rate of change of the slope of the tangent to the curve at \( \theta = \frac{\pi}{3} \) is:
\[ \frac{2k}{3} \]

Since this value does not match any of the provided options, the answer is (d) none of these.