Finding a vector given a tangent vector

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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]

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

 

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  • #2
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.
 
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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).
 
  • #4
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 [itex]-\sin(t)e^{\cos t}[/itex] instead of what you have written?
 
  • #5
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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 [itex]-\sin(t)e^{\cos t}[/itex] 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.:blushing:
 
  • #6
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.:blushing:
And now that the problem is stated clearly, you see how to solve it, right?
 
  • #7
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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?
 
  • #8
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.
 

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