# Find the minimum distance (Calculus II)

• April 19th 2009, 08:22 PM
bkarpuz
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.
• April 20th 2009, 02:22 AM
NonCommAlg
Quote:

Originally Posted by bkarpuz

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! (Clapping) it can be solved geometrically without using methods of multivariable calculus. here's my answer, which i think is correct: (Nod)

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 $x^2+y^2+z^2 \geq \frac{1}{2}.$
• April 20th 2009, 07:16 AM
bkarpuz
Quote:

Originally Posted by NonCommAlg
this is a nice problem! (Clapping) it can be solved geometrically without using methods of multivariable calculus. here's my answer, which i think is correct: (Nod)

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? :p
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 $\ell=\sqrt{2}/2$.
• April 20th 2009, 01:30 PM
NonCommAlg
Quote:

Originally Posted by bkarpuz
I was looking for such a nice solution, thanks NonCommAlg.
I guess you wonder if I have a solution or not? :p
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.