Unit tangent vector to a curve at a point

[memish]

Homework Statement

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

Homework Equations

none

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 I am not sure which part I went wrong on
help?

THANKS

 

[Dick]

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

 

[memish]

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

 

[memish]

ohhhh i have to use quotient rule?!

 

[memish]

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

 

[Dick]

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).

 

[Office_Shredder]

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