MATHEMATICS COLLEGE

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.

Answers

Answer 1

Answer:

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 = $a=(52)/(2)=26\ \text{feet}$

Semi-minor axis = $b=13\ \text{feet}$

The formula for ellipse is

$(x^2)/(a^2)+(y^2)/(b^2)=1\n\Rightarrow (x^2)/(26^2)+(y^2)/(13^2)=1$

At 10 feet is the width of the truck.

The truck is passing through the center so the $x=(10)/(2)=5$

$(5^2)/(26^2)+(y^2)/(13^2)=1\n\Rightarrow y=\sqrt{(1-(5^2)/(26^2))* 13^2}\n\Rightarrow y=12.757\ \text{feet}$

Now $y=12.757\ \text{feet}>12\ \text{feet}$

So, the height of the archway 5 feet away from the center is 12.757 feet.

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Answer 2

Answer:

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

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Answers

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Answer:

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Answers

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Answers

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