Figure is missing, so i have attached it.
Answer:
it will clear the arch because the height of the archway of the bridge 5 feet from the center is approximately 12.76 ft
Step-by-step explanation:
The standard form of equation of an ellipse is;
x²/a² + y²/b² = 1
From the figure in the image attached, we can see that the radius is; a = 52/2 = 26 ft
While the value of b = 13 ft
Thus;
x²/26² + y²/13² = 1
x²/676 + y²/169 = 1
We want to find the height of the archway of the bridge 5 feet from the center.
Thus, we will plug in 5 for x to get;
5²/676 + y²/169 = 1
(25/676) + (y²/169) = 1
Multiply through by 676 to get;
25 + 4y² = 676
4y² = 676 - 25
y² = 651/4
y² = 162.75
y = 12.76 ft
Thus height of the truck is 12 ft and so it will clear the arch because the height of the archway of the bridge 5 feet from the center is approximately 12.76 ft
Height of the archway 5 feet from the center is required.
The height of the archway 5 feet away from the center is 12.757 feet.
Semi-major axis = [tex]a=\dfrac{52}{2}=26\ \text{feet}[/tex]
Semi-minor axis = [tex]b=13\ \text{feet}[/tex]
The formula for ellipse is
[tex]\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\\\Rightarrow \dfrac{x^2}{26^2}+\dfrac{y^2}{13^2}=1[/tex]
At 10 feet is the width of the truck.
The truck is passing through the center so the [tex]x=\dfrac{10}{2}=5[/tex]
[tex]\dfrac{5^2}{26^2}+\dfrac{y^2}{13^2}=1\\\Rightarrow y=\sqrt{(1-\dfrac{5^2}{26^2})\times 13^2}\\\Rightarrow y=12.757\ \text{feet}[/tex]
Now [tex]y=12.757\ \text{feet}>12\ \text{feet}[/tex]
So, the height of the archway 5 feet away from the center is 12.757 feet.
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