Answer:
Can give you the example then you have to figure out.
Step-by-step explanation:
We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: \displaystyle \frac{{x}^{2}}{{a}^{2}}-\frac{{y}^{2}}{{b}^{2}}=1
a
2
x
2
−
b
2
y
2
=1, where the branches of the hyperbola form the sides of the cooling tower. We must find the values of \displaystyle {a}^{2}a
2
and \displaystyle {b}^{2}b
2
to complete the model.
First, we find \displaystyle {a}^{2}a
2
. Recall that the length of the transverse axis of a hyperbola is \displaystyle 2a2a. This length is represented by the distance where the sides are closest, which is given as \displaystyle \text{ }65.3\text{ } 65.3 meters. So, \displaystyle 2a=602a=60. Therefore, \displaystyle a=30a=30 and \displaystyle {a}^{2}=900a
2
=900.
To solve for \displaystyle {b}^{2}b
2
, we need to substitute for \displaystyle xx and \displaystyle yy in our equation using a known point. To do this, we can use the dimensions of the tower to find some point \displaystyle \left(x,y\right)(x,y) that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y-axis bisects the tower, our x-value can be represented by the radius of the top, or 36 meters. The y-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,
x
2
a
2
−
y
2
b
2
=
1
Standard form of horizontal hyperbola
.
b
2
=
y
2
x
2
a
2
−
1
Isolate
b
2
=
(
79.6
)
2
(
36
)
2
900
−
1
Substitute for
a
2
,
x
,
and
y
≈
14400.3636
Round to four decimal places
The sides of the tower can be modeled by the hyperbolic equation
\displaystyle \frac{{x}^{2}}{900}-\frac{{y}^{2}}{14400.3636 }=1,\text{or}\frac{{x}^{2}}{{30}^{2}}-\frac{{y}^{2}}{{120.0015}^{2} }=1
900
x
2
−
14400.3636
y
2
=1,or
30
2
x
2
−
120.0015
2
y
2
=1
The shape of hyperbolic cooling tower is a conic section that can be described with a formula equation.
a = 28.5, b ≈ 52.65
[tex]Equation \ of \ hyperbola;\ \dfrac{x^2}{28.5^2} - \dfrac{(y-155)^2}{56.62^2} = 1[/tex]
Reasons:
The parameters of the hyperbolic cooling tower are;
Minimum diameter = 57 feet
The height at which the diameter occurs = 155 ft.
The diameter of the tower = 175 feet
Total height of the tower = 200 feet
Height at which the center of the hyperbola occurs = 155 feet
Solution:
The equation of a hyperbola is; [tex]\dfrac{(x - h)^2}{a^2} - \dfrac{(y- k)^2}{b^2} = 1[/tex]
The center of the hyperbola = (0, 155)
[tex]a = \dfrac{57}{2} = 28.5[/tex]
At y = 155,
Therefore;
[tex]\dfrac{x^2}{28.5^2} - \dfrac{(y -155)^2}{b^2} = 1[/tex]
At y = 200 feet [tex]x = \dfrac{75 \ feet}{2} = 37.5 \ feet[/tex]
Therefore;
[tex]\dfrac{37.5^2}{28.5^2} - \dfrac{(200 -155)^2}{b^2} = 1[/tex]
b² = 2769.0340909
b ≈ 52.62
The equation of the hyperbola is therefore;
[tex]\dfrac{x^2}{28.5^2} - \dfrac{(y-155)^2}{56.62^2} = 1[/tex]
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