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Math Help - Find the Osculating Circle

  1. #1
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    Find the Osculating Circle

    Cheers everyone.

    I have to find the osculating circle of r(t) = <2cos(t),3sin(t)> at t = pi/2.

    I find r'(t)=<-2sin(t),3cos(t)> and |r'(t)|=(4sin^2(t)+9cos^2(t))^(1/2). I converted that into |r'(t)| = (4 + 5cos^2(t))^(1/2). I am stuck on taking the derivative of the Unit Tangent vector to find the Principal Normal Unit vector in order to find the curvature. The derivative is getting horrible. Am I on the right track?
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  2. #2
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    Quote Originally Posted by palmaas View Post
    Cheers everyone.

    I have to find the osculating circle of r(t) = <2cos(t),3sin(t)> at t = pi/2.

    I find r'(t)=<-2sin(t),3cos(t)> and |r'(t)|=(4sin^2(t)+9cos^2(t))^(1/2). I converted that into |r'(t)| = (4 + 5cos^2(t))^(1/2). I am stuck on taking the derivative of the Unit Tangent vector to find the Principal Normal Unit vector in order to find the curvature. The derivative is getting horrible. Am I on the right track?
    I'm not sure what method you are using here. The curve r(t) = (2cos(t),3sin(t)) is an ellipse centred at the origin, with its axes parallel to the coordinate axes. When t = \pi/2, r(t) is the point (0,3) on the y-axis, so the centre of curvature will also be on the y-axis. To locate it, you just need to know the radius of curvature at that point.

    With x = 2\cos t and y = 3\sin t, the curvature is given by \kappa = \frac{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}{(\dot{x}^2 + \dot{y}^2)^{3/2}} = \frac{6\sin^2t+6\cos^2t}{(4\sin^2t+9\cos^2t)^{3/2}}. Evaluate that at t = \pi/2, then the radius of curvature is 1/\kappa, and that will tell you where the centre of curvature is.
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