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Determine the parametric equations of the tangent line to the curve r(t) =< cos 2t; sin 3t; sin 4t > when t = pi/4
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we want the time derivative of r(t) r'(t)=<-2sin(2t), 3cos(3t), 4cos(4t)> now find r'(pi/4) r'(pi/4)=<-2, -3sqrt(2)/2, -4> so we have a direction vector, now we want a point on the line. We can use r(pi/4) r(pi/4)=<0, sqrt(2)/2, 0> so the equation of the tangent line is T=<-2s, sqrt(2)/2-(3sqrt(2)/2)*s, -4s> where s is an arbitrary parameter.
I took the derivative but was unsure where to go from there, thanks!
in the last equation why the variables swapped?
so it's not confused with t, since it doesn't need to be the same variable.
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