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Integration of e

  • Thread starter olivia333
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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, e5t, e-5t>
0≤t≤1

Homework Equations



L = ∫|r'(t)|

The Attempt at a Solution



r'(t) = √((5√(2))2+(5e5t)2+(-5e-5t)2)

|r'(t)| = 5√(2+e10t+e-10t)

Now this is where I get confused.

∫(5√(2+e10t+e-10t)) = ((e10t-1)*√(2+e10t+e-10t))/(e10t+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.
 
Last edited:

Answers and Replies

  • #2
Dick
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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.
 
  • #3
Ray Vickson
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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, e5t, e-5t>
0≤t≤1

Homework Equations



L = ∫|r'(t)|

The Attempt at a Solution



r'(t) = √((5√(2))2+(5e5t)2+(-5e-5t)2)

|r'(t)| = 5√(2+e10t+e-10t)

Now this is where I get confused.

∫(5√(2+e10t+e-10t)) = ((e10t-1)*√(2+e10t+e-10t))/(e10t+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 [itex] 5\sqrt{2 + e^{10t} + e^{-10t}}[/itex] can be re-written, using the identity [itex] \cosh(10t) =2 \cosh(5t)^2 - 1,[/itex] to give a much simpler integral.

RGV
 
Last edited:

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