## 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)

1. amistre64

wouldnt that be a caternary and not a parabola?

2. amistre64

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

3. amistre64

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

4. amistre64

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

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

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

7. amistre64

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

8. amistre64

otherwise you get a-a=0 right?

9. anonymous

2 times that

10. amistre64

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

11. amistre64

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

12. anonymous

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

13. amistre64

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

14. amistre64

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

15. anonymous

ok cool. i got 6.97 ft seem about right?

16. amistre64

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

17. amistre64

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

18. amistre64

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

19. anonymous

ok thanks! :)

20. amistre64

youre welcome :)