From a Scalar Formula to a Curve
A scalar field \(f\) assigns a number to each point in its domain. Its formula can describe how the value changes across the plane, but a useful way to examine that information is to ask: where does the field take one chosen value? The answer is a level set. When the points lie in the plane, such a set is often a line or a curve.
To find one, choose a constant \(c\), set \(f(x,y)=c\), and solve for the points \((x,y)\) that satisfy the equation. The constant is the field value on the entire set; it is not a coordinate or a new variable.
Why Fixing the Field Value Produces a Level Curve
The graph of a scalar field is the surface \(z=f(x,y)\). A horizontal plane at height \(c\) has equation \(z=c\). A point on the graph lies in that plane exactly when its height satisfies \(f(x,y)=c\). Thus the intersection is described by the two equations \(z=f(x,y)\) and \(z=c\), which together require \(f(x,y)=c\).
This gives a direct way to understand the construction: solving \(f(x,y)=c\) identifies the horizontal locations of the points where the graph meets the plane \(z=c\). If the graph is a bowl, for example, the intersection at a positive height can be a circle. If the graph is a tilted plane, the intersection is a line.
A Reliable Procedure for Constructing Levels
For each selected value, use the same sequence. The algebra may be short, but checking the allowed values and interpreting the resulting equation are essential parts of the construction.
Write the selected constant as \(c\). Use several distinct values when you want to see how the family of curves changes.
Begin with \(f(x,y)=c\), keeping any domain restrictions in view.
Rearrange the equation into a recognizable line, circle, or other curve. Check whether the chosen value permits any solutions.
Mark features such as a line's slope, a circle's center, or a radius. Label each curve with its field value.
Worked Example: Level Lines of an Affine Field
Worked Example: Level Lines of an Affine Field
Find and describe the level sets of \(f(x,y)=2x-y+1\) for \(c=-1\), \(c=1\), and \(c=3\).
The field is defined for every \((x,y)\in\mathbb{R}^2\). For a general chosen value \(c\), set the formula equal to \(c\):
Isolate \(y\) by subtracting \(2x+1\) from both sides:
For \(c=-1\), substitute that value into the equation:
At \(x=0\), this line gives \(y=2\), so the point is \((0,2)\). Checking it in the original field gives \(f(0,2)=2(0)-2+1=-1\), as required.
For \(c=1\):
At \(x=0\), the point is \((0,0)\), and \(f(0,0)=2(0)-0+1=1\).
For \(c=3\):
At \(x=0\), the point is \((0,-2)\), and \(f(0,-2)=2(0)-(-2)+1=3\).
Each level is a line with slope \(2\). Their vertical intercepts are \(2\), \(0\), and \(-2\), respectively, so the three lines are parallel. To sketch them, plot the checked point on each line and use slope \(2\), meaning that a move of \(1\) unit to the right corresponds to a move of \(2\) units up.
Worked Example: Level Curves of a Bowl
Worked Example: Level Curves of a Bowl
Find the level sets of \(f(x,y)=x^2+y^2\) for \(c=-1\), \(c=0\), \(c=1\), and \(c=4\).
For any selected value, the equation is \(x^2+y^2=c\). Since squares are nonnegative, \(x^2+y^2\geq 0\) for every point. Therefore there are no points when \(c=-1\). This is an empty level set.
When \(c=0\), both squares must be zero: \(x^2=0\) and \(y^2=0\). Hence \(x=0\) and \(y=0\). The level set consists of the single point \((0,0)\), not a circle of positive radius. Substitution checks the value: \(f(0,0)=0^2+0^2=0\).
When \(c=1\), the level equation is
This is a circle centered at \((0,0)\) with radius \(1\). For example, \((1,0)\) lies on it, and \(f(1,0)=1^2+0^2=1\).
When \(c=4\), the level equation is
This is a circle centered at \((0,0)\) with radius \(2\). The point \((0,2)\) lies on it, and \(f(0,2)=0^2+2^2=4\).
The sketch therefore has a point at the origin for level \(0\), a circle of radius \(1\) for level \(1\), and a larger concentric circle of radius \(2\) for level \(4\). There is no curve for level \(-1\). Each point on a given circle is the same distance from the origin, and its squared distance gives the field value.
