The Central Idea
A field answers a location-based question:
\[ \boxed{\text{What quantity is assigned at this point?}} \]For a scalar field,
\[ f(x,y)=\text{one scalar value}. \]For a vector field,
\[ \mathbf{F}(x,y)=\text{one vector}. \]Scalar fields tell us how much. Vector fields can tell us both how much and which way.
Location Comes First
Every field value is attached to a location.
If
\[ f(2,3)=7, \]then \((2,3)\) is the point and \(7\) is the scalar value at that point.
If
\[ \mathbf{F}(2,3)=\langle 4,-1\rangle, \]then the same point \((2,3)\) carries the vector \(\langle 4,-1\rangle\).
Scalar Field Values
In a scalar field, evaluating the function produces one number.
Suppose
\[ T(x,y)=20+x-2y. \]At \((4,3)\),
\[ T(4,3)=20+4-2(3)=18. \]The scalar value at that location is
\[ \boxed{18}. \]Vector Field Values
In a vector field, evaluation produces an ordered vector.
Let
\[ \mathbf{F}(x,y)=\langle x+y,2x-y\rangle. \]At \((3,1)\),
\[ \mathbf{F}(3,1) = \langle 3+1,2(3)-1\rangle = \langle 4,5\rangle. \]The complete field value is therefore
\[ \boxed{\langle 4,5\rangle}. \]Components Carry Directional Information
A two-dimensional vector
\[ \langle a,b\rangle \]has a horizontal component \(a\) and a vertical component \(b\).
Positive \(a\) points toward increasing \(x\), while negative \(a\) points toward decreasing \(x\).
Positive \(b\) points toward increasing \(y\), while negative \(b\) points toward decreasing \(y\).
Reading Direction from Signs
The signs of vector components give a quick directional description.
| Vector | General direction |
|---|---|
| \(\langle +,+\rangle\) | right and up |
| \(\langle -,+\rangle\) | left and up |
| \(\langle -,-\rangle\) | left and down |
| \(\langle +,-\rangle\) | right and down |
Magnitude Is Not Direction
A vector's magnitude measures its size:
\[ \|\langle a,b\rangle\| = \sqrt{a^2+b^2}. \]For
\[ \mathbf{v}=\langle 3,4\rangle, \]the magnitude is
\[ \|\mathbf{v}\|=5. \]The number \(5\) does not by itself tell us the direction.
Different Vectors Can Have the Same Magnitude
Consider
\[ \langle 3,4\rangle \]and
\[ \langle -3,-4\rangle. \]Both have magnitude
\[ 5, \]but they point in opposite directions.
Magnitude therefore describes strength or size, not orientation.
Unit Direction
For any nonzero vector \(\mathbf{v}\), dividing by its magnitude produces a unit vector in the same direction:
\[ \boxed{ \widehat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|} }. \]If
\[ \mathbf{v}=\langle 3,4\rangle, \]then
\[ \widehat{\mathbf{v}} = \left\langle \frac35,\frac45\right\rangle. \]Its magnitude is \(1\), so only direction remains.
Direction Angles in the Plane
A nonzero planar vector
\[ \mathbf{v}=\langle a,b\rangle \]can also be described by an angle \(\theta\) measured from the positive \(x\)-axis.
When \(a\neq 0\),
\[ \tan\theta=\frac{b}{a}. \]The component signs determine the correct quadrant.
Example: Direction Angle
Let
\[ \mathbf{v}=\langle 1,\sqrt{3}\rangle. \]Then
\[ \tan\theta=\sqrt{3}. \]Both components are positive, so the vector lies in the first quadrant. Therefore,
\[ \boxed{\theta=60^\circ=\frac{\pi}{3}}. \]Three-Dimensional Direction
In three dimensions,
\[ \mathbf{v}=\langle a,b,c\rangle \]combines motion or orientation in three coordinate directions.
Its magnitude is
\[ \|\mathbf{v}\| = \sqrt{a^2+b^2+c^2}. \]The unit direction is
\[ \widehat{\mathbf{v}} = \frac{\langle a,b,c\rangle} {\sqrt{a^2+b^2+c^2}}. \]The Zero Vector Has No Direction
The vector
\[ \mathbf{0}=\langle 0,0\rangle \]has magnitude \(0\).
