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Thread: a(t)*a'(t)=0?

  1. #1
    Oct 2010


    how do I prove that a(t)*a'(t)=0; given that a(t) is a parametrized curve where the image of the curve lies on a sphere in R3?
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  2. #2
    MHF Contributor
    Prove It's Avatar
    Aug 2008
    You could solve the DE...

    \displaystyle a\,\frac{da}{dt} = 0

    \displaystyle \int{a\,\frac{da}{dt}\,dt} = \int{0\,dt}

    \displaystyle \int{a\,da} = C_1

    \displaystyle \frac{a^2}{2} + C_2 = C_1

    \displaystyle a^2 = C where \displaystyle C = 2(C_1 - C_2).
    Last edited by Prove It; Feb 16th 2011 at 06:15 AM.
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  3. #3
    MHF Contributor

    Apr 2005
    I don't believe Prove It's proof is valid. He appears to be assuming that a is a real valued function when it should be vector valued. Also, he is trying to prove the converse of the statement- that if the dot product is 0 then the curve lies in a sphere.

    In fact, I don't believe the theorem, as stated, is true. Consider the curve \vec{a}(\theta)= Rcos(\theta)\vec{i}+ (Rsin(\theta)- R)\vec{k}. At any point on that curve, x^2+ y^2+ (z- R)^2= R^2 cos^2(\theta)+ R^2 sin^2(\theta)= R^2 so the curve lies on the surface of the sphere with center at (0, 0, R) and radius R. But \vec{a}'(\theta)= -Rsin(\theta)\vec{i}+ Rcos(\theta)\vec{k}.

    \vec{a}\cdot\vec{a'}= -R^2sin(\theta)cos(\theta)+ R^2sin(\theta)cos(\theta)- R^2cos(\theta)= -R^2cos(\theta)\ne 0.

    What is true is that if \vec{a} is a curve lieing on the surface of a sphere with center at the origin, then \vec{a}\cdot\vec{a}'= 0.

    There are two ways to prove that

    Analytically: Any curve lieing on the surface of a sphere with radius R and center at the origin can be written as \vec{a}(t)= Rcos(\theta(t))sin(\phi(t))\vec{i} + Rsin(\theta(t))sin(\phi(t))\vec{j}+ Rcos(\phi(t))\vec{k}
    for some functions \theta(t) and \phi(t)

    Then \vec{a}'(t)= (-Rsin(\theta(t))sin(\phi(t))\theta'(t)+ Rcos(\theta(t))cos(\phi(t))\phi'(t))\vec{i} + (Rcos(\theta(t))sin(\phi(t))\theta'(t)+ Rsin(\theta(t))cos(\phi(t))\phi'(t))\vec{j}+ (-Rsin(\phi(t))\phi'(t)\ve{k}
    Take the dot product of those two vectors.

    Geometrically: \vec{a(t)}, the position vector, for each t, is a vector from the origin, the center of the sphere, to the sphere. It is, of course, perpendicular to the tangent plane at that point. And \vec{a}'(t), the tangent vector to a curve lieing in the surface of the sphere, is itself in the tangent plane.
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