Distance-Based Fields Produce Circles
A distance formula measures how far \((x,y)\) is from a fixed center \((a,b)\). The distance is \(\sqrt{(x-a)^2+(y-b)^2}\), so a level equation sets that distance equal to the selected value \(c\). Because distance cannot be negative, only \(c\geq 0\) can produce a level set. For positive \(c\), the equation describes a circle centered at \((a,b)\) with radius \(c\); at \(c=0\), it gives the center alone.
Squaring both sides converts the distance equation into the standard circle equation, but the nonnegativity condition must not be discarded. The equivalent circle equation for \(c\geq 0\) is \((x-a)^2+(y-b)^2=c^2\).
Worked Example: Levels of a Distance Field
Worked Example: Levels of a Distance Field
Let \(f(x,y)=\sqrt{(x-1)^2+(y+2)^2}\). Find and sketch the levels for \(c=-1\), \(c=0\), and \(c=3\).
The formula is defined for all real \(x,y\), because the expression inside the square root is a sum of squares and is therefore nonnegative. Also \(f(x,y)\geq 0\), so the level \(c=-1\) is empty.
For \(c=0\), set the field equal to zero:
A square root is zero only when its radicand is zero. Since the radicand is a sum of squares, both squares must be zero: \(x-1=0\) and \(y+2=0\). Thus the level is the single point \((1,-2)\). Checking gives \(f(1,-2)=\sqrt{0^2+0^2}=0\).
For \(c=3\), start with the level equation and square both sides. This is valid here because both sides are nonnegative:
This is a circle with center \((1,-2)\) and radius \(3\). Four points that help locate it are \((4,-2)\), \((-2,-2)\), \((1,1)\), and \((1,-5)\). For \((4,-2)\), the distance is \(\sqrt{(4-1)^2+(-2+2)^2}=\sqrt{9+0}=3\). For \((-2,-2)\), it is \(\sqrt{(-3)^2+0^2}=3\). For \((1,1)\), it is \(\sqrt{0^2+3^2}=3\). For \((1,-5)\), it is \(\sqrt{0^2+(-3)^2}=3\). Each point checks in the original field, so all four lie on the level circle.
To sketch the results, mark the center \((1,-2)\), plot the four radius-\(3\) points, and draw the circle through them. The level \(0\) is the center point, and the level \(-1\) has no points.
A Common Mistake: Ignoring the Allowed Values
It is tempting to manipulate a level equation without tracking which constants can occur. For the distance field in the previous example, a mistaken solution might claim that the level \(c=-1\) is a circle of radius \(-1\), or square the equation and call the result a circle of radius \(-1\). That conclusion is incorrect: a radius is nonnegative, and the original field is a square root, so \(f(x,y)\geq 0\) everywhere. The level \(f=-1\) has no solutions.
The correction is to check the range restriction before interpreting a squared equation. In general, when \(f(x,y)=\sqrt{(x-a)^2+(y-b)^2}\), a level \(c\) exists only if \(c\geq 0\). For \(c>0\), the radius is \(c\); for \(c=0\), the set is one point; and for \(c<0\), the set is empty. The same kind of check matters for the bowl \(x^2+y^2\), whose levels cannot have negative values.
Sketching and Checking the Result
Once the equation is solved, sketching should reflect the geometry rather than just the algebra. For a line, identify its slope and intercept or plot two points. For a circle, identify its center and radius, then mark points at that radius in the horizontal, vertical, or diagonal directions. Label each curve with its chosen value \(c\) so that the geometry remains tied to the field.
A quick substitution provides a useful check: choose a point on the proposed curve and evaluate the original formula there. The result should equal the level value. This check can catch a sign error in a line equation, a mistaken center, or an incorrect radius. It does not replace solving the level equation, but it verifies that the sketch and formula agree at the checked point.
Check Your Understanding
For each question, begin by setting the scalar-field formula equal to the stated value and consider whether the selected value is allowed.
- For \(f(x,y)=3x+y\), what equation describes the level \(c=2\), and what kind of curve is it?
- For \(f(x,y)=x^2+y^2\), what are the level sets at \(c=0\) and \(c=9\)?
- For \(f(x,y)=\sqrt{(x+2)^2+(y-1)^2}\), identify the center and radius of the level at \(c=4\).
- Why is the level \(c=-2\) empty for a distance field defined by a square root of a sum of squares?
- What point is the level set of \(f(x,y)=(x-3)^2+(y+1)^2\) at \(c=0\)?