Because division by zero is undefined, it cannot be normalized:
\[ \frac{\mathbf{0}}{\|\mathbf{0}\|} \]is undefined.
Therefore, the zero vector has no unique direction.
Comparing Scalar Field Values
For a scalar field, comparing two points means comparing two numbers.
Let
\[ f(x,y)=x^2+y^2. \]Then
\[ f(1,1)=2 \]and
\[ f(3,4)=25. \]The second point has the larger scalar field value.
Comparing Vector Field Values
Vector values cannot generally be ordered as larger or smaller in the same way as ordinary numbers.
For example,
\[ \langle 5,0\rangle \]and
\[ \langle 0,5\rangle \]have equal magnitude but different directions.
Comparing vector fields may require comparing components, magnitudes, directions, or all three.
Comparing Magnitudes at Different Points
Suppose
\[ \mathbf{F}(x,y)=\langle x,y\rangle. \]At \((1,1)\),
\[ \|\mathbf{F}(1,1)\|=\sqrt{2}. \]At \((3,4)\),
\[ \|\mathbf{F}(3,4)\|=5. \]The vector field is stronger in magnitude at \((3,4)\).
Comparing Directions at Different Points
In the radial field
\[ \mathbf{F}(x,y)=\langle x,y\rangle, \]the vector at \((2,0)\) is
\[ \langle 2,0\rangle, \]while the vector at \((0,2)\) is
\[ \langle 0,2\rangle. \]The magnitudes are equal, but the directions differ by \(90^\circ\).
Parallel Field Vectors
Two field vectors point in the same or opposite line direction when one is a scalar multiple of the other.
For example,
\[ \langle 2,4\rangle \]and
\[ \langle 1,2\rangle \]point in the same direction because
\[ \langle 2,4\rangle=2\langle 1,2\rangle. \]Opposite Directions
The vectors
\[ \mathbf{u} \]and
\[ -\mathbf{u} \]have the same magnitude but opposite directions.
For example,
\[ \langle 3,-1\rangle \]and
\[ \langle -3,1\rangle \]differ only by multiplication by \(-1\).
Perpendicular Field Vectors
Two vectors are perpendicular when their dot product is zero.
If
\[ \mathbf{u}=\langle a,b\rangle \]and
\[ \mathbf{v}=\langle c,d\rangle, \]then perpendicularity requires
\[ ac+bd=0. \]This provides an algebraic test for directional relationships inside a field.
Example: Testing Perpendicular Directions
Consider
\[ \mathbf{u}=\langle 2,1\rangle \]and
\[ \mathbf{v}=\langle -1,2\rangle. \]Their dot product is
\[ \mathbf{u}\cdot\mathbf{v} = 2(-1)+1(2)=0. \]Therefore, the vectors are perpendicular.
Direction Relative to Position
A useful question is whether the field vector points along, against, or across the position vector.
The position vector is
\[ \mathbf{r}=\langle x,y\rangle. \]Comparing \(\mathbf{F}(x,y)\) with \(\mathbf{r}\) can reveal radial, inward, or tangential behavior.
Outward Direction
In
\[ \mathbf{F}(x,y)=\langle x,y\rangle, \]the field vector equals the position vector.
Every nonzero vector therefore points directly away from the origin.
Its magnitude is
\[ \sqrt{x^2+y^2}, \]so the field also becomes stronger with distance.
Inward Direction
In
\[ \mathbf{F}(x,y)=\langle -x,-y\rangle, \]the field vector is the negative of the position vector.
Every nonzero vector points directly toward the origin.
Tangential Direction
In the field
\[ \mathbf{F}(x,y)=\langle -y,x\rangle, \]the field vector is perpendicular to the position vector:
\[ \langle x,y\rangle\cdot\langle -y,x\rangle=0. \]Thus the vectors point tangent to circles centered at the origin.
Direction Can Be Described Locally
A vector field may not have one global direction.
Instead, each point has its own local vector direction.
In
\[ \mathbf{F}(x,y)=\langle -y,x\rangle, \]the direction continuously changes as the point moves around the origin.
Scalar Values Do Not Contain Direction
If a temperature field satisfies
\[ T(2,3)=30, \]the value \(30\) gives no inherent direction.
To learn which way temperature increases or decreases, we must compare nearby field values.
This distinction motivates the later study of gradients.
