schoolSo, we've seen planes and spheres as far as surfaces go. Now we're gonna wrap up the surfaces with the last bit, starting with the simpler cylinders and then generalizing to all quadric surfaces.
Now, one thing you need to do is remove the idea that a cylinder is two circles connected. It's not. A cylinder is a surface that consists of all lines (or rulings) that are parallel to a given line and pass through a given plane curve.
If your given plane curve is a flat circle, and we take every line that can pass through it that is parallel to a given line perpendicular to the circle, you have your circle-based cylinder!
But consider this! Consider the graph of z = x2. You draw your XYZ axis and notice that since the function z does not involve y at all, every line is parallel to the y axis and runs through the ENTIRE y axis. And then all you have left to determine is the shape of the graph, which, if you simply do an XZ plane where X is the horizontal, you see a parabola. Extending this infinitely through the y-axis, you get a shape that looks exactly like a folded piece of paper. This is called a parabolic cylinder.
So, you can clearly see cylinders don't even need to be closed!
If you have x2 + y2 = 1 through the XYZ, then you see it's a circular cylinder, the circle being on the XY plane and then going up and down infinitely, the circles' radii = 1. So, drawing your XYZ axis with Z pointing up, it's just a vertical, circular cylinder. On the other hand, y2 + z2 = 1 is a circular cylinder that goes in the direction of the X axis.
Now you should know what cylinders generally look like. Consider the parabolic cylinder example again. Mathematically, we are saying that any vertical plane with the equation y = k (parallel to the XZ plane) intersects the graph in a curve with an equation z = x2.
And now that that's in your head, let's move on to the general quadric surface.
A quadric surface is the graph of any second-degree equation in three variables, and the most general form is
Ax2 + By2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0
where A,B,C,D,E,F,G,H,I,J are all constants.
This is gonna get kinda tricky to explain how to draw, so bear with me.
Usually, quadric surfaces can be rotated and translated about until they reach one of two standard forms:
Ax2 + By2 + Cz2 + J = 0
Ax2 + By2 + Iz = 0
A note: they do not HAVE to be in THAT standard form. Look at them. You see all three variables x,y,z, and if they are all squared, a constant. The first one will always look like that, but the second could be Ax2 + Cz2 + Hy = 0 instead. Just so long as one variable is not squared.
In fact, there are six common shapes with formulas associated with them. Here they are--
ellipsoid: x2/a2 + y2/b2 + z2/c2 = 1. All traces (we'll talk about traces in a moment) are ellipses. If a = b = c, then the ellipsoid is a sphere. Looks like a 3D ellipse.
elliptic paraboloid: z/c = x2/a2 + y2/b2. Horizontal traces are ellipses. Vertical traces are parabolas. Whatever variable is in the first power (in this case, z) indicates the axis (or direction) of the paraboloid. Looks like a rounded funnel.
cone: z2/c2 = x2/a2 + y2/b2. Horizontal traces are ellipses. Vertical traces in the planes x = k, y = k are hyperbolas if k =/= 0, but pairs of lines if k = 0 (we'll discuss this "k" stuff later). Looks like two funnels attached at the tiny ends.
hyperboloid of one sheet: x2/a2 + y2/b2 - z2/c2 = 1. Horizontal traces are ellipses. Vertical traces are hyperbolas. Axis of symmetry corresponds to the variable whose coefficient is negative (goes in the direction of this axis). Looks like two lampshades attached at the smaller ends.
hyperboloid of two sheets: -x2/a2 - y2/b2 + z2/c2 = 1. Horizontal traces in z = k are ellipses if k > c or k < -c. Vertical traces are hyperbolas. The two minus signs indicate two sheets. Looks like a bump down on the ground, and then you mirror this so you have an upside-down bump floating in the sky.
hyperbolic paraboloid: z/c = x2/a2 - y2/b2. Horizontal traces are hyperbolas. Vertical traces are parabolas. Has varying shapes (just Google Image this sucker).
