Nuclear power plant some nuclear power plants utilize "natural draft" cooling towers in the shape of a hyperboloid, a solid obtained by rotating a hyperbola about its conjugate axis. suppose that such a cooling tower has a base diameter of 400 feet and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. if the diameter at the top of the tower is 300 feet, how tall is the tower?

  592.4 feet

Step-by-step explanation:

The cross section of the tower can be modeled by a hyperbola with its vertex at (x, y) = (100, 360) and its center at (0, 360). The point (x, y) = (200, 0) helps define the equation of it.

Basic form:

  ((x -h)/a)^2 -((y -k)/b)^2 = 1

Filling in (h, k) = (0, 360) and a=100, we can find b using the given point:

  (200/100)^2 -((0-360)/b)^2 = 1

  (360/b)^2 = 3 . . . . . . . . . . . . . . . rearrange, simplify

  b^ = 360^2/3 = 43200

So, the equation of the hyperbola is ...

  x^2/10000 - (y-360)^2/43200 = 1

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Then for x = 150 (the distance from center at the top, we can find y to be ...

  (150/100)^2 - (y -360)^2/43200 = 1

  1.25 = (y -360)^2/43200

  y = 360 +√(1.25·43200) ≈ 592.379

The height of the tower is about 592.4 feet.

answer: 1 + 1 = 11

step-by-step explanation: