identify the quadric surface of x^2+y^2+z^2+127=0
the left hand side is always positive, the right hand side is zero they can't be equal for any real numbers x,y and z.
so you mean that this has no quadric surface?
yes. no triple (x,y,z) will satisfy the equation.
ah so whats the best reason that i can say to my profeesor if ever i was ask in by her?
it's a 3D counterpart of a degenerate curve. in analytic geometry, it is possible that an equation, say of a circle \[x^2 + y^2 + Dx + ey + F =0\] to have no solution, from which we say the circle is degenerate. this idea extends to surfaces and curves in 3D, too.
here's a clear comparison \[x^2+y^2+z^2=64\] sphere; \(C(0,0,0), r=8\) \[x^2+y^2+z^2=0\] point sphere at \(C(0,0,0)\) \[x^2+y^2+z^2=-127\] degenerate sphere
@Wencester do you see why? Whenever you square a real number, the result is always positive... there's no way to square three numbers and add them for a neg. result.
thnks for that but what if the equation will be \[x ^{2}+y ^{2}+z ^{2}=127\]
Then you have a sphere, of radius sqrt(127).
@agent0smith i agree
\[x^2+y^2+z^2=r^2 \]^ equation of a sphere at the origin.
@Wencester have you read the central and non-central quadric surfaces?
not yet sir... whats up on that?
you should read them, in addition to asking questions.
ah ok tnx ill read on it
type it here, @Wencester
how will i know the rate of chnge in atmospheric pressure using a contour plt...if the distance between the two location is 1400 miles
it based on the contour pot wherein the level of curves are given
do you have an equation?
nothing
how about the plot?
its in my photocopy..its hard to draw
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