An arch has the shape of a semi ellipse that is 10 ft. long and 3 ft. high at the center. How high is the arch at a point 2 ft. from the center? *
We are told that 2a = 10 and 2b = 3
so the equation of the ellipse is
x^2/25 + y^2/2.25 = 1 , using x^2/a^2 + y^2/b^2 = 1
so when x = 2
4/25 + y^2/2.25 = 1
y^2/2.25 = 21/25
y^2 = 2.23(21/25) = 1.89
y = √1.89 = 1.375 ft
we have a = 5, b = 3
x^2/25 + y^2/9 = 1
now find y when x = 2
Agree with oobleck.
I read the height to be the whole ellipse not the semi.
To find the height of the arch at a point 2 ft. from the center, you can use the equation of a semi ellipse:
𝑦 = √(𝑏^2 − 𝑏^2(𝑥/𝑎)^2),
where 𝑦 is the height, 𝑥 is the horizontal distance from the center, 𝑎 is the length of the semi ellipse, and 𝑏 is the height at the center.
In this case, 𝑎 (length of the semi ellipse) is given as 10 ft., 𝑏 (height at the center) is given as 3 ft., and 𝑥 (the horizontal distance from the center) is 2 ft.
Plugging in these values into the equation, you can find the height at that particular point:
𝑦 = √(3^2 − 3^2(2/10)^2),
= √(9 − 9(2/10)^2),
= √(9 − 9(4/100)),
= √(9 − 9(0.04)),
= √(9 − 0.36),
= √8.64,
≈ 2.94 ft.
Therefore, the height of the arch at a point 2 ft. from the center is approximately 2.94 ft.