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

Options

A. 2a
B. `asqrt2`
C. `a//2`
D. `a//sqrt2`

Answer

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

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

Given:
- \( x = a(t + \sin t) \)
- \( y = a(1 - \cos t) \)

We differentiate both equations with respect to \( t \):

\[ \frac{dx}{dt} = a(1 + \cos t) \]

\[ \frac{dy}{dt} = a \sin t \]

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

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

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{a \sin t}{a(1 + \cos t)} = \frac{\sin t}{1 + \cos t} \]

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

Substituting \( t = \frac{\pi}{2} \):

\[ \sin\left(\frac{\pi}{2}\right) = 1 \quad \text{and} \quad \cos\left(\frac{\pi}{2}\right) = 0 \]

Thus,

\[ \frac{dy}{dx} \bigg|_{t = \frac{\pi}{2}} = \frac{1}{1 + 0} = 1 \]

### Step 4: Find the slope of the normal

The slope of the normal is the negative reciprocal of the slope of the tangent:

\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -1 \]

### Step 5: Find the coordinates of the point on the curve at \( t = \frac{\pi}{2} \)

Substituting \( t = \frac{\pi}{2} \) into the equations for \( x \) and \( y \):

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

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

So, the point is \( \left(a\left(\frac{\pi}{2} + 1\right), a\right) \).

### Step 6: Find the equation of the normal

Using the point-slope form of the line:

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

This simplifies to:

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

### Step 7: Find the length of the normal

The length of the normal \( L \) can be calculated using the formula:

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

Substituting \( y = a \) and \( \frac{dy}{dx} = 1 \):

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

### Final Answer

The length of the normal to the curve at \( t = \frac{\pi}{2} \) is:

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