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)

`(x^2)/(a^2)-(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=(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} (50^2)/(24^2)-((-84)^2)/(b^2)=1 \\ (50^2)/(24^2)-(7056)/(b^2)=1 \\ (50^2)/(24^2)-1=(7056)/(b^2) \end{gathered}`
`\begin{gathered} 3.340277778=(7056)/(b^2) \\ 3.340277778b^2=7056 \\ b^2=(7056)/(3.340277778) \\ b^2=(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} \\ (x^2)/(24^2)-(36^2)/(2112.399168)=1 \\ (x^2)/(24^2)=1+(36^2)/(2112.399168) \\ (x^2)/(24^2)=(1265)/(784) \\ 784x^2=728640 \\ x^2=(728640)/(784) \\ x^2=(45540)/(49) \\ x=\sqrt[]{(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