Tangent Vector for Vector Function?

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SUMMARY

The discussion focuses on finding the unit tangent vector for the vector function r(t) = <8√2t, e^(-8t), e^(8t)>. The user correctly identifies the derivative r'(t) = <8√2, -8e^(-8t), 8e^(8t)> but struggles with calculating the magnitude |r'(t)|. A critical error is noted in the calculation of e^(8t) squared, which should yield e^(16t) instead of e^(64t). The user ultimately resolves the issue by applying the identity (e^(8t) + e^(-8t))^2 = e^(16t) + 2 + e^(-16t).

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  • Knowledge of calculating magnitudes of vectors
  • Familiarity with exponential functions and their properties
  • Ability to simplify algebraic expressions involving exponentials
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Students studying calculus, particularly those focusing on vector functions and their applications. This discussion is beneficial for anyone needing assistance with derivatives and magnitudes in vector calculus.

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Tangent function vector?

Homework Statement



i know the formula for this problem. My homework was to like 5 of these and got the other 4. now stuck on the algebra part. please take a look and let me know where to go from here.
problem is
consider the vector function given below:
r(t) = <8sqrt(2)t, e^(-8t), e^(8t)> only the 2 in the x cordinate is square root
find the unit tangen vector

it will be r'(t) / |r'(t)|

derivative of r(t) = 8sqrt(2) , -8e^(-8t) , 8e^(8t)

then i get stuck on the magnitude of this. i know it is each component squared then square root of all them added togeather.
i get

128 + 64e^(-64t) + 64e^(64t)
then square root this whole thing.

im not sure how to simplify the whole e thing.
any help would be great!
thanks
 
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I didn't see it at first, but |r'(t)| is incorrect. When you square e8t you don't get e64t; you get e16t.

Also, you might want to keep in mind that (e8t + e-8t)2 = e16t + 2 + e-16t
 


ok thanks i got it now!
 

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