The length of normal to the curve `x=a(theta+sin theta), y=a(1-cos theta),` at `theta=(pi)/(2)` is

Options

A. 2a
B. a unit from origin
C. `sqrt2a`
D. `2sqrt2a`

Answer

To find the length of the normal to the curve defined by the parametric equations \( x = a(\theta + \sin \theta) \) and \( y = a(1 - \cos \theta) \) at the point where \( \theta = \frac{\pi}{2} \), we can follow these steps:

### Step 1: Find \( \frac{dx}{d\theta} \) and \( \frac{dy}{d\theta} \)

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

We differentiate \( x \) and \( y \) with respect to \( \theta \):

\[ \frac{dx}{d\theta} = a(1 + \cos \theta) \]
\[ \frac{dy}{d\theta} = a \sin \theta \]

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

Using the chain rule, we can find \( \frac{dy}{dx} \) as follows:

\[ \frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{a \sin \theta}{a(1 + \cos \theta)} = \frac{\sin \theta}{1 + \cos \theta} \]

### Step 3: Evaluate \( \frac{dy}{dx} \) at \( \theta = \frac{\pi}{2} \)

Now, we substitute \( \theta = \frac{\pi}{2} \):

\[ \frac{dy}{dx} = \frac{\sin\left(\frac{\pi}{2}\right)}{1 + \cos\left(\frac{\pi}{2}\right)} = \frac{1}{1 + 0} = 1 \]

### Step 4: Find the coordinates \( (x, y) \) at \( \theta = \frac{\pi}{2} \)

Now, we calculate the coordinates of the point on the curve when \( \theta = \frac{\pi}{2} \):

\[ x = a\left(\frac{\pi}{2} + \sin\left(\frac{\pi}{2}\right)\right) = a\left(\frac{\pi}{2} + 1\right) \]
\[ y = a\left(1 - \cos\left(\frac{\pi}{2}\right)\right) = a(1 - 0) = a \]

### Step 5: Calculate the length of the normal

The length of the normal \( L \) at the point \( (x, y) \) is given by the formula:

\[ L = y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \]

Substituting the values we found:

\[ L = a \sqrt{1 + 1^2} = a \sqrt{2} \]

### Final Answer

Thus, the length of the normal to the curve at \( \theta = \frac{\pi}{2} \) is:

\[ \boxed{a \sqrt{2}} \]