Question

Find an equation of the tangent to the curve x=e^t sinπt,y=e^(2t) at the point where t=0

Answer 1

dy/dx = (dy/dt) / (dx/dt) = (2e^(2t))/(e^t(sinπt + πcosπt)) = 2e^t / (sinπt + πcosπt)

at t=0, dy/dx = 2/π
so now you have a point (1,1) and a slope, so the line is
y-1 = 2/π (x-1)