Comparing Nearby Scalar Values
Let
\[ f(x,y)=x+2y. \]At \((1,1)\),
\[ f(1,1)=3. \]One unit to the right,
\[ f(2,1)=4. \]One unit upward,
\[ f(1,2)=5. \]The field rises more across the second comparison than across the first.
Direction of Increasing Scalar Values
A scalar field may increase differently in different spatial directions.
For
\[ f(x,y)=x+2y, \]increasing \(x\) by \(1\) changes the field by \(1\), while increasing \(y\) by \(1\) changes the field by \(2\).
This elementary comparison previews the idea that spatial change itself can have direction.
Direction of Decreasing Scalar Values
Suppose
\[ f(x,y)=10-x-3y. \]Moving toward increasing \(x\) lowers the field value, and moving toward increasing \(y\) lowers it even more rapidly per coordinate unit.
Scalar fields therefore contain directional information indirectly through how values vary from point to point.
Contours Help Reveal Directional Change
A contour joins points that have equal scalar field values.
If nearby contours correspond to rapidly changing values, the scalar field is changing quickly across that region.
If contours are farther apart, the change is more gradual.
The dedicated lesson on level sets and contour maps develops this visual relationship in detail.
Vector Length in Field Diagrams
When arrow lengths are drawn proportionally, longer arrows indicate larger vector magnitudes.
If one point has
\[ \|\mathbf{F}\|=2 \]and another has
\[ \|\mathbf{F}\|=8, \]the second location has a vector quantity four times as large in magnitude.
Normalized Field Diagrams
Sometimes every arrow is displayed with the same length so that direction is easier to see.
Mathematically, this resembles replacing each nonzero field vector by
\[ \frac{\mathbf{F}(x,y)} {\|\mathbf{F}(x,y)\|}. \]Such a diagram preserves direction but suppresses magnitude information.
Magnitude as Its Own Scalar Field
Every vector field generates a scalar field through its magnitude:
\[ M(x,y)=\|\mathbf{F}(x,y)\|. \]If
\[ \mathbf{F}(x,y)=\langle x,y\rangle, \]then
\[ M(x,y)=\sqrt{x^2+y^2}. \]Thus vector strength can be studied using scalar-field tools.
Example: Magnitude Field
Let
\[ \mathbf{F}(x,y)=\langle 2x,-2y\rangle. \]Its magnitude field is
\[ M(x,y) = \sqrt{(2x)^2+(-2y)^2}. \]Therefore,
\[ \boxed{ M(x,y)=2\sqrt{x^2+y^2} }. \]The vector magnitude depends only on distance from the origin.
Component Fields Can Be Studied Separately
For
\[ \mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle, \]the functions \(P\) and \(Q\) may be analyzed as ordinary scalar fields.
One component may grow while the other decreases. One may vanish along a curve while the other remains nonzero.
The final vector direction results from their combination.
Where a Component Vanishes
Suppose
\[ \mathbf{F}(x,y)=\langle x-y,x+y\rangle. \]The horizontal component vanishes where
\[ x-y=0, \]or
\[ y=x. \]Along that line, the vectors have no horizontal component.
Where Another Component Vanishes
For the same field,
\[ \mathbf{F}(x,y)=\langle x-y,x+y\rangle, \]the vertical component vanishes where
\[ x+y=0, \]or
\[ y=-x. \]Along that line, the vectors are purely horizontal.
Finding Where the Whole Vector Vanishes
A vector field equals the zero vector only when every component is zero.
For
\[ \mathbf{F}(x,y)=\langle x-y,x+y\rangle, \]solve
\[ x-y=0 \]and
\[ x+y=0. \]The unique solution is
\[ \boxed{(0,0)}. \]Direction Can Reverse across a Boundary
Consider
\[ \mathbf{F}(x,y)=\langle x,0\rangle. \]When \(x>0\), vectors point right.
When \(x<0\), vectors point left.
At \(x=0\), the field is zero.
A change in component sign can therefore mark a directional reversal.
Direction and Physical Interpretation
The meaning of a vector direction depends on what the field represents.
For a velocity field, the direction indicates instantaneous motion.
For a force field, it indicates the direction of force.
For an acceleration field, it indicates the direction of acceleration.
The same mathematical vector may carry different physical meanings in different models.
Units of Field Values
Field values should be interpreted with units.
A scalar temperature field may have units of degrees Celsius.
A velocity vector field may have components measured in meters per second.
