Evaluating Scalar and Vector Fields at Points

Introduction

A scalar field assigns a number to each point in its domain. A vector field assigns a vector to each point in its domain. In either case, evaluating the field at a specified point means finding the value that the assignment gives at that point.

The calculation usually involves substituting the point’s coordinates into a formula. The main care required is to substitute every coordinate in the right place, preserve signs and parentheses, and—in a vector field—keep the components in their stated order.

Key idea: To evaluate a field at a point, replace each coordinate variable in its formula with the corresponding coordinate of the point. For a vector field, do this separately for every component function.

Evaluating a Scalar Field

A scalar field in two variables can be written \(f:D\to\mathbb{R}\), where \(D\) is a set of points in the plane. If \((a,b)\) is a point in \(D\), then \(f(a,b)\) is the number assigned to that point. The first coordinate replaces \(x\), and the second replaces \(y\).

For a scalar field in three variables, the same principle applies: \(f(a,b,c)\) is found by replacing \(x\) with \(a\), \(y\) with \(b\), and \(z\) with \(c\). Parentheses are especially helpful when a coordinate is negative.

Write the point’s coordinates in order.
For \((a,b)\), the first coordinate corresponds to \(x\), and the second corresponds to \(y\).

Substitute each coordinate.
Use parentheses around substituted values when needed, particularly for negative numbers.

Simplify and report the value.
The result for a scalar field is a number.

Worked Example: Evaluating a Scalar Field in Two Variables

Let \(f(x,y)=x^2-2xy+3y^2\). Evaluate \(f(2,-1)\).

Substitute \(x=2\) and \(y=-1\) into the formula:

\begin{aligned} f(2,-1) &=(2)^2-2(2)(-1)+3(-1)^2\\ &=4+4+3\\ &=11. \end{aligned}

Thus the value assigned to \((2,-1)\) is \(11\). To check, evaluate the three terms in the original expression: \(x^2=4\), \(-2xy=-2(2)(-1)=4\), and \(3y^2=3(1)=3\). Their sum is \(4+4+3=11\), as obtained above.

Worked Example: Evaluating a Scalar Field in Three Variables

Let \(g(x,y,z)=2x-yz+z^2\). Evaluate \(g(3,-1,2)\).

Substitute \(x=3\), \(y=-1\), and \(z=2\), retaining parentheses around the negative value:

\begin{aligned} g(3,-1,2) &=2(3)-(-1)(2)+(2)^2\\ &=6-(-2)+4\\ &=6+2+4\\ &=12. \end{aligned}

For a check, the product \(yz\) at this point is \((-1)(2)=-2\), so the term \(-yz\) is \(-(-2)=2\). The remaining terms are \(2x=6\) and \(z^2=4\). Adding \(6+2+4\) gives \(12\), confirming the result.

Evaluating a Vector Field

A vector field assigns a vector to each point. In two dimensions, a vector field can be expressed using two component functions, for example \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\). At a point \((a,b)\), its value is the vector whose first component is \(P(a,b)\) and whose second component is \(Q(a,b)\).

In three dimensions, a vector field can be written \(\mathbf{G}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle\). Its value at \((a,b,c)\) is \(\langle P(a,b,c),Q(a,b,c),R(a,b,c)\rangle\). The order matters: the first computed component stays first, the second stays second, and the third stays third.

Definition: If \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\), then \(\mathbf{F}(a,b)=\langle P(a,b),Q(a,b)\rangle\). For a three-dimensional field \(\mathbf{G}(x,y,z)=\langle P,Q,R\rangle\), its value is \(\mathbf{G}(a,b,c)=\langle P(a,b,c),Q(a,b,c),R(a,b,c)\rangle\).

Worked Example: Evaluating a Two-Dimensional Vector Field

Let \(\mathbf{F}(x,y)=\langle y-x,\ x^2+2y\rangle\). Find \(\mathbf{F}(1,-2)\).

Evaluate the first component using \(x=1\) and \(y=-2\):

$$ y-x=(-2)-1=-3. $$

Evaluate the second component at the same point:

$$ x^2+2y=(1)^2+2(-2)=1-4=-3. $$

Place the computed values in the original component order:

\mathbf{F}(1,-2)=\langle -3,-3\rangle.

