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theEric

  • 2 years ago

Calculus 3 - Find work done along curve in force field. How do I set up the integral? The curve is the upper half of a circle with radius 3, centered at the origin, from (-3,0) to (3,0).

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  1. theEric
    • 2 years ago
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    |dw:1367800601926:dw|

  2. theEric
    • 2 years ago
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    |dw:1367800740264:dw|

  3. theEric
    • 2 years ago
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    So I think it works nicely if I chose the parametric equations for the curve as\[x(t)=cos(t)\]\[y(t)=sin(t)\]and so\[\frac{dx(t)}{dt}=-sin(t)\]and\[\frac{dy(t)}{dt}=cos(t)\]

  4. theEric
    • 2 years ago
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    \[\text {So then the parameter, }t\text{, goes from }\pi \text{ to }0 \text{.}\] \[\text{Is it okay to write:} \int_\pi ^0 F(t)dt\text{, or must it be }-\int_0 ^ \pi F(t)dt \text{?}\] Don't they equal the same thing?

  5. Zarkon
    • 2 years ago
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    those two integrals are the same

  6. Zarkon
    • 2 years ago
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    you might want to look at your parametric equations again

  7. theEric
    • 2 years ago
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    \[\text{Thanks.. I've considered rewriting them to go from }0\text{ to }\pi \text{. In which case:}\] \[x(t)=-cos(t)\]\[y(t)=sin(t)\] Is this what you mean? Or were my original equations incorrect?

  8. Zarkon
    • 2 years ago
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    neither are correct

  9. Zarkon
    • 2 years ago
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    "circle with radius 3"

  10. theEric
    • 2 years ago
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    Oh... Yes.. Thank you! I tend to not carry all of the math into the next step. I tend to carry over my immediate thoughts and stop there...

  11. theEric
    • 2 years ago
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    \[x(t)=3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t \text{ goes from }0\text{ to }\pi \text{.}\] or\[x(t)=-3cos(t)\]\[y(t)=3sin(t)\]\[\text{where }t\text{ goes from }0 \text{ to } \pi \text{.}\]

  12. theEric
    • 2 years ago
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    The first one is from \[\pi\] to 0, I mean.

  13. Zarkon
    • 2 years ago
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    ok

  14. theEric
    • 2 years ago
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    Those work, then?

  15. theEric
    • 2 years ago
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    And so they're the same?

  16. Zarkon
    • 2 years ago
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    What is your force field \(F\)?

  17. Zarkon
    • 2 years ago
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    \[W=\int\limits_{a}^{b}F(c(t))\cdot c'(t) dt\]

  18. Zarkon
    • 2 years ago
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    \(c(t)\) is the particular path

  19. theEric
    • 2 years ago
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    To continue the problem I'm given,\[F(x,y)=(2y+x^2)i + (x^2 - 2x)j\]

  20. theEric
    • 2 years ago
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    yielding... \[W=-\int _0 ^\pi ((6sin(t)+9cos(t))(-3sint)+(9cos^2(t)-6cos(t))(3cos(t)))dt\]

  21. theEric
    • 2 years ago
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    Which is ugly.. I've simplified it a lot, and I think I've solved it, but I'm not sure. I'll just check with Wolfram Alpha. Thank you for clearing up the integration!

  22. Zarkon
    • 2 years ago
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    np

  23. theEric
    • 2 years ago
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    Take care!

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