Question

A curve is defined parametrically by x=e^(t) and y=2e^(-t). An equation of the tangent line to the curve at t=ln2 is

Answer 1

vector form of the line:\[\vec r(t)=<e^t,2e^{-t}>\]and the point is\[r(\ln2)=<2,1>\]now we need the derivative of the vector function\[\vec r'(t)=<e^t,-2e^{-t}>\]and the derivative at that point is\[r'(
\ln2)=<2,-1>\]the parametric formula for a line parallel to a vector u from a point is\[\vec v(t)<x_0,y_0>+t\vec u=<2,1>+t<2,-1>=<2+2t,1-t>\]with the components written individually we have\[x=2+2t\]\[y=1-t\]