Integrating vector functions

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  • #1
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Homework Statement


See figure


Homework Equations





The Attempt at a Solution



For an integral like this am I allowed to evaluate the cross product between the to vectors,

[tex]\vec{u}[/tex] and [tex]\vec{v}[/tex]?

If so then,

[tex]\vec{p} = \vec{u}\times\vec{v}[/tex]

and,

[tex]\int f\vec{p} = -3\int t\vec{p} [/tex]

Does this work?
 

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Answers and Replies

  • #2
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under what conditions should you think you cannot?
 
  • #3
HallsofIvy
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Yes, it is always true that [itex]\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)[/itex].

(But [itex]\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)[/itex].)
 
  • #4
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Yes, it is always true that [itex]\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)[/itex].

(But [itex]\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)[/itex].)

So,

[tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \int \left(-3t^{3} - 3t\right)dt\hat{i} + \int \left(-6t^{2} + 3t\right)dt\hat{j} + \int \left(6t + 3t^{4} \right)dt\hat{k} [/tex]

then,

[tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \left(\frac{-3}{4}t^{4} - \frac{3}{2}t^{2} \right)\hat{i} + \left( -2t^{3} + \frac{3}{2}t^{2} \right)\hat{j} + \left( 3t^{2} + \frac{3}{5}t^{5} \right)\hat{k} + C [/tex]

Is this correct?
 
  • #5
lanedance
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Yes, it is always true that [itex]\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)[/itex].

(But [itex]\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)[/itex].)

what about when a = 1...

may be missing somethig, but I'd say [itex]\vec{u}\times\left(a\vec{v}\right)= a\left(\vec{u}\times\vec{v}\right)[/itex], if you change the order of u & v thats another story
 
Last edited:
  • #6
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Bump, still looking for a quick check on my answer.
 
  • #7
HallsofIvy
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Yes, it is always true that [itex]\left(a\vec{u}\right)\times\vec{v}= a\left(\vec{u}\times\vec{v}\right)[/itex].

(But [itex]\vec{u}\times\left(a\vec{v}\right)= -a\left(\vec{u}\times\vec{v}\right)[/itex].)

what about when a = 1...

may be missing somethig, but I'd say [itex]\vec{u}\times\left(a\vec{v}\right)= a\left(\vec{u}\times\vec{v}\right)[/itex], if you change the order of u & v thats another story
Yes, you are right. For some reason, I was thinking [itex]\vec{u}\times(a\vec{v})= -((a\vec{v})\times\vec{u}= -a(\vec{v}\times\vec{u})[/itex] but now we have to swap u and v back again!

[itex](a\vec{u})\times\vec{v}= \vec{u}\times(a\vec{v})= a(\vec{u}\times\vec{v})[/itex]
 
  • #8
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Bump, still looking to see if the route I took is the right one.
 
  • #9
HallsofIvy
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What part of "yes", in the second response, do you have trouble with?
 
  • #10
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What part of "yes", in the second response, do you have trouble with?

I was just curious whether or not I had done the integration correctly, that's all.
 
  • #11
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So,

[tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \int \left(-3t^{3} - 3t\right)dt\hat{i} + \int \left(-6t^{2} + 3t\right)dt\hat{j} + \int \left(6t + 3t^{4} \right)dt\hat{k} [/tex]

then,

[tex]\int f\vec{p}dt = -3\int t\vec{p}dt = \left(\frac{-3}{4}t^{4} - \frac{3}{2}t^{2} \right)\hat{i} + \left( -2t^{3} + \frac{3}{2}t^{2} \right)\hat{j} + \left( 3t^{2} + \frac{3}{5}t^{5} \right)\hat{k} + C [/tex]

Is this correct?

Yep, that's right. Just make sure your C is a constant vector, not a real number constant.
 

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