Restricting a Scalar Field to a Curve

From a Field to a Function of One Parameter

A scalar field assigns a number to each point in its domain. A parametrized curve, in turn, describes a moving point. When we evaluate the field at the point on the curve corresponding to parameter \(t\), the result is a function of \(t\). This step turns a spatial expression into an ordinary one-variable integrand.

This composition is central to computing a scalar line integral. The field supplies the value at each point; the parametrization tells us which point is being visited. Once those coordinates are substituted and simplified, the arc-length factor can be included using the scalar line integral formula.

Definition: Let \(f:U\to\mathbb{R}\) be a scalar field and let \(\mathbf{r}:[a,b]\to U\) be a parametrized curve. The restriction of \(f\) to this parametrized curve is the function \(g:[a,b]\to\mathbb{R}\) defined by \(g(t)=f(\mathbf{r}(t))\). If \(\mathbf{r}(t)=(x(t),y(t),z(t))\), then \(g(t)=f(x(t),y(t),z(t))\).

The word “restriction” here means that we evaluate the field only at points on the curve, rather than at every point in its domain. The resulting \(g\) is an ordinary function of the parameter. It is not a new scalar field on all of space.

For a continuously differentiable curve \(\mathbf{r}:[a,b]\to U\) and a continuous scalar field \(f\) on the curve, the scalar line integral is computed by

\int_C f\,ds = \int_a^b f(\mathbf{r}(t))\,\|\mathbf{r}'(t)\|\,dt.

Thus the composition \(f(\mathbf{r}(t))\) is the field-value factor in the parameter integral. The speed \(\|\mathbf{r}'(t)\|\) converts a small parameter change into a small arc-length change. The focus here is on forming and simplifying the composition accurately.

Why Composition Gives the Correct Field Value

At parameter value \(t\), the point on the curve is \(\mathbf{r}(t)\). The field value at that point is, by definition, \(f(\mathbf{r}(t))\). This explains why all coordinate formulas for the curve must be substituted into the field, not merely one coordinate or an endpoint.

The scalar line integral formula can also be understood by approximating the curve with short pieces. On a sufficiently short piece near parameter \(t_i\), continuity makes the field value approximately \(f(\mathbf{r}(t_i))\). The piece's length is approximately \(\|\mathbf{r}'(t_i)\|\Delta t\). Therefore its contribution to the sum approximating the integral is approximately

f(\mathbf{r}(t_i))\,\|\mathbf{r}'(t_i)\|\,\Delta t.

Adding these contributions and taking the limit as the parameter partition gets finer gives the Riemann integral of \(f(\mathbf{r}(t))\|\mathbf{r}'(t)\|\). This is why the field must first be evaluated along the curve, while the speed factor remains as a separate part of the line-integral integrand.

Key idea: To restrict a scalar field to a parametrized curve, replace each spatial coordinate in \(f\) by its corresponding coordinate function from \(\mathbf{r}(t)\), then simplify the resulting expression before integrating.

Worked Examples

Worked Example: Restricting a Polynomial Field to a Line

Let \(f(x,y,z)=x^2+xy+z^2\) and \(\mathbf{r}(t)=(t,2t,0)\), for \(0\le t\le1\). First compose the field with the curve:

f(\mathbf{r}(t)) =f(t,2t,0) =t^2+(t)(2t)+0^2 =t^2+2t^2 =3t^2.

For the scalar line integral, differentiate the parametrization and calculate its speed:

\mathbf{r}'(t)=(1,2,0), \qquad \|\mathbf{r}'(t)\| =\sqrt{1^2+2^2+0^2} =\sqrt{5}.

Therefore, the parameter integral is

\int_C f\,ds =\int_0^1 3t^2\sqrt{5}\,dt =3\sqrt{5}\left[\frac{t^3}{3}\right]_0^1 =\sqrt{5}(1-0) =\sqrt{5}.

