If the parametric equation of a curve is given by $x=e^{t}cos~t$, $y=e^{t}sin~t$, then the tangent to the curve at the point $t=\frac{\pi}{4}$ makes with the axis of x the angle:

Question

Accepted Answer

To determine the angle that the tangent to the curve makes with the x-axis at the point $ t = \frac{\pi}{4} $, we need to find the derivative $\frac{dy}{dx}$ at this point. The parametric equations given are:

$$ x = e^t \cos t $$
$$ y = e^t \sin t $$

First, we find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$:

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

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

Now, the derivative $\frac{dy}{dx}$ is given by:

$$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t (\sin t + \cos t)}{e^t (\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t} $$

At $ t = \frac{\pi}{4} $, we have:

$$ \sin \frac{\pi}{4} = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

Substituting these values into the expression for $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0} $$

The expression $\frac{\sqrt{2}}{0}$ indicates that the slope is undefined, which means the tangent line is vertical. A vertical line makes an angle of $\frac{\pi}{2}$ radians with the x-axis.

Therefore, the correct answer is Option 3: $\frac{\pi}{2}$.

**Explanation for Incorrect Options:**

1. **Option 1 (0):** This would imply a horizontal tangent line, which occurs when the slope $\frac{dy}{dx} = 0$. However, in this case, the slope is undefined, indicating a vertical line, not horizontal.

2. **Option 2 ($\pi$):** An angle of $\pi$ radians would also imply a horizontal line, but in the opposite direction. Since the slope is undefined, the tangent is vertical, not horizontal.

3. **Option 4 ($\frac{\pi}{4}$):** This angle would imply a line with a slope of 1, which is not the case here. The slope is undefined, indicating a vertical line, not one with a slope of 1.

Thus, the only correct option is Option 3, where the tangent line is vertical, making an angle of $\frac{\pi}{2}$ with the x-axis.