From Sampled Values to a Field Sketch
A two-dimensional vector field assigns a vector to each point in its domain. A sketch makes those assignments visible: at each selected point, draw an arrow whose components come from the field’s formula. The arrow’s location matters just as much as its components. It begins at the point where the vector field is being evaluated, not automatically at the origin.
A field is usually sampled at a manageable set of grid points. A sketch then shows the values at those points, rather than every value in the domain. To make the picture readable, choose a regular grid, keep the coordinate scales consistent, and use the same plotting scale for every vector.
The first component tells how far the arrow moves in the horizontal coordinate direction, and the second tells how far it moves in the vertical coordinate direction. If a scale factor is used, it multiplies both components. It does not change the base point, and it must be applied the same way to every sampled vector.
How to Place and Scale the Arrows
Suppose the value at a sample point \((a,b)\) is \(\langle u,v\rangle\). With scale factor \(s\), the plotted displacement is \(\langle su,sv\rangle\). Therefore, the arrow tip is found by adding that displacement to the base point:
This endpoint rule follows directly from the meaning of vector components: \(\langle su,sv\rangle\) describes a move of \(su\) units horizontally and \(sv\) units vertically. Starting that move at \((a,b)\) gives the endpoint \((a+su,b+sv)\). Thus the rule preserves both pieces of information in the field value: the vector is drawn at the correct point, and its component ratio is preserved by multiplying both components by the same factor.
The plotted scale factor is a drawing choice, not a change to the field itself. A factor smaller than \(1\) shortens all arrows, which can prevent overlap on a dense grid. A factor greater than \(1\) lengthens them. If one factor is used throughout, comparisons between sampled vectors remain meaningful. Choosing a different factor for each arrow can make the sketch misleading.
Use equal coordinate spacing for equal coordinate differences, and use the same physical scale on the horizontal and vertical axes. For example, if one unit horizontally occupies twice as much space as one unit vertically, an arrow’s apparent slope will not represent the ratio of its vertical and horizontal components. A clear sketch labels the axes and marks the sample points before adding arrows.
Worked Example: Plotting a Field on a Three-by-Three Grid
Worked Example: Plotting a Field on a Three-by-Three Grid
Sketch sampled arrows for \(\mathbf{F}(x,y)=\langle x+y,x-y\rangle\) at the grid points with \(x,y\in\{-1,0,1\}\). Use scale factor \(s=\frac12\).
At each point, substitute its coordinates into both components. Multiply the resulting vector by \(\frac12\) to get the plotted displacement, then add that displacement to the base point to get the tip. The table gives all three calculations for each sample.
| Base point | \(\mathbf{F}(x,y)\) | Plotted displacement \(\frac12\mathbf{F}(x,y)\) | Arrow tip |
|---|---|---|---|
| \((-1,-1)\) | \(\langle-1-1,-1-(-1)\rangle=\langle-2,0\rangle\) | \(\langle-1,0\rangle\) | \((-1,-1)+(-1,0)=(-2,-1)\) |
| \((-1,0)\) | \(\langle-1+0,-1-0\rangle=\langle-1,-1\rangle\) | \(\langle-\frac12,-\frac12\rangle\) | \((-1,0)+(-\frac12,-\frac12)=(-\frac32,-\frac12)\) |
| \((-1,1)\) | \(\langle-1+1,-1-1\rangle=\langle0,-2\rangle\) | \(\langle0,-1\rangle\) | \((-1,1)+(0,-1)=(-1,0)\) |
| \((0,-1)\) | \(\langle0-1,0-(-1)\rangle=\langle-1,1\rangle\) | \(\langle-\frac12,\frac12\rangle\) | \((0,-1)+(-\frac12,\frac12)=(-\frac12,-\frac12)\) |
| \((0,0)\) | \(\langle0+0,0-0\rangle=\langle0,0\rangle\) | \(\langle0,0\rangle\) | \((0,0)+(0,0)=(0,0)\) |
| \((0,1)\) | \(\langle0+1,0-1\rangle=\langle1,-1\rangle\) | \(\langle\frac12,-\frac12\rangle\) | \((0,1)+(\frac12,-\frac12)=(\frac12,\frac12)\) |
| \((1,-1)\) | \(\langle1-1,1-(-1)\rangle=\langle0,2\rangle\) | \(\langle0,1\rangle\) | \((1,-1)+(0,1)=(1,0)\) |
| \((1,0)\) | \(\langle1+0,1-0\rangle=\langle1,1\rangle\) | \(\langle\frac12,\frac12\rangle\) | \((1,0)+(\frac12,\frac12)=(\frac32,\frac12)\) |
| \((1,1)\) | \(\langle1+1,1-1\rangle=\langle2,0\rangle\) | \(\langle1,0\rangle\) | \((1,1)+(1,0)=(2,1)\) |
Plot each arrow from the base point in the first column to its tip in the last column. For instance, the arrow at \((1,0)\) ends at \((\frac32,\frac12)\). The row at \((0,0)\) gives a zero vector, so it is represented by a point rather than a visible arrow of positive length.
