The distance, from the origin, of the normal to the curve, `x = 2 cos t + 2t sin t, y = 2 sin t - 2t "cos t at t" = (pi)/(4)`, is :

Options

A. 2
B. 4
C. `sqrt(2)`
D. `2 sqrt(2)`

Answer

To find the distance from the origin to the normal to the curve defined by the parametric equations \( x = 2 \cos t + 2t \sin t \) and \( y = 2 \sin t - 2t \cos t \) at \( t = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Calculate the coordinates of the point on the curve at \( t = \frac{\pi}{4} \).

Substituting \( t = \frac{\pi}{4} \) into the equations for \( x \) and \( y \):

\[ x = 2 \cos\left(\frac{\pi}{4}\right) + 2\left(\frac{\pi}{4}\right) \sin\left(\frac{\pi}{4}\right) = 2 \cdot \frac{1}{\sqrt{2}} + 2 \cdot \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} + \frac{\pi}{2\sqrt{2}} = \sqrt{2} + \frac{\pi}{2\sqrt{2}} = \frac{2 + \pi}{2\sqrt{2}} \]

\[ y = 2 \sin\left(\frac{\pi}{4}\right) - 2\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{4}\right) = 2 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{\pi}{2\sqrt{2}} = \sqrt{2} - \frac{\pi}{2\sqrt{2}} = \frac{2 - \pi}{2\sqrt{2}} \]

Thus, the point on the curve at \( t = \frac{\pi}{4} \) is:

\[ \left( \frac{2 + \pi}{2\sqrt{2}}, \frac{2 - \pi}{2\sqrt{2}} \right) \]

### Step 2: Calculate the derivatives \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

\[ \frac{dx}{dt} = -2 \sin t + 2t \cos t \]
\[ \frac{dy}{dt} = 2 \cos t - 2t \sin t \]

At \( t = \frac{\pi}{4} \):

\[ \frac{dx}{dt} = -2 \cdot \frac{1}{\sqrt{2}} + 2 \cdot \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = -\frac{2}{\sqrt{2}} + \frac{\pi}{2\sqrt{2}} = \frac{-2 + \pi}{2\sqrt{2}} \]

\[ \frac{dy}{dt} = 2 \cdot \frac{1}{\sqrt{2}} - 2 \cdot \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} - \frac{\pi}{2\sqrt{2}} = \frac{2 - \pi}{2\sqrt{2}} \]

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

The slope of the tangent is given by \( \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \).

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

\[ \text{slope of normal} = -\frac{\frac{dx}{dt}}{\frac{dy}{dt}} = -\frac{-\frac{2 - \pi}{2\sqrt{2}}}{\frac{-2 + \pi}{2\sqrt{2}}} = \frac{2 - \pi}{2 - \pi} = 1 \]

### Step 4: Write the equation of the normal line.

Using the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = 1 \), \( (x_1, y_1) = \left( \frac{2 + \pi}{2\sqrt{2}}, \frac{2 - \pi}{2\sqrt{2}} \right) \):

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

Rearranging gives:

\[ y = x + \frac{2 - \pi}{2\sqrt{2}} - \frac{2 + \pi}{2\sqrt{2}} = x - \frac{2 + \pi - 2 + \pi}{2\sqrt{2}} = x - \frac{2\pi}{2\sqrt{2}} = x - \frac{\pi}{\sqrt{2}} \]

### Step 5: Find the distance from the origin to the normal line.

The equation of the normal line is:

\[ y = x - \frac{\pi}{\sqrt{2}} \]

This can be rearranged to the standard form \( Ax + By + C = 0 \):

\[ x - y - \frac{\pi}{\sqrt{2}} = 0 \]

Here, \( A = 1, B = -1, C = -\frac{\pi}{\sqrt{2}} \).

The distance \( d \) from the origin \( (0, 0) \) to the line \( Ax + By + C = 0 \) is given by:

\[ d = \frac{|C|}{\sqrt{A^2 + B^2}} = \frac{\left| -\frac{\pi}{\sqrt{2}} \right|}{\sqrt{1^2 + (-1)^2}} = \frac{\frac{\pi}{\sqrt{2}}}{\sqrt{2}} = \frac{\pi}{2} \]

### Final Answer:

The distance from the origin to the normal to the curve at \( t = \frac{\pi}{4} \) is:

\[ \frac{\pi}{2} \]