# How do you find the unit tanget vector to the curve ?

## Homework Statement

r(t) = <cosh(t), -sinh(t), t>
not given a value for t.

r'(t)
N(t)

## The Attempt at a Solution

T = r'(t)/distance[r'(t)]

how do you find unit tangent vector and the unit normal vector without being given a value for t?

r'(t) = <sinh(t),- cosh(t),t>

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your r'(t) is not right. The tangent vector is r'(t), an the unit tangent vector is r'(t)/||r'(t)||

oh and to answer your question, the answers are in terms of t. since you are not given a value of t(it is a variable so there is no set value), your answer is in terms of t. this allows for the answer that you get to be 'generalized'. this means that you can substitute any value of t you want into your equation and get the answer. it looks to me like you will be dealing with these t variables exclusively for the rest of your calculus career

so when i get the unit tangent vector it is <sinh(t)/(sinh(t)+cosh(t)),-cosh(t)/(sinh(t)+cosh(t)),0> how do i go about taking the derivative of that in order to find the unit normal vector?

i believe the derivative of cosh(x) is sinh(x) and vice versa, you should double check that. and what level of calculus are you at? there are different methods for finding the normal vector and i want to know which one will suit your level. this is mostly because i can only remember the higher level one because its more recent in my memory.

calc 3. the derivatives where given to me. and the equation we have been using for N(t) is T'/dist(T').
thank you

and using that formula you are still having issues? I don't quite understand that notation. is dist(T') = ||T'||? or the magnitude of T'

ideasrule
Homework Helper
It's supposed to be the magnitude of T'. T' is just the derivative of the tangent vector.