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August 22, 2014

Homework Help: Calculus

Posted by COFFEE on Wednesday, June 13, 2007 at 5:58pm.

Graph the curve and find its exact length.

x = e^t + e^-t, y = 5 - 2t, from 0 to 3

Length = Integral from 0 to 3 of:

Sqrt[(dx/dt)^2 + (dy/dt)^2]

dx/dt = e^t - e^-t, correct?
dy/dt = -t^2 - 5t, correct?

So: Integral from 0 to 3 of

Sqrt[(e^t - e^-t)^2 + (-t^2 - 5t)^2]

Then what do I do? Thanks.


ok on dx/dt

dy/dt= -2

INT sqrt(4+e^t-e^-t)dt

sqrt (4x + (e^t + e^-t)/loge)

dy/dt = -2 as Bob pointed out.

Next, use that:

Sqrt[(e^t - e^-t)^2 + 4] =

Sqrt[4 + 4 Sinh^2(t)] = =

2 Sqrt[1 + Sinh^2(t)] =

2 Cosh(t)

thanks!!!

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