From a Formula to a Picture
A field assigns a value to each allowed point in space. A formula tells you how to calculate those values, but a useful picture should show more than a collection of computations: it should make the field’s type, domain, and behavior visible.
The first decision is what kind of value the formula returns. A scalar field assigns one number to each point, as a temperature model might. A vector field assigns a vector to each point, as a velocity or force model might. These different outputs call for different visualizations.
A scalar field \(f(x,y)\) can be visualized as a surface \(z=f(x,y)\), or with a contour map that marks points where \(f\) has a fixed value. A vector field is usually shown with arrows: each arrow begins at an input point and indicates the vector assigned there. The length and direction of the arrow represent the vector’s magnitude and direction.
A Reliable Translation Process
Use the formula in a consistent order. First identify the input and output: two input variables with one real-valued output indicate a scalar field on the plane; two input variables with two component functions indicate a planar vector field. Next determine which inputs are allowed by the formula. Then evaluate at a small set of points that reveal how the values change. Finally, choose a visualization that matches the output and use the calculated values to check the picture.
Count the components of the output: one real number is scalar; an ordered list of components is a vector.
Check for operations such as division or square roots that restrict which input points are allowed.
Substitute selected points into every component, keeping the input coordinates in the correct order.
Use a surface or contours for scalar values and arrows for vectors; compare the picture with the computed samples.
Worked Example: A Rotating Vector Field
Worked Example: A Rotating Vector Field
Consider the planar vector field \(\mathbf{F}(x,y)=\langle -y,x\rangle\). Identify its type and domain, evaluate it at four points, and describe an appropriate arrow sketch.
Type and domain. The formula returns two components, so it is a vector field \(\mathbf{F}:\mathbb{R}^2\to\mathbb{R}^2\). Both components are defined for every real \(x\) and \(y\), so its domain is all of \(\mathbb{R}^2\).
Sample values. Substitute each point into both components:
At the point \((1,0)\), the arrow points upward; at \((0,1)\), it points left; at \((-1,0)\), it points down; and at \((0,-1)\), it points right. Thus, on the four points of the unit circle, the arrows point counterclockwise around the origin. At the origin, \(\mathbf{F}(0,0)=\langle 0,0\rangle\), so the vector has zero length there.
Check and sketch. For a general point \((x,y)\), the dot product with its position vector is
This calculation verifies that every nonzero arrow is perpendicular to the line from the origin to its base point. Its magnitude is
So the arrows grow longer as their base points move farther from the origin. A correct sketch has a zero vector at the origin, arrows tangent to circles centered there, and counterclockwise direction. The four sample values agree with that pattern.
Why the Geometric Check Works
The dot-product check in the example illustrates how algebra can justify a visual feature. Two vectors are perpendicular when their dot product is zero. Here the position vector is \(\langle x,y\rangle\), while the field vector is \(\langle -y,x\rangle\); their dot product cancels exactly. Therefore, wherever the field vector is nonzero, its arrow is perpendicular to the radial direction. The sample arrows determine which of the two possible tangent directions occurs: at \((1,0)\), the arrow points upward, establishing counterclockwise rotation.
This method is useful beyond this particular formula. A few computed arrows suggest a pattern, while a calculation such as a dot product or a magnitude formula can check whether that pattern holds throughout the domain. A sketch should be guided by both: use samples to establish specific values and use algebra to identify general behavior.
Worked Example: A Scalar Field and Its Contours
Worked Example: A Scalar Field and Its Contours
Let \(f(x,y)=4-x^2-y^2\). Identify its type and domain, evaluate it at two points, and describe its graph and contour map.
Type and domain. The formula returns one real number, so \(f:\mathbb{R}^2\to\mathbb{R}\) is a scalar field. It is a polynomial expression, which is defined for every real \(x\) and \(y\). Its domain is therefore \(\mathbb{R}^2\).
Sample values. At \((1,1)\),
At \((0,2)\),
Substitution back into the original expression confirms the values: \(4-1-1=2\) and \(4-0-4=0\). The scalar outputs can be plotted as heights on the surface \(z=4-x^2-y^2\).
Contours. A contour at level \(c\) consists of the input points satisfying \(f(x,y)=c\). Rearranging this equation gives
When \(c<4\), this is a circle centered at the origin with radius \(\sqrt{4-c}\). At \(c=4\), it is just the origin. When \(c>4\), there are no real points on that contour because \(x^2+y^2\) cannot be negative. For example, level \(c=0\) gives \(x^2+y^2=4\), a circle of radius \(2\); level \(c=3\) gives \(x^2+y^2=1\), a circle of radius \(1\). The point \((0,2)\) lies on the level-zero circle, consistent with the calculated value \(f(0,2)=0\).
