
#1
Sep2309, 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 



#2
Sep2309, 09:40 PM

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




#3
Sep2309, 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?




#4
Sep2309, 09:45 PM

P: 15

Unit tangent vector to a curve at a point
ohhhh i have to use quotient rule?!




#5
Sep2309, 09:47 PM

P: 15

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




#6
Sep2309, 09:48 PM

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#7
Sep2309, 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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