Okay, wait. Where the hell did this "k" come from, and what are these "traces"? Well, first off, they're related so we hit them both in the same explanation. Second of all, we use them to actually determine how we would draw this sucker.
First off, let's talk about these "traces." Traces are our attempts at taking a 3D object and making it 2D at various angles. To be more specific, instead of graphing something in the XYZ space, we're going to graph something in the XY, YZ, and XZ planes. After that, it's up to our ability to "see in 3D" that will helps us connect these three planes into the 3D space.
Let's consider the surface z = y2 - x2. From our table, we can see this is a hyperbolic paraboloid, where a,b,c = 1 (the table says x - y, but our problem has y - x. This is okay because it basically means we're going in a different direction, is all. The hyperbolic paraboloid will just have one part negative in this format).
So, how do we draw this? First, we need to draw 3 planes: ZY (make Y the horizontal), ZX (make X the horizontal), and YX (make X the horizontal).
Now, let's start with the ZY plane. From z = y2 - x2, we need to get this in terms of ONLY z and y (since ZY plane does not care about x). This means we need to remove the x2. Well, suppose x is no longer a variable, but is now some constant k. x = k, so x2 = k2, and we have z = y2 - k2. You should make the connection that this looks VERY similar to y = x2 - c, which is a parabola. So, we're drawing parabolas, and we do so by picking arbitrary values for k. If k = 0, we have z = y2. If k = 1, we have z = y2 - 1. If k = 2, we have z = y2 - 4, etc.
As you can tell, because it's k2, no matter what value we put in, it will always turn negative because of the minus sign in front of the k2. This means we only look at the parabolas whose z-intercepts are 0 and lower. So you draw them. These are called "traces in x = k."
Now, let's look at the ZX plane. We need to remove y such that we only have a function of z and x, and we do so by, again, letting y = k. This gives us z = k2 - x2. By letting k = anything, we see that these are upside-down (negative) parabolas whose Z intercepts now start with 0 and go to positive infinity, so we can draw them out just like we did the last one. And these are our traces in y = k.
Lastly, it's the YX plane. We gotta get rid of z, so we let z = k. This gives us k = y2 - x2. Suppose k = 0, then we have y2 = x2, or + y = + x. This is two lines (y = x, y = -x), so we draw out those two lines and basically make an X on our YX plane. Now, if we let k = 1, we get 1 = y2 - x2. Manipulating, we get x2 = y2 - 1, or +x = + sqrt(y2 - 1), which is a hyperbola that opens up and down. By letting k = -1, we get the same thing, only this time the hyperbola opens left and right By continuing to change k, we get hyperbolas that keep filling up the space left in that plane, and these are our traces in z = k.
Now, the hard part is trying to put this all together in the XYZ space. First, you definitely need to draw the XYZ axes. Now, taking one at a time, say, starting with x = k, you need to sort of "lightly draw" each parabola. In the 2D traces, you drew parabolas above and under each other because k changed the z-intercepts. Here, you'll also be changing the front-back (assuming you draw a standard XYZ axis system, where X points front and back) to accommodate x = k. It's not really THAT hard. You know how to T Table stuff: plug stuff in for, say, X, and then plug something in for Y and find Z, and just keep doing this.
Doing the same for the y = k traces, you simply also account for left-right (assuming standard XYZ coordinates where the Y axis goes left and right) changes, too, as y changes with k.
And then you do the same for the z = k traces, accounting for up-down movement as z changes (standard XYZ coordinates have Z axis going up-down).
And then you can MAYBE see the picture.
In all seriousness, you will probably never be asked to manually graph such an oblique shape. It's really too complicated. Ellipsoids, spheres, cones, and paraboloids are easy, but hyperbolic paraboloids are really just ugly looking when trying to do them manually. You will probably only graph them in software (like Mathematica, MathCAD, MATLAB, etc.), but you will most likely be expected to graph the traces. And, really, that's the important part. As long as you can see what's going on in each of the planes, that's more important than seeing the overall picture because in physical analysis, you will be a LOT more concerned with 2D mappings than you will ever be with the entire 3D object. So this is fine for those of us who can't think in 3D (I know I can't....).
