Logical error in Spivak's Calculus?

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SUMMARY

The discussion centers on a potential logical error in Michael Spivak's treatment of conic sections in his calculus text. The user identifies a discrepancy in the assertion that the intersection of a cone defined by the equation \( z = \pm m \sqrt{x^2 + y^2} \) and a plane described by \( z = Mx + B \) implies that any point satisfying \( Mx + B = \pm m \sqrt{x^2 + y^2} \) lies on both the cone and the plane. The user argues that this is not necessarily true, as the converse does not hold without additional assumptions about the point's coordinates. The conclusion suggests that Spivak's exposition lacks clarity regarding the conditions under which the intersection holds.

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nimon
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Yes I am another plucky young fool who decided to self study Spivak. I think I have found an error in his section on conic sections, but Spivak is seldom wrong and I want to be sure I'm thinking straight.

Let $C$ be a cone generated by a line of gradient $m$ which goes through the origin. Then $(x,y,z)$ is on $C$ if $$(1) \qquad z = \pm m \sqrt{ x^{2} + y^{2} }.$$

Let $P$ be a plane which intersects with the cone and whose intersection with the $xy$-plane is a line parallel to the $y-$ plane. Thus, the intersection of $P$ with the $xz$-plane is a line: $L$, say. Supposing $L$ to have gradient $M$ and $z$-intercept $B$, the line $L$ can be described by the equation $$ (2) \qquad z = Mx+B.$$

All is right and well. But then he says 'combining $(1)$ and $(2)$, we see that (x,y,z) is in the intersection of the cone and the plane if and only if $$Mx+B = \pm m \sqrt{ x^{2} + y^{2} }.$$

I understand why, if $(x,y,z)$ is in the intersection, then $(1) = (2),$ but why, is the converse true? Surely we can find an infinite number of points where the equations are equal, but $z$ could be any number and the point not on either plane.

He doesn't first assume that the point is already on $C$ or $P,$ just that it is in $\mathbb{R}^{3},$ and I haven't missed anything in his argument out. Am I just being thick, or do I have a point?
 
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the point is, if you "eliminate $z$" you are left with a quadratic equation of $x,y$ in the $xy$-plane. what happens in the 3 cases:

$M^2-m^2 > 0$
$M^2-m^2 = 0$
$M^2-m^2 < 0$?

not "all points" $(x,y)$ are going to satisfy:

$\{(x,y) \in \Bbb R^2: Mx + B = \pm m\sqrt{x^2 + y^2}\}$

when $M,m,B$ are fixed before-hand.

if we pick such an $(x,y)$, this completely determines $(x,y,Mx+B)$ yes?
 
Thank you for your post. Whilst I don't disagree with anything you have said, I still have a problem accepting the proposition: $$Mx+B = \pm m \sqrt{ x^{2} + y^{2} } \Rightarrow (x,y,z) \in P \cap C. $$ In fact, suppose $(x,y,z) \in P \cap C,$ then by that very proposition it follows that $(x,y,z+1) \in P \cap C$ which, among other things, contradicts the assumption that $C$ is a cone.

I think I have concluded that this is really an error, at least in exposition. He should have made it clear that it is assumed that $(x,y,z)$ has the form $(x,y,Mx+B)$ in which case the proposition definitely holds.
 

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