On page 80 of Spivak's Calculus, 4th Edition, he writes:

One of the simplest subsets of this three-dimensional space is the (infinite) cone illustrated in Figure 2; this cone may be produced by rotating a "generating line," of slope C say, around the third axis.

For any given first two coordinates x and y, the point (x,y,0) in the horizontal plane has distance (x^2+y^2)^(1/2) from the origin, and thus

(1) (x,y,z) is on the cone if and only if z=±C(x^2+y^2)^(1/2)

I don't understand how (1) follows from the preceding paragraph. Can someone please shed some light?

Question

On page 80 of Spivak's Calculus, 4th Edition, he writes:

One of the simplest subsets of this three-dimensional space is the (infinite) cone illustrated in Figure 2; this cone may be produced by rotating a "generating line," of slope C say, around the third axis.

For any given first two coordinates x and y, the point (x,y,0) in the horizontal plane has distance (x^2+y^2)^(1/2) from the origin, and thus

(1) (x,y,z) is on the cone if and only if z=±C(x^2+y^2)^(1/2)

I don't understand how (1) follows from the preceding paragraph. Can someone please shed some light?

unklerhaukus · Answer

a point rotating about an axis will make  circle

annur · Answer

I still don't see how if z=±C(x^2+y^2)^(1/2) from that...

unklerhaukus · Answer

well the equation of a circle radius r is 

x^2+y^2=r^2

i.e.
±√(x^2+y^2)=r

annur · Answer

okay...things are starting to be clear. How does C (the slope of the generating line) come into the picture?

unklerhaukus · Answer

i'm not really sure about the detail there, it looks like it should be right