Graph the curve and find its exact length.

Question

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!!!

Answer 1

So the integral for the length becomes:

Length = Integral from 0 to 3 of 2 Cosh(t) dt

Now, to solve this integral, we know that the derivative of sinh(t) is cosh(t). Thus,

Length = 2 Sinh(t) evaluated from 0 to 3

Length = 2[Sinh(3) - Sinh(0)]

Since Sinh(0) = 0,

Length = 2 Sinh(3)

So, the exact length of the curve is 2 Sinh(3).