Check both components directly in the defining formula: \(y-x=-2-1=-3\), and \(x^2+2y=1+2(-2)=-3\). The ordered pair is therefore \(\langle-3,-3\rangle\).

Worked Example: Evaluating a Three-Dimensional Vector Field

Let \(\mathbf{G}(x,y,z)=\langle xy+z,\ x-2z,\ y^2-z^2\rangle\). Evaluate \(\mathbf{G}(-1,2,1)\).

Compute each component using \(x=-1\), \(y=2\), and \(z=1\):

\begin{aligned} xy+z&=(-1)(2)+1=-2+1=-1,\\ x-2z&=(-1)-2(1)=-1-2=-3,\\ y^2-z^2&=(2)^2-(1)^2=4-1=3. \end{aligned}

Keep these results in their original order:

\mathbf{G}(-1,2,1)=\langle -1,-3,3\rangle.

To check, substitute the point into the original component list again: the first expression gives \(-2+1=-1\), the second gives \(-1-2=-3\), and the third gives \(4-1=3\). Hence the reported vector has the correct three components.

Why Componentwise Substitution Works

The component functions specify exactly how the assigned vector is built. A point supplies values for the coordinate variables; each component function uses those same values to produce one number. The definition of the vector field then places those numbers into an ordered vector.

More explicitly, suppose \(\mathbf{F}(x,y)=\langle P(x,y),Q(x,y)\rangle\). At \((a,b)\), the first component must be \(P(a,b)\), because that is the value of the first component function at the point. The second component must be \(Q(a,b)\) for the same reason. By the definition of an ordered vector field value, the result is therefore \(\langle P(a,b),Q(a,b)\rangle\). The three-component case follows identically, with one evaluated number for each of \(P\), \(Q\), and \(R\).

\begin{aligned} \mathbf{F}(a,b) &=\langle P(a,b),Q(a,b)\rangle,\\ \mathbf{G}(a,b,c) &=\langle P(a,b,c),Q(a,b,c),R(a,b,c)\rangle. \end{aligned}

This explains why it is valid to calculate the components separately and assemble the results afterward. It also explains why changing the order is not harmless: an ordered vector records which computed value belongs to each component position.

A Common Substitution Mistake

A frequent error is to lose the negative sign when substituting a negative coordinate into an expression containing a square or a product. For example, in the scalar field \(g(x,y,z)=2x-yz+z^2\), the value at \((3,-1,2)\) is not \(6-2+4=8\). That incorrect calculation treats the term \(-yz\) as if it were \(-2\), despite \(yz=(-1)(2)=-2\).

The correct substitution keeps the sign within the product and then applies the minus sign already present in the formula:

-yz=-\bigl((-1)(2)\bigr)=-(-2)=2, \qquad g(3,-1,2)=6+2+4=12.

Writing parentheses makes the operations visible. In vector fields, another related mistake is to calculate all component values correctly but report them in a different order. Preserve the component order from the original formula when forming the final vector.

How to Report an Evaluation Clearly

A clear solution makes the correspondence between the point and the variables easy to follow. State the point, show the substitution, simplify, and identify the result as a scalar or as an ordered vector. For a vector field, showing each component on its own line can make a sign or arithmetic error easier to detect.

Match coordinates and variables by position: \(a\) replaces \(x\), \(b\) replaces \(y\), and \(c\) replaces \(z\).
Use parentheses around negative coordinates when substituting them into products, powers, or sums.
Evaluate every component of a vector field at the same point.
Retain the component order specified by the field formula.
Check the result against the original expression before reporting it.

These steps apply whenever the specified point is one at which the field formula is defined. The resulting scalar is a number, while the resulting vector is reported with its components in order.

Check Your Understanding

For each question, substitute the coordinates carefully and show enough work to make the result checkable.

Let \(h(x,y)=3x-y^2\). Find \(h(-2,3)\).
Let \(k(x,y,z)=xy+2z\). Find \(k(2,-3,1)\).
Let \(\mathbf{A}(x,y)=\langle x+y,xy\rangle\). Find \(\mathbf{A}(-1,4)\).
Let \(\mathbf{B}(x,y,z)=\langle z-x,\ x+y,\ yz\rangle\). Find \(\mathbf{B}(2,1,-2)\).
In your own words, why must the component order be preserved when reporting a vector-field value?