Adding and Scaling Scalar and Vector Fields

Combining Fields Point by Point

A scalar field assigns a number to each point in its domain. Examples include a temperature assigned to every location in a room or a concentration assigned to every point in a solution. A vector field assigns a vector to each point; a velocity field, for example, assigns a velocity vector at each location.

Once fields have been defined, we can form new fields by adding their values or multiplying their values by a constant. These operations are carried out at each point. The location stays fixed while the field values at that location are combined.

Definition: Let $f$ and $g$ be scalar fields with the same domain $D$. Their sum $f+g$ and constant multiple $cf$, where $c$ is a constant, are the scalar fields defined by $$ (f+g)(\mathbf{x})=f(\mathbf{x})+g(\mathbf{x}), \qquad (cf)(\mathbf{x})=c f(\mathbf{x}) $$ for every $\mathbf{x}\in D$. The same pointwise definitions apply to vector fields with values in the same vector space.

Two fields can be added only where both are defined, and their values must be compatible for addition. Thus scalar fields can be added to scalar fields, and vector fields can be added to vector fields of the same dimension. If their domains differ, the sum is defined on the intersection of their domains. In applications, the quantities being added must also be compatible: for instance, temperatures can be combined as temperatures, but a temperature and a length do not make a meaningful sum.

Scalar-Field Formulas

If $f$ and $g$ are scalar fields on a common domain $D$, their sum at a point is simply the sum of the two numbers assigned there. Multiplying a scalar field by a constant multiplies each of its values by that constant.

(f+g)(x,y,z)=f(x,y,z)+g(x,y,z), \qquad (cf)(x,y,z)=c f(x,y,z).

The same formulas work when the domain has two coordinates or another specified form: use the coordinates of a point in that domain. The operation does not change which point is being evaluated. It changes the value assigned to that point.

Worked Example: Adding and Scaling Two Scalar Fields

Let $f(x,y)=x+2y$ and $g(x,y)=3x-y$, defined on $\mathbb{R}^2$. Find the formula for $h=f+g$, then evaluate $h$ and $-2f$ at $(1,2)$.

At a general point, add the field values:

$$ h(x,y)=f(x,y)+g(x,y) =(x+2y)+(3x-y)=4x+y. $$

At $(1,2)$, $f(1,2)=1+2(2)=5$ and $g(1,2)=3(1)-2=1$. Therefore, $h(1,2)=5+1=6$. Substituting into the formula for $h$ gives $h(1,2)=4(1)+2=6$, as required.

For the scaled field, $(-2f)(x,y)=-2(x+2y)=-2x-4y$. At $(1,2)$, $(-2f)(1,2)=-2(5)=-10$. Its formula gives $-2(1)-4(2)=-2-8=-10$, confirming the value.

Vector-Field Formulas

A vector field in two dimensions has two component functions. Write $\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle$ and $\mathbf{G}(x,y)=\langle R(x,y),S(x,y)\rangle$. At every point, vector addition adds corresponding components, and multiplication by a constant multiplies each component by that constant.

(\mathbf{F}+\mathbf{G})(x,y) = \langle P(x,y)+R(x,y),\;Q(x,y)+S(x,y)\rangle, \qquad (c\mathbf{F})(x,y) = \langle cP(x,y),\;cQ(x,y)\rangle.

For a three-dimensional field, write $\mathbf{F}(x,y,z)=\langle P,Q,R\rangle$ and $\mathbf{G}(x,y,z)=\langle U,V,W\rangle$, where each component is evaluated at $(x,y,z)$. The corresponding formulas are

(\mathbf{F}+\mathbf{G})(x,y,z) = \langle P+U,\;Q+V,\;R+W\rangle, \qquad (c\mathbf{F})(x,y,z) = \langle cP,\;cQ,\;cR\rangle.

These formulas combine vector values at a common location; they do not add or scale the coordinates of the location itself. For example, to find $(\mathbf{F}+\mathbf{G})(1,2)$, first evaluate both fields at $(1,2)$, then add the resulting vectors component by component.

Worked Example: Adding Two Vector Fields

Let $\mathbf{F}(x,y)=\langle x-y,2x\rangle$ and $\mathbf{G}(x,y)=\langle y,1-x\rangle$. Find $\mathbf{H}=\mathbf{F}+\mathbf{G}$, and calculate the resulting vector at $(1,2)$.

Add corresponding component functions:

\mathbf{H}(x,y) = \langle (x-y)+y,\;2x+(1-x)\rangle = \langle x,\;x+1\rangle.

At $(1,2)$, $\mathbf{F}(1,2)=\langle 1-2,2(1)\rangle=\langle-1,2\rangle$, and $\mathbf{G}(1,2)=\langle 2,1-1\rangle=\langle2,0\rangle$. Their sum is $\langle-1+2,2+0\rangle=\langle1,2\rangle$. Substitution into the combined formula gives $\mathbf{H}(1,2)=\langle1,1+1\rangle=\langle1,2\rangle$, confirming the result.

Worked Example: Scaling a Vector Field

Let $\mathbf{V}(x,y)=\langle 2x,y-1\rangle$. Find $-3\mathbf{V}$, and evaluate it at $(1,3)$.

