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
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)] =
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