The length of the normal at t on the curve `x=a(t+sint), y=a(1-cos t),` is

Options

A. `a sin t`
B. `2a "sin"^(3)(t)/(2)"sec"(t)/(2)`
C. `2a "sin"(t)/(2)"tan"(t)/(2)`
D. `2a "sin"(t)/(2)`

Answer

To find the length of the normal at \( t \) on the curve defined by the parametric equations \( x = a(t + \sin t) \) and \( y = a(1 - \cos t) \), we will follow these steps:

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

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

We need to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

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

Calculating \( \frac{dy}{dt} \):
\[ \frac{dy}{dt} = a \sin t \]

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

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

### Step 3: Find the slope of the normal

The slope of the normal \( m \) is the negative reciprocal of \( \frac{dy}{dx} \):
\[ m = -\frac{1 + \cos t}{\sin t} \]

### Step 4: Use the formula for the length of the normal

The length of the normal \( L \) at a point on the curve is given by:
\[ L = y \sqrt{1 + m^2} \]

### Step 5: Substitute \( y \) and \( m \) into the formula

Substituting \( y = a(1 - \cos t) \) and \( m = -\frac{1 + \cos t}{\sin t} \):
\[ m^2 = \left(-\frac{1 + \cos t}{\sin t}\right)^2 = \frac{(1 + \cos t)^2}{\sin^2 t} \]

Now, substituting into the length formula:
\[ L = a(1 - \cos t) \sqrt{1 + \frac{(1 + \cos t)^2}{\sin^2 t}} \]

### Step 6: Simplify the expression under the square root

We know that:
\[ 1 + \frac{(1 + \cos t)^2}{\sin^2 t} = \frac{\sin^2 t + (1 + \cos t)^2}{\sin^2 t} \]

Expanding \( (1 + \cos t)^2 \):
\[ (1 + \cos t)^2 = 1 + 2\cos t + \cos^2 t \]

Thus:
\[ \sin^2 t + (1 + \cos t)^2 = \sin^2 t + 1 + 2\cos t + \cos^2 t = 2 + 2\cos t = 2(1 + \cos t) \]

So:
\[ 1 + \frac{(1 + \cos t)^2}{\sin^2 t} = \frac{2(1 + \cos t)}{\sin^2 t} \]

### Step 7: Final expression for the length of the normal

Now substituting back into the length formula:
\[ L = a(1 - \cos t) \sqrt{\frac{2(1 + \cos t)}{\sin^2 t}} = a(1 - \cos t) \cdot \frac{\sqrt{2(1 + \cos t)}}{\sin t} \]

### Conclusion

Thus, the length of the normal at \( t \) on the curve is:
\[ L = a(1 - \cos t) \cdot \frac{\sqrt{2(1 + \cos t)}}{\sin t} \]