Answer
$y = \frac{2}{\pi}x+1$
Work Step by Step
$x = e^t~sin~\pi t$
$y = e^{2t}$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2e^{2t}}{e^t~sin~\pi t+\pi~e^t~cos~\pi t} = \frac{2e^t}{sin~\pi t+\pi~cos~\pi t}$
When $t=0$:
$x = e^0~sin~(\pi \cdot 0) = 0$
$y = e^{(2)\cdot (0)} = 1$
$\frac{dy}{dx} = \frac{2e^0}{sin~(\pi\cdot 0 )+\pi~cos~(\pi\cdot 0)} = \frac{2}{0+\pi} = \frac{2}{\pi}$
We can find the equation of the tangent line:
$(y-1) = \frac{2}{\pi}(x-0)$
$y = \frac{2}{\pi}x+1$
