Exercises — Functions of several variables — graphs, level curves, level surfaces

What we need: the inside of a real square root must be $\ge 0$ — you cannot take the real root of a negative number.

• Require $16-x^2-y^2\ge 0$, i.e. $x^2+y^2\le 16$.
• What it looks like: $x^2+y^2\le 16$ is a filled disk of radius $4$ centred at the origin (see Figure s01). So domain $=\{(x,y):x^2+y^2\le 16\}$.
• Range: $16-x^2-y^2$ runs from $0$ (on the rim) up to $16$ (at the centre). Its square root runs from $0$ to $4$. So range $=[0,4]$.

Set each equal to a constant $c$ and read the standard shape:

• (a) $3x+2y=c\Rightarrow y=\frac{c-3x}{2}$ — straight lines (all parallel, slope $-\frac{3}{2}$).
• (b) $x^2+y^2=c$ — circles of radius $\sqrt{c}$ (for $c>0$).
• (c) $x^2-y^2=c$ — hyperbolas.
• (d) $y-x^2=c\Rightarrow y=x^2+c$ — parabolas shifted vertically.

Why set $f=c$? By definition a level curve collects inputs of equal output.

• $\sqrt{x^2+y^2}=c$. Since a square root is never negative, we need $c\ge 0$.
• Square both sides (legal because both sides are $\ge 0$): $x^2+y^2=c^2$.
• What it looks like: a circle of radius $c$ (not $\sqrt{c}$!). So $c=1,2,3$ give circles of radius $1,2,3$; $c=0$ gives the single point at the origin.
• Spacing: here radius $=c$, so equal $c$-steps give equally spaced circles — this cone climbs at a constant steepness. Contrast with $x^2+y^2$ whose radius $=\sqrt{c}$ crowds outward. (Figure s02 puts the two side by side.)

What we need: the input of a logarithm must be strictly positive — $\ln$ of zero or a negative number is undefined.

• Require $x-y^2>0$, i.e. $x>y^2$.
• What it looks like: the curve $x=y^2$ is a sideways parabola opening to the right. The condition $x>y^2$ is the region to the right of (inside) that parabola, boundary not included (strict inequality).
• Domain $=\{(x,y):x>y^2\}$.

• Set $2x+3y+6z=c$. One linear equation in three unknowns removes one degree of freedom, leaving a 2D object — a plane.
• These are all parallel planes (same left-hand side, different $c$). Their common normal direction is $(2,3,6)$.
• As $c$ increases in equal steps the planes shift by equal distances, because the equation is linear (no crowding). Distance between $c$ and $c+\Delta c$ planes $=\frac{\Delta c}{\sqrt{2^2+3^2+6^2}}=\frac{\Delta c}{7}$.

• Radius $=\sqrt{c}$: for $c=1,4,9,16$ we get radii $1,2,3,4$.
• Gaps (outer minus inner radius): $2-1=1$, $3-2=1$, $4-3=1$. The radius gaps happen to be equal here because we chose perfect squares — but note the $c$-jumps were $3,5,7$, i.e. unequal. Equal radius-steps came from ever-larger height jumps.
• Turn the reading around: for equal height jumps (say $c=1,2,3,4$) the radii would be $1,\sqrt{2},\sqrt{3},2\approx 1,1.41,1.73,2$, whose gaps $0.41,0.32,0.27$ shrink. Circles crowd outward → the outer bowl is steepest. Crowded level curves $\iff$ steep terrain.

• $c>0$: $x^2-y^2=c\Rightarrow \frac{x^2}{c}-\frac{y^2}{c}=1$ — hyperbolas opening left/right (they cross the $x$-axis).
• $c<0$: write $c=-k$, $k>0$: $x^2-y^2=-k\Rightarrow \frac{y^2}{k}-\frac{x^2}{k}=1$ — hyperbolas opening up/down (they cross the $y$-axis).
• $c=0$: $x^2=y^2\Rightarrow y=\pm x$ — the two crossing diagonal lines. This is the degenerate hyperbola: it has collapsed onto its own asymptotes. (Figure s03.)
• The nested hyperbolas separated by the crossing lines are the fingerprint of a saddle.

