1. The problem statement, all variables and given/known data Find the unit tangent vector T(t) to the curve r(t) at the point with the given value of the parameter, t. r(t)=<e^(2t), t^(-2), 1/(3t)> t=1 2. Relevant equations none 3. The attempt at a solution So first I took the derevative to get r'(t) which I got to be <2*e^(2x), -2t^(-3), -3t^(-2)> and plugged in the paramter, 1, so I got <2(e^2), -2,-3> and then I think I should divide that by its own magnitude, which I got to be the square root of (4(e^4) + 13) Buttt that's not working and Im not sure which part I went wrong on help? THANKS
Ohhh ok... can you kinda explain that? Isn't 1/(3t) basically (3t)^(-1) and then you can use the power rule?
Ok. d/dt (3t)^(-1)=(-1)*(3t)^(-2)*(d/dt (3t)). The last part comes from the chain rule. I think that's what your are forgetting. That's -3/(3t)^2=(-1)/(3t^2).
Yes, but then you need to use the chain rule [tex] \frac{d}{dt} (3t)^{-1} = -1*(3t)^{-2}*\frac{d}{dt}(3t) = -1*(3t)^{-2}*3[/tex]