Asked by Angie
A railroad tunnel is shaped like a semi-eclipse. The height of the tunnel is 28ft and the total width is 50ft. Find the vertical clearance at a point 13ft to the right from the center. Please show all work that leads to your answer. Round your final answer to the nearest tenth.
Answers
Answered by
Steve
Let the ellipse be modeled by
x^2/a^2 + y^2/b^2 = 1
Since the base has width 50, and the height is 28, we have
b = 28
a = 25
and the ellipse is
x^2/25^2 + y^2/28^2 = 1
When x=13, y=23.9
x^2/a^2 + y^2/b^2 = 1
Since the base has width 50, and the height is 28, we have
b = 28
a = 25
and the ellipse is
x^2/25^2 + y^2/28^2 = 1
When x=13, y=23.9
Answered by
Bosnian
Standard form equation of an ellipse:
x² / a² + y² / b² = 1
where:
a is the semimajor axis
b is the semiminor axis
In this case:
a = total width / 2 = 50ft / 2 = 25ft
b = The height = 28ft
x² / a² + y² / b² = 1
x² / 25² + y² / 28² = 1
The vertical clearance at a point 13ft mean:
x = 13
x² / 25² + y² / 28² = 1
13² / 25² + y² / 28² = 1
Subtract 13² / 25² to both sides
13² / 25² + y² / 28² *- 13² / 25² = 1 - 13² / 25²
y² / 28² = 1 - 13² / 25²
y² / 28² = 25² / 25² - 13² / 25²
y² / 28² = ( 25² - 13² ) / 25²
Multiply both sides by 28²
y² = 28² ∙ ( 25² - 13² ) / 25²
Take the square root of both sides
y = ± √ [ 28² ∙ ( 25² - 13² ) / 25² ]
y = ± √ 28² ∙ √ ( 25² - 13² ) / √ 25²
y = ± 28 ∙ √ ( 25² - 13² ) / 25
y = ± 28 ∙ √ ( 625 - 169 ) / 25
y = ± 28 ∙ √ 456 / 25
y = ± 28 ∙ √ ( 4 ∙ 114 ) / 25
y = ± 28 ∙ √ 4 ∙ √ 114 / 25
y = ± 28 ∙ 2 ∙ √ 114 / 25
y = ± 56 ∙ √ 114 / 25
y = ± 56 ∙ 10.677078252 / 25
y = ± 597.916382112 / 25
y = ± 23.91665528448
y = ± 23.9 ft
rounded to the nearest tenth
x² / a² + y² / b² = 1
where:
a is the semimajor axis
b is the semiminor axis
In this case:
a = total width / 2 = 50ft / 2 = 25ft
b = The height = 28ft
x² / a² + y² / b² = 1
x² / 25² + y² / 28² = 1
The vertical clearance at a point 13ft mean:
x = 13
x² / 25² + y² / 28² = 1
13² / 25² + y² / 28² = 1
Subtract 13² / 25² to both sides
13² / 25² + y² / 28² *- 13² / 25² = 1 - 13² / 25²
y² / 28² = 1 - 13² / 25²
y² / 28² = 25² / 25² - 13² / 25²
y² / 28² = ( 25² - 13² ) / 25²
Multiply both sides by 28²
y² = 28² ∙ ( 25² - 13² ) / 25²
Take the square root of both sides
y = ± √ [ 28² ∙ ( 25² - 13² ) / 25² ]
y = ± √ 28² ∙ √ ( 25² - 13² ) / √ 25²
y = ± 28 ∙ √ ( 25² - 13² ) / 25
y = ± 28 ∙ √ ( 625 - 169 ) / 25
y = ± 28 ∙ √ 456 / 25
y = ± 28 ∙ √ ( 4 ∙ 114 ) / 25
y = ± 28 ∙ √ 4 ∙ √ 114 / 25
y = ± 28 ∙ 2 ∙ √ 114 / 25
y = ± 56 ∙ √ 114 / 25
y = ± 56 ∙ 10.677078252 / 25
y = ± 597.916382112 / 25
y = ± 23.91665528448
y = ± 23.9 ft
rounded to the nearest tenth
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