## anonymous 4 years ago 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

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$