Surfaces (…well, “Quadrics”)

1. Let {w=f(x_{1},x_{2},\dots,x_{n})} where {x_{1}}, {x_{2}}, …, {x_{n}} are all independent variables. Then the “Domain” of {f} is the set of {n}-tuples {{\mathbb R}^{n}}. Note that an ordered pair is
(1)\displaystyle  (x,y)=\mbox{2-tuple}.
The set of corresponding values is the “Range” (or Codomain) of the function. So we have
(2)\displaystyle  f\colon\underbrace{{\mathbb R}^{n}}_{\text{domain}}\rightarrow\underbrace{{\mathbb R}}_{\text{range}}
Sometimes we write {\mathrm{dom}(f)} for the domain of {f}, and {\mathrm{ran}(f)} or {\mathrm{cod}(f)} for the range (or codomain) of {f}. We DO NOT write {f({\mathbb R}^{2})} for the range, because this is the collection of all points mapped by {f}.

Example 1. Consider {f\colon{\mathbb R}^{2}\rightarrow{\mathbb R}} defined by
(3)\displaystyle  f(x,y) = \sqrt{x^{2}+y^{2}}
Notice that {f(x,y)\geq0} for any {x,y\in{\mathbb R}}. So
(4)\displaystyle  f({\mathbb R}^{2})=\{f(x,y)\in{\mathbb R} : x,y\in{\mathbb R}\} = \{ u\in{\mathbb R} : u\geq0\}\not={\mathbb R}.
This is not the codomain! It’s contained in the codomain, though. The image is always a subset of the codomain.

Example 2. Let {g(x,y)=\ln(x^{2}-y)}. For {g} to be defined, we need
(5)\displaystyle  x^{2}-y>0\quad\mbox{or}\quad x^{2}>y.
The boundary of the domain is {x^{2}=y}, which we can doodle:

Since the boundary is not in the domain, then the domain of {g} is open.

Example 3. Consider
(6)\displaystyle  h(x,y)=\sqrt{25-x^{2}-y^{2}}.
The domain is the set of {(x,y)} such that
(7)\displaystyle  x^{2}+y^{2}\leq25.
We can doodle this:

Observe that the boundary is the circle; the disc is the circle and everything enclosed in it.

2. So we have just discussed domains and codomains, but we have not discussed the graph of the function {z=f(x,y)}. What would this look like? Well, when we plot {y=g(x)}, it’s on the plane {{\mathbb R}^{2}}. So plotting {z=f(x,y)} would be on the 3-space {{\mathbb R}^{3}}. Lets start considering examples of what this looks like.

Example 4. Consider the graph given by
(8)\displaystyle  z = f(x,y) = 9 - x^{2}-y^{2}.
How can we draw this? Well, the first trick is to draw when {x=0} and {y=0}:
(9)\displaystyle  \begin{aligned} z &= f(0,y) = 9-y^{2}\\ z &= f(x,0) = 9-x^{2} \end{aligned}
These are parabolas in the {yz} and {xz} planes, respectively. Now we can start drawing “Level Curves”, i.e., curves where we fix {z} to be some constant.

For example, when {z=0}, we have a circle
(10)\displaystyle  x^{2}+y^{2}=9
Observe then that {z} controls the radius of the circles: for nonzero {z}, we have
(11)\displaystyle  x^{2}+y^{2}=9-z.
The left hand side must be non-negative, and can be zero only when {z=9}. So we get a surface that looks like

We call this surface a “Paraboloid” as we have parabolas along the {x}– and {y}-axes.

This is the general scheme for picturing a surface: draw level curves {f(x,y)=c} for some constant {c}, which produces the level curve for {f} corresponding to {c}.

Example 5. Consider {x^{2}+y^{2}-z^{2}=0}. What does this surface look like?

Solution: First we observe the curves along the {x}-axis, i.e., when {y=0} is {x^{2}=z^{2}}. Similarly when {x=0} we have {y^{2}=z^{2}}. What sort of curves are these? Well, {(\pm t,0,t)} and {(0,\pm t,t)} are the curves, for {t\in{\mathbb R}}.

Also note that {x^{2}+y^{2}} describes a circle, whose radius happens to be {z^{2}}. This tells us the level curves are simply circles. So we have a cone:

This surface is precisely a “Cone”.

Example 6. What if we deform the previous example, writing
(12)\displaystyle  x^{2}+y^{2}-z^{2}=1.
What surface does this describe?

Solution: Well, we see that the radius of the level curves deform {z^{2}\rightarrow z^{2}+1}. So the resulting surface looks like a cone, but along the {x}-axis we don’t have a straight-line: we have a hyperbola {x^{2}-z^{2}=1}. Similarly along the {y}-axis we have another hyperbola {y^{2}-z^{2}=1}. Thus our surface is doodled as:

This “deformed cone” is called a “One Sheeted Hyperboloid”.

Example 7. Another variation, consider the surface
(13)\displaystyle  x^{2}+y^{2}-z^{2}=-1
What does it look like?

Solution: Well, we see that {x^{2}+y^{2}\geq0} always, whereas {z^{2}-1\geq0} only when {|z|\geq1}. So we have two “surfaces” or “Sheets” here. The level curves are again circles, and this enables us to doodle the graph:

Along the {x}-axis and {y}-axis, we have hyperbolas. For this reason, we call the surface a “Two-Sheeted Hyperbaloid”.

Example 8. Suppose we have a surface given by {z=x^{2}-y^{2}}. What does it look like?

Solution: Observe the level curves for {z>0} gives us hyperbolas, and for {z<0} we again have hyperbolas (the same as before but reflected about the line {x=y}). For {z=0} we have a cone.

When we take {y=\pm1,0} we see {z=x^{2}-C} for some constant {C}. These curves look like smiles. For {x=\pm1,0}, we have {z=C-y^{2}} for some constant {C}. These curves look like frowns.

This surface is a saddle, and we can doodle it as

Since the curves when we hold {z} constant form hyperbolas, and the curves when we hold {x} or {y} constaant are parabolas, people sometimes call this saddle a “Hyperbolic Paraboloid”.


About Alex Nelson

I like math. I like programming. Most of all, I love puzzles.
This entry was posted in Calculus, Geometry, Vector Calculus. Bookmark the permalink.

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