If
\[ \mathbf{v}(x,y)=\langle 3,4\rangle\ \text{m/s}, \]then its magnitude is
\[ 5\ \text{m/s}. \]Field Values Can Depend on Time
If
\[ \mathbf{F}(x,y,t), \]then fixing \((x,y)\) and changing \(t\) shows how the field vector at one location evolves with time.
Fixing \(t\) instead gives a snapshot of the field across space.
These are different ways of slicing the same space-time model.
Comparing the Same Point at Different Times
Suppose
\[ \mathbf{v}(x,y,t)=\langle x+t,y-t\rangle. \]At the point \((1,2)\),
\[ \mathbf{v}(1,2,0)=\langle 1,2\rangle, \]while
\[ \mathbf{v}(1,2,3)=\langle 4,-1\rangle. \]The location stayed fixed, but the field value changed with time.
Comparing Different Points at the Same Time
For the same field,
\[ \mathbf{v}(x,y,t)=\langle x+t,y-t\rangle, \]fix \(t=2\).
Then
\[ \mathbf{v}(0,0,2)=\langle 2,-2\rangle \]and
\[ \mathbf{v}(3,1,2)=\langle 5,-1\rangle. \]This comparison shows spatial variation at one instant.
Common Error: Confusing Coordinates with Components
In
\[ \mathbf{F}(2,3)=\langle 5,-1\rangle, \]the point \((2,3)\) is the input location.
The numbers \(5\) and \(-1\) are components of the output vector.
They do not describe a second location unless the context explicitly says so.
Common Error: Comparing Vectors Like Ordinary Numbers
Statements such as
\[ \langle 3,4\rangle > \langle 2,5\rangle \]do not have the ordinary numerical meaning of \(7>4\).
Vector comparison must specify what is being compared: magnitude, direction, a component, or another defined criterion.
Common Error: Using Magnitude to Recover Direction
Knowing only
\[ \|\mathbf{F}\|=5 \]does not determine the vector.
Infinitely many planar vectors have magnitude \(5\), including
\[ \langle 5,0\rangle,\quad \langle 0,5\rangle,\quad \langle 3,4\rangle,\quad \langle -3,-4\rangle. \]Direction requires additional information.
Common Error: Assigning Direction to a Scalar
A scalar value such as
\[ T=30^\circ\text{C} \]has no inherent spatial direction.
A direction of increasing temperature can be derived from the surrounding scalar field, but that direction is not contained in the isolated number \(30\).
Common Error: Normalizing the Zero Vector
The unit-vector formula
\[ \widehat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|} \]works only when
\[ \|\mathbf{v}\|\neq 0. \]The zero vector cannot be normalized because division by zero would be required.
A Reliable Procedure for Reading a Scalar Field Value
- Identify the location.
- Check that the point is in the domain.
- Evaluate the scalar field at that point.
- Attach units when relevant.
- Compare with nearby or reference values only when needed.
- Do not assign a direction to the scalar value itself.
A Reliable Procedure for Reading a Vector Field Value
- Identify the location.
- Evaluate every component of the field.
- Keep the components in the correct coordinate order.
- Read the signs to understand broad directional behavior.
- Compute the magnitude when strength or size matters.
- Normalize the vector when only direction is needed.
- Interpret the vector according to the physical or geometric model.
Example: Full Scalar Interpretation
Let
\[ T(x,y)=25-2x+y. \]At \((3,4)\),
\[ T(3,4)=25-6+4=23. \]Therefore,
\[ \boxed{T(3,4)=23}. \]The point \((3,4)\) is the location. The number \(23\) is the scalar field value there. No direction has yet been specified.
Example: Full Vector Interpretation
Let
\[ \mathbf{F}(x,y)=\langle 2x-y,x+2y\rangle. \]At \((2,1)\),
\[ \mathbf{F}(2,1) = \langle 4-1,2+2\rangle = \langle 3,4\rangle. \]Its magnitude is
\[ \|\mathbf{F}(2,1)\| = 5. \]Its unit direction is
\[ \left\langle \frac35,\frac45\right\rangle. \]Thus one evaluation gives the complete vector, its strength, and its direction.
Example: Compare Two Directions
Suppose
\[ \mathbf{F}(1,0)=\langle 2,0\rangle \]and
\[ \mathbf{F}(0,1)=\langle 0,2\rangle. \]Both vectors have magnitude \(2\).
