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Evaluating a Scalar Function Along a Line Path

Learn to turn a scalar function evaluated along a line path into a function of one parameter.

Beginner 8 min read

What You'll Learn

  • Identify the starting point and direction vector in a line path.
  • Substitute each coordinate expression into the matching input of a scalar function.
  • Simplify the result to express the function's values in terms of the parameter.
  • Check a substituted formula by evaluating both forms at selected parameter values.
  • Recognize how a zero direction vector gives a constant path.

From a Line Path to a One-Variable Formula

A scalar function of several variables assigns a number to each allowed input point. If the input point moves along a line, the scalar function's value may change as the point moves. The line path describes each coordinate in terms of a parameter, so substituting those coordinates into the scalar function produces a formula in that parameter alone.

This is a direct way to track a function's values along a specified path. The key is to preserve the correspondence between each coordinate and its input slot: the expression for \(x\) goes into the \(x\)-input, the expression for \(y\) goes into the \(y\)-input, and, when present, the expression for \(z\) goes into the \(z\)-input.

Definition: A line path in the plane has coordinate functions \(x=a+bt\) and \(y=c+dt\), where \(a,b,c,d\) are fixed real numbers and \(t\) is the parameter. A line path in space also has \(z=e+ft\), where \(e,f\) are fixed real numbers. Evaluating a scalar function along the path means substituting these coordinate functions into the function.

The point at \(t=0\) is the starting point \((a,c)\) in the plane or \((a,c,e)\) in space. The coefficients of \(t\), namely \(b,d\) or \(b,d,f\), give the direction in which the coordinates change as the parameter changes. The parameter \(t\) is not itself one of the spatial coordinates; it tells us which point on the path to use.

The Substitution Rule

Suppose \(F\) is a scalar function of two variables and the path is \(x=a+bt,\ y=c+dt\). Define a new function \(g\) of the parameter by evaluating \(F\) at the path's point:

$$ g(t)=F(a+bt,c+dt). $$

For a scalar function \(H\) of three variables and a path \(x=a+bt,\ y=c+dt,\ z=e+ft\), the corresponding expression is

$$ g(t)=H(a+bt,c+dt,e+ft). $$

In each case, the result is a one-variable formula. It describes the scalar function's value at the point reached when the parameter has value \(t\). The formula is understood for parameter values whose path points are inputs at which the original scalar function is defined.

The substitution can be organized as a short procedure:

1
Write the path coordinates.
Record the expression for each coordinate as a function of \(t\).
2
Match each expression to its input.
Replace \(x\) with its path expression, \(y\) with its path expression, and \(z\) with its path expression if the function has a third input.
3
Simplify and check.
Expand or collect terms as needed, then compare the result with the original function at a convenient value of \(t\).

Worked Example: A Quadratic Function Along a Plane Line

Worked Example: A Quadratic Function Along a Plane Line

Let \(F(x,y)=x^2+2y\), and evaluate it along the line path \(x=1+2t,\ y=3-t\).

Replace \(x\) in the function with \(1+2t\), and replace \(y\) with \(3-t\):

$$ g(t)=F(1+2t,3-t)=(1+2t)^2+2(3-t). $$

Expand each part:

$$ (1+2t)^2=1+4t+4t^2, \qquad 2(3-t)=6-2t. $$

Combine like terms:

$$ g(t)=1+4t+4t^2+6-2t=4t^2+2t+7. $$

Check at \(t=0\). The path gives \((x,y)=(1,3)\), so the original function gives \(F(1,3)=1^2+2(3)=1+6=7\). The new formula gives \(g(0)=4(0)^2+2(0)+7=7\). Both evaluations agree.

Check also at \(t=1\). The path gives \((x,y)=(3,2)\), and \(F(3,2)=3^2+2(2)=9+4=13\). The parameter formula gives \(g(1)=4+2+7=13\), as required.

Why the Substitution Works

The rule follows from what it means to evaluate a function at a point. At a chosen parameter value \(t\), the path specifies one point. The value along the path is, by definition, the value of the original scalar function at that point. Writing the coordinates of that point as expressions in \(t\) gives exactly the substitution formula.

Derivation: Let the plane path be \(\mathbf{r}(t)=(a+bt,c+dt)\). The scalar function \(F\) takes an ordered pair as input. At parameter value \(t\), the point on the path has first coordinate \(a+bt\) and second coordinate \(c+dt\). Therefore the value of \(F\) at that point is \(F(a+bt,c+dt)\). Defining \(g(t)\) to be this value gives \(g(t)=F(a+bt,c+dt)\). The three-variable case follows by the same argument: the path point has coordinates \((a+bt,c+dt,e+ft)\), so its value is \(H(a+bt,c+dt,e+ft)\). Thus the formulas are not additional rules about scalar functions; they express evaluation at the point selected by the path.

This reasoning also explains why input order matters. If a function is specified as \(F(x,y)\), its first input is the \(x\)-coordinate and its second input is the \(y\)-coordinate. The path's ordered coordinates must be inserted in that same order.

Worked Example: Keeping the Inputs in Order

Worked Example: Keeping the Inputs in Order

Let \(F(x,y)=xy+y^2\), and use the line path \(x=-1+t,\ y=2+3t\).

