How do I integrate this?

dt/[t(1+t^2)] = A/t*dt + (Bt+C)/(1+t^2)*dt ---integrate----> A*ln|t| + B/2*ln|1+t^2| + C*arctan(t)
 
This is a variation of the brachistochrone problem in my mechanics book.

I worked out the integration using partial fractions, but I don't see how to get the catenary equation.
 
I'm not sure what you mean. What exactly is the objective here? If y' is the derivative of another function, perhaps you need to find y? I know little of the catenary equation and nothing of the "brachistochrone problem".

edit:OK, I did some googling, and I have seen this before, but I haven't studied this in detail. Maybe post what the original problem was or just more details about how you got to this point in your work and what is needed to be done.
 
daon said:
I'm not sure what you mean. What exactly is the objective here? If y' is the derivative of another function, perhaps you need to find y? I know little of the catenary equation and nothing of the "brachistochrone problem".

edit:OK, I did some googling, and I have seen this before, but I haven't studied this in detail. Maybe post what the original problem was or just more details about how you got to this point in your work and what is needed to be done.
Click to expand...

Sorry. Yes, we do need to get to y. This is a calculus of variations problem.

http://i111.photobucket.com/albums/n149 ... 1287846510
 
I have played around with the brachistochrine in the past. So, I dug out my old notes.

Note, that \(\displaystyle \sqrt{1+y'^{2}}\) is the arc length and \(\displaystyle \sqrt{2gy}\) is the rate. So, we know from way back that

t=d/r. I will use the parametric aspect.

Think of a small length of arc to be ds. Then, \(\displaystyle ds=\sqrt{x'^{2}+y'^{2}}\) and \(\displaystyle v=\sqrt{2gy}\)

and \(\displaystyle dt=\frac{ds}{v}=\frac{\sqrt{x'^{2}+y'^{2}}}{\sqrt{2gy}}\)

But, in the brachistochrone, \(\displaystyle x=a(t-sin(t)), \;\ y=a(1-cos(t))\)

\(\displaystyle x'=a(1-cos(t)), \;\ y'=a\cdot sin(t)\)

This gives:

\(\displaystyle dt=\sqrt{\frac{2a^{2}(1-cos(t))}{2ga(1-cos(t))}}=\sqrt{\frac{a}{g}}\)

Now,integrate:

\(\displaystyle \int_{0}^{\pi}dt={\pi}\sqrt{\frac{a}{g}}\)

This is how I looked at it.
 
Wow. I like your method better. Let me see if I can reproduce it myself.

Yeah, I got the same thing. Thanks.

Minimum point is 2a and this implies theta equals pi.
 
Wow, you're digging up some stuff I have not looked at in a while. That's cool.

A surface of revolution is given by \(\displaystyle 2{\pi}\int_{x_{1}}^{x_{2}}y\sqrt{1+y'^{2}}dy\)

Using Euler's equation, \(\displaystyle y\sqrt{1+y'^{2}}\)

\(\displaystyle \frac{d}{dx}\left(\frac{yy'}{\sqrt{1+y'^{2}}}\right)-\sqrt{1+y'^{2}}=0\)

\(\displaystyle 1+y'^{2}-yy''=0\)..........[1]

Let \(\displaystyle y'=u\), and \(\displaystyle y''=\frac{du}{dx}=\frac{du}{dy}\cdot \frac{dy}{dx}=u\frac{du}{dy}\)

Separate variables and integrate:

Then, [1] becomes \(\displaystyle \int \frac{u}{1+u^{2}}du=\int\frac{dy}{y}\)

\(\displaystyle \frac{1}{2}ln(1+u^{2})=ln(y)+C\)

\(\displaystyle u=\frac{dy}{dx}=\frac{\sqrt{y^{2}-c^{2}}}{c}\)

Separate and integrate:

\(\displaystyle \int\frac{dy}{\sqrt{y^{2}-c^{2}}}=\int\frac{dx}{c}\)

\(\displaystyle cosh^{-1}(\frac{y}{c})=\frac{x+K}{c}\)

\(\displaystyle y=c\cdot cosh^{-1}(\frac{x+K}{c})\)

Which is the form for a catenary.

I should mention that this does not prove that the surface is actually a minimum. That would require some more grunt work and analyses.

But it does derive the catenary....the shape that gives the minimum. Rather anal, but I thought I would mention it.
 
This is part of my classical mechanics course. He told us the first day of class he would give us the hardest problems in the textbook - we get ten each chapter. Virtually all of the problems require at least one mathematical trick to solve them. I'm sure I'll need help on the rest of the problem set. It's due Monday. I've completely finished three, and this is the fourth one.

How would I proceed with the integration from where I stopped? I have about a page's worth of work, so I probably shouldn't abandon it right now. lol. My equations are a bit different than yours.
 
I bet you also have the one that asks to find the curve of length L that encloses the max area?.

If you do, it's a circle.
 
6-10 reads like an optimization problem from calc I.

The surface area of a cylinder is minimized when the diameter and height are equal.
 
galactus said:
6-10 reads like an optimization problem from calc I.

The surface area of a cylinder is minimized when the radius and height are equal.
Click to expand...

Now we're using calculus of variations and the various equations I posted.

The answer in the back of the book is R= (1/2)H.
 
Yes, that is correct. I should have said the surface is minimum when the DIAMETER and the height are equal.

I mistakenly said when the radius and height are equal.

\(\displaystyle S=2{\pi}r^{2}+2{\pi}rh\)

\(\displaystyle V={\pi}r^{2}h\)

\(\displaystyle h=\frac{V}{{\pi}r^{2}}\)

\(\displaystyle S=2{\pi}r^{2}+\frac{2V}{r}, \;\ r>0\)

\(\displaystyle \frac{dS}{dr}=4{\pi}r-\frac{2V}{r^{2}}\)

\(\displaystyle \frac{dS}{dr}=0\) when \(\displaystyle r=\sqrt[3]{\frac{V}{2\pi}}\)

then \(\displaystyle h=\frac{V}{{\pi}r^{2}}=\frac{V}{{\pi}r^{3}}\cdot r=2r\)
 
galactus said:
Yes, that is correct. I should have said the surface is minimum when the DIAMETER and the height are equal.

I mistakenly said when the radius and height are equal.
Click to expand...

Can you shed any light on the problem using the calculus of variations techniques?

This of course is all new to me. Next chapter we're going over Lagrangian and Hamiltonian dynamics.
 
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