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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)
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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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0wouldnt that be a caternary and not a parabola?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0we can find the "parabola" by wrapping our curve around the 3 points established

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(0,0) (1.5,3) (1.5,3)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02.25a 1.5b = 3 2.25a +1.5b = 3  4.5a = 6 a = 6/4.5 = 60/45 = 20/15 = 4/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0techinically 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
 5 years ago
Best ResponseYou've already chosen the best response.0b = 3....yeah, i spose there aint a b lol...its centered at the orign.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you need to integrate it between [0,1.5] then double the results

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0otherwise you get aa=0 right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0woops \[2\int\limits_{0}^{1.5}\sqrt{1(64x^{2}/9)}\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and make sure you got that + in there correctly instead of the 

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok cool. i got 6.97 ft seem about right?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[2\int\limits_{0}^{1.5}\sqrt{1+\frac{64x^2}{9}}dx\]

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, it should be a little over 6 ft...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt do the calculations, but sounds about right :)
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