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!!!

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).

To find the length of the curve, you need to integrate the expression

√[(dx/dt)^2 + (dy/dt)^2]

where dx/dt = e^t - e^-t and dy/dt = -2.

First, let's substitute the values of dx/dt and dy/dt into the expression:

√[(e^t - e^-t)^2 + (-2)^2]

Simplify the expression inside the square root:

√[(e^2t - 2 + e^-2t) + 4]

This simplifies to:

√[e^2t + e^-2t + 2]

Now we have the integrand. The integral will be with respect to t, from 0 to 3:

∫[0 to 3] √[e^2t + e^-2t + 2] dt

To evaluate this integral, there is no known elementary antiderivative for the function inside the square root. Therefore, we need to use numerical methods to approximate the integral. One common method is to use Simpson's rule or any other numerical integration method.

Alternatively, if you have access to mathematical software, you can input the integral directly into the software to obtain the exact value of the length.