Restricting a Function to a Parametrized Boundary

From a Boundary Curve to One Variable

A function \(f(x,y)\) depends on two coordinates, but along a specified boundary curve the coordinates are linked. If the curve is described by a parameter \(t\), then both \(x\) and \(y\) become functions of \(t\). Substituting those expressions into \(f\) produces a one-variable function.

This is useful when studying a function on a region with a boundary. The restriction records the function's values along a particular boundary piece, using the parameter to keep track of position. The essential steps are to choose a parametrization that covers the intended piece and then substitute both coordinate functions into \(f\).

The expression \(g(t)\) gives the value of \(f\) at the point on the curve corresponding to \(t\). It does not change the original function; it describes its values only along the curve traced by \(\mathbf{r}\).

Why the Substitution Captures the Curve

Suppose a parametrization \(\mathbf{r}:I\to\mathbb{R}^2\) traces exactly the boundary piece \(C\), meaning that \(\mathbf{r}(I)=C\). Each parameter value \(t\) gives a point of \(C\), and every point of \(C\) is reached by at least one parameter value. It follows that the values of the restricted function are exactly the values of \(f\) on \(C\).

To see both inclusions, first take any \(t\in I\). Since \(\mathbf{r}(t)\in C\), we have \(g(t)=f(\mathbf{r}(t))\), which is a value of \(f\) on \(C\). Conversely, take any \((x,y)\in C\). Because \(\mathbf{r}(I)=C\), there is some \(t\in I\) with \(\mathbf{r}(t)=(x,y)\). Then \(f(x,y)=g(t)\). Thus the two sets of values are equal.

This argument explains why the parameter interval matters as much as the substitution. If the parametrization misses part of \(C\), the resulting one-variable function describes only the part it reaches. If it traces points outside the intended piece, the restriction includes values from the wrong curve.

Parametrizing a Line Segment

Let the endpoints of a line segment be \(P=(x_0,y_0)\) and \(Q=(x_1,y_1)\). A standard parametrization is the linear interpolation \(\mathbf{r}(t)=(1-t)P+tQ\), for \(0\le t\le 1\). In coordinates, this is

At \(t=0\), the point is \(P\); at \(t=1\), it is \(Q\). For values between 0 and 1, both coordinates move linearly between their endpoint values, so the parametrization traces the segment. The interval must be retained: allowing \(t\) to range over all real numbers traces the entire line through the endpoints, not just the segment.

Parametrizing a Circle

A circle of radius \(R>0\), centered at \((h,k)\), has the standard parametrization

For a full traversal, take \(0\le t\le 2\pi\). This traces every point on the circle, with the starting point repeated at the final parameter value. Repeating a point is harmless: the purpose is to cover the circle, not necessarily to assign a unique parameter to each point. For a circle centered at the origin, set \(h=k=0\). An arc uses a smaller interval chosen to trace just that arc.

The Chain Rule Along a Parametrized Curve

If \(f\) has continuous first partial derivatives near the curve and \(x(t)\) and \(y(t)\) are differentiable, then \(g(t)=f(x(t),y(t))\) is differentiable. The multivariable chain rule gives

This rule follows by tracking how changes in \(t\) change each coordinate. To first order, a small change \(\Delta t\) produces coordinate changes \(x'(t)\Delta t\) and \(y'(t)\Delta t\). The corresponding first-order change in \(f\) is \(f_x\) times the change in \(x\), plus \(f_y\) times the change in \(y\). Dividing by \(\Delta t\) and taking the limit gives the displayed formula. The formula applies where the stated differentiability conditions hold; the substitution \(g(t)=f(x(t),y(t))\) itself does not require differentiability.

For example, in the line-segment calculation, \(f_x=2x+y\), \(f_y=x+4y\), \(x'(t)=2\), and \(y'(t)=3\). Along the segment, \(x=1+2t\) and \(y=-1+3t\). Therefore

Direct differentiation of the restricted function \(g(t)=2-7t+28t^2\) also gives \(g'(t)=-7+56t\), confirming the chain-rule calculation.

Common Parametrization Errors

A frequent line-segment error is to use endpoint coordinates as increments. For the segment from \((1,-1)\) to \((3,2)\), one might incorrectly write \(x=1+3t\), \(y=-1+2t\), with \(0\le t\le1\). At \(t=1\), this gives \((4,1)\), not the required endpoint \((3,2)\). The correction is to subtract the starting coordinates from the ending coordinates: the increments are \(3-1=2\) and \(2-(-1)=3\). Thus the correct parametrization is \(x=1+2t\), \(y=-1+3t\).

Another error is to use a correct circle formula with an interval that traces the wrong portion. For example, \(x=3\cos t\), \(y=3\sin t\), \(0\le t\le\pi\) describes the upper semicircle, not the full circle. To trace the full circle, use \(0\le t\le2\pi\). Always check both the equations of the curve and the parameter range.

Once a boundary piece has been converted into a one-variable function, it can be studied using one-variable methods. The boundary restriction is the setup: its correctness depends on the curve, the interval, and the substitution all matching the intended boundary piece.