# Finding a vector given a tangent vector

## Homework Statement

Find a tangent vector r that satisfies r(0)= (e^(1),0) given T(t) = (-e^(cos(t)sin(t)),cos(t)), where t is an element of [0,2π]

## Homework Equations

Tangent vector T = r'(t)/(norm(r'(t))

## The Attempt at a Solution

I was thinking that r(t) = ∫r'(t), and that the norm of r(t) = 1; but i am having a hard time identifying a function compatible with the tangent vector that also has a norm of 1. I also attempted to find a value for t that would force the exponential aspect of -e (cos(t)sin(t)) to equal 1, while also allowing cos(t) = 0, but this did not work. Now i'm stuck second guessing myself.

LCKurtz
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## Homework Statement

Find a tangent vector r that satisfies r(0)= (e^(1),0) given T(t) = (-e^(cos(t)sin(t)),cos(t)), where t is an element of [0,2π]
This question doesn't make any sense to me. Your T isn't a unit vector in the first place.

Sorry, maybe i've omitted something important: for each (t), the function T provides a tangent vector to an assortment of curves. The curve r exists in this assortment; find r that satisfies r(0) = (e^1, 0).

LCKurtz
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Sorry, maybe i've omitted something important: for each (t), the function T provides a tangent vector to an assortment of curves. The curve r exists in this assortment; find r that satisfies r(0) = (e^1, 0).
Are you trying to find a vector function r(t) satisfying r(0) =<e,0> and such that r'(t) = T(t)? Is that it? And if so, are you sure the first component of T isn't suppsed to be $-\sin(t)e^{\cos t}$ instead of what you have written?

Are you trying to find a vector function r(t) satisfying r(0) =<e,0> and such that r'(t) = T(t)? Is that it? And if so, are you sure the first component of T isn't suppsed to be $-\sin(t)e^{\cos t}$ instead of what you have written?
Yes i am!

It is given in wolfram mathematica form T(t) = (-exp(cos(t))sin(t), cos(t)); so now that you think about it, i could see it being T(t) = -sin(t)e^(cos(t)), cos(t).

I apologize for my errors.

LCKurtz
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Yes i am!

It is given in wolfram mathematica form T(t) = (-exp(cos(t))sin(t), cos(t)); so now that you think about it, i could see it being T(t) = -sin(t)e^(cos(t)), cos(t).

I apologize for my errors.
And now that the problem is stated clearly, you see how to solve it, right?

And now that the problem is stated clearly, you see how to solve it, right?
Would i need to integrate it in order to get a function of 0, r(0) = (e,0)?

rdr = e^(cos(t))+constant, ∫rdr = sin(t)+constant; if t=0, the function turns into the desired form.

Is this on the right track?

LCKurtz
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Would i need to integrate it in order to get a function of 0, r(0) = (e,0)?

rdr = e^(cos(t))+constant, ∫rdr = sin(t)+constant; if t=0, the function turns into the desired form.

Is this on the right track?
Don't forget you are integrating a vector -- both components and each component gets its own constant. And your answer for r will be a vector. But, yes, it is that easy.