Unit Tangent Vectors for Position Vectors: Finding T(t) for Given Values of t

AI Thread Summary
To find the unit tangent vector T(t) for the given position vectors, the derivative r'(t) must be calculated and then divided by its magnitude |r'(t)|. For r(t) = (cos(5t), sin(5t)), the user struggles with T(pi/4) and seeks clarification on their calculations. For r(t) = (t^2, t^3), the user reports obtaining (1, 1) for T(1) but is uncertain about the correctness of their results. The discussion highlights confusion over the steps involved in deriving the unit tangent vectors and the need for accurate calculations to avoid errors. Overall, the focus is on correctly applying the derivative and magnitude to find T(t) for various functions.
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positon vectors r(t) find the unit tangent vectors T(t) for the given value of t

r(t) = (cos5t, sin5t)
T(pi/4) = ( , )

r(t) = (t^2, t^3)
T(1) = ?

r(t) = e^5t i + e^-1t j + t k
T(2) = ? i+ ? j+ ? k

now the to find it i use r'(t)/lr'(t)l
I did that, but i get wrong answers i don't know what I am doing wrong because all i c is that is how u find the unit tangent vectors... someone please help me and explain what u did.. thanks alot!
 
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the derivative is the tangent vector.
then divide it by its magnitude to get unit tangent vector.
 
I did that and the damn computer say i got the wrong answers... like the 1st one i took it derivative then i divid it by its magnitude which is squaring everything and sq rt it right that's what i did..
 
give the answer that you have found for the first one.
 
well i worked out the second one i got (1, 1) i really I am doing something stupid
 
i get 2/sq rt of 13, 3/ sq rt of 13
what is wrong with your steps?
 

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