# Simple Derivation has me stumped

1. Jul 22, 2012

### Nano-Passion

I have no idea how eq 20.10b to eq 20.10c.

Hartle goes from the equation for the arc length to $$\sqrt{\frac{R^2}{R^2-x^2}}$$

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2. Jul 22, 2012

### PAllen

compute dy/dx and plug into 2.10.b. It leads directly to 2.10c. I certainly couldn't see it by eye - but it is a pretty small calculation.

Last edited: Jul 23, 2012
3. Jul 22, 2012

### Nano-Passion

Sorry, I'm not able to read between the line here. What do you mean by compute dy/dx? Compute dy/dx of what?

4. Jul 22, 2012

### PAllen

You have x^2+y^2=R^2. Solve for y and take dy/dx. Plug this into where dy/dx is in 2.10.b. You can then simplify to 2.10.c.

5. Jul 23, 2012

### Staff: Mentor

d(x2 + y2) = 2xdx + 2ydy = dR2 = 0

dy/dx = -x/y

6. Jul 23, 2012

### Nano-Passion

Oh, alright thanks. ^.^

7. Jul 23, 2012

### Nano-Passion

One more question, do you know how he got -1, 1 as the limits of integration? It looks as if he pulled that number randomly.

8. Jul 23, 2012

### George Jones

Staff Emeritus
A change of integration variable was made such that the new integration variable is dimensionless, i.e., $\xi = x/R$.

9. Jul 23, 2012

### Nano-Passion

Okay, I also don't know how that integral will give you Pi. Something is missing in my knowledge-base.

10. Jul 23, 2012

### PAllen

That integral should be covered in any first course in calculus; or looked up in even the smallest table of integrals; or recognize that it is the circumference of a unit semi-circle. If you're reading this book, you should have calculus book, and can review integration of trigonometric forms.

11. Jul 23, 2012

### Nano-Passion

Thank you, I should have realized that. What I also should have done is just evaluated the integral to see that it equals pi.

I suppose I should jump back to classical mechanics, I need to test out of classical mechanics I anyways. I'll come back to this book later.