The surface is a downward-opening paraboloid: its maximum height is \(4\) at the origin, and its height decreases as \(x^2+y^2\) increases. The circular contours display this same radial pattern in the plane. These conclusions follow from the formula and the contour equation, not from guessing the shape of a graph.
Worked Example: A Vector Field with an Excluded Point
Worked Example: A Vector Field with an Excluded Point
Consider \(\mathbf{G}(x,y)=\left\langle \frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right\rangle\). Find its domain, evaluate it at three points, and describe a suitable sketch.
Type and domain. The formula returns two components, so this is a planar vector field. Both components have denominator \(x^2+y^2\). That denominator is zero exactly when \(x=0\) and \(y=0\). Division by zero is undefined, so the domain is \(\mathbb{R}^2\setminus\{(0,0)\}\).
Sample values. At \((1,0)\), the denominator is \(1^2+0^2=1\), so
At \((0,2)\), the denominator is \(0^2+2^2=4\), so
At \((-1,-1)\), the denominator is \((-1)^2+(-1)^2=2\), so
Each result can be checked by multiplying its components by the original denominator: at \((1,0)\), this gives \(\langle 1,0\rangle\); at \((0,2)\), it gives \(4\langle 0,\frac12\rangle=\langle 0,2\rangle\); and at \((-1,-1)\), it gives \(2\langle-\frac12,-\frac12\rangle=\langle-1,-1\rangle\).
Sketch. At every allowed point, the components are a positive scalar multiple of \(\langle x,y\rangle\), so arrows point radially outward. Their magnitude is
Thus arrows are longer near the origin and shorter farther away. The origin must be left out of the arrow sketch: the formula assigns no vector there. The computed values agree with the pattern—rightward at \((1,0)\), upward at \((0,2)\), and diagonally down-left at \((-1,-1)\).
A Common Mistake: Filling in a Missing Point
For the field \(\mathbf{G}\) above, a tempting but incorrect substitution at the origin is
The fact that both numerators are zero does not make either quotient equal to zero. Division by zero is undefined, so \(\mathbf{G}(0,0)\) does not exist and the origin is excluded from the domain. The corrected description is a radial vector field defined on \(\mathbb{R}^2\setminus\{(0,0)\}\), with no arrow placed at the origin. A diagram that includes an arrow there would claim a value not supplied by the formula.
Choosing a Visualization That Fits
A formula does not determine just one possible picture; the output type determines which pictures preserve the relevant information. For a scalar field of two variables, a surface graph shows heights, while contours show locations of equal value. For a vector field of two variables, an arrow sketch shows the vector assigned to each sampled point. For a field with three input variables, the same type check applies, though drawing many arrows in three-dimensional space can make a diagram crowded; selected sample points may be clearer.
| Field formula | Output | Useful visualization |
|---|---|---|
| \(f(x,y)\) | One real number at each point | Surface \(z=f(x,y)\) or contours \(f(x,y)=c\) |
| \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\) | A planar vector at each point | Arrow sketch with vectors based at sample points |
| \(\mathbf{F}(x,y,z)\) | A three-dimensional vector at each point | Three-dimensional arrow sketch or selected samples |
For any of these choices, the domain belongs in the picture as well as in the description. A scalar contour map should not suggest values at excluded inputs, and a vector sketch should not place arrows where the vector formula is undefined. In practice, a small number of carefully chosen samples is often more informative than a crowded plot: include points that test symmetry, reveal direction, and check what happens near any excluded location.
Check Your Understanding
For each question, use the formula to support your description rather than relying on an imagined plot.
- Is \(h(x,y)=x+2y\) a scalar field or a vector field, and what is its domain?
- Evaluate \(\mathbf{H}(x,y)=\langle y,-x\rangle\) at \((1,0)\) and \((0,1)\). What arrow directions do these values suggest?
- Describe the contour \(x^2+y^2=9\) for the scalar field \(f(x,y)=x^2+y^2\). What is the contour level?
- For \(\mathbf{K}(x,y)=\left\langle\frac{x}{x^2+y^2},\frac{y}{x^2+y^2}\right\rangle\), which point is excluded from the domain, and why?
- What kind of visualization would you choose for a scalar field of two variables, and what kind would you choose for a planar vector field?