If the parametric equation of a curve is given by \( x=e^{t} \cos t \) and \( y=e^{t} \sin t \) then the tangent to the curve at the point \( \mathrm{t}=\frac{\pi}{4} \) makes the angle with the axis of x is
Step-by-step Solution:
Slope of tangent for a parametric curve
\[
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]
Given
\[
x=e^t\cos t,\qquad y=e^t\sin t
\]
Differentiate
\[
\frac{dx}{dt}=e^t(\cos t-\sin t),\qquad
\frac{dy}{dt}=e^t(\sin t+\cos t)
\]
\[
\Rightarrow
\frac{dy}{dx}
=\frac{\sin t+\cos t}{\cos t-\sin t}
\]
At (t=\frac{\pi}{4})
\[
\sin\frac{\pi}{4}=\cos\frac{\pi}{4}
\]
\[
\frac{dy}{dx}
=\frac{\sin t+\cos t}{\cos t-\sin t}
=\frac{\sqrt2}{0}
=\infty
\]
So the tangent is vertical.
Angle with x-axis
\[
\theta=\frac{\pi}{2}
\]