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Thread: Find the force F = < 2x, e^y + z cos y,sin y > of a particle with line integrals

  1. #1

    Find the force F = < 2x, e^y + z cos y,sin y > of a particle with line integrals

    Given F = < 2x, e^y + z cos y,sin y >
    Find the work done by the force in moving a particle from P(1, 0, 1) to Q(1, 2, −3) along the curved path given by C : r(t) =< 1 + sin πt, 2 sin(πt/2), 1 − 4t >, 0 ≤ t ≤ 1.

    I tried plugging r into F but ended getting an extremely long and complex vector function to take the integral of. The hint said to think, so I assume I'm not supposed to brute force the integral. Is there any property of either F or r that will allow me to simplify my work?

  2. #2
    Elite Member
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    Please demonstrate.

    I guess that "n"-looking thing is a [tex]\pi[/tex].

    How sure are you that you are to evaluate it? Maybe the problem statement says just "set it up"?
    Last edited by tkhunny; 10-21-2017 at 05:07 PM.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  3. #3
    Quote Originally Posted by tkhunny View Post
    Please demonstrate.

    I guess that "n"-looking thing is a [tex]\pi[/tex].

    How sure are you that you are to evaluate it? Maybe the problem statement says just "set it up"?
    The problem says to evaluate it. However, if I find the curl of F, it is zero, so then the line integral would be zero?

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    That would work. I didn't get 0. I may have missed something.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

  5. #5
    Quote Originally Posted by tkhunny View Post
    That would work. I didn't get 0. I may have missed something.
    How did you set up your integral?

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    Elite Member stapel's Avatar
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    Cool

    Quote Originally Posted by Superyoshiom View Post
    How did you set up your integral?
    Please reply showing your work. Thank you!

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    Quote Originally Posted by Superyoshiom View Post
    How did you set up your integral?
    Yeah, that's not how this works. You first.
    "Unique Answers Don't Care How You Find Them." - Many may have said it, but I hear it most from me.

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