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Question 274145: 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 such a cooling tower has a base diameter of 400 feet, and the diameter at its narrowest point, 360 feet above the round, is 200 feet. If the diameter at the top of the tower is 300 feet, how tall is the tower?
Answer by CharlesG2(834)   (Show Source):
You can put this solution on YOUR website! 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 such a cooling tower has a base diameter of 400 feet, and the diameter at its narrowest point, 360 feet above the round, is 200 feet. If the diameter at the top of the tower is 300 feet, how tall is the tower?
(200,0) and (100,360) and (150,y)
x distance from central axis, y height above ground
center of hyperboloid is (0,360)
(x-0)^2 / a^2 - (y-360)^2 / b^2 = 1
we know (100,360) is on the hyperbola
100^2 / a^2 = 1
100^2 = a^2
100 = a
we also know (200,0) is on hyperbola
200^2 / 100^2 - (-360)^2 / b^2 = 1
40000/10000 - 129600/b^2 = 1
4 - 129600/b^2 = 1
129600/b^2 = 3
129600 = 3b^2
43200 = b^2
4*108*100 = 4 * 4 * 27 * 100 = 3 * 4 * 4 * 9 * 100 = b^2
2 * 2 * 3 * 10 * sqrt(3) = b
120sqrt(3) = b
(x-0)^2 / a^2 - (y-360)^2 / b^2 = 1 becomes
x^2 / 10000 - (y-360)^2 / 43200 = 1
now plug in (150,y)
150^2 / 10000 - (y-360)^2 / 43200 = 1 and solve for y
22500/10000 - (y-360)^2/43200 = 1
2.25 - (y-360)^2/43200 = 1
(y-360)^2/43200 = 1.25
(y-360)^2 = 54000
y-360 = sqrt(6*9*10*100) = 30sqrt(60) = 30sqrt(4*15) = 60sqrt(15)
y = 360 + 60sqrt(15)
y = 360 + 60 * 3.872983
y = 360 + 232.37898
y = 592.37898
y is approx 592 feet
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