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anonymous

  • 5 years ago

Need help finding the distance traveled by a particle that travels along these parametric curves: x=9-3cos^(2)(6t) y=5sin^(2)(6t) from −2pi less than or equal to t less than or equal to 3pi

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  1. anonymous
    • 5 years ago
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    I get \[\sqrt{4896}\int\limits_{-2\pi}^{3\pi}\left| \sin(6t)\cos(6t) \right|dt\]

  2. anonymous
    • 5 years ago
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    Did I set it up wrong?

  3. anonymous
    • 5 years ago
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    I would use the arc length formula here -> the integral of sqrt( 1 + (dy/dx)^2) from -2pi to 3pi

  4. anonymous
    • 5 years ago
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    Because it's parametric how would I do that?

  5. anonymous
    • 5 years ago
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    right, it needs to be modified a bit for parametric curves. It should be the integral of the square root of (dx/dt)^2 + (dy/dt)^2 from -2pi to 3pi

  6. anonymous
    • 5 years ago
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    L=\[\int\limits_{a}^{b}\sqrt{(dx/dt)^2 + (dy/dt)^2 dt}\]

  7. anonymous
    • 5 years ago
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    yes, like that

  8. anonymous
    • 5 years ago
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    Well, I get the previous answer I posted ^

  9. anonymous
    • 5 years ago
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    Idk, maybe my calculator is rounding it to an answer that doesn't work.

  10. anonymous
    • 5 years ago
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    I end up getting 351.5278214

  11. anonymous
    • 5 years ago
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    You should get the equation: integral from -2pi to 3pi of sqrt((6sin(12t))^2 + (30sin(12t))^2)

  12. anonymous
    • 5 years ago
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    I get 305.941

  13. anonymous
    • 5 years ago
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    Oh.. I kept getting something else. I have no idea. But Thanks, honestly.

  14. anonymous
    • 5 years ago
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    no problem

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