# Equations of two concentric paraboloids

• MCB277
In summary: I don't think you will have much luck with that in general. For example, if you take the parabola ##y = x^2## and look for the curve that three units away and parallel "above" it, that is too high to fit above it near the vertex and you get a graph that looks like this:

## Homework Statement

I need to obtain the equation of 2 paraboloids separated by a distance L.

## Homework Equations

I think that the equations should be:

z_1=x^2+y^2
z_2=x^2+y^2-L

## The Attempt at a Solution

The problem is that when I plot the region between two inequations,
x^2+y^2>=z and z+L>=x^2+y^2
the region is not of size L

Thanks[/B]

What you have done is to construct two paraboloids that are separated by a distance ##L## in the ##z##-direction. Being separated by ##L## in the ##z##-direction is not the same thing as the closest point on the second surface to a given point on the first being a distance ##L## away. In order to solve the problem, you need to specify which of these is your intended problem and solve that problem.

Edit: It should also be pointed out that the curve that is everywhere a distance ##L## away from a paraboloid is not a paraboloid in itself.

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So, if I want to build a paraboloid with a certain thickness L, how would it be done?

MCB277 said:
So, if I want to build a paraboloid with a certain thickness L, how would it be done?
You cannot. A paraboloid is a hypersurface of one dimension lower than the space you are constructing it in. It does not have a thickness.

you are right there is no such thing, I want to find a surface parallel to the paraboloid given in that equation.

So for a given point on the paraboloid, what is the normal vector to the paraboloid?

MCB277
Thanks you !

MCB277 said:
Thanks you !
Do you think you could follow up here or via message how you proceeded from this point?

I made an attempt to do the analogous problem in one variable, starting with y = x^2 and its normal vectors, but I got buried in algebra trying to convert the resulting equations for "the point (x1, y1) which is distance L from (x, x^2) along the normal" to a closed form y1 = f(x1).

RPinPA said:
Do you think you could follow up here or via message how you proceeded from this point?

I made an attempt to do the analogous problem in one variable, starting with y = x^2 and its normal vectors, but I got buried in algebra trying to convert the resulting equations for "the point (x1, y1) which is distance L from (x, x^2) along the normal" to a closed form y1 = f(x1).

Since this isn't the OP's question, I will give a bit more suggestion to you. Let's write the parabola in vector form ##\vec r(t) = \langle t, t^2\rangle##. A tangent vector is ##r'(t) = \langle 1,2t \rangle## and a unit normal pointed down is ##\hat n(t) =\frac 1 {\sqrt{1+4t^2}}\langle 2t,-1\rangle##. Now let's say you want a parallel curve ##h## distance from and below the parabola, use$$\vec q(t) = \vec r(t) +h\hat n(t)$$Here's a picture with ##h=1##:

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Pretty much what I did, giving me ##<q_1(x), q_2(x)>## parametrically in terms of ##x##. But I was hoping to eliminate the ##x## so that I could analyze the result to describe exactly what kind of curve it is, if not a parabola.

RPinPA said:
Pretty much what I did, giving me ##<q_1(x), q_2(x)>## parametrically in terms of ##x##. But I was hoping to eliminate the ##x## so that I could analyze the result to describe exactly what kind of curve it is, if not a parabola.
I don't think you will have much luck with that in general. For example, if you take the parabola ##y = x^2## and look for the curve that three units away and parallel "above" it, that is too high to fit above it near the vertex and you get a graph that looks like this:

The equation for that is$$\vec R(t) =\langle t - \frac{6t}{\sqrt{1+4t^2}}, t^2 - \frac{3}{\sqrt{1+4t^2}} \rangle$$I haven't tried it, but looking at the image I'm guessing if you manage to put this in the ##y = f(x)## form, it won't be either simple or and some "standard" form you would recognize. It gets even more interesting if you do the same thing to an ellipse and put the second curve too far on the inside.

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• par2.jpg
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## 1. What is an equation of two concentric paraboloids?

An equation of two concentric paraboloids is a mathematical equation that represents two parabolic surfaces that share the same axis of symmetry. This means that the two paraboloids have the same vertex, but different curvatures.

## 2. How can I graph an equation of two concentric paraboloids?

To graph an equation of two concentric paraboloids, you can use a 3D graphing calculator or software, such as Desmos or Geogebra. You can also plot points by hand and connect them to create a visual representation of the paraboloids.

## 3. What is the significance of two concentric paraboloids in real-life applications?

Two concentric paraboloids are commonly used in optics and astronomy to represent the shape of satellite dishes, reflectors, and mirrors. They are also used in engineering and architecture to design structures with curved surfaces.

## 4. How do I find the focus and directrix of an equation of two concentric paraboloids?

The focus and directrix of an equation of two concentric paraboloids can be found by solving the equation for the variable that represents the focus and directrix. In general, the focus lies on the axis of symmetry and the directrix is a line perpendicular to the axis of symmetry passing through the vertex.

## 5. Can an equation of two concentric paraboloids have a negative coefficient?

Yes, an equation of two concentric paraboloids can have a negative coefficient for the squared term. This indicates that the paraboloids open downwards instead of upwards. However, the coefficient for the linear term must be equal to zero in order to have two concentric paraboloids.