Will a truck that is 10 feet wide carrying a load that reaches 12 feet above the ground clear the semielliptical arch on the one-way road that passes under the bridge shown in the figure on the right? 13 ft --- 52 ft It will clear the arch because the height of the archway of the bridge 5 feet from the center is approximately (Round to two decimal places as needed.) feet.

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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