Asked by Dom
Find an equation of the tangent to the curve x=e^t sinπt,y=e^(2t) at the point where t=0
Answers
Answered by
oobleck
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)
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)
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