Scale each component function by $-3$:

(-3\mathbf{V})(x,y) = \langle -3(2x),-3(y-1)\rangle = \langle -6x,-3y+3\rangle.

First, $\mathbf{V}(1,3)=\langle2(1),3-1\rangle=\langle2,2\rangle$, so scaling that vector gives $-3\langle2,2\rangle=\langle-6,-6\rangle$. Direct substitution into the scaled formula gives $\langle-6(1),-3(3)+3\rangle=\langle-6,-9+3\rangle=\langle-6,-6\rangle$. Both calculations agree.

Why the Component Rules Work

The formulas follow from the definition of vector addition and scalar multiplication. In two dimensions, adding $\langle a,b\rangle$ and $\langle d,e\rangle$ means adding their first components and their second components, giving $\langle a+d,b+e\rangle$. Multiplying $\langle a,b\rangle$ by $c$ means multiplying each component, giving $\langle ca,cb\rangle$. Apply these definitions to the vectors assigned by the fields at an arbitrary fixed point $(x,y)$.

Evaluate both fields at the same point.
Write $\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle$ and $\mathbf{G}(x,y)=\langle R(x,y),S(x,y)\rangle$.

Apply vector addition.
By componentwise addition, the result is $\langle P(x,y)+R(x,y),Q(x,y)+S(x,y)\rangle$.

Apply scalar multiplication.
By componentwise scaling, multiplying $\mathbf{F}(x,y)$ by $c$ gives $\langle cP(x,y),cQ(x,y)\rangle$.

This reasoning also verifies that the pointwise sum is a vector field: at each point of the common domain it assigns a vector with the same number of components as the original fields. The scaled field does too. The argument works in three dimensions by applying the same definitions to three components.

How Scaling Changes Values and Arrows

For a scalar field, multiplying by a positive constant preserves the sign of every value and multiplies its magnitude by that constant. Multiplication by a negative constant reverses the sign as well as scaling the magnitude. Multiplication by zero gives the zero scalar field, which has value zero at every point.

For a vector field, a positive scaling factor changes arrow lengths by the factor $c$ without changing their directions. A negative factor reverses each nonzero arrow and changes its length by the factor $|c|$. Scaling by zero makes every arrow the zero vector. These statements follow from the corresponding effects of scalar multiplication on an individual vector; they apply separately at each point.

Adding vector fields is different from scaling one field. At each point, the two arrows are added as vectors, so the result depends on both their directions and their lengths. The magnitude of a sum is not generally the sum of the magnitudes. For example, opposite vectors can cancel: $\langle1,0\rangle+\langle-1,0\rangle=\langle0,0\rangle$.

Worked Example: Adding and Scaling in Three Dimensions

Let $\mathbf{A}(x,y,z)=\langle x,y,z\rangle$ and $\mathbf{B}(x,y,z)=\langle1,-1,2\rangle$. Find $\mathbf{C}=2\mathbf{A}+\mathbf{B}$, then evaluate $\mathbf{C}$ at $(-1,2,1)$.

First scale each component of $\mathbf{A}$, then add the components of $\mathbf{B}$:

\mathbf{C}(x,y,z) = \langle2x,2y,2z\rangle+\langle1,-1,2\rangle = \langle2x+1,2y-1,2z+2\rangle.

At $(-1,2,1)$, $\mathbf{A}(-1,2,1)=\langle-1,2,1\rangle$. Thus $2\mathbf{A}(-1,2,1)=\langle-2,4,2\rangle$, while $\mathbf{B}(-1,2,1)=\langle1,-1,2\rangle$. Adding gives $\langle-2+1,4-1,2+2\rangle=\langle-1,3,4\rangle$. Substitution into the formula for $\mathbf{C}$ gives $\langle2(-1)+1,2(2)-1,2(1)+2\rangle=\langle-1,3,4\rangle$, confirming the result.

A Common Mistake

A common error is to add the input coordinates instead of adding the field values. Consider the fields in the earlier worked example at the point $(1,2)$. The incorrect calculation might replace $(1,2)$ by the coordinate pair $(1+1,2+2)=(2,4)$, treating that as the location for the sum. But the sum field is defined at the original point $(1,2)$, and its value there is $\mathbf{F}(1,2)+\mathbf{G}(1,2)=\langle1,2\rangle$, not a vector obtained by changing the location.

The reliable procedure is: keep the point fixed, evaluate each field there, and then add or scale the resulting values. When working from formulas, derive the combined component formula first and substitute the location afterward. This makes it easier to see which quantities are inputs and which are field values.

Key takeaway: Field addition and constant scaling are pointwise operations. The location is unchanged; scalar values are combined as numbers, and vector values are combined or scaled component by component.

Check Your Understanding

Answer each question using the pointwise definitions.

If $f(x,y)=2x-y$ and $g(x,y)=x+3y$, what is the formula for $f+g$, and what is its value at $(1,2)$?
For $\mathbf{F}(x,y)=\langle x,2y\rangle$, write the component formula for $-\mathbf{F}$. What does the negative sign do to each nonzero arrow?
Why must two vector fields have the same number of components to be added?
At a fixed point, can the magnitude of the sum of two vectors always be found by adding their magnitudes? Explain with an example or counterexample.
What is the difference between adding two field values at a point and adding the coordinates of that point?