Curvature of Ellipse

  1. Dec 10, 2008 #1
    Find Curvature of Ellipse given x=3*cos(t) and y=4*sin(t) at the points (3,0) and (0,4)

    Relevant equations: curvature at r(s) is k(s)=||dT/ds|| when r(s) is arc length parametrization and T is the unit tangent vector
    I usually use the formula k(t)= (||r'(t) x r''(t)||)/||r'(t)||^3

    So, r(t)=<3cost,4sint> and r'(t)= <-3sint,4cost> and r''(t)=<-3cost,-4sint>
    do I just plug them in for k(t)? And I'm clueless about how to use the given points. Someone help please. Thank you
     
  2. jcsd
  3. Dec 10, 2008 #2

    Dick

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    Yes, you can just plug them into k(t). You'll need to do a cross product, so take the z component of the vectors to be zero. When you get k as a function of t, then you just need to go back and set (3*cos(t),4*sin(t))=(3,0) and figure out what t is for the first point. Ditto for the second.
     
  4. Dec 10, 2008 #3
    Thanks, but I'm not sure I completely understand

    This is what I got so far, please correct me if I'm wrong:
    r'(t) x r''(t)= 12sint^2+12cost^2 = 12
    ||r'(t) x r''(t)||= sqrt(144) =12
    ||r'(t)||^3 = (9sint^2+16cost^2)^(3/2)

    k(t)= 12/(9sint^2+16cost^2)^(3/2)

    from what you've said, (3*cos(t),4*sin(t))=(3,0), so 3cost=3 => t= 0 or 2pi and 4sint=0 => t=0 or 2pi
    similarly (3*cos(t),4*sin(t))=(0,4), so 3cost=0 => t=pi/2 or 3pi/2 and 4sint=4 => t=pi/2

    Sry if this is a stupid question, but how do I apply that to k(t)?
     
  5. Dec 10, 2008 #4

    Dick

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    That looks ok to me. So the question is just asking you what is k(0) and k(pi/2), right? Those are the curvatures of the ellipse at the two given points.
     
  6. Dec 10, 2008 #5
    Oh I see.. okay, thank you very much!
     
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