$$z = \sqrt{x^2 + y^2 - 1}$$
How can I graph something like this without using graphing devices?
I equal $z = 0$ to find the graph on the xy plane. So I got a circle, $1=x^2 + y^2.$ But when I equal 0 for either the $x$ or the $y,$ I get $z = \sqrt{x^2 - 1}$, but what's the graph of that? Different graphing websites were telling me different answers...
Please don't show some crazy and complicated methods to graph this. I just want simple steps just as plugging $x,y,z$ as zeroes and etc.
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$\begingroup$What I usually do is break a three-dimensional graph up into three separate planes, X-Y, X-Z, Y-Z, and I draw them individually and try to visualize how they fit together.
To graph the X-Y plane you set Z = 0 and plot the function as you normally would, so $$z = \sqrt(x^2 + y^2 - 1) == 0 = \sqrt(x^2 + y^2 - 1)$$ $$\text {Therefore:} x^2 + y^2 = 1$$ is your X-Y axis graph, which is just a circle of radius 1 centered at the origin.
Now onto the X-Z plane, we do this by setting y = 0. $$z = \sqrt(x^2 - 1)$$ Which should be fairly trivial to plot (at least qualitatively).
And finally the Y-Z plane which will look exactly like the X-Z plane. $$z = \sqrt(y^2 - 1)$$
So here are the three graphs (for visualization purposes... this is much easier on paper, though):
$x^2 + y^2 = 1$:
$z = \sqrt(x^2 - 1)$:
And finally: $z = \sqrt(y^2 - 1)$
Now the tricky part is to arrange these three planes so that the shape of the object can be inferred, here is how these three planes are arranged:
So all that's left to do is to imagine the graph on each of those axes and interpolate what the shape should be.
This is the graph Wolfram Alpha gives:
Notice the circle of radius 1 on the X-Y plane and the hyperbolic curves on both the Y-Z plane and X-Z plane.
$\endgroup$ 2 $\begingroup$Your first step of checking the contour line for $z=0$ should give you a hint what to do otherwise: The contour line for $z=c$ (with $c>0$) turns out to be a circle as well: $z=c\iff \sqrt{x^2+y^2-1}=c\iff x^2+y^2=c^2+1$. That is a circle of radius $\sqrt{c^2+1}$. Of course, $z<0$ does not occur because the square root is always non-negative. And also there is no $z$ value corresponding to $(x,y)$ with $x^2+y^2<1$, i.e. for points inside the unit disk. The overall appearence of this graph is therefore like a cone with a widened hole at its apex.
$\endgroup$ $\begingroup$If $z = \sqrt{x^2+y^2-1}$ then $z^2 = x^2 + y^2 - 1$. Although the solution set of the first equation only gives part of the solution set of the second because $z^2=x^2 + y^2 - 1$ gives $z=\pm \sqrt{x^2+y^2-1}$. We actually get the set $z^2 = x^2 + y^2 - 1$ where $z\ge 0$. Let's consider $z^2 = x^2 + y^2 -1$.
Consider how the surface meets the axes. The $x$-axis has $y=z=0$. When $y=z=0$ we have $x^2-1=0$ and so $x = \pm 1$. The $y$-axis has $x=z=0$. When $x=z=0$ we have $y^2-1=0$ and so $y = \pm 1$. The $z$-axis has $x=y=0$. When $x=y=0$ we have $z^2=-1$ and there are no real solutions. Hence we have a quadric surface which meets the $x$- and $y$-axis at two points and misses the $z$ axis. We have a hyperboloid of a single sheet (like a squashed cylinder).
For $z = \sqrt{x^2+y^2-1}$ we take the above hyperboloid and throw away the part with $z < 0$.
There is a more complicated method for not-so-simple examples. If you had lower-order and mixed terms, e.g. $xy$ or $2z$, then you complete the square to find the translation component. Then you find the matrix of the resulting quadratic form. The eigenvalues of this matrix tell you all about the surface and the way that it's oriented in three-space.
$\endgroup$ 1 $\begingroup$This particular equation is very easy to understand, because it’s a surface of revolution. Note that if we put $r=\sqrt{x^2+y^2}$, the distance of a point from the $z$-axis, then your equation is $z = \sqrt{r^2-1}$. This means that no matter what plane $\Pi$ through the $z$-axis you draw, the intersection of your graph with $\Pi$ will look the same. And what’s that? It’s just half of a hyperbola, because your equation was not $z^2=r^2-1$, the more “geometric” form. So take the part of the hyperbola $z^2=x^2-1$ that’s above the $x,y$-plane, and rotate it through the $z$ axis. That’s your graph.
If I may say so, this particular problem falls much more easily to pure thought than to any graphing calculator.
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