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.
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
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
To find the vertical clearance at a point 13ft to the right from the center, we need to find the equation of the ellipse that represents the shape of the tunnel.
First, let's find the length of the semi-major axis (a). Since the total width of the tunnel is 50ft, the length of the semi-major axis is half of that, which is 25ft.
Next, we need to find the length of the semi-minor axis (b). To determine the length of b, we can use the height of the tunnel, which is 28ft. Since the tunnel is shaped like a semi-eclipse, the height of the tunnel is equal to the length of the semi-minor axis. Therefore, b = 28ft.
The equation of an ellipse with the center at the origin (0,0) and the semi-major axis (a) along the x-axis and the semi-minor axis (b) along the y-axis is given by:
(x^2/a^2) + (y^2/b^2) = 1
Plugging in the values we found, the equation becomes:
(x^2/25^2) + (y^2/28^2) = 1
To find the vertical clearance at a point 13ft to the right from the center, we need to find the corresponding y-value on the ellipse.
Plugging in x = 13 into the equation, we have:
(13^2/25^2) + (y^2/28^2) = 1
Simplifying the equation, we get:
(169/625) + (y^2/784) = 1
Rearranging the equation to solve for y, we have:
y^2/784 = 1 - (169/625)
y^2/784 = (625 - 169)/625
y^2/784 = 456/625
Cross-multiplying, we get:
625y^2 = 784 * 456
625y^2 = 354,624
Dividing both sides by 625, we have:
y^2 = 354,624/625
y^2 = 567.3984
Taking the square root of both sides, we get:
y = √567.3984
y ≈ 23.8
Therefore, the vertical clearance at a point 13ft to the right from the center is approximately 23.8ft.