Topic: Quadric surfaces.
I have two homework problems, and they are quite a doozy for me. So the point is, I am given details of a quadric surface on the Cartesian coordinate system, and thus have to use the data to construct an equation through it.
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Find an equation for the surface consisting of all points P for which the distance from P to the x-axis is 4 times the distance from P to the yz-plane.
(1) So this a cone, so its equation will be the form of (x^2/a^2)+(y^2/b^2)-(z^2/c^2)=0.
Unfortunately, this is about as far as I can get. Same situation with the next one.
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A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet. The diameter at the base is 280 m and the minimum diameter, 500 m above the base, is 180 m. Find an equation for the tower. (Assume the center is at the origin with axis the z-axis and the minimum diameter is at the center.)
(1) A hyperboloid of one sheet with horizontal circles will have the equation of (x^2/a^2)+(y^2/a^2)-(z^2/b^2)=1.
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Honestly, the only attempts I have made have been on the second problem. My strategy was at first to set z=0 and find that the horizontal traces should construct a circle with radius of 180 m. But now I get stuck -- don't hyperboloids of one sheet extend infinitely? How do I know if the height is 1,000 m? I'm pretty sure I don't have to worry about the bases extending to a radius of 280 m since those will extend with height, but determining the height is a different animal to me.
The first problem, I am a little lost in the language.
Insight is appreciated. I do not expect you to hand me a silver platter answer.
Thanks.
For P:(x,y,z)
distance from P to the x-axis: √(y^2+z^2)
distance from P to the yz-plane: x
So,
√(y^2+z^2) = 4x
y^2+z^2 = 16x^2
x^2/1 - y^2/16 - z^2/16 = 0
I'll get back to you on the rest.
For the first problem, you are asked to find an equation for the surface consisting of all points P for which the distance from P to the x-axis is 4 times the distance from P to the yz-plane. Let's break it down.
To construct the equation of the surface, we need to understand the given information and translate it into mathematical terms.
1. The distance from P to the x-axis is 4 times the distance from P to the yz-plane.
Let d1 be the distance from P to the x-axis, and d2 be the distance from P to the yz-plane. We can express this relationship as:
d1 = 4d2
2. The equation for a quadric surface is typically of the form:
(x^2/a^2) + (y^2/b^2) + (z^2/c^2) = 1 or 0, depending on the specific surface.
Now, let's proceed with finding the equation for this particular quadric surface:
1. Start by expressing the distance from P to the x-axis and the yz-plane in terms of the coordinates of P. Given that the yz-plane is defined by x = 0, we can express d1 and d2 as follows:
d1 = |x|
d2 = √(y^2 + z^2)
2. Since d1 is four times d2, we can rewrite the equation as:
|x| = 4√(y^2 + z^2)
3. Square both sides to remove the square root:
x^2 = 16(y^2 + z^2)
4. Rearrange the equation to isolate the variables and manipulate it into the standard form of a quadric surface:
x^2 - 16y^2 - 16z^2 = 0
Therefore, the equation for the surface consisting of all points P for which the distance from P to the x-axis is 4 times the distance from P to the yz-plane is x^2 - 16y^2 - 16z^2 = 0.
Let's move on to the second problem, the cooling tower for a nuclear reactor in the shape of a hyperboloid of one sheet:
1. Given that the tower is a hyperboloid of one sheet with horizontal circles, we can use the equation:
(x^2/a^2) + (y^2/b^2) - (z^2/c^2) = 1
2. You correctly determined that the base has a diameter of 280m, which means the radius is 140m. Thus, we have a = 140.
3. The minimum diameter, 500m above the base, is 180m, which means the radius is 90m. This gives us b = 90.
4. Since the minimum diameter is at the center and the center is at the origin, we have c = height of the tower = 500m.
5. Now we can substitute the values of a, b, and c into the equation to get the final equation for the tower:
(x^2/140^2) + (y^2/90^2) - (z^2/500^2) = 1
So the equation for the cooling tower, in the shape of a hyperboloid of one sheet, is (x^2/140^2) + (y^2/90^2) - (z^2/500^2) = 1.
It's essential to note that hyperboloids of one sheet do extend infinitely, and the equation we derived represents the surface at any given height within the tower.
I hope this explanation helps you solve these problems on your own. Let me know if you have any further questions!