Question

A cooling tower, such as the one shown in the figure

Answer 1

Step 1

State the standard form of a hyperbola with the horizontal transverse axis with center at (0,0)

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Step 2

For a hyperbola with the horizontal transverse axis with center at (0,0)

$\begin{gathered} Length\text{ of horizontal transverse axis }=48=2a \\ \text{Therefore} \\ a=\frac{48}{2}=24 \end{gathered}$

Points to work with include;

$\begin{gathered} one\text{ corner of the base -(50,-84)} \\ one\text{ corner of the top- (}x,36) \\ At\text{ the center- (}24,0) \end{gathered}$

Step 3

Using points (50,-84) to find b²

We have;

$\begin{gathered} \frac{50^2}{24^2}-\frac{(-84)^2}{b^2}=1 \\ \frac{50^2}{24^2}-\frac{7056}{b^2}=1 \\ \frac{50^2}{24^2}-1=\frac{7056}{b^2} \end{gathered}$$\begin{gathered} 3.340277778=\frac{7056}{b^2} \\ 3.340277778b^2=7056 \\ b^2=\frac{7056}{3.340277778} \\ b^2=\frac{7056}{3.340277778} \\ b^2=2112.399168 \end{gathered}$

Step 4

Solve for the x-coordinate of the right corner of the top

$\begin{gathered} The\text{ equation now becomes} \\ \frac{x^2}{24^2}-\frac{36^2}{2112.399168}=1 \\ \frac{x^2}{24^2}=1+\frac{36^2}{2112.399168} \\ \frac{x^2}{24^2}=\frac{1265}{784} \\ 784x^2=728640 \\ x^2=\frac{728640}{784} \\ x^2=\frac{45540}{49} \\ x=\sqrt[]{\frac{45540}{49}} \\ x=30.48586156 \end{gathered}$

But the diameter at the top = 2x

Therefore,

$\begin{gathered} 2(30.48586156)=60.97172312 \\ \approx60.97\text{meters} \end{gathered}$

Hence approximately to 2, decimal places the diameter at the top = 60.97 meters