Finding a vector given a tangent vector

[tsamocki]

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.

Homework Statement

 

[LCKurtz]

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.

 

[tsamocki]

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]

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?

 

[tsamocki]

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]

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?

 

[tsamocki]

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]

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.