Curvature of Ellipse

  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. Dick

    Dick 25,735
    Science Advisor
    Homework Helper

    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. 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. Dick

    Dick 25,735
    Science Advisor
    Homework Helper

    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. Oh I see.. okay, thank you very much!
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