So, let's do some problems.
---------
1) a. What does the equation y = x2 represent as a curve in R2? It represents a parabola whose absolute minimum is 0 and has an infinite domain.
b. What does it represent as a surface in R3? It represents a parabolic cylinder whose absolute minimum is 0 and infinitely extends along the Z-axis, opening in the positive Y direction.
c. What does the equation z = y2 represent? In 2D, this represents a parabola whose absolute minimum is 0 and has an infinite domain on the ZY plane. In 3D, it represents another parabolic cylinder whose absolute minimum is 0 and infinitely extends along the X-axis, opening in the positive Z direction.
2) Reduce the equation to one of the standard forms and classify the surface: 4y2 + z2 - x - 16y - 4z + 20 = 0.
Oh, boy. Well, first, let's group things up and move that 20 out of the way: 4y2 - 16y + z2 - 4z - x = -20.
Now, let's complete the square for the y. We look at 4y2 - 16y. We pull out the 4, 4(y2 - 4y), take half the coefficient of the one-degree term, 4/2 = 2, then square this, 22 = 4, and we put that back in there and add to both sides of the equation.
4(y2 - 4y + 4) = -20 + 16 (we're ignoring the z and x parts for now, and it's +16 because of that 4*4 on the left side). That gives us 4((y - 2)2) = -4.
Now we gotta do the same thing for z2 - 4z, which actually turns out to be the very same except we don't have to deal with the obscure 4, so we get (z - 2)2 = -4 + 4 = 0.
Because we ALREADY manipulated the -4 from completing the square with Y, we simply pull in the Y terms, put them with the Z terms, and bring back that X term we had to do nothing to--
4((y - 2)2) + (z - 2)2 - x = 0. Now we just divide by 4 to get rid of that coefficient.
(y - 2)2 + ((z - 2)2)/4 = x/4. This is an elliptic paraboloid because the two squared terms are positive, and there is one term that is not squared. This moves in the direction of the X-axis because the X is not squared.
3) Find an equation for the surface consisting of all points P for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.
First, let's let P = (x,y,z), that is, just some arbitrary point. We need the distance from P to the x-axis. The x-axis, of course, is simply any arbitrary (x,0,0). So, the distance between the two is just D = sqrt((x - x)2 + (y - 0)2 + (z - 0)2). That's just D = sqrt(y2 + z2).
Now we need the distance from some P to the YZ plane. The YZ plane is just some (0,y,z). So, we find the distance between P and this plane, which is D = sqrt((x - 0)2 + (y - y)2 + (z - z)2) = sqrt(x2) = |x|. We have our two distances, BUT our condition was that the distance from P to x-axis is TWICE the distance from P to the YZ plane, so instead of just
D = sqrt(y2 + z2) = |x|, we do D = sqrt(y2 + z2) = 2*|x|. And then we square all sides to get rid of the radical:
y2 + z2 = 4x2. This takes the general form of a cone, and because not all axes are equal in length (radius of 1, radius of 1, and radius of 2 from that 4x2), this is in fact an elliptical cone. So, our answer is...
y2 + z2 = 4x2, elliptical cone
------
Why do you care? Ever use Solid Works or AutoCad to build 3D stuff? Know those neat little features that'll automatically cover the entire surface in one little design you created, like pinholes or ridges or even fillets? You get that "instant copy and paste" thing by these sorts of equations since ALL solid objects are engineered through these types of equations.
And if you're any good at physics, if you have such a nice design, you know just how to increase a number oh-so-slightly to turn those two cone-shaped funnel things attached at the small ends into a little bit of a wider attachment and keep going until you can maximize what you need done (like shortening the gap in order to increase velocity if it were an air tube, per se, or increasing the gap in order to give it more structural support).
