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anonymous

  • 5 years ago

A person 6 ft tall wants to buy a jump rope. If, when the rope is at its lowest point, their hands will be 3 ft apart and 3 ft above the ground and if the rope will take on the shape of a parabola just barely hitting the ground, how long must the rope be? Consider the rope to be described by the equation y=ax^2 with the origin being a spot on the ground where the jump rope touches. arc length = sqrt(1+(dy/dx)^2)

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  1. amistre64
    • 5 years ago
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    wouldnt that be a caternary and not a parabola?

  2. amistre64
    • 5 years ago
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    we can find the "parabola" by wrapping our curve around the 3 points established

  3. amistre64
    • 5 years ago
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    (0,0) (-1.5,3) (1.5,3)

  4. amistre64
    • 5 years ago
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    2.25a -1.5b = 3 2.25a +1.5b = 3 ---------------- 4.5a = 6 a = 6/4.5 = 60/45 = 20/15 = 4/3

  5. anonymous
    • 5 years ago
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    techinically yes. but so far ive graphed it with the points (-1.5,3), (0,0) and (1.5,3). I've found a to be 4/3 and the equation y=(4/3)x^2. deriving that i got dy=(8/3)x dx and squaring that I get (64/9)x^2 I plugged that into the arc length equation and got something that did not match what I got on my calculator

  6. amistre64
    • 5 years ago
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    b = 3-....yeah, i spose there aint a b lol...its centered at the orign.

  7. amistre64
    • 5 years ago
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    you need to integrate it between [0,1.5] then double the results

  8. amistre64
    • 5 years ago
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    otherwise you get a-a=0 right?

  9. anonymous
    • 5 years ago
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    2 times that

  10. amistre64
    • 5 years ago
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    \[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

  11. amistre64
    • 5 years ago
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    \[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

  12. anonymous
    • 5 years ago
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    woops \[2\int\limits_{0}^{1.5}\sqrt{1-(64x^{2}/9)}\]

  13. amistre64
    • 5 years ago
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    and make sure you got that + in there correctly instead of the -

  14. amistre64
    • 5 years ago
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    r = sqrt(x^2 + y^2)

  15. anonymous
    • 5 years ago
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    ok cool. i got 6.97 ft seem about right?

  16. amistre64
    • 5 years ago
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    \[2\int\limits_{0}^{1.5}\sqrt{1+\frac{64x^2}{9}}dx\]

  17. amistre64
    • 5 years ago
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    yeah, it should be a little over 6 ft...

  18. amistre64
    • 5 years ago
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    i didnt do the calculations, but sounds about right :)

  19. anonymous
    • 5 years ago
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    ok thanks! :)

  20. amistre64
    • 5 years ago
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    youre welcome :)

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