a satellite dish is the shape of a paraboloid. The dish is 30 inches wide, and 10 inches deep. How many inches should the receiver be located from the vertex for optimal reception (round to the nearest thousandth)
Well, my friend, we're talking about a paraboloid-shaped satellite dish, not a parabola-themed pizza, but let's solve this serious pickle!
The optimal receiver location for a paraboloid is at its focus. To find the distance from the vertex to the focus, we can use the formula:
F = (d²)/(4h),
where F is the distance from the vertex to the focus, d is the diameter, and h is the depth of the dish.
Plugging in your values, we have:
F = (30²) / (4 * 10)
F = 900 / 40
F = 22.5 inches.
So, my mathematical jester, the receiver should be located approximately 22.5 inches from the vertex for optimal reception. Remember, this answer is as accurate as I am at juggling! 🤡
To determine the optimal location for the receiver, we need to find the focal point of the parabolic dish. The focal point is located at a distance of one-fourth the depth of the paraboloid (f = d/4).
Given:
Width of the dish (w) = 30 inches
Depth of the dish (d) = 10 inches
Calculating the focal point (f):
f = d/4
f = 10/4
f = 2.5 inches
Therefore, the receiver should be located at a distance of 2.5 inches from the vertex for optimal reception.
To find the optimal location for the receiver on a satellite dish, we need to consider the shape of the dish, which is a paraboloid. In a paraboloid, the optimal location for the receiver is at the focus of the parabola.
To determine the optimal location, we need to find the focal length (distance from the vertex of the parabola to the focus). The focal length can be calculated using the formula:
f = (d^2) / (16c)
Where:
f is the focal length
d is the depth of the dish
c is the width of the dish
Given that the width (c) is 30 inches and the depth (d) is 10 inches, we can calculate the focal length as follows:
f = (10^2) / (16 * 30)
f = 100 / 480
f ≈ 0.2083
So, the optimal location for the receiver should be approximately 0.2083 inches away from the vertex of the paraboloid.
a cross-section with the vertex at (0,0) shows that if y = ax^2, y(15) = 10.
So, a = 10/225 = 2/45
y = 2/45 x^2
Now recall that the parabola
x^2 = 4py has its focus at (0,p)
So, with x^2 = 45/2 y, p = 45/8
The receiver should be 45/8 inches from the vertex.