Chapter 10.4, Problem 77AYU

Nuclear Power Plaut 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?

Source: Bay Area Air Quality Management District

To determine

How tall is the tower if the diameter at the top of the tower is 300 feet.

Answer to Problem 77AYU

$592.4$ feet approx.

Explanation of Solution

Given:

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. The cooling tower has a base diameter of 400 feet and the diameter at its narrowest point, 360 feet above the ground, is 200 feet. The diameter at the top of the tower is 300 feet.

Formula used:

Equation of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $b^2=c^2-a^2$.

Calculation:

Let the equation of the hyperbola at the origin be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.

The diameter at its narrowest point, which is the center diameter is 200 feet. So we have $a=\frac{200}{2}=100$. We also know that the base diameter is 400 feet.

Since the center of hyperbola is at the origin and 360 feet above the ground, the points $\left(200,-360\right)$ and $\left(-200,-360\right)$ must be on the graph of hyperbola.

Therefore,

$\frac{{\left(200\right)}^2}{{100}^2}-\frac{{\left(-360\right)}^2}{b^2}=1$

$4-\frac{{360}^2}{b^2}=1$

$3=\frac{{360}^2}{b^2}$

$b^2=43,200$

$b=120\sqrt{3}$

The equation of the hyperbola is

$\frac{{150}^2}{10000}-\frac{y^2}{43200}=1$

$y^2=54000$

$y\approx 232.4$

The height of the tower is $232.4+360=592.4$ feet approx.