Check the endpoint calculations by subtracting each base point from its listed tip. For the first row, \((-2,-1)-(-1,-1)=(-1,0)\), which is the listed displacement. For the row at \((0,1)\), \((\frac12,\frac12)-(0,1)=(\frac12,-\frac12)\). Applying the same subtraction to every row recovers its displacement column; multiplying each displacement by \(2\) recovers the field value in that row. Thus the plotted arrows use the stated scale and the correct base points.
A Practical Sketching Procedure
A reliable sketch can be made in a short sequence. The order matters: computing the field values first reduces the risk of confusing a base point with an arrow tip while drawing.
Use a regular grid over the region of interest, with enough points to show how the assigned vectors vary across the region.
At each sample point, substitute its \(x\)- and \(y\)-coordinates into \(P\) and \(Q\) to obtain \(\langle P(x,y),Q(x,y)\rangle\).
If arrows would be too long or crowded, select a convenient positive factor \(s\) and multiply every vector by it.
Use \((x,y)\) as the tail and \((x,y)+s\mathbf{F}(x,y)\) as the tip. Draw a point when the vector is zero.
Keep equal units equally spaced on both axes, and check that each arrow’s horizontal and vertical changes match its scaled components.
Worked Example: Choosing a Smaller Scale for a Sparse Grid
Worked Example: Choosing a Smaller Scale for a Sparse Grid
For \(\mathbf{G}(x,y)=\langle 2x+y,x-y\rangle\), plot the values at \((-1,-1),(-1,1),(1,-1),(1,1)\), and \((0,0)\). Use \(s=\frac13\).
At \((-1,-1)\), the components are \(2(-1)+(-1)=-3\) and \((-1)-(-1)=0\). Thus \(\mathbf{G}(-1,-1)=\langle-3,0\rangle\), the plotted displacement is \(\langle-1,0\rangle\), and the tip is \((-1,-1)+(-1,0)=(-2,-1)\).
At \((-1,1)\), the components are \(2(-1)+1=-1\) and \((-1)-1=-2\). Thus \(\mathbf{G}(-1,1)=\langle-1,-2\rangle\), the displacement is \(\langle-\frac13,-\frac23\rangle\), and the tip is \((-1,1)+(-\frac13,-\frac23)=(-\frac43,\frac13)\).
At \((1,-1)\), the components are \(2(1)+(-1)=1\) and \(1-(-1)=2\). Thus \(\mathbf{G}(1,-1)=\langle1,2\rangle\), the displacement is \(\langle\frac13,\frac23\rangle\), and the tip is \((1,-1)+(\frac13,\frac23)=(\frac43,-\frac13)\).
At \((1,1)\), the components are \(2(1)+1=3\) and \(1-1=0\). Thus \(\mathbf{G}(1,1)=\langle3,0\rangle\), the displacement is \(\langle1,0\rangle\), and the tip is \((1,1)+(1,0)=(2,1)\).
At \((0,0)\), both components are zero, so the vector and displacement are \(\langle0,0\rangle\), and the tip remains \((0,0)\).
