The tangent to the curve given by `x = e^(t) cos t y = e^(t) " sint at t " =(pi)/(4)` makes with x-axis an angle of

Options

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

Answer

To find the angle that the tangent to the curve makes with the x-axis at \( t = \frac{\pi}{4} \), we can follow these steps:

### Step 1: Find the parametric equations
The parametric equations given are:
\[ x = e^t \cos t \]
\[ y = e^t \sin t \]

### Step 2: Differentiate \( x \) and \( y \) with respect to \( t \)
We need to find \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \).

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 3: Find \( \frac{dy}{dx} \)
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t} \]

### Step 4: Evaluate \( \frac{dy}{dx} \) at \( t = \frac{\pi}{4} \)
Substituting \( t = \frac{\pi}{4} \):
\[ \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus,
\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}} = \frac{\frac{2}{\sqrt{2}}}{0} \]
This indicates that the slope is undefined, which means the tangent line is vertical.

### Step 5: Determine the angle with the x-axis
For a vertical line, the angle \( \theta \) with the x-axis is:
\[ \theta = \frac{\pi}{2} \text{ (or } 90^\circ\text{)} \]

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

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