# Unit tangent vector to a curve at a point

by memish
Tags: curve, point, tangent, unit, vector
 P: 15 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)= 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
 Sci Advisor HW Helper Thanks P: 24,975 The derivative of 1/(3t) is -(1/3)*t^(-2). Not -3t^(-2).
 P: 15 Ohhh ok... can you kinda explain that? Isn't 1/(3t) basically (3t)^(-1) and then you can use the power rule?
P: 15

## Unit tangent vector to a curve at a point

ohhhh i have to use quotient rule?!
 P: 15 ahhh yay i got it thanks!!! picking up calc againafter the summer sucks...
 Mentor P: 4,499 Yes, but then you need to use the chain rule $$\frac{d}{dt} (3t)^{-1} = -1*(3t)^{-2}*\frac{d}{dt}(3t) = -1*(3t)^{-2}*3$$