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Why is it obvious that the plane $z=0$ is tangent to the surface $z=x^{2}+y^{2}$ I don't quite understand, is this obvious? I have a problem with the background knowledge, I don't even know how to deal with the surface $z=x^{2}+y^{2}$.

user10354138
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2018 xyz
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    Just visualise a paraboloid sitting on the $xy$-plane. – Angina Seng Jun 22 '19 at 10:34
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    In general, if you have to ask "why is X obvious?", it isn't really. Then again, "obvious" is very often only "obvious in hindsight" – Hagen von Eitzen Jun 22 '19 at 10:43
  • I think the key point is to know how to visualize the graph of $z = x^2 + y^2$. And for that, maybe it would help to first restrict attention to points on the graph that satisfy $y=0$ (so $z = x^2$). Do you see a parabola opening upwards? With some more thought, you can convince yourself that the graph of $z=x^2 + y^2$ is a paraboloid opening upwards, and you can picture it clearly. – littleO Jun 22 '19 at 10:53

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Define the surface $\;f(x,y,z)=x^2+y^2-z=0\;$ , so that

$$\frac{\partial f}{\partial x}=2x\,,\;\;\frac{\partial f}{\partial y}=2y\,,\;\;\frac{\partial f}{\partial z}=-1$$

Then the tangent plane to $\;f\;$ at $\;(0,0)\;$ is given by

$$\frac{\partial f(0,0)}{\partial x}(x-0)+\frac{\partial f(0,0)}{\partial y}(y-0)+\frac{\partial f(0,0)}{\partial z}(z-0)=0\implies-z=0$$

DonAntonio
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An explanation is that for an algebraic surface with equation $p(x,y,z)=0$, $p$ being a polynomial such that $p(0,0,0)=0$, the tangent plane at the origin has as an equation the linear approximation of $p(x,y,z)$ in a neighbourhood of $(0,0,0)$, i.e. the linear part of the polynomial $p(x,y,z)$.

Bernard
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If you want to imagine why this is true, think about how $x^2 + y^2 = z$ changes as $z$ changes by looking at the level curves. The curve $x^2 + y^2 = 1$, for example, is a circle of radius 1, and $x^2 + y^2 = 0$ is a point. So, as $z$ grows, the radius of the circle grows, and the surface looks like a big circular bowl sitting on the $xy$-plane (aka $z=0$). The surface just meets the plane at the point $(0,0)$.

Erin Carmody
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The surface is a parabolic bowl, sitting on the plane $z=0$, with the $z$ axis as its axis of symmetry. A typical slice of it, cut perpendicular to its axis, is a circle centred on the $z$ axis of radius $\surd z$. Cutting it another way—say with the $y=0$ plane—gives a parabola $x^2=z$ in the plane, and similarly with any other plane through the $z$ axis.

John Bentin
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In fact at $z=0$ the equation $x^2+y^2=0$ defines only $1$ point, which is $(0,0)$.

So the plane $z=0$ and the surface intersect in a single point.

But since the surface is smooth (the equation is polynomial), it cannot have singular points, and the touching point is automatically a tangent point.

zwim
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