Check the tips by subtracting their bases: \((-2,-1)-(-1,-1)=(-1,0)\); \((-\frac43,\frac13)-(-1,1)=(-\frac13,-\frac23)\); \((\frac43,-\frac13)-(1,-1)=(\frac13,\frac23)\); \((2,1)-(1,1)=(1,0)\); and \((0,0)-(0,0)=(0,0)\). These are exactly the five scaled vectors. Draw the arrows only after marking their five distinct base points.
Worked Example: Plotting a Field with Nonlinear Components
Worked Example: Plotting a Field with Nonlinear Components
For \(\mathbf{H}(x,y)=\langle x^2-y,x+y^2\rangle\), plot the values on the four points with \(x,y\in\{0,1\}\). Use \(s=\frac12\).
At \((0,0)\), the components are \(0^2-0=0\) and \(0+0^2=0\). The field value and displacement are both \(\langle0,0\rangle\), so the tip is \((0,0)\).
At \((1,0)\), the components are \(1^2-0=1\) and \(1+0^2=1\). The field value is \(\langle1,1\rangle\), the displacement is \(\langle\frac12,\frac12\rangle\), and the tip is \((1,0)+(\frac12,\frac12)=(\frac32,\frac12)\).
At \((0,1)\), the components are \(0^2-1=-1\) and \(0+1^2=1\). The field value is \(\langle-1,1\rangle\), the displacement is \(\langle-\frac12,\frac12\rangle\), and the tip is \((0,1)+(-\frac12,\frac12)=(-\frac12,\frac32)\).
At \((1,1)\), the components are \(1^2-1=0\) and \(1+1^2=2\). The field value is \(\langle0,2\rangle\), the displacement is \(\langle0,1\rangle\), and the tip is \((1,1)+(0,1)=(1,2)\).
For a check, subtract each base point from its tip. The differences are \((0,0)\), \((\frac12,\frac12)\), \((-\frac12,\frac12)\), and \((0,1)\), respectively. Doubling those differences recovers the four field values. The calculation is complete; the sketch consists of these four arrows, each drawn with its tail at the corresponding grid point.
A Common Mistake: Starting Every Arrow at the Origin
A vector can be drawn as a free arrow to show its components, but a vector-field sketch assigns a vector to a particular point. Moving the tail to the origin loses that location information.
For example, in the third worked example, \(\mathbf{H}(1,0)=\langle1,1\rangle\). With scale factor \(\frac12\), the plotted displacement is \(\langle\frac12,\frac12\rangle\). A mistaken calculation might draw the arrow from \((0,0)\) to \((\frac12,\frac12)\). That endpoint correctly represents the displacement from the origin, but the field value was sampled at \((1,0)\), not at the origin.
The correction is to add the displacement to the actual base point:
So the arrow must begin at \((1,0)\) and end at \((\frac32,\frac12)\). As a final check, the tip minus the base is \((\frac32,\frac12)-(1,0)=(\frac12,\frac12)\), the required scaled vector.
Why Careful Plotting Matters
A vector-field sketch is a compact way to see the values assigned at several points at once. It is useful only when the positions and plotted displacements are trustworthy. If the arrows are drawn from the wrong points, use inconsistent scales, or are distorted by unequal axis scales, the picture can suggest relationships that are not present in the formula.
A sketch also represents sampled values, not every value between the sample points. Adding more grid points gives more information about the field’s assignments, but the arrows should still be computed from the component functions at their own base points. If a component formula is undefined at a location, it provides no vector there, so do not draw an arrow at that point.
Check Your Understanding
For each question, distinguish the field value from the plotted displacement and from the arrow’s endpoint.
- If a field value at \((2,-1)\) is \(\langle3,2\rangle\) and the plotting factor is \(\frac12\), what are the plotted displacement and arrow tip?
- Why must the plotting factor be applied to both components of every sampled vector?
- For \(\mathbf{F}(x,y)=\langle x+y,x-y\rangle\), evaluate the field at \((0,1)\). With scale factor \(\frac12\), find the displacement and tip.
- What information is lost if every arrow is drawn with its tail at the origin?
- Why can unequal physical scales on the two coordinate axes make a sketch misleading?