Now, one thing you need to do is remove the idea that a cylinder is two circles connected. It's not. A cylinder is a surface that consists of all lines (or rulings) that are parallel to a given line and pass through a given plane curve.
If your given plane curve is a flat circle, and we take every line that can pass through it that is parallel to a given line perpendicular to the circle, you have your circle-based cylinder!
But consider this! Consider the graph of z = x2. You draw your XYZ axis and notice that since the function z does not involve y at all, every line is parallel to the y axis and runs through the ENTIRE y axis. And then all you have left to determine is the shape of the graph, which, if you simply do an XZ plane where X is the horizontal, you see a parabola. Extending this infinitely through the y-axis, you get a shape that looks exactly like a folded piece of paper. This is called a parabolic cylinder.
So, you can clearly see cylinders don't even need to be closed!
If you have x2 + y2 = 1 through the XYZ, then you see it's a circular cylinder, the circle being on the XY plane and then going up and down infinitely, the circles' radii = 1. So, drawing your XYZ axis with Z pointing up, it's just a vertical, circular cylinder. On the other hand, y2 + z2 = 1 is a circular cylinder that goes in the direction of the X axis.
Now you should know what cylinders generally look like. Consider the parabolic cylinder example again. Mathematically, we are saying that any vertical plane with the equation y = k (parallel to the XZ plane) intersects the graph in a curve with an equation z = x2.
And now that that's in your head, let's move on to the general quadric surface.
A quadric surface is the graph of any second-degree equation in three variables, and the most general form is
Ax2 + By2 + Cz2 + Dxy + Eyz + Fxz + Gx + Hy + Iz + J = 0
where A,B,C,D,E,F,G,H,I,J are all constants.
This is gonna get kinda tricky to explain how to draw, so bear with me.
Usually, quadric surfaces can be rotated and translated about until they reach one of two standard forms:
Ax2 + By2 + Cz2 + J = 0
Ax2 + By2 + Iz = 0
A note: they do not HAVE to be in THAT standard form. Look at them. You see all three variables x,y,z, and if they are all squared, a constant. The first one will always look like that, but the second could be Ax2 + Cz2 + Hy = 0 instead. Just so long as one variable is not squared.
In fact, there are six common shapes with formulas associated with them. Here they are--
ellipsoid: x2/a2 + y2/b2 + z2/c2 = 1. All traces (we'll talk about traces in a moment) are ellipses. If a = b = c, then the ellipsoid is a sphere. Looks like a 3D ellipse.
elliptic paraboloid: z/c = x2/a2 + y2/b2. Horizontal traces are ellipses. Vertical traces are parabolas. Whatever variable is in the first power (in this case, z) indicates the axis (or direction) of the paraboloid. Looks like a rounded funnel.
cone: z2/c2 = x2/a2 + y2/b2. Horizontal traces are ellipses. Vertical traces in the planes x = k, y = k are hyperbolas if k =/= 0, but pairs of lines if k = 0 (we'll discuss this "k" stuff later). Looks like two funnels attached at the tiny ends.
hyperboloid of one sheet: x2/a2 + y2/b2 - z2/c2 = 1. Horizontal traces are ellipses. Vertical traces are hyperbolas. Axis of symmetry corresponds to the variable whose coefficient is negative (goes in the direction of this axis). Looks like two lampshades attached at the smaller ends.
hyperboloid of two sheets: -x2/a2 - y2/b2 + z2/c2 = 1. Horizontal traces in z = k are ellipses if k > c or k < -c. Vertical traces are hyperbolas. The two minus signs indicate two sheets. Looks like a bump down on the ground, and then you mirror this so you have an upside-down bump floating in the sky.
hyperbolic paraboloid: z/c = x2/a2 - y2/b2. Horizontal traces are hyperbolas. Vertical traces are parabolas. Has varying shapes (just Google Image this sucker).
Okay, wait. Where the hell did this "k" come from, and what are these "traces"? Well, first off, they're related so we hit them both in the same explanation. Second of all, we use them to actually determine how we would draw this sucker.
