Answer
$\displaystyle y=\frac{2}{\pi}x+1$
Work Step by Step
$\qquad\qquad\qquad\qquad\qquad\qquad$Find the point at which $t = 0$
$x|_{t=0}=e^0\sin(0)=1(0)=0$
$y|_{t=0}=e^{2(0)}=1$
Point: $(0,1)$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad$Now find $\displaystyle \frac{dy}{dx}$
$\displaystyle \frac{dy}{dt}=2e^{2t}$
$\displaystyle \frac{dx}{dt}=e^t[π\cos(πt)]+e^t\sin(πt)=e^t[\pi\cos(\pi t)+\sin(πt)]$
$\displaystyle \frac{dy}{dx}=\frac{2e^{2t}}{e^t[\pi\cos(\pi t)+\sin(πt)]}=\frac{2e^t}{\pi\cos(\pi t)+\sin(πt)}$
$\qquad\qquad\qquad\qquad$Now plug $t=0$ in for $t$ and find the slope of the line:
$\displaystyle \frac{dy}{dx}|_{t=0}=\frac{2e^0}{\pi\cos(0)+\sin(0)}=\frac{2(1)}{\pi(1)+0}=\frac{2}{\pi}$
$\qquad\qquad\qquad\qquad$Now use the point-slope form of a line to
$\qquad\qquad\qquad\qquad$find the equation tangent to the parametric curve.
$y-y_1=m(x_1-x)\quad$
$\displaystyle y-1=\frac{2}{π}(x-0)$
$\displaystyle y=\frac{2}{\pi}x+1$