• Set $100-x^2-4y^2=64\Rightarrow x^2+4y^2=36$.
• Divide by $36$: $\frac{x^2}{36}+\frac{y^2}{9}=1$. What it looks like: an ellipse with semi-axis $6$ along $x$ and $3$ along $y$.
• Interpretation: all points on this ellipse sit at exactly $64^{\circ }$. The hottest point is the centre ($T=100$); isotherms are nested ellipses shrinking toward it.

• Parallel lines $y=2x+c$: rearrange to $c=y-2x$. So $f(x,y)=y-2x$ has exactly these level curves — set $f=c$ and you recover $y=2x+c$. ✅
• Circles centred at $(1,-2)$: a circle of that centre is $(x-1{)}^2+(y+2{)}^2=r^2$. To make that a level set, take $f(x,y)=(x-1{)}^2+(y+2{)}^2$; then $f=c$ (for $c>0$) is a circle of radius $\sqrt{c}$ centred at $(1,-2)$. ✅

• Cone: level curve $\sqrt{x^2+y^2}=c\Rightarrow$ radius $=c$. Equal height-steps → equally spaced circles → constant steepness everywhere.
• Paraboloid: radius $=\sqrt{c}$. Equal height-steps → circles crowd outward → steepness increases without bound as you move out.
• Conclusion: far from the axis the paraboloid is steeper (its circles are more crowded). Near the axis the paraboloid is actually gentler (its circles start out widely spaced). The crossover is where $\sqrt{c}=c$, i.e. $c=1$ (radius $1$).

• Set $x^2+y^2+z^2-2z=c$. Complete the square in $z$ (add and subtract $1$): $x^2+y^2+(z-1{)}^2-1=c$, so $x^2+y^2+(z-1{)}^2=c+1$.
• What it looks like: spheres centred at $(0,0,1)$ with radius $\sqrt{c+1}$, valid when $c+1>0$ i.e. $c>-1$. At $c=-1$ the surface is the single point $(0,0,1)$; for $c<-1$ it is empty.
• So $g$'s level surfaces are the same spheres as $f$'s, merely shifted up by $1$ and re-labelled.

Write $A{x}^2+B{y}^2=c$ and reason by signs.

• $A>0,B>0$: left side $\ge 0$.
  • $c>0$: ellipse $\frac{x^2}{c/A}+\frac{y^2}{c/B}=1$ (a circle iff $A=B$).
  • $c=0$: only $x=y=0$ works → the single point origin.
  • $c<0$: sum of non-negatives can't be negative → empty.
• $A<0,B<0$: mirror image. $c<0$: ellipse; $c=0$: point; $c>0$: empty.
• $A>0,B<0$ (opposite signs), i.e. a saddle-type:
  • $c\ne 0$: hyperbola (opening along $x$ if $c>0$, along $y$ if $c<0$).
  • $c=0$: $A{x}^2=-B{y}^2$, i.e. $y=\pm \sqrt{A/|B|}x$ → two crossing lines (degenerate hyperbola).
• One coefficient zero, say $B=0,A>0$: $A{x}^2=c$.
  • $c>0$: $x=\pm \sqrt{c/A}$ → two parallel vertical lines.
  • $c=0$: $x=0$ → the single line (the $y$-axis).
  • $c<0$: empty. Every case is now covered: ellipse, point, empty, hyperbola, crossing lines, parallel lines, single line.

• Idea: the equation $f(x_1,\dots ,x_n)=c$ is one constraint on $n$ free numbers. Generically one equation removes exactly one degree of freedom, so the surviving object has dimension $n-1$.
• $n=1$: $f(x)=c$ picks out isolated points (dimension $0$) — e.g. $x^2=4\Rightarrow x=\pm 2$.
• $n=2$: $f(x,y)=c$ is a curve (dimension $1$) — level curve.
• $n=3$: $f(x,y,z)=c$ is a surface (dimension $2$) — level surface.
• "Generically" flags the degenerate exceptions we met (a level set can shrink to a point or vanish), but the typical dimension is always $n-1$.

L1 name the shape from the formula · L2 turn a formula into an actual equation of a circle/plane/parabola · L3 read steepness off the crowding of curves · L4 run it backwards — invent an $f$ with the level set you want · L5 classify every case by sign, and know why dimension drops by exactly one.