Minimum Distance from an Ellipsoid to a Paraboloid

Calculus Level 5

In the x y z xyz coordinate system, an ellipsoid has its center at the point ( 0 , 0 , 0 ) (0,0,0) , and semi-axes coinciding with the vectors ( 1 , 2 , 3 ) (1,2,3) , ( 5 , 2 , 3 ) (-5,-2,3) , and ( 6 , 9 , 4 ) (6,-9,4) .

A paraboloid has the general shape of the surface z = x 2 + y 2 z = x^2 + y^2 , but its vertex location and orientation are different. The paraboloid's vertex is at the point ( 3 , 7 , 4 ) (3,7,4) , and it opens up in the direction of the vector ( 9 , 3 , 8 ) (9,-3,-8) in the same way in which the surface z = x 2 + y 2 z = x^2 + y^2 opens up in the direction of the positive z z axis.

We want to find the minimum distance d d^* , between a point P = ( x 1 , y 1 , z 1 ) P =(x_1, y_1, z_1 ) on the ellipsoid and a point Q = ( x 2 , y 2 , z 2 ) Q = (x_2, y_2, z_2 ) on the paraboloid. If the minimum occurs with P = ( x 1 , y 1 , z 1 ) P^* = ({x_1}^*,{y_1}^*,{z_1}^*) , and Q = ( x 2 , y 2 , z 2 ) Q^* = ( {x_2}^*,{y_2}^*,{z_2}^*) , then report the value of 100 ( x 1 + y 1 + z 1 + x 2 + y 2 + z 2 + d ) \lfloor 100({x_1}^*+{y_1}^*+{z_1}^* + {x_2}^*+{y_2}^*+{z_2}^* + d^* ) \rfloor , where \lfloor \cdot \rfloor is the floor function; for example, 7.6 = 7 \lfloor 7.6 \rfloor = 7 .

Inspiration


The answer is 1728.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

0 solutions

No explanations have been posted yet. Check back later!

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...