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Spivak: Conic Sections appendix, problem 1

  • #1
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10

Homework Statement


"Consider a cylinder with a generator perpendicular to the horizontal plane; the only requirement for a point ##(x,y,z)## to lie on this cylinder is that ##(x,y## lies on a circle: ##x^2+y^2=C^2##.
Show that the intersection of a plane with this cylinder can be described by an equation of the form ##(αx+β)^2+y^2=C^2##"

Homework Equations


##x^2+y^2=C^2##.
##z=Mx+B##
##(αx+β)^2+y^2=C^2##

The Attempt at a Solution


If we assume the plane in question is perpendicular to the x,z plane (Which the book does seem to assume, going by the explanations preceding this question), it is of the form ##ax+bz=c##. Thus a point ##(x,y,z)## is on it iff ##z=Mx+B## for some ##M, B## (This is how Spivak described the plane so I'm going with this).

So a point ##(x,y,z)## belongs to the intersection iff
1) ##x^2+y^2=C^2##.
2) ##z=Mx+B##

Obviously right from the start I'm misunderstanding the question since all the equation ##(αx+β)^2+y^2=C^2## does is takes the original cylinder, squeezes (or stretches) it along the x axis and transports it along the x axis. Besides, since we can insert any z coordinate we want into a point described by this equation, it can't describe a 2D intersection.

I'm guessing that this equation is for after we "eliminate" the z axis and describe the intersection using a 2d coordinate system (of the intersecting plane), but I don't really know how to do that. I know that the new coordinate system would be:
##x = αx'+β, y=y'##
By this, it looks like all we need to do is insert these new coordinates in to the cylinder equation and call it a day, but like I said, something more is needed. What am I missing here?
 

Answers and Replies

  • #2
tnich
Homework Helper
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I think there is an error in the problem statement. If you compare ##(αx+β)^2+y^2=C^2## with ##x^2+y^2=C^2##, you can see that it implies ##αx+β = \pm x##, which can only be true for two values of ##x##. I think what Spivak meant to say was ##(αz+β)^2+y^2=C^2##.
 
  • #3
123
10
I think there is an error in the problem statement. If you compare ##(αx+β)^2+y^2=C^2## with ##x^2+y^2=C^2##, you can see that it implies ##αx+β = \pm x##, which can only be true for two values of ##x##. I think what Spivak meant to say was ##(αz+β)^2+y^2=C^2##.
I know there isn't an error since in the answer book Spivak reaches the same result. This problem follows directly from a segment where he explains conic sections and does basically the same thing, so I can't really show what he means here without quoting those 3 pages. So I'll just add them to show what he means here (I think that's copyright infringement but that's about all I can do, I'll delete them when I figure this out...).

https://ibb.co/kAKyVU
https://ibb.co/jXVSPp
https://ibb.co/kwRdVU

The solution goes as follows:
"The point ##(x,y,z)## is in the cylinder if and only if
##x^2+y^2=C^2##.
Choosing the coordinates in the plane P as on page 81 ,we see that the points in the intersection of P and the cylinder are those satisfying
##(αx+β)^2+y^2=C^2##.
The possibilities are ∅, a straight line, 2 parallel straight lines or an ellipse (or circle)."
The problem starts following the pages linked, so I guess in order to find out what he wants here I need to figure out what he meant there...
 
Last edited:
  • #4
tnich
Homework Helper
1,048
336
I know there isn't an error since in the answer book Spivak reaches the same result. This problem follows directly from a segment where he explains conic sections and does basically the same thing, so I can't really show what he means here without quoting those 3 pages. So I'll just add them to show what he means here (I think that's copyright infringement but that's about all I can do, I'll delete them when I figure this out...).

https://ibb.co/kAKyVU
https://ibb.co/jXVSPp
https://ibb.co/kwRdVU

The solution goes as follows:
"The point ##(x,y,z)## is in the cylinder if and only if
##x^2+y^2=C^2##.
Choosing the coordinates in the plane P as on page 81 ,we see that the points in the intersection of P and the cylinder are those satisfying
##(αx+β)^2+y^2=C^2##.
The possibilities are ∅, a straight line, 2 parallel straight lines or an ellipse (or circle)."
The problem starts following the pages linked, so I guess in order to find out what he wants here I need to figure out what he meant there...
Look at how he defined ##\alpha x + \beta## on p. 81. The author is essentially saying that ##x = \alpha x + \beta##. In that context, his solution for the cylinder makes sense, although it is confusing and in my opinion rather sloppy to use the same label ##x## to represent two different variables in the same problem. The normal way of avoiding problems like this is to use different labels, ##x## and ##w##, or ##x## and ##x'## for example. This makes it possible to use the rules of algebra without generating spurious results.
 
  • #5
123
10
I get that there is a new ##x## being measured from the new coordinate system (I'll just call it ##x'##), what I don't get is how the solution was received. At first we have the equation ##x^2+y^2=C^2##, which describes a cylinder. Then all we do is express this equation in terms of a new coordinate system ##(αx'+β)^2+y'^2=C^2## and suddenly it describes only the points on the cylinder which are also on the plain. It's as if he gets rid of the 3rd dimension without ever explaining why, he does it with the conic section too:

In the (*) equation (page 81) he claims "A point ##(x,y,z)## is in the intersection of the cone and the plane if and only if ##Mx+B=±C\sqrt{x^2+y^2}##." But one can find infinite points that follow this equation with some random z coordinate that are not in the intersection. All points in the intersection follow this equation but the opposite is not necessarily true. The proper equation to my understanding would be ##Mx+B=±C\sqrt{x^2+y^2}=z##

I feel like both the misunderstandings I have about the cylinder and the cone stem from the same thing, something about the transition to the new coordinate system that suddenly turns equations in the 3d space to equations in the 2d plane...
 
  • #6
tnich
Homework Helper
1,048
336
I get that there is a new ##x## being measured from the new coordinate system (I'll just call it ##x'##), what I don't get is how the solution was received. At first we have the equation ##x^2+y^2=C^2##, which describes a cylinder. Then all we do is express this equation in terms of a new coordinate system ##(αx'+β)^2+y'^2=C^2## and suddenly it describes only the points on the cylinder which are also on the plain. It's as if he gets rid of the 3rd dimension without ever explaining why, he does it with the conic section too:

In the (*) equation (page 81) he claims "A point ##(x,y,z)## is in the intersection of the cone and the plane if and only if ##Mx+B=±C\sqrt{x^2+y^2}##." But one can find infinite points that follow this equation with some random z coordinate that are not in the intersection. All points in the intersection follow this equation but the opposite is not necessarily true. The proper equation to my understanding would be ##Mx+B=±C\sqrt{x^2+y^2}=z##

I feel like both the misunderstandings I have about the cylinder and the cone stem from the same thing, something about the transition to the new coordinate system that suddenly turns equations in the 3d space to equations in the 2d plane...
That's exactly what he is trying to do, get the equation of the intersection of the cylinder and the plane. He is trying to express that equation in a coordinate system embedded in the plane. He already showed you how that worked for the plane and the cone, and you didn't seem to have trouble with that. The only difference here is that the variable ##z## does not show up in the equation of the cylinder, so there is not need to express ##z## in terms of ##x##, and no need to substitute ##\alpha x' + \beta## for ##x## in the expression for ##z## (because there is no expression for ##z##).
 

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