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Simple Derivation has me stumped

  1. Jul 22, 2012 #1
    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 [tex]\sqrt{\frac{R^2}{R^2-x^2}}[/tex]

    Attached Files:

  2. jcsd
  3. Jul 22, 2012 #2


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    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
  4. Jul 22, 2012 #3
    Sorry, I'm not able to read between the line here. What do you mean by compute dy/dx? Compute dy/dx of what?
  5. Jul 22, 2012 #4


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    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.
  6. Jul 23, 2012 #5
    d(x2 + y2) = 2xdx + 2ydy = dR2 = 0

    dy/dx = -x/y
  7. Jul 23, 2012 #6
    Oh, alright thanks. ^.^
  8. Jul 23, 2012 #7
    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.
  9. Jul 23, 2012 #8

    George Jones

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    A change of integration variable was made such that the new integration variable is dimensionless, i.e., [itex]\xi = x/R[/itex].
  10. Jul 23, 2012 #9
    Okay, I also don't know how that integral will give you Pi. Something is missing in my knowledge-base.
  11. Jul 23, 2012 #10


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    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.
  12. Jul 23, 2012 #11
    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.
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