Chapter 10.4, Problem 45E

(a)

To determine

To write: An equation of the hyperbola used to generate the hyperboloid, if the cooling tower is 150 feet wide at its narrowest point.

Answer to Problem 45E

  $\frac{x^2}{{75}^2}-\frac{y^2}{{100}^2}=1$

Explanation of Solution

Given information: Nuclear Power: A nuclear cooling tower is a hyperboloid , that is, a hyperbola rotated around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the cooling tower has an eccentricity of $\frac{5}{3}$ .

  

Calculation:

  

If its 150 feet wide at the narrowest point (conjugate axis), then a = 75.

And,

  $e=\frac{c}{75}=\frac{5}{3}\to c=125$

If: $a^2+b^2=c^2\to b^2={125}^2-{75}^2=10,000$

This gives us:

  $\frac{x^2}{{75}^2}-\frac{y^2}{{100}^2}=1$

(b)

To determine

To find: The radius of the top and the base of the tower, if the tower is 450 feet tall, the top is 100 feet above the center of the hyperbola and the base is 350 feet below the center.

Answer to Problem 45E

  $\begin{array}{l}\text{Top radius}=106.06\text{ feet}\\ \text{Base radius}=273.004\text{ feet}\end{array}$

Explanation of Solution

Given information: Nuclear Power: A nuclear cooling tower is a hyperboloid , that is, a hyperbola rotated around its conjugate axis. Suppose the hyperbola used to generate the hyperboloid modeling the shape of the cooling tower has an eccentricity of $\frac{5}{3}$ .

  

Calculation:

  

Given:

  $\begin{array}{l}\frac{{\left(r_1\right)}^2}{{75}^2}-\frac{{100}^2}{{100}^1}=1\to \frac{{\left(r_1\right)}^2}{{75}^2}=2\\ r_1=\sqrt{2\times {75}^2}=75\sqrt{2}=106.06\text{ feet}\end{array}$

  $\begin{array}{l}\frac{{\left(r_2\right)}^2}{{75}^2}-\frac{{\left(-350\right)}^2}{{100}^1}=1\to \frac{{\left(r_2\right)}^2}{{75}^2}=1+\frac{49}{4}\\ \frac{{\left(r_2\right)}^2}{{75}^2}=\frac{53}{4}\\ r_2=\frac{75\sqrt{53}}{2}\\ r_2=273.004\text{}\text{feet}\end{array}$