# Unit tangent vector to a curve at a point

1. ### memish

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)=<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

2. ### Dick

25,810
The derivative of 1/(3t) is -(1/3)*t^(-2). Not -3t^(-2).

3. ### memish

15
Ohhh ok... can you kinda explain that? Isn't 1/(3t) basically (3t)^(-1) and then you can use the power rule?

4. ### memish

15
ohhhh i have to use quotient rule?!

5. ### memish

15
ahhh yay i got it thanks!!! picking up calc againafter the summer sucks...

6. ### Dick

25,810
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).

7. ### Office_Shredder

4,500
Staff Emeritus
Yes, but then you need to use the chain rule
$$\frac{d}{dt} (3t)^{-1} = -1*(3t)^{-2}*\frac{d}{dt}(3t) = -1*(3t)^{-2}*3$$