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Thread: Find the minimum distance (Calculus II)

  1. April 19th 2009, 08:22 PM #1
    bkarpuz
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    Arrow Find the minimum distance (Calculus II)

    Let S(x,y,z):=(x-y)^{2}-z^{2}-1=0.
    Find the minimum distance of the points on the surface S to the origin.
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  2. April 20th 2009, 02:22 AM #2
    NonCommAlg
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    Quote Originally Posted by bkarpuz View Post

    Let S(x,y,z):=(x-y)^{2}-z^{2}-1=0. Find the minimum distance of the points on the surface S to the origin.
    this is a nice problem! it can be solved geometrically without using methods of multivariable calculus. here's my answer, which i think is correct:

    Spoiler:
    i believe the minimum distance is \frac{\sqrt{2}}{2}, which is attained at two points: x=\pm \frac{1}{2}, \ y = \mp \frac{1}{2}, \ z=0. the reason is that the lines x-y-1=0 and x-y+1=0 are tangent to the circle

    x^2+y^2 = \frac{1}{2}. therefore (x-y-1)(x-y+1)=z^2 \geq 0 implies that x^2 + y^2 \geq \frac{1}{2} and thus x^2+y^2+z^2 \geq \frac{1}{2}.
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  3. April 20th 2009, 07:16 AM #3
    bkarpuz
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    Quote Originally Posted by NonCommAlg View Post
    this is a nice problem! it can be solved geometrically without using methods of multivariable calculus. here's my answer, which i think is correct:

    Spoiler:
    i believe the minimum distance is \frac{\sqrt{2}}{2}, which is attained at two points: x=\pm \frac{1}{2}, \ y = \mp \frac{1}{2}, \ z=0. the reason is that the lines x-y-1=0 and x-y+1=0 are tangent to the circle

    x^2+y^2 = \frac{1}{2}. therefore (x-y-1)(x-y+1)=z^2 \geq 0 implies that x^2 + y^2 \geq \frac{1}{2} and thus x^2+y^2+z^2 \geq \frac{1}{2}.
    I was looking for such a nice solution, thanks NonCommAlg.
    I guess you wonder if I have a solution or not?
    Here it follows.
    Spoiler:
    Rotating a surface along a point does not changes the distance of the surface to that point.
    Therefore, rotating S with \pi/4 radians along O, we have S(x,y,z)=\widehat{S}(t,s,z)=2t^{2}-z^{2}-1=0, by making the substitution x(t,s)=\sqrt{2}t/2+\sqrt{2}s/2 and y(t,s)=-\sqrt{2}t/2+\sqrt{2}s/2.
    Clearly \widehat{S} is a hyperbola of two parts (I made this term up, and if possible teach me the correct one).
    Hence, the distance \ell attains it smallest value at the peak points (\pm\sqrt{2}/2,0,0), it is clear that \ell=\sqrt{2}/2.
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  4. April 20th 2009, 01:30 PM #4
    NonCommAlg
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    Quote Originally Posted by bkarpuz View Post
    I was looking for such a nice solution, thanks NonCommAlg.
    I guess you wonder if I have a solution or not?
    Here it follows.
    Spoiler:
    Rotating a surface along a point does not changes the distance of the surface to that point.
    Therefore, rotating S with \pi/4 radians along O, we have S(x,y,z)=\widehat{S}(t,s,z)=2t^{2}-z^{2}-1=0, by making the substitution x(t,s)=\sqrt{2}t/2+\sqrt{2}s/2 and y(t,s)=-\sqrt{2}t/2+\sqrt{2}s/2.
    Clearly \widehat{S} is a hyperbola of two parts (I made this term up, and if possible teach me the correct one).
    Hence, the distance \ell attains it smallest value at the peak points (\pm\sqrt{2}/2,0,0), it is clear that \ell=\sqrt{2}/2.
    \widehat{S} is actually a cylinder in tsz cooridinates, along s axis, with hyperbolic base . by the way, the Lagrange multipliers method will also solve the problem easily.
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