Their dot product is
\[ \langle 2,0\rangle\cdot\langle 0,2\rangle=0. \]Therefore, their directions are perpendicular.
Practice 1: Scalar Field Value
Let
\[ f(x,y)=4x-y+2. \]Find \(f(3,5)\).
Practice 2: Vector Field Value
Let
\[ \mathbf{F}(x,y)=\langle x+2y,3x-y\rangle. \]Find \(\mathbf{F}(2,-1)\).
Practice 3: Magnitude
Find the magnitude of
\[ \mathbf{v}=\langle 8,15\rangle. \]Practice 4: Unit Direction
Find a unit vector in the direction of
\[ \mathbf{v}=\langle 5,12\rangle. \]Practice 5: Read the Signs
Describe the broad planar direction of
\[ \langle -4,7\rangle. \]Practice 6: Compare Magnitudes
Which vector has greater magnitude?
\[ \mathbf{u}=\langle 6,8\rangle \]or
\[ \mathbf{v}=\langle 9,0\rangle. \]Practice 7: Parallel or Perpendicular?
Determine the relationship between
\[ \mathbf{u}=\langle 2,3\rangle \]and
\[ \mathbf{v}=\langle -3,2\rangle. \]Practice 8: Find Where a Component Vanishes
For
\[ \mathbf{F}(x,y)=\langle x+2y,3x-y\rangle, \]find the line where the first component is zero.
Practice 9: Find the Zero Vector Location
Find where
\[ \mathbf{F}(x,y)=\langle x-1,y+4\rangle \]equals \(\langle 0,0\rangle\).
Practice 10: Scalar versus Direction
A temperature field satisfies
\[ T(2,5)=18. \]Explain what the number \(18\) tells us and what it does not tell us.
Practice Answers
Practice 1
\[ f(3,5)=4(3)-5+2=9. \]Therefore,
\[ \boxed{f(3,5)=9}. \]Practice 2
\[ \mathbf{F}(2,-1) = \langle 2+2(-1),3(2)-(-1)\rangle. \]Therefore,
\[ \boxed{\mathbf{F}(2,-1)=\langle 0,7\rangle}. \]Practice 3
\[ \|\mathbf{v}\| = \sqrt{8^2+15^2} = \sqrt{289} = \boxed{17}. \]Practice 4
\[ \|\mathbf{v}\| = \sqrt{5^2+12^2} = 13. \]Therefore,
\[ \boxed{ \widehat{\mathbf{v}} = \left\langle \frac5{13},\frac{12}{13}\right\rangle }. \]Practice 5
The first component is negative and the second is positive, so the vector points broadly
\[ \boxed{\text{left and up}}. \]Practice 6
\[ \|\mathbf{u}\| = \sqrt{6^2+8^2} = 10, \] \[ \|\mathbf{v}\|=9. \]Therefore, \(\mathbf{u}\) has the greater magnitude.
Practice 7
\[ \mathbf{u}\cdot\mathbf{v} = 2(-3)+3(2)=0. \]Therefore,
\[ \boxed{\mathbf{u}\perp\mathbf{v}}. \]Practice 8
Set the first component equal to zero:
\[ x+2y=0. \]Therefore,
\[ \boxed{y=-\frac{x}{2}}. \]Practice 9
Set both components equal to zero:
\[ x-1=0, \qquad y+4=0. \]Therefore,
\[ \boxed{(x,y)=(1,-4)}. \]Practice 10
The number \(18\) is the scalar temperature assigned to the location \((2,5)\). It does not by itself tell us which direction is warmer, cooler, or changing most rapidly.
What to Remember
A field value must always be interpreted together with its location.
- a scalar field assigns one number to each point;
- a vector field assigns one vector to each point;
- vector components encode coordinate-direction contributions;
- vector magnitude measures size or strength;
- a unit vector isolates direction;
- the zero vector has no unique direction;
- vectors can be compared by magnitude, components, or directional relationships;
- scalar fields contain directional-change information only through comparisons across space.
The essential discipline is to keep location, field value, magnitude, and direction conceptually separate.
Where We Go Next
Field values tell us what is assigned at individual locations. The next step is to organize locations that share the same scalar value.
The next lesson, Level Sets and Contour Maps, shows how equal-value sets turn scalar fields into visible geometric patterns.