Why Radial Contours Are Special
A function of two variables can have contours with many different shapes. But when its formula depends on \(x\) and \(y\) only through \(x^2+y^2\), the contour equation has a useful structure: every point on a contour must have the same distance from the origin, or belong to one of several such distances.
The key is to replace \(x^2+y^2\) with a single variable. This turns the contour problem into solving an equation in one nonnegative number. The solutions then tell us exactly which circles, points, or empty sets make up the contour.
The condition \(s\ge 0\) is essential: \(x^2+y^2\) cannot be negative for real \(x\) and \(y\). Every admissible solution \(s\) corresponds to the distance \(\sqrt{s}\) from the origin.
From a One-Variable Solution to a Contour
Suppose the equation \(g(s)=c\) has an admissible solution \(s_0\). The points on the level set that correspond to this solution satisfy \(x^2+y^2=s_0\). If \(s_0>0\), this is a circle centered at the origin with radius \(\sqrt{s_0}\). If \(s_0=0\), then \(x^2+y^2=0\), which holds only at \((0,0)\). If there are no admissible solutions, there are no points on the level set.
This description allows for more than one circle. If \(g\) takes the same value at several nonnegative inputs, each input gives a separate circle, except that the input \(0\) gives a point. The contours are therefore concentric, but need not consist of exactly one circle.
Why the Description Holds
The result follows directly from the definition of a level set and the identity \(s=x^2+y^2\). The argument also explains why no other shapes can occur.
If \((x,y)\) is on the level set at height \(c\), then \(g(x^2+y^2)=c\). Set \(s=x^2+y^2\). Since squares are nonnegative, \(s\ge 0\), and \(g(s)=c\).
For a fixed solution \(s\), the points associated with it satisfy \(x^2+y^2=s\). When \(s>0\), this is the circle centered at the origin with radius \(\sqrt{s}\). When \(s=0\), both squares must be zero, so \(x=0\) and \(y=0\).
Every point on the contour produces a nonnegative solution of \(g(s)=c\) by Step 1. Conversely, every point satisfying \(x^2+y^2=s\) for a solution \(s\ge 0\) has \(f(x,y)=g(s)=c\). Thus the contour is exactly the union stated in the theorem.
This proves both directions: every contour point belongs to one of the listed circles or is the origin, and every point in those listed sets really is on the contour.
Worked Example: A Basic Quadratic Radial Function
Worked Example: A Basic Quadratic Radial Function
Find the contours at heights \(c=-1\), \(c=0\), and \(c=4\) for \(f(x,y)=x^2+y^2\).
Here \(g(s)=s\). The contour equation is \(s=c\), with \(s=x^2+y^2\ge 0\).
For \(c=-1\), the equation gives \(s=-1\), which is inadmissible. Equivalently, \(x^2+y^2=-1\) has no real solutions, so this contour is empty.
For \(c=0\), we have \(x^2+y^2=0\). Since both squares are nonnegative, their sum is zero only when \(x=0\) and \(y=0\). The contour is the single point \((0,0)\).
For \(c=4\), we have \(x^2+y^2=4\), a circle of radius \(\sqrt{4}=2\).
Check: At the origin, \(f(0,0)=0\). At every point of the radius-\(2\) circle, \(x^2+y^2=4\), so \(f(x,y)=4\). There are no real points with \(f(x,y)=-1\), because a sum of squares cannot be negative.
Worked Example: A Square-Root Radial Function
Worked Example: A Square-Root Radial Function
Find the contours at heights \(c=2\), \(c=0\), and \(c=-1\) for \(f(x,y)=\sqrt{x^2+y^2}\).
Write \(g(s)=\sqrt{s}\), defined for \(s\ge 0\). The contour equation is \(\sqrt{s}=c\), where \(s=x^2+y^2\ge 0\).
For \(c=2\), the equation \(\sqrt{s}=2\) gives \(s=4\) after squaring both sides. This solution satisfies \(s\ge 0\), so the contour is \(x^2+y^2=4\), a circle of radius \(2\).
