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anonymous

  • 5 years ago

Hey everyone, need help finding the arc length of the curve y^2=x^3 from (0,0) to (1/4,1/8).

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  1. anonymous
    • 5 years ago
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    I know the formula and I think y'=(3x^2)/(2y)

  2. anonymous
    • 5 years ago
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    let me try this, see if it works. \[y=x^\frac{3}{2}\] yes?

  3. anonymous
    • 5 years ago
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    ok right, do we only take the +x^3/2?

  4. anonymous
    • 5 years ago
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    \[y'=\frac{3}{2}x^\frac{1}{2}\]

  5. anonymous
    • 5 years ago
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    y = x^(3/2) y' = 3/2*x^1/2 (y')^2 = 9/4*x integrate sqrt(1+9/4*x) from 1/8 to 1/4 use u-sub if you have to

  6. anonymous
    • 5 years ago
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    Rim would it not be from 0 to 1/4?

  7. anonymous
    • 5 years ago
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    oh sorry yes 0 to 1/4

  8. anonymous
    • 5 years ago
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    we are going from (0,0) to (1/8,1/4) so it is ok yes? maybe i am wrong, but i will graph it. in any case we have to integrate \[\int_0^{\frac{1}{8}} \sqrt{1-\frac{9}{4}x} dx\]

  9. anonymous
    • 5 years ago
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    oops you are right it is 1/4

  10. anonymous
    • 5 years ago
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    anti derivative is \[-\frac{(4-9x)^\frac{3}{2}}{27}\]

  11. anonymous
    • 5 years ago
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    and integral is \[\frac{8}{27}-\frac{7\sqrt{7}}{216}\]

  12. anonymous
    • 5 years ago
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    did you guys get .2824?

  13. anonymous
    • 5 years ago
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    let me get a calculator i will check

  14. anonymous
    • 5 years ago
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    no in fact i got .21055...

  15. anonymous
    • 5 years ago
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    ok, but isnt it also sqrt(1+(9/4)x)?

  16. anonymous
    • 5 years ago
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  17. anonymous
    • 5 years ago
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    yes it is .2824

  18. anonymous
    • 5 years ago
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    well i guess i messed up hold on

  19. anonymous
    • 5 years ago
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    you are right and i am wrong (0ften the case)

  20. anonymous
    • 5 years ago
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    take advantage of Wolfram alpha it is a very useful tool

  21. anonymous
    • 5 years ago
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    i put - where i should have had +. sorry

  22. anonymous
    • 5 years ago
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    no problem thanks for the help guys

  23. anonymous
    • 5 years ago
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    no problem

  24. anonymous
    • 5 years ago
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    Just why do we pick positive sqrt all the time?

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