First off, let's talk about these "traces." Traces are our attempts at taking a 3D object and making it 2D at various angles. To be more specific, instead of graphing something in the XYZ space, we're going to graph something in the XY, YZ, and XZ planes. After that, it's up to our ability to "see in 3D" that will helps us connect these three planes into the 3D space.
Let's consider the surface z = y2 - x2. From our table, we can see this is a hyperbolic paraboloid, where a,b,c = 1 (the table says x - y, but our problem has y - x. This is okay because it basically means we're going in a different direction, is all. The hyperbolic paraboloid will just have one part negative in this format).
So, how do we draw this? First, we need to draw 3 planes: ZY (make Y the horizontal), ZX (make X the horizontal), and YX (make X the horizontal).
Now, let's start with the ZY plane. From z = y2 - x2, we need to get this in terms of ONLY z and y (since ZY plane does not care about x). This means we need to remove the x2. Well, suppose x is no longer a variable, but is now some constant k. x = k, so x2 = k2, and we have z = y2 - k2. You should make the connection that this looks VERY similar to y = x2 - c, which is a parabola. So, we're drawing parabolas, and we do so by picking arbitrary values for k. If k = 0, we have z = y2. If k = 1, we have z = y2 - 1. If k = 2, we have z = y2 - 4, etc.
As you can tell, because it's k2, no matter what value we put in, it will always turn negative because of the minus sign in front of the k2. This means we only look at the parabolas whose z-intercepts are 0 and lower. So you draw them. These are called "traces in x = k."
Now, let's look at the ZX plane. We need to remove y such that we only have a function of z and x, and we do so by, again, letting y = k. This gives us z = k2 - x2. By letting k = anything, we see that these are upside-down (negative) parabolas whose Z intercepts now start with 0 and go to positive infinity, so we can draw them out just like we did the last one. And these are our traces in y = k.
Lastly, it's the YX plane. We gotta get rid of z, so we let z = k. This gives us k = y2 - x2. Suppose k = 0, then we have y2 = x2, or + y = + x. This is two lines (y = x, y = -x), so we draw out those two lines and basically make an X on our YX plane. Now, if we let k = 1, we get 1 = y2 - x2. Manipulating, we get x2 = y2 - 1, or +x = + sqrt(y2 - 1), which is a hyperbola that opens up and down. By letting k = -1, we get the same thing, only this time the hyperbola opens left and right By continuing to change k, we get hyperbolas that keep filling up the space left in that plane, and these are our traces in z = k.
Now, the hard part is trying to put this all together in the XYZ space. First, you definitely need to draw the XYZ axes. Now, taking one at a time, say, starting with x = k, you need to sort of "lightly draw" each parabola. In the 2D traces, you drew parabolas above and under each other because k changed the z-intercepts. Here, you'll also be changing the front-back (assuming you draw a standard XYZ axis system, where X points front and back) to accommodate x = k. It's not really THAT hard. You know how to T Table stuff: plug stuff in for, say, X, and then plug something in for Y and find Z, and just keep doing this.
Doing the same for the y = k traces, you simply also account for left-right (assuming standard XYZ coordinates where the Y axis goes left and right) changes, too, as y changes with k.
And then you do the same for the z = k traces, accounting for up-down movement as z changes (standard XYZ coordinates have Z axis going up-down).
And then you can MAYBE see the picture.
In all seriousness, you will probably never be asked to manually graph such an oblique shape. It's really too complicated. Ellipsoids, spheres, cones, and paraboloids are easy, but hyperbolic paraboloids are really just ugly looking when trying to do them manually. You will probably only graph them in software (like Mathematica, MathCAD, MATLAB, etc.), but you will most likely be expected to graph the traces. And, really, that's the important part. As long as you can see what's going on in each of the planes, that's more important than seeing the overall picture because in physical analysis, you will be a LOT more concerned with 2D mappings than you will ever be with the entire 3D object. So this is fine for those of us who can't think in 3D (I know I can't....).
So, let's do some problems.