For \(c=0\), \(\sqrt{s}=0\) gives \(s=0\). Thus \(x^2+y^2=0\), and the contour consists only of \((0,0)\).
For \(c=-1\), there is no solution: a square root is nonnegative, so it cannot equal \(-1\). The contour is empty.
Check: On the radius-\(2\) circle, \(\sqrt{x^2+y^2}=\sqrt{4}=2\). At the origin, the function value is \(\sqrt{0}=0\). The function is nonnegative everywhere in its domain, confirming that the level \(-1\) is unattainable.
Worked Example: One Height, Two Circles
Worked Example: One Height, Two Circles
Find the contour at height \(c=9\) for \(f(x,y)=(x^2+y^2-4)^2\).
Set \(s=x^2+y^2\), so \(s\ge 0\). The level equation becomes \((s-4)^2=9\). Taking both square roots gives \(s-4=3\) or \(s-4=-3\). Therefore \(s=7\) or \(s=1\). Both values are admissible.
The solution \(s=1\) gives \(x^2+y^2=1\), a circle of radius \(1\). The solution \(s=7\) gives \(x^2+y^2=7\), a circle of radius \(\sqrt{7}\). The contour is the union of these two concentric circles.
Check: When \(x^2+y^2=1\), the function value is \((1-4)^2=(-3)^2=9\). When \(x^2+y^2=7\), it is \((7-4)^2=3^2=9\). Both circles therefore lie on the stated contour.
Admissibility and a Common Mistake
Solving the one-variable equation is not quite enough: every solution must also satisfy \(s\ge 0\). This condition can remove algebraic solutions before they become contours. A negative value of \(s\) cannot be turned into a real radius, because no real \(x,y\) can satisfy \(x^2+y^2=s<0\).
Common Mistake: Keeping a Negative Value of \(s\)
Consider again \(f(x,y)=(x^2+y^2-4)^2\), but now seek the contour at height \(c=25\). The equation in \(s\) is \((s-4)^2=25\), so \(s-4=5\) or \(s-4=-5\). Thus \(s=9\) or \(s=-1\).
Incorrect calculation: Keeping both algebraic solutions and writing \(x^2+y^2=-1\) as a second contour. This is not a real circle or any other real point set: \(x^2+y^2\) cannot equal \(-1\).
Correction: Retain only \(s=9\). It gives the circle \(x^2+y^2=9\), of radius \(3\). The other candidate is inadmissible and contributes nothing.
Check: On the radius-\(3\) circle, \(x^2+y^2=9\), so \(f(x,y)=(9-4)^2=25\). There is no real point corresponding to \(s=-1\).
How the Shape Depends on \(g\)
The radial form restricts the possible contour components, while the one-variable function \(g\) determines which radii occur. If \(g(s)=c\) has exactly one positive solution, the contour is one circle. If its only nonnegative solution is zero, the contour is the origin. If it has no nonnegative solution, the contour is empty. If it has several positive solutions, the contour consists of several concentric circles; a solution at zero can add the origin as well.
For a strictly increasing or strictly decreasing \(g\) on its domain, each level equation has at most one solution, so there can be at most one circle or point at a given height. Without that one-to-one behavior, multiple circles are possible, as the previous example shows.
The practical procedure is short: introduce \(s=x^2+y^2\), solve \(g(s)=c\), discard any solutions with \(s<0\), and translate each remaining solution into its corresponding circle or point. This reduction is useful because it separates the algebra of the function from the geometric interpretation of its contours.
Check Your Understanding
For each question, remember to restrict \(s\) to nonnegative values before describing the contour.
- For \(f(x,y)=x^2+y^2\), what is the contour at height \(c=9\), and what is its radius?
- For \(f(x,y)=\sqrt{x^2+y^2}\), what are the contours at heights \(0\) and \(-2\)?
- For \(f(x,y)=(x^2+y^2-1)^2\), solve for the values of \(s\) on the contour at height \(c=4\). Which solutions are admissible?
- Why does \(x^2+y^2=0\) give a point rather than a circle of positive radius?
- What does a positive solution of \(g(s)=c\) contribute to the contour of \(f(x,y)=g(x^2+y^2)\)?