Introduction
A field diagram places information at locations in space. Reading it correctly requires keeping two things separate: where a field is evaluated and the value the field assigns there. In a vector-field diagram, an arrow represents the value, while its base marks the location.
That distinction sounds simple, but several common errors follow from overlooking it. A viewer may treat an arrow tip as the location where the field is evaluated, mistake arrow length for the location’s distance from the origin, draw a scalar value as though it were a vector, or assume that a field is defined at a point where its formula is not defined.
Location and Value Are Different Objects
A two-dimensional vector field assigns a vector to each point in a domain \(D\) in the plane. In function notation, it is a map \(\mathbf{F}:D\to\mathbb{R}^2\). If \(\mathbf{p}=(a,b)\) is in \(D\), then \(\mathbf{F}(\mathbf{p})\) is the vector value assigned at that point.
For example, if \(\mathbf{F}(a,b)=(u_1,u_2)\), then \((a,b)\) is the location and \((u_1,u_2)\) is the vector value. A standard diagram draws an arrow with its base at \((a,b)\). If the drawing uses the same coordinate scale for vector components as for position, the arrow’s displacement from its base to its tip is \((u_1,u_2)\). If the diagram uses a drawing scale factor \(s\), that displayed displacement is \(s(u_1,u_2)\), not necessarily the vector value itself.
The distinction can be derived directly. Let the arrow base be \(\mathbf{p}=(a,b)\) and let the field value be \(\mathbf{F}(\mathbf{p})=(u_1,u_2)\). With drawing scale factor \(s\), the displayed tip is \(\mathbf{q}=(a+su_1,b+su_2)\). Subtracting the base from the tip gives the displayed displacement:
Thus the base records the input point, while the displacement from base to tip records a scaled drawing of the output vector. The tip’s coordinates are not, in general, another field value or the point at which the original arrow is based.
Worked Example: Keeping the Base Point Separate from the Tip
Worked Example: Keeping the Base Point Separate from the Tip
Consider \(\mathbf{F}(x,y)=(x,-y)\) and evaluate it at \(\mathbf{p}=(2,1)\). Find the vector value and the displayed tip when the drawing scale factor is \(s=\tfrac12\). Then check the displayed displacement.
First evaluate each component at the base point:
The input location remains \((2,1)\). With scale factor \(s=\tfrac12\), the displayed displacement is
Add this displayed displacement to the base to obtain the tip:
Check by subtracting the base from the tip:
The check confirms that the drawn arrow has the correct scaled displacement. Its base is \((2,1)\), its vector value is \((2,-1)\), and its tip is \((3,\tfrac12)\). These are three distinct pieces of information.
Do Not Read an Arrow Tip as Its Location
A frequent mistake is to treat the arrow tip as the point where the field value is evaluated. In the example above, the tip \((3,\tfrac12)\) is not the location of the value \(\mathbf{F}(2,1)\). That value belongs at the base \((2,1)\). The tip simply shows the end of the drawn arrow.
This error can create a false picture of the field. If every arrow is shifted to its own tip, then the diagram no longer shows which vector is assigned to which point. For a field in which vectors point toward or away from a center, that shift may make the pattern appear to have a different center or direction.
Arrow Length Is Not Position
The location of an arrow is determined by its base, not by its length. Length represents the magnitude of the vector value, subject to the diagram’s drawing scale. If a common scale factor \(s\) is used, a vector of magnitude \(\|\mathbf{F}(\mathbf{p})\|\) is drawn with length \(s\|\mathbf{F}(\mathbf{p})\|\). Changing the scale changes the displayed length, not the field value or the base point.
It is also important not to compare lengths across a diagram if the drawing does not use a consistent scale. A diagram may shorten arrows to prevent overlap. In that case, their directions and general relative magnitudes may be emphasized, but their displayed lengths need not equal the numerical magnitudes.
Worked Example: Distinguishing Position from Magnitude
Let \(\mathbf{F}(x,y)=(x,y)\), and compare the field values at \(\mathbf{p}_1=(1,0)\) and \(\mathbf{p}_2=(2,0)\). Use drawing scale \(s=\tfrac12\). Find the values, their magnitudes, and the displayed arrow lengths.
At the first point, substitute \(x=1\) and \(y=0\):
The displayed length is \(s\|\mathbf{F}(1,0)\|=\tfrac12\cdot1=\tfrac12\). The arrow’s base is still \((1,0)\).
At the second point, substitute \(x=2\) and \(y=0\):
The displayed length is \(s\|\mathbf{F}(2,0)\|=\tfrac12\cdot2=1\). The arrow’s base is \((2,0)\).
Check the two drawn displacements by scaling each value:
Their lengths are \(\tfrac12\) and \(1\), as computed. The second arrow is longer because its vector value has greater magnitude, not because its base is farther along the diagram in some sense encoded by arrow length. The positions are read from the bases: \((1,0)\) and \((2,0)\).
