How Does Hartle Derive the Equation from Arc Length to a Trigonometric Integral?

[Nano-Passion]

Please click the attached image.

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}}

 

[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.

 

[Nano-Passion]

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

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

 

[PAllen]

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

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.

 

[Chestermiller]

d(x² + y²) = 2xdx + 2ydy = dR² = 0

dy/dx = -x/y

 

[Nano-Passion]

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.

Oh, alright thanks. ^.^

 

[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.

 

[George Jones]

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.

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

 

[Nano-Passion]

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

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

 

[PAllen]

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

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.

 

[Nano-Passion]

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.

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.