Clarification on this derivation

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We recently started calculus of variations in my classical mechanics course earlier this week, just before our midterm which was today. Unfortunately, I did very poorly on the exam. However, I'm going to dedicate more time to the class as well as my modern physics 2 class. It's unfortunate that I work 20 hours during the business week which is really at least 24 hours including lunch, commute, etc.

At any rate, I'm not following 23 and 24. This is for the brachistochrone problem. Also, I'm not exactly sure what the general approach is. Of course, 21 is the function (time) that you want minimized, so it looks like they did in a roundabout way by minimizing the distance y.

http://i111.photobucket.com/albums/n149/camarolt4z28/2010-10-15214417.jpg?t=1287197386
 
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The 23 is the 22 manipulated.

<br /> \frac{{y}&#039;^2}{x(1+{y}&#039;^2)} = \frac{1}{2a} \\<br />

<br /> \frac{x(1+{y}&#039;^2)}{{y}&#039;^2} = 2a \\<br />

<br /> \frac{1}{{y}&#039;^2} = \frac{2a-x}{x} \\<br />

<br /> {y}&#039;^2 = \frac{x}{2a-x} \\<br />

<br /> {y}&#039; = \sqrt{\frac{x}{2a-x}} \\<br />

<br /> {y}&#039; = \frac{x}{\sqrt{x(2a-x)}} \\<br />

<br /> {y}&#039; = \frac{x}{\sqrt{2ax-x^2}} \\<br />
 
Yeah, I worked that out a little bit ago. Thanks. I was mainly concerned with 23 to 24. Is that a typical trigonometric substitution?
 
mmm I don't think it's typical.
It's a passage you make when you already know the solution by other ways, or you know it exist.
 
Quinzio said:
mmm I don't think it's typical.
It's a passage you make when you already know the solution by other ways, or you know it exist.

What is the little trick they used?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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