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Arc lengths of polar curves

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the total length of the cardioid r=a(1-cos theta)


    2. Relevant equations

    ds2=r2dtheta2+dr2

    ds= integral from beta to alpha sqrt[r2 + (dr/d theta)2]dtheta

    3. The attempt at a solution

    dr=a(sin theta)d theta

    ds2=a2(1-cos theta)2d theta2 + a2sin2theta (d theta2), which reduces to:

    ds=21/2a(1-cos theta)1/2(d theta)

    I'm good up to this point. My book says that the above equation for ds simplifies to:

    2a|sin1/2theta|d theta

    I don't see how to arrive at this simplification. Can someone please explain? Many thanks.
     
  2. jcsd
  3. Apr 5, 2010 #2
    I have a separate, basic question related to arc lengths of polar curves. I'm trying to derive the formula ds2=r2dtheta2+dr2, which enables me to compute arc lengths of polar curves by integration, from the rectangular equation for the differential element of arc length, namely ds2=dx2 + dy2.

    I know that I need to use the transformation equations:

    x=r cos theta and y= r sin theta

    My book says that I need to differentiate with respect to theta using the product rule to obtain:

    dx/dtheta = -r sin theta + cos theta dr/dtheta

    When I differentiate, I get dx/dtheta = -r sin theta. I'm not sure how the product rule fits in, which I presume is where does the second half come from (+ cos theta dr/dtheta). I think I'm missing something basic here.
     
  4. Apr 5, 2010 #3

    Dick

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    It's a trig trick. 1-cos(theta)=2*cos^2(theta/2). Half angle formula.
     
  5. Apr 5, 2010 #4
    I'm not so strong in trig, but did you mean to say 1-cos(theta)=2*sin2theta/2?

    If 1-cos2theta = 2sin2theta, then
    1-cos theta = 2sin2theta/2 because we halved the angle cos2theta on the left side?
     
  6. Apr 5, 2010 #5

    Dick

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    Sure. Sorry. That's what I meant to say. Must be getting late.
     
  7. Apr 5, 2010 #6
    Thanks, Dick.

    Any ideas on my proof question from above (my 2nd post)?
     
  8. Apr 6, 2010 #7

    Dick

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    I would use the product rule on the differentials. dx=d(r*cos(t))=dr*cos(t)+r*d(cos(t))=dr*cos(t)-r*sin(t)dt. Do the same with dy and evaluate dx^2+dy^2.
     
  9. Apr 6, 2010 #8

    Dick

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    I would use the product rule on the differentials. dx=d(r*cos(t))=dr*cos(t)+r*d(cos(t))=dr*cos(t)-r*sin(t)dt. Do the same with dy and evaluate dx^2+dy^2.
     
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