Question

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)

Answer 1

wouldnt that be a caternary and not a parabola?

Answer 2

we can find the "parabola" by wrapping our curve around the 3 points established

Answer 3

(0,0) (-1.5,3) (1.5,3)

Answer 4

2.25a -1.5b = 3
2.25a +1.5b = 3
----------------
4.5a = 6

a = 6/4.5 = 60/45 = 20/15 = 4/3

Answer 5

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

Answer 6

b = 3-....yeah, i spose there aint a b lol...its centered at the orign.

Answer 7

you need to integrate it between [0,1.5] then double the results

Answer 8

otherwise you get a-a=0 right?

Answer 9

2 times that

Answer 10

\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

Answer 11

\[\int\limits_{0}^{1.5} \sqrt{1 + [f'(x)]^2}dx\] right? and double it

Answer 12

woops \[2\int\limits_{0}^{1.5}\sqrt{1-(64x^{2}/9)}\]

Answer 13

and make sure you got that + in there correctly instead of the -

Answer 14

r = sqrt(x^2 + y^2)

Answer 15

ok cool. i got 6.97 ft seem about right?

Answer 16

\[2\int\limits_{0}^{1.5}\sqrt{1+\frac{64x^2}{9}}dx\]

Answer 17

yeah, it should be a little over 6 ft...

Answer 18

i didnt do the calculations, but sounds about right :)

Answer 19

ok thanks! :)

Answer 20

youre welcome :)