The tangent to the curve given by `x=e^t.cost, y=e^t.sin t` at `t=pi/4` makes with X-axis an angle:

Options

A. 0
B. `pi/4`
C. `pi/3`
D. `pi/2`

Answer

To find the angle that the tangent to the curve \( x = e^t \cos t \) and \( y = e^t \sin t \) at \( t = \frac{\pi}{4} \) makes with the X-axis, we will follow these steps:

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

Given:
- \( x = e^t \cos t \)
- \( y = e^t \sin t \)

We need to differentiate both \( x \) and \( y \) with respect to \( t \).

1. **Differentiate \( x \)**:
\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t + e^t (-\sin t) = e^t (\cos t - \sin t) \]

2. **Differentiate \( y \)**:
\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

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

Using the chain rule, the slope of the tangent line is given by:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t} \]

### Step 3: Evaluate the slope at \( t = \frac{\pi}{4} \)

Now, we substitute \( t = \frac{\pi}{4} \):
\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0} \]

### Step 4: Determine the angle with the X-axis

Since the slope \( \frac{dy}{dx} \) is undefined (infinity), this indicates that the tangent line is vertical. The angle \( \theta \) that a vertical line makes with the X-axis is:
\[ \theta = \frac{\pi}{2} \text{ radians or } 90^\circ \]

### Final Answer

The angle that the tangent to the curve at \( t = \frac{\pi}{4} \) makes with the X-axis is \( \frac{\pi}{2} \) radians.

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