A Scalar Value Is Not a Vector Arrow
A scalar field and a vector field assign different kinds of values. A scalar field is a function \(f:D\to\mathbb{R}\); its value at a point is a single number. A vector field is a function \(\mathbf{F}:D\to\mathbb{R}^2\) in the plane; its value is an ordered pair. The diagram must represent the appropriate kind of output.
A scalar value such as \(3\) is not, by itself, the vector \((3,0)\). The latter is a particular vector pointing in a particular direction, while the scalar \(3\) has no direction. Scalar fields can be represented with numbers, colors, or heights, depending on the diagram. Those conventions do not turn the scalar output into a vector.
Worked Example: Reading a Scalar Value Correctly
Let \(f(x,y)=x+y\), a scalar field, and evaluate it at \((1,2)\). Explain why interpreting the result as a horizontal arrow is an error.
Substitute the coordinates into the formula:
The output is the scalar number \(3\). It is not an ordered pair. In particular, writing \((3,0)\) would introduce an extra component and a direction that the scalar field did not assign.
Check the input and output types: \((1,2)\in D\subseteq\mathbb{R}^2\) is a point, and \(f(1,2)=3\in\mathbb{R}\) is a scalar. By contrast, a vector field at that point would have an output in \(\mathbb{R}^2\). The type check confirms that no vector arrow follows from the scalar value alone.
Every Field Has a Domain
A formula does not necessarily define a field at every point in the plane. The domain is the set of inputs where the field is defined. A diagram should not assign an arrow at a point excluded from that domain. In particular, a point where the formula requires division by zero cannot be treated as a point with vector value zero; the value is undefined there.
A plotted diagram may show only selected sample points, so the absence of an arrow does not always prove that the field is undefined there. But if the field’s domain is specified and excludes a point, that exclusion is meaningful. It must be respected when interpreting the picture.
Worked Example: Respecting an Excluded Point
Consider the vector field
Determine its domain, evaluate it at \((1,0)\), and decide whether an arrow can be assigned at \((0,0)\).
The denominator is \(x^2+y^2\). For the formula to be defined, require \(x^2+y^2\ne0\). Since each square is nonnegative, their sum is zero exactly when \(x=0\) and \(y=0\). Thus the domain is
At \((1,0)\), the denominator is \(1^2+0^2=1\), so
At \((0,0)\), the denominator is \(0^2+0^2=0\), and each component would require division by zero. Therefore \(\mathbf{F}(0,0)\) is undefined, and no field-value arrow belongs at the origin.
Check the domain conclusion: \((1,0)\ne(0,0)\), so it is included and the computed value is valid. The origin is excluded, so assigning it a zero vector would contradict the formula rather than evaluate it.
A Reliable Reading Procedure
When a field diagram seems confusing, use a fixed order of checks rather than trying to infer everything from the overall pattern.
Determine whether the output is scalar or vector. A scalar value has no direction; a vector value does.
Confirm that the point being read belongs to the field’s domain before assigning it a value.
For a vector field, use the arrow’s base to identify the input point where the vector is assigned.
Use the arrow’s orientation and, when the scale is consistent, its length to interpret the vector value.
Do not assume displayed displacement equals the vector value unless the diagram uses the same scale. With scale \(s\), displayed displacement is \(s\mathbf{F}(\mathbf{p})\).
Common Mistakes and Their Corrections
Here are four errors to watch for. Each confuses two different aspects of the diagram.
| Mistake | Why it is wrong | Correction |
|---|---|---|
| Treating the arrow tip as the input point | The tip is determined by the base, the vector value, and the drawing scale. | Use the arrow’s base to identify where the field is evaluated. |
| Using arrow length as the location | Length represents magnitude, subject to drawing scale; location is given by the base. | Read position from the base and vector size from the arrow. |
| Drawing a scalar as a vector | A scalar has no direction and is not an ordered pair. | Represent scalar and vector outputs according to their types. |
| Adding an arrow at an excluded point | The field has no value there if the point is outside its domain. | Check the domain before plotting or interpreting a value. |
Why These Distinctions Matter
Field diagrams compress many values into a single picture. That makes them useful, but it also makes it easy to confuse the location of a value with the way that value is drawn. The distinctions are not cosmetic: changing a base point changes where the field value is assigned; changing the vector changes the field value; changing the drawing scale changes only the picture’s size.
Domain information is equally important. A diagram that includes an arrow at an excluded point asserts the existence of a value the field does not have. Conversely, a missing arrow at an ordinary point may simply mean that the diagram samples the field sparsely. Read the stated domain and the diagram together, rather than treating every visible gap as evidence of a singularity or every drawn point as valid.
Check Your Understanding
Use the distinctions in this tutorial to answer the following questions.
- If a vector arrow is based at \((2,1)\), what does the base tell you, and what does the arrow represent?
- A field value is \((2,-1)\), and the diagram uses scale factor \(\tfrac12\). What is the displayed displacement from base to tip?
- Why is it incorrect to interpret the scalar value \(4\) automatically as the vector \((4,0)\)?
- For the field in the excluded-point example, why is the origin not assigned the zero vector?
- If two arrows have different bases but the same length, what can you conclude about their locations? What, if the drawing scale is consistent, can you conclude about their vector magnitudes?