Trouble Calculating Length of a Curve Segment (Surface Integral?)

• May 13th 2011, 09:27 PM
johnnyjohnny
Trouble Calculating Length of a Curve Segment (Surface Integral?)
So, I am trying to find the length of a segment of a curve that is given by an x equation and a y equation and a limit placed on the 3rd variable t. Sorry in advance for the lack of actual images. Apparently I can't use that extension here? The initial problem is:

x = (e^t) + (e^-t) y = -2t + 5 0 <= t <= 3

Since the equation for finding the length of a curve segment described like this is: { [ (dx/dt)^2 ] + [ (dy/dt)^2 ] }^(1/2)
Using a pic from wolfram in the hopes of being a little clearer the formula is: http://www4a.wolframalpha.com/Calcul...=61&w=106&h=28

Then I calculate:

dx/dt = (e^t) - (e^-t)
dy/dt = -2t + 5

Which finally leads me to: { ( [ (e^t) - (e^-t) ]^2 ) + 4 }^(1/2) or http://www4a.wolframalpha.com/Calcul...=61&w=300&h=47

Now, I know that the answer is clearly in that last image but really I'm not so concerned with that as I am with the actual steps to solve this problem. By "problem" I am referring to that final integral calculation. I am led to believe that that is just true similar to how \int (1/x) = ln x. Even that equality can be stretched out to include some steps though if I remember correctly. I am looking for the steps required for this problem. Anyone?
• May 13th 2011, 09:36 PM
TheEmptySet
It is mostly algebra lets focus on the radicand to start

$\displaystyle (e^t-e^{-t})^2+(-2)^2=e^{2t}-2+e^{-2t}+4=e^{2t}+\frac{1}{e^{2t}}+2$

$\displaystyle \frac{(e^{2t})^2+1+2e^{2t}}{e^{2t}}=\frac{(e^{2t}) ^2+2e^{2t}+1}{e^{2t}}=\frac{(e^{2t}+1)^2}{e^{2t}}$

Now if you take the square root we get

$\displaystyle \frac{e^{2t}+1}{e^{t}}=e^{t}+e^{-t}$

Now just integrate

$\displaystyle \int_{0}^{3} e^{t}+e^{-t} dt$

You should be able to finish from here.
• May 13th 2011, 09:42 PM
johnnyjohnny
Thank you! I really appreciate this. Now, it's time for sleep.. I should be able to now that I can stop thinking about this :p
• May 14th 2011, 04:53 AM
HallsofIvy
You might find it simpler to write $\displaystyle x= 2cosh(t)$, $\displaystyle y= -2t+ 5$ so that $\displaystyle \frac{dx}{dt}= 2sinh(t)$, $\displaystyle \frac{dy}{dt}= -2dt$ and then $\displaystyle \sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}= 2\sqrt{sinh^2(t)+ 1}= 2cosh(t)$