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Show that a curve is regular

  1. Sep 26, 2015 #1
    Hello !

    1. The problem statement, all variables and given/known data
    Consider a parametrized curve
    C(θ)=( (R+r)*cos(θ) - d*cos(θ(R+r)/r) ; (R+r)*sin(θ) - d*sin(θ(R+r)/r) )
    Show that C is regular for d<r. Is it regular if d=r ?

    2. Relevant equations


    3. The attempt at a solution
    C'(θ)=( -(R+r)*sin(θ) +d*(R+r)/r*sin(θ(R+r)/r) ; (R+r)*cos(θ) - d*(R+r)/r*cos(θ(R+r)/r) )
    I don't know how to show that it is regular for d<r, i am supposed to show that C'(θ)≠0 but I don't know how to introduce the condition to prove that

    Thanks
     
  2. jcsd
  3. Sep 26, 2015 #2

    Svein

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    You have found [itex]\frac{dx}{d\theta}, \frac{dy}{d\theta} [/itex], which are the components of the tangent. Now find the length of that vector.
     
  4. Sep 26, 2015 #3
    By using sin²x+cos²x=1 and using formulas for cos(a)*cos(b) and sin(a)*sin(b) I ended up with
    d²(R+r)²/r²+(R+r)²-2d(R+r)²/r * cos(θ[(R+r)/r -(R+r)])
     
  5. Sep 26, 2015 #4

    Svein

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    So what are the maximum and minimum value of that expression?
     
  6. Sep 26, 2015 #5
    Max when cos(θ[(R+r)/r -(R+r)])=0 we have d²(R+r)²/r²+(R+r)²
    and Min when cos(θ[(R+r)/r -(R+r)])=1 we have d²(R+r)²/r²+(R+r)²-2d(R+r)²/r
     
  7. Sep 26, 2015 #6
    There's a mistake in my equations it is
    d²(R+r)²/r²+(R+r)²-2d(R+r)²/r * cos(θ[(R+r)/r -1])

    Max when cos(θ[R/r])=0 we have d²(R+r)²/r²+(R+r)²
    and Min when cos(θ[R/r])=1 we have d²(R+r)²/r²+(R+r)²-2d(R+r)²/r
     
    Last edited: Sep 26, 2015
  8. Sep 26, 2015 #7

    Svein

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    So, the maximum is >0. The minimum is (R+r)²(d²/r² -2d/r + 1) = (R+r)²(d/r -1)2 which is ≥0. When is the minimum = 0?
     
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