---------
1) a. What does the equation y = x2 represent as a curve in R2? It represents a parabola whose absolute minimum is 0 and has an infinite domain.
b. What does it represent as a surface in R3? It represents a parabolic cylinder whose absolute minimum is 0 and infinitely extends along the Z-axis, opening in the positive Y direction.
c. What does the equation z = y2 represent? In 2D, this represents a parabola whose absolute minimum is 0 and has an infinite domain on the ZY plane. In 3D, it represents another parabolic cylinder whose absolute minimum is 0 and infinitely extends along the X-axis, opening in the positive Z direction.
2) Reduce the equation to one of the standard forms and classify the surface: 4y2 + z2 - x - 16y - 4z + 20 = 0.
Oh, boy. Well, first, let's group things up and move that 20 out of the way: 4y2 - 16y + z2 - 4z - x = -20.
Now, let's complete the square for the y. We look at 4y2 - 16y. We pull out the 4, 4(y2 - 4y), take half the coefficient of the one-degree term, 4/2 = 2, then square this, 22 = 4, and we put that back in there and add to both sides of the equation.
4(y2 - 4y + 4) = -20 + 16 (we're ignoring the z and x parts for now, and it's +16 because of that 4*4 on the left side). That gives us 4((y - 2)2) = -4.
Now we gotta do the same thing for z2 - 4z, which actually turns out to be the very same except we don't have to deal with the obscure 4, so we get (z - 2)2 = -4 + 4 = 0.
Because we ALREADY manipulated the -4 from completing the square with Y, we simply pull in the Y terms, put them with the Z terms, and bring back that X term we had to do nothing to--
4((y - 2)2) + (z - 2)2 - x = 0. Now we just divide by 4 to get rid of that coefficient.
(y - 2)2 + ((z - 2)2)/4 = x/4. This is an elliptic paraboloid because the two squared terms are positive, and there is one term that is not squared. This moves in the direction of the X-axis because the X is not squared.
3) Find an equation for the surface consisting of all points P for which the distance from P to the x-axis is twice the distance from P to the yz-plane. Identify the surface.
First, let's let P = (x,y,z), that is, just some arbitrary point. We need the distance from P to the x-axis. The x-axis, of course, is simply any arbitrary (x,0,0). So, the distance between the two is just D = sqrt((x - x)2 + (y - 0)2 + (z - 0)2). That's just D = sqrt(y2 + z2).
Now we need the distance from some P to the YZ plane. The YZ plane is just some (0,y,z). So, we find the distance between P and this plane, which is D = sqrt((x - 0)2 + (y - y)2 + (z - z)2) = sqrt(x2) = |x|. We have our two distances, BUT our condition was that the distance from P to x-axis is TWICE the distance from P to the YZ plane, so instead of just
D = sqrt(y2 + z2) = |x|, we do D = sqrt(y2 + z2) = 2*|x|. And then we square all sides to get rid of the radical:
y2 + z2 = 4x2. This takes the general form of a cone, and because not all axes are equal in length (radius of 1, radius of 1, and radius of 2 from that 4x2), this is in fact an elliptical cone. So, our answer is...
y2 + z2 = 4x2, elliptical cone
------
Why do you care? Ever use Solid Works or AutoCad to build 3D stuff? Know those neat little features that'll automatically cover the entire surface in one little design you created, like pinholes or ridges or even fillets? You get that "instant copy and paste" thing by these sorts of equations since ALL solid objects are engineered through these types of equations.
And if you're any good at physics, if you have such a nice design, you know just how to increase a number oh-so-slightly to turn those two cone-shaped funnel things attached at the small ends into a little bit of a wider attachment and keep going until you can maximize what you need done (like shortening the gap in order to increase velocity if it were an air tube, per se, or increasing the gap in order to give it more structural support).
Quick Links
Latest Comment
Re: I have never - hey, at least you're cooking... I burned the chicken... WITH the broth in it... so yeah......
| Terms of Service
| Privacy Policy