anonymous
  • anonymous
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)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
wouldnt that be a caternary and not a parabola?
amistre64
  • amistre64
we can find the "parabola" by wrapping our curve around the 3 points established
amistre64
  • amistre64
(0,0) (-1.5,3) (1.5,3)

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amistre64
  • amistre64
2.25a -1.5b = 3 2.25a +1.5b = 3 ---------------- 4.5a = 6 a = 6/4.5 = 60/45 = 20/15 = 4/3
anonymous
  • anonymous
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
amistre64
  • amistre64
b = 3-....yeah, i spose there aint a b lol...its centered at the orign.
amistre64
  • amistre64
you need to integrate it between [0,1.5] then double the results
amistre64
  • amistre64
otherwise you get a-a=0 right?
anonymous
  • anonymous
2 times that
amistre64
  • amistre64
\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it
amistre64
  • amistre64
\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it
anonymous
  • anonymous
woops \[2\int\limits_{0}^{1.5}\sqrt{1-(64x^{2}/9)}\]
amistre64
  • amistre64
and make sure you got that + in there correctly instead of the -
amistre64
  • amistre64
r = sqrt(x^2 + y^2)
anonymous
  • anonymous
ok cool. i got 6.97 ft seem about right?
amistre64
  • amistre64
\[2\int\limits_{0}^{1.5}\sqrt{1+\frac{64x^2}{9}}dx\]
amistre64
  • amistre64
yeah, it should be a little over 6 ft...
amistre64
  • amistre64
i didnt do the calculations, but sounds about right :)
anonymous
  • anonymous
ok thanks! :)
amistre64
  • amistre64
youre welcome :)

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