AP Calc

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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 -

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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