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Thread: Find K and R = 1/K

  1. #1
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    Find K and R = 1/K

    Find the curvature and radius of curvature.

    y = 2x^2 + 3, x = -1

    r (t) = < t, 2t^2 + 3, 0 >

    r'(t) = < 1, 4t, 0 >

    The magnitude of r'(t) = root {16t^2 + 1}.

    Is this correct so far?
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  2. #2
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    Re: Find K and R = 1/K

    Wky did you title your post "Find K and R = 1/K" ?
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    Re: Find K and R = 1/K

    Quote Originally Posted by DenisB View Post
    Wky did you title your post "Find K and R = 1/K" ?
    I wanted the title to stand out. Am I right so far?
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  4. #4
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    Re: Find K and R = 1/K

    Am I correct here?
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    Re: Find K and R = 1/K

    yes
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  6. #6
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    Re: Find K and R = 1/K

    I suspect that the "K" in the title was supposed to be "kappa", \kappa, the curvature, and R the radius of curvature. The first, and easiest, step of one method of finding those would be to find the length of the tangent vector.
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  7. #7
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    Re: Find K and R = 1/K

    I will continue with this problem today. If I get stuck or cannot finish, I will post that as well.
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  8. #8
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    Re: Find K and R = 1/K

    Quote Originally Posted by romsek View Post
    yes
    Let's see how far I can go with this problem.
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    Re: Find K and R = 1/K

    Quote Originally Posted by HallsofIvy View Post
    I suspect that the "K" in the title was supposed to be "kappa", \kappa, the curvature, and R the radius of curvature. The first, and easiest, step of one method of finding those would be to find the length of the tangent vector.
    Yes, K = curvature and R = radius of curvature.
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  10. #10
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    Re: Find K and R = 1/K

    If a curve is given by a vector function, \vec{r}(t)= x(t)\vec{i}+ y(t)\vec{j}+ z(t)\vec{k}, then the tangent vector at any point is the derivative: \vec{T}(t)= x'(t)\vec{i}+ y'(t)\vec{j}+ z'(t)\vec{k}. If, in particular, the parameter, t, is the arc-length, s, then that tangent vector will have length 1 and is the "unit tangent vector". If t is not the arc-length then we still have relation s= \int |T|dt.

    With arc-length as parameter, so that the tangent vector has unit length, the derivative of the tangent vector is perpendicular to the tangent vector (Since |\vec{T|}= \sqrt{\vec{T}\cdot \vec{T}}= 1, \vec{T}\cdot \vec{T}= 1, a constant, so \frac{d\vec{T}\cdot \vec{T}}{ds}= 2\vec{T}\cdot\frac{d\vec{T}}{ds}= 0) and is called the "normal vector" to the curve at that point. The "curvature" is the length of that normal vector. The "radius of curvature" is 1 over the curvature.

    Where t is not the arc-length, we have, by the chain rule, \frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{dt}{ds}. And, since s= \int \left| \frac{d\vec{T}}{dt}\right| dt, \frac{ds}{dt}= |\vec{T}|, \frac{d\vec{T}}{ds}= \frac{d\vec{T}}{dt}\frac{ds}{dt}.
    Last edited by HallsofIvy; Jan 6th 2017 at 06:31 AM.
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  11. #11
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    Re: Find K and R = 1/K

    Cool.
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