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AmTran_Bus

  • one year ago

Find the arc length of ellipse x=4sintheta and y=3costheta

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  1. AmTran_Bus
    • one year ago
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    So I know you can say\[\int\limits_{a}^{b}\sqrt{1+(dy/dx)^2}\]

  2. AmTran_Bus
    • one year ago
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    and also \[\frac{ x }{ 4 }=\sin \theta , \frac{ y }{ 3 }= \cos \theta\]

  3. AmTran_Bus
    • one year ago
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    so (x/4)^2 +(y/3)^2 =1

  4. AmTran_Bus
    • one year ago
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    Taking the derivative, you got 4 cos theta and -3 sin theta

  5. AmTran_Bus
    • one year ago
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    @amistre64

  6. amistre64
    • one year ago
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    id use: \[\int \sqrt{(x')^2+(y')^2}~dt\]

  7. AmTran_Bus
    • one year ago
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    That isnt the formula in my book? How did you get it and why does it work? Just from the trig identity?

  8. amistre64
    • one year ago
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    its the general setup for the arclength "s", which is just the pythag thrm s, the section, squared; is equal to sum of the square of the parts |dw:1438546179447:dw| s^2 = x^2 + y^2, or simply s = sqrt(x^2 + y^2) as s and x and y get small ...

  9. amistre64
    • one year ago
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    |dw:1438546232411:dw|

  10. amistre64
    • one year ago
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    ds = sqrt(dx^2 + dy^2) now when all of these are "with respect to x", then dx=1 ds = sqrt(1 + (y')^2) is the initial introducion

  11. amistre64
    • one year ago
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    if it is with respect to "t", dx = x'

  12. AmTran_Bus
    • one year ago
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    Ok, well thank you then. So all I have is\[\int\limits_{0}^{2\pi}\sqrt{16\cos^2 \theta+9\sin^2 \theta }\]

  13. AmTran_Bus
    • one year ago
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    *d theta

  14. amistre64
    • one year ago
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    yep

  15. AmTran_Bus
    • one year ago
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    Ok. And that is an answer choice.

  16. AmTran_Bus
    • one year ago
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    Thanks.

  17. amistre64
    • one year ago
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    lets hope memory serves us right then lol

  18. AmTran_Bus
    • one year ago
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    Yea, for real. I apparently do not get this real well, cause its still fuzzy why every other example they have said use the (dy/dx)^2 thing.

  19. AmTran_Bus
    • one year ago
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    That is, the very first one I posted.

  20. amistre64
    • one year ago
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    we could define y in terms of x if you are working with parameters of course .. but its just extra work

  21. AmTran_Bus
    • one year ago
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    Ok.

  22. amistre64
    • one year ago
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    the thing about arclength is, we dont want to be confined to a coordinant system ...

  23. amistre64
    • one year ago
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    anyways, thats my perspective :)

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