A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet. What is the height of the tunnel 5 feet from the edge? Round your answer to the hundredths place. O9.99 feet O 13.82 feet O24.65 feet O 25.01 feet
A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet. What is the height of the tunnel 5 feet from the edge? Round your answer to the hundredths place. O9.99 feet O 13.82 feet O24.65 feet O 25.01 feet
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Semieliptical Arch Tunnel Problem
#### Problem Statement
A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet. What is the height of the tunnel 5 feet from the edge? Round your answer to the hundredths place.
#### Options:
- 0.99 feet
- 13.82 feet
- 24.85 feet
- 25.01 feet
#### Solution Explanation
To solve this problem, we would need to use the equation for a semiellipse. The general form of an ellipse equation is:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
where:
- \(2a\) is the width of the tunnel (distance along the x-axis), so \(a = \frac{60}{2} = 30\) feet.
- \(b\) is the height of the ellipse (distance along the y-axis), so \(b = 25\) feet.
We need to find the height (y) at \(x = 30 - 5 = 25\) feet from the center of the ellipse (5 feet from the edge).
Plug \(x = 25\) into the ellipse equation:
\[\frac{25^2}{30^2} + \frac{y^2}{25^2} = 1\]
Solve for \(y\):
\[\frac{625}{900} + \frac{y^2}{625} = 1\]
Simplify fractions and isolate \( y^2 \):
\[ \frac{625}{900} + \frac{y^2}{625} = 1 \]
\[ \frac{625}{900} + \frac{y^2}{625} = \frac{900}{900} \]
\[ \frac{625}{900} + \frac{y^2}{625} = \frac{900}{900} \]
\[ \frac{y^2}{625} = \frac{900}{900} - \frac{625}{900} \]
\[ \frac{y^2}{625} = \frac{275}{900} \]
\[ y^2 = 625 \ast \frac{275}{900} \]
\[ y = \sqrt{625](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3b617bc-0d34-47a7-bf50-98a510ba10e9%2Fb283111d-ed13-4126-a224-9517a002f493%2F3iqy67_processed.png&w=3840&q=75)
Transcribed Image Text:### Semieliptical Arch Tunnel Problem
#### Problem Statement
A tunnel is constructed with a semielliptical arch. The width of the tunnel is 60 feet, and the maximum height at the center of the tunnel is 25 feet. What is the height of the tunnel 5 feet from the edge? Round your answer to the hundredths place.
#### Options:
- 0.99 feet
- 13.82 feet
- 24.85 feet
- 25.01 feet
#### Solution Explanation
To solve this problem, we would need to use the equation for a semiellipse. The general form of an ellipse equation is:
\[\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\]
where:
- \(2a\) is the width of the tunnel (distance along the x-axis), so \(a = \frac{60}{2} = 30\) feet.
- \(b\) is the height of the ellipse (distance along the y-axis), so \(b = 25\) feet.
We need to find the height (y) at \(x = 30 - 5 = 25\) feet from the center of the ellipse (5 feet from the edge).
Plug \(x = 25\) into the ellipse equation:
\[\frac{25^2}{30^2} + \frac{y^2}{25^2} = 1\]
Solve for \(y\):
\[\frac{625}{900} + \frac{y^2}{625} = 1\]
Simplify fractions and isolate \( y^2 \):
\[ \frac{625}{900} + \frac{y^2}{625} = 1 \]
\[ \frac{625}{900} + \frac{y^2}{625} = \frac{900}{900} \]
\[ \frac{625}{900} + \frac{y^2}{625} = \frac{900}{900} \]
\[ \frac{y^2}{625} = \frac{900}{900} - \frac{625}{900} \]
\[ \frac{y^2}{625} = \frac{275}{900} \]
\[ y^2 = 625 \ast \frac{275}{900} \]
\[ y = \sqrt{625
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