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Unit tangent vector to a curve at a point

by memish
Tags: curve, point, tangent, unit, vector
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memish
#1
Sep23-09, 09:21 PM
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)=<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
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Dick
#2
Sep23-09, 09:40 PM
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The derivative of 1/(3t) is -(1/3)*t^(-2). Not -3t^(-2).
memish
#3
Sep23-09, 09:43 PM
P: 15
Ohhh ok... can you kinda explain that? Isn't 1/(3t) basically (3t)^(-1) and then you can use the power rule?

memish
#4
Sep23-09, 09:45 PM
P: 15
Unit tangent vector to a curve at a point

ohhhh i have to use quotient rule?!
memish
#5
Sep23-09, 09:47 PM
P: 15
ahhh yay i got it thanks!!! picking up calc againafter the summer sucks...
Dick
#6
Sep23-09, 09:48 PM
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Quote Quote by memish View Post
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
#7
Sep23-09, 09:49 PM
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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]


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