Substitute the \(x\)-expression into the first input and the \(y\)-expression into the second:

$$ g(t)=F(-1+t,2+3t)=(-1+t)(2+3t)+(2+3t)^2. $$

Expand the product in the first term:

$$ (-1+t)(2+3t) =-2-3t+2t+3t^2 =-2-t+3t^2. $$

Expand the square in the second term:

$$ (2+3t)^2=4+12t+9t^2. $$

Add the two expressions and collect like terms:

$$ g(t)=(-2-t+3t^2)+(4+12t+9t^2) =2+11t+12t^2. $$

For a check at \(t=0\), the path point is \((-1,2)\), and the original function gives \(F(-1,2)=(-1)(2)+2^2=-2+4=2\). The simplified formula gives \(g(0)=2\). At \(t=-1\), the path point is \((-2,-1)\), so \(F(-2,-1)=(-2)(-1)+(-1)^2=2+1=3\). The formula gives \(g(-1)=2-11+12=3\). The checks agree.

Worked Example: A Scalar Function Along a Space Line

Worked Example: A Scalar Function Along a Space Line

Let \(H(x,y,z)=x^2+yz+2z\), and use the path \(x=2-t,\ y=1+2t,\ z=-1+t\).

Substitute each coordinate expression into its matching input:

$$ g(t)=H(2-t,1+2t,-1+t) =(2-t)^2+(1+2t)(-1+t)+2(-1+t). $$

Expand the square, product, and final term separately:

$$ (2-t)^2=4-4t+t^2, $$ $$ (1+2t)(-1+t)=-1+t-2t+2t^2=-1-t+2t^2, $$ $$ 2(-1+t)=-2+2t. $$

Now add the terms and combine coefficients:

$$ g(t)=(4-4t+t^2)+(-1-t+2t^2)+(-2+2t) =1-3t+3t^2. $$

At \(t=0\), the path point is \((2,1,-1)\). The original function gives \(H(2,1,-1)=2^2+(1)(-1)+2(-1)=4-1-2=1\). The parameter formula gives \(g(0)=1\). At \(t=1\), the point is \((1,3,0)\), and \(H(1,3,0)=1^2+(3)(0)+2(0)=1\). The formula gives \(g(1)=1-3+3=1\). Both checks confirm the result.

Reading the Path and Its Parameter

The fixed terms in a path specify the point at parameter value zero; the coefficients of \(t\) specify how the coordinates vary. For example, in \(x=1+2t,\ y=3-t\), the starting point is \((1,3)\). Increasing \(t\) by one increases \(x\) by two and decreases \(y\) by one. This information helps you keep the path coordinates straight, but the function value is still found by substitution into the function.

A path with all direction coefficients equal to zero does not move. For instance, \(x=2,\ y=-1\) gives the same point for every \(t\). Along such a path, \(F(x,y)\) has the constant value \(F(2,-1)\), whenever that input is defined. More generally, a zero coefficient for just one coordinate means only that coordinate stays fixed; the other coordinates may still vary.

The parameter can be any real number for which the path point is an allowed input to the scalar function. If a particular problem specifies a parameter interval, use only values in that interval. Substitution gives the formula along the path; it does not change which input points are permitted by the function's definition.

Common Mistake: Swapping Coordinate Expressions

A frequent error is to substitute the right expressions in the wrong order. This changes the point at which the function is evaluated, and can therefore change the result.

Worked Example: Correcting a Coordinate Swap

Let \(F(x,y)=x+2y\), with path \(x=1+2t,\ y=3-t\). A mistaken substitution might reverse the expressions and write

$$ F(3-t,1+2t)=(3-t)+2(1+2t)=5+3t. $$

That calculation is algebraically correct for the point \((3-t,1+2t)\), but that is not the specified path point. The path says that \(x=1+2t\) and \(y=3-t\), so the correct substitution is

$$ g(t)=F(1+2t,3-t) =(1+2t)+2(3-t) =1+2t+6-2t =7. $$

Check at \(t=0\). The specified path point is \((1,3)\), so \(F(1,3)=1+2(3)=7\). The corrected formula gives \(g(0)=7\). The mistaken expression gives \(5\) at \(t=0\), which is the value at the incorrectly swapped point \((3,1)\), not at \((1,3)\).

Why Evaluating Along a Line Is Useful

A multivariable function describes values across a set of points, while a line path selects a particular sequence of points from that set. Substitution reduces the evaluation along that path to a one-variable expression. This makes it possible to compare values at different positions on the line, identify where the value is constant, and prepare for later questions about how the value changes. For the present task, the essential step is simply to form the substituted expression correctly and simplify it without losing the coordinate order.

Check Your Understanding

For each question, write the parameterized scalar-function value before simplifying, and use a selected parameter value to check your result.

  1. For \(F(x,y)=x^2-y\) along \(x=2+t,\ y=1-2t\), what is the resulting formula in \(t\)?
  2. For \(F(x,y)=3x+y\) along \(x=-1+t,\ y=4+t\), what is the value at \(t=0\), and what is the full formula along the path?
  3. For \(H(x,y,z)=xy+z\) along \(x=t,\ y=2-t,\ z=1+3t\), find the formula in \(t\).
  4. In the path \(x=5-2t,\ y=-1+4t\), what is the starting point, and by how much does each coordinate change when \(t\) increases by one?
  5. If \(x=2,\ y=3\) for every \(t\), what can be said about \(F(x,y)\) along the path, provided \(F(2,3)\) is defined?

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