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Thread: minimising the square of a distance

  1. #1
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    minimising the square of a distance

    Well because of an exercise involving a curve and a point not in that curve I found out that the minimum of the distance between the curve and the point is found at a point such that the derivative of the curve vector is perpendicular to the vector (curve-external point vector). I could find it cause being the curve X(t) and the point P the distance between the curve and the point is abs(X(t)-Q) and setting the derivative to zero to get the critical point we get that to have a critical point X'(t) is perpendicular to X(t)-Q but i cant prove thats a minimum and not a maximum... how can i do it?

    help would be apreciated...thanks
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  2. #2
    Super Member Failure's Avatar
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    Quote Originally Posted by Mppl View Post
    Well because of an exercise involving a curve and a point not in that curve I found out that the minimum of the distance between the curve and the point is found at a point such that the derivative of the curve vector is perpendicular to the vector (curve-external point vector). I could find it cause being the curve X(t) and the point P the distance between the curve and the point is abs(X(t)-Q) and setting the derivative to zero to get the critical point we get that to have a critical point X'(t) is perpendicular to X(t)-Q but i cant prove thats a minimum and not a maximum... how can i do it?

    help would be apreciated...thanks
    I'm not exactly sure about your notation. I assume that you have a curve $\displaystyle \vec{x}(t)$ parametrized by $\displaystyle t$, and a point $\displaystyle P$ with position vector $\displaystyle \vec{p}:= \vec{OP}$. If so, why don't you just consider the square of the euclidian distance of $\displaystyle \vec{x}(t)$ and $\displaystyle \vec{p}$, that is
    $\displaystyle d^2(t) := (\vec{x}(t)-\vec{p})^2$

    Its first derivative is $\displaystyle \frac{d}{dt}d^2(t)=2\cdot (\vec{x}(t)-\vec{p})\cdot \dot{\vec{x}}(t)$

    Obviously, the condition that this derivative be equal to 0 is equivalent to your condition that the vector $\displaystyle \vec{x}(t)-\vec{p}$ be perpendicular to the tangent vector $\displaystyle \dot{\vec{x}}(t)$.

    To distinguish between points $\displaystyle \vec{x}(t)$ of shortest (local) distance to $\displaystyle P$ from those of greatest (local) distance, you can now check the sign of the second derivative of $\displaystyle d^2(t)$
    Last edited by Failure; Jul 2nd 2010 at 10:01 AM. Reason: Factor 2 added to the derivative of d^2(t)
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  3. #3
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    I already knew that is just that I'm wondering there must be a way of proving that in the general case that this very crit point we found is a minimum and not a maxmimum, and as I compute my second derivative I can't find a reason for it to be always positive,is it possible to prove it on the general case?
    thank you
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by Mppl View Post
    I already knew that is just that I'm wondering there must be a way of proving that in the general case that this very crit point we found is a minimum and not a maxmimum, and as I compute my second derivative I can't find a reason for it to be always positive,is it possible to prove it on the general case?
    thank you
    No, because those critical points $\displaystyle \vec{x}(t)$ can be points of locally minimal or locally maximal distance from $\displaystyle P$, as the case may be.

    For example, let $\displaystyle \vec{x}(t)$ be the parametrization of a circle and let $\displaystyle P$ be a point outside the circle. There are two points where the derivative of $\displaystyle d^2(t)$ is zero: one has minimal distance and the other has maximal distance from $\displaystyle P$.
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