Posted by **Vanna** on Thursday, March 12, 2009 at 3:28pm.

Cooling towers for nuclear reactors are often constructed as hyperboloids of one sheet because of the structural stability of that surface. Suppose all horizontal cross sections are circular, with a minimum radius of 200 feet occurring at a height of 600 feet. The tower is to be 800 feet tall with a maximum cross-sectional radius of 300 feet. Find the equation of the surface.

- Math -
**drwls**, Thursday, March 12, 2009 at 4:33pm
Make the axis of symmetry the y axis and the minimum cross section be the y = 0 plane. The general equation for the hyperbola that generates the surface is

x^2/a^2 - y^2/b^2 = 1

where a is the minimum radial distance from the y axis, which in this case is 200 feet. To get b, require that x = 300 when y = 200 feet. (That is, y = 200 feet above the "throat" where x = 200.

(300/200)^2 - (200/b)^2 = 1

(200/b)^2 = 1.25

b/200 = sqrt(4/5)

b = 178.9 ft

(x/200)^2 - (y/178.9)^2 = 1

The domain of y is -600 to +200

- Math -
**Vanna**, Thursday, March 12, 2009 at 5:29pm
However, the equation is for a hyperboloid of one sheet, not a hyperbola. I need to find a equation that follows the general equation of a hyperboloid of one sheet: ax^2 + by^2 - cz^2 where a,b,c are > 0.

- Math -
**drwls**, Thursday, March 12, 2009 at 8:04pm
The hyperbola that I described, when rotated about the y axis, in an x,y,z coordinate system. becomes the hyperboloid of one sheet. I leave the rest up to you

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