Posted by **Anon** on Thursday, July 31, 2014 at 7:32pm.

When a chain hangs under its own weight, its shape is catenary. The equation of a catenary with a vertex on the y-axis is y=e^x+e^-x. Find the length of the chain from x=-2 to x=2. Then use the same technique from Problem #8(Previous AP calc problem I posted) to find the length of the parabola with the same vertex and endpoints.

I go the first part that asks to find the length to be 12.225, but I don't understand what the next part is asking or how to do it!

- AP Calc -
**Damon**, Thursday, July 31, 2014 at 8:13pm
when x = 0, middle, y = e^0 + e^-0 = 2

when x = 2 or -2 (end) y = e^2 + 1/e^2

= 7.524

so parabola has vertex at (0, 2)

and goes through (-2, 7.524 ) and (-2,7.524)

we might as well move this down weo so the vertex is at the origin

vertex at (0,0)

and

(2, 5.524) and (-2, 5.524)

so form is

y = a x^2 + b x

b is 0 because symmetric about y axis

(even function)

y = a x^2

when x = 2, y = 5.524

5.524 = a (4)

a = 1.381

so our parabola is

y = 1.381 x^2

now use your technique (I do not know which you are using) to find the length of y = 1.381 x^2 from 0 to 2 and double that for -2 to + 2

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