Arclength integral of r(t) = <5√(2)t, e^(5t), e^(-5t)> from 0 to 1

[olivia333]

Homework Statement

I have the problem and answer, I'm just confused on the last step, but I'll put down everything anyways.

Find the arclength of the curve r(t) = <5√(2)t, e^5t, e^-5t>
0≤t≤1

Homework Equations

L = ∫|r'(t)|

The Attempt at a Solution

r'(t) = √((5√(2))²+(5e^5t)²+(-5e^-5t)²)

|r'(t)| = 5√(2+e^10t+e^-10t)

Now this is where I get confused.

∫(5√(2+e^10t+e^-10t)) = ((e^10t-1)*√(2+e^10t+e^-10t))/(e^10t+1)

I just don't understand how to integrate my answer to get that. If someone could just go through the steps that'd be great. Thanks!

Then obviously sub in for 1 and 0, getting 148.4064 which is correct so I know that the integration is correctly done.

 

[Dick]

That's an odd form for the resulting integral. To see how to do it more easily what is (e^(5t)+e^(-5t))^2? Expand it out.

 

[Ray Vickson]

Homework Statement

I have the problem and answer, I'm just confused on the last step, but I'll put down everything anyways.

Find the arclength of the curve r(t) = <5√(2)t, e^5t, e^-5t>
0≤t≤1

Homework Equations

L = ∫|r'(t)|

The Attempt at a Solution

r'(t) = √((5√(2))²+(5e^5t)²+(-5e^-5t)²)

|r'(t)| = 5√(2+e^10t+e^-10t)

Now this is where I get confused.

∫(5√(2+e^10t+e^-10t)) = ((e^10t-1)*√(2+e^10t+e^-10t))/(e^10t+1)

I just don't understand how to integrate my answer to get that. If someone could just go through the steps that'd be great. Thanks!

Then obviously sub in for 1 and 0, getting 148.4064 which is correct so I know that the integration is correctly done.

Your function $5\sqrt{2 + e^{10t} + e^{-10t}}$ can be re-written, using the identity $\cosh(10t) =2 \cosh(5t)^2 - 1,$ to give a much simpler integral.

RGV