If the parametric equation of a curve is given by `x=e^tcost,y=e^tsint,` then the tangent to the curve at the point `t =pi//4` makes with the axis of x the angle

Options

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

Answer

To solve the problem, we need to find the angle that the tangent to the curve at the point where \( t = \frac{\pi}{4} \) makes with the x-axis. The parametric equations of the curve are given by:

\[ x = e^t \cos t \]
\[ y = e^t \sin t \]

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

We need to find \( \frac{dy}{dx} \) which can be computed using the chain rule:

\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

#### Finding \( \frac{dx}{dt} \):

Using the product rule:

\[ \frac{dx}{dt} = \frac{d}{dt}(e^t \cos t) = e^t \cos t - e^t \sin t = e^t (\cos t - \sin t) \]

#### Finding \( \frac{dy}{dt} \):

Using the product rule:

\[ \frac{dy}{dt} = \frac{d}{dt}(e^t \sin t) = e^t \sin t + e^t \cos t = e^t (\sin t + \cos t) \]

### Step 2: Evaluate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \) at \( t = \frac{\pi}{4} \)

#### For \( \frac{dx}{dt} \):

\[ \frac{dx}{dt} \bigg|_{t=\frac{\pi}{4}} = e^{\frac{\pi}{4}} \left(\cos\left(\frac{\pi}{4}\right) - \sin\left(\frac{\pi}{4}\right)\right) = e^{\frac{\pi}{4}} \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right) = 0 \]

#### For \( \frac{dy}{dt} \):

\[ \frac{dy}{dt} \bigg|_{t=\frac{\pi}{4}} = e^{\frac{\pi}{4}} \left(\sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right)\right) = e^{\frac{\pi}{4}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right) = e^{\frac{\pi}{4}} \cdot \sqrt{2} \]

### Step 3: Calculate \( \frac{dy}{dx} \)

Now we can find \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^{\frac{\pi}{4}} \cdot \sqrt{2}}{0} \]

Since \( \frac{dx}{dt} = 0 \), \( \frac{dy}{dx} \) is undefined, which indicates a vertical tangent line.

### Step 4: Determine the angle \( \theta \)

The slope \( m = \frac{dy}{dx} \) being undefined means the tangent line is vertical. The angle \( \theta \) that a vertical line makes with the positive x-axis is:

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

### Conclusion

Thus, the angle that the tangent to the curve at the point \( t = \frac{\pi}{4} \) makes with the x-axis is:

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