Hey guys and gals,
I just wanted to see if any one can explain to me how to derive the arc length of y=e^x; 0<x<1.
I know the formula to finding the arc length is the square root of 1 + (dy/dx)^2 in the integrand. my question is how do you go from √(1 + u²)/u du to √(1 + u²)/u². Can some one help out?
thanks in advance,
Jose
sorry mr fantastic,
first of all let me thank you for taking the time in helping me out, I appreciate a lot.
second,
I am trying to derive the arc length of y = e^x from 0 to 1
first step is to get dy/dx
that is simply e^x dx
next I have ∫√(1+(e^x)²) dx
then I use u substitution, where u=e^x and du=e^x dx
next I plug the values in and I get ∫√(1+(u)²) du/u or ∫(√(1+(u)²))/u du
now, I have been helped at school and was told that the next step to this would be to go from ∫(√(1+(u)²))/u du to ∫(√(1+(u)²))/u² du,
my question is what did they do to get there and what are the next steps?
hey mr fantastic,
thanks again,
that substitution really makes it easier. Just to clarify though, when you substitute u=√(1+e^2x), I see how you have to square u to get u^2=1+e^2x what I dont understand is how you isolated du to be equal to u/u^2-1
Can you clarify?
Ok, Fantastic,
I got a break through,
feels good too!! anyways, so I see how you got u^2-1. that is, you set e^2x=u^2-1 right?
got it!!!
So, let me ask you this, this would be the same for y=e^3x or y=e^9x? I mean the technique and substitution we used for y=e^x?
Please tell me it is so!
Thanks in advance for your efforts,
Jose