an arch is in the form of a semi ellipse it is 50m wide at the base and has height of 20m how wide is the arch at the height of 10m above the base?
an astronaut is to be fired into an ellipitical orbit about the earth having a minimum altitude of 800km and a maximum altitude of 5400km find the equation of the curve followed by the astronaut consider the radius of the eathr to be 6400km
For the ellipse, you can see that
a = 50/2 = 25
b = 20
So, put that into standard form, then find 2x when y=10
To find the width of the arch at a height of 10m above the base, we can use the formula for the equation of a semi-ellipse.
The equation of a semi-ellipse in standard form is given by:
(x^2/a^2) + (y^2/b^2) = 1
where a is the semi-major axis (half of the width) and b is the semi-minor axis (half of the height).
Given that the width of the arch at the base is 50m and the height is 20m, we have:
a = 50/2 = 25m
b = 20/2 = 10m
Now, we need to solve the equation of the semi-ellipse for the width (x) at a height (y) of 10m above the base.
Plugging in the values into the equation, we get:
(x^2/25^2) + (10^2/10^2) = 1
Simplifying the equation:
(x^2/625) + 1 = 1
(x^2/625) = 0
x^2 = 0
The width of the arch at a height of 10m above the base is 0m.
Please note that there seems to be an issue with the given measurements or the shape described. The width cannot be 0m at any height above the base for a semi-ellipse.
To find the width of the arch at a certain height above the base, we can use the formula for the equation of a semi ellipse:
x^2 / (a^2) + y^2 / (b^2) = 1
where "x" is the width at a given height above the base, "y" is the height, "a" is the width at the base, and "b" is the height at the highest point of the arch.
In this case, the base width (a) is 50m and the height (b) is 20m. Let's use these values to find the width at 10m above the base.
Substituting the values into the equation, we have:
x^2 / (50^2) + 10^2 / (20^2) = 1
Simplifying the equation, we have:
x^2 / 2500 + 100 / 400 = 1
x^2 / 2500 + 1 / 4 = 1
Now, we can solve for x.
Multiply the equation by 2500 to eliminate the fractions:
x^2 + 625 = 2500
Subtract 625 from both sides:
x^2 = 1875
Take the square root of both sides:
x = √1875
Now, let's calculate the width of the arch:
x ≈ 43.30
Therefore, the arch width at a height of 10m above the base is approximately 43.30m.