Find the width of the tower at a height of 11 feet, to 1 decimal place.

Question

The sides of a nuclear power plant cooling tower form a hyperbola. The diameter of the bottom of the tower is 318 feet. The smallest diameter of the tower is 182 which is 356 feet above the ground. The tower is 512 feet tall.

512 ft

182 ft

318 ft

356 ft

Photo remixed from work by Jiří Sedláček, CC-BY-SA

Find the width of the tower at a height of 11 feet, to 1 decimal place.

 feet.

Answer 1

I wish I can put my graph but I'm only limited to 10,000 character (or probably 10,000 bytes).

We use this standard form of the equation of hyperbola because the transverse axis is horizontal:

x²/a² -y²/b² = 1

2a = distance between two vertices

2a = 182

a=91

2x₁ = 318

x₁ = 159 ft. (We need the radius not the diameter)

356 ft. is from transverse axis down. Therefore: y₁=-356

Use (159,-356) for (x,y) to solve for b:

159²/91² - (-356)²/b² = 1

b ≈ 248.466

Let W = the width of the tower at 11 ft.

Therefore the coordinates in the hyperbola at 11 ft. are (W/2,-356+11) and (-W/2,-356+11)

Solve for W:

(W/2)²/91² - (-356+11)²/(248.466...)² = 1

W ≈ 311.4 ft.