To check the composition, substitute the curve coordinates into the original field once more: \(x^2=t^2\), \(xy=2t^2\), and \(z^2=0\), so their sum is \(3t^2\), as obtained. To check the integration, the antiderivative of \(3t^2\sqrt{5}\) is \(\sqrt{5}t^3\), whose derivative is \(3t^2\sqrt{5}\). The endpoint evaluation gives \(\sqrt{5}\).

Worked Example: Restricting an Exponential Field

Let \(f(x,y,z)=e^x(1+y)\) and let \(\mathbf{r}(t)=(t,t,0)\), for \(0\le t\le1\). Substitution gives

f(\mathbf{r}(t)) =f(t,t,0) =e^t(1+t).

The coordinate functions \(x(t)=t\) and \(y(t)=t\) both depend on \(t\). In particular, the factor \(1+y\) becomes \(1+t\), not a constant. The curve's derivative and speed are

\mathbf{r}'(t)=(1,1,0), \qquad \|\mathbf{r}'(t)\| =\sqrt{1^2+1^2+0^2} =\sqrt{2}.

Now integrate the restricted field with respect to arc length:

\int_C f\,ds =\sqrt{2}\int_0^1 e^t(1+t)\,dt.

Since \(\frac{d}{dt}(te^t)=e^t+te^t=e^t(1+t)\), an antiderivative is \(te^t\). Hence

\int_C f\,ds =\sqrt{2}\left[te^t\right]_0^1 =\sqrt{2}\bigl(1\cdot e^1-0\cdot e^0\bigr) =\sqrt{2}e.

The composition checks directly: at the point \((t,t,0)\), \(e^x=e^t\) and \(1+y=1+t\), so their product is \(e^t(1+t)\). The antiderivative check follows from the derivative above, and the endpoint values produce the stated result.

Worked Example: Restricting a Trigonometric Field to a Space Curve

Let \(f(x,y,z)=\sin z+xy\) and \(\mathbf{r}(t)=(\cos t,\sin t,t)\), for \(0\le t\le\frac{\pi}{2}\). Substitute all three coordinates:

f(\mathbf{r}(t)) =\sin(t)+(\cos t)(\sin t) =\sin t+\sin t\cos t.

The derivative and speed are

\mathbf{r}'(t)=(-\sin t,\cos t,1), \qquad \|\mathbf{r}'(t)\| =\sqrt{\sin^2t+\cos^2t+1} =\sqrt{2}.

Using \(\int_0^{\pi/2}\sin t\,dt=1\) and \(\sin t\cos t=\frac12\frac{d}{dt}(\sin^2t)\), the line integral becomes

\begin{aligned} \int_C f\,ds &=\sqrt{2}\int_0^{\pi/2}(\sin t+\sin t\cos t)\,dt\\ &=\sqrt{2}\left(1+\frac12[\sin^2t]_0^{\pi/2}\right)\\ &=\sqrt{2}\left(1+\frac12(1-0)\right) =\frac{3\sqrt{2}}{2}. \end{aligned}

For a check, \(\sin z\) becomes \(\sin t\), while \(xy\) becomes \((\cos t)(\sin t)\); both terms match the composition. An antiderivative of the parenthesized expression is \(-\cos t+\frac12\sin^2t\), whose derivative is \(\sin t+\sin t\cos t\). Its values at \(\frac{\pi}{2}\) and \(0\) differ by \(\frac32\), giving \(\frac{3\sqrt{2}}{2}\) after multiplication by the speed.

A Reliable Substitution Process

When a field has several terms or a curve has several nonconstant coordinates, writing each substitution explicitly reduces errors. A dependable calculation follows this order:

Record the coordinate functions.
Write \(\mathbf{r}(t)=(x(t),y(t),z(t))\), including the parameter interval.

Substitute into every field coordinate.
Form \(f(x(t),y(t),z(t))\), preserving products, powers, and functions such as exponentials and trigonometric functions.

Simplify the one-variable expression.
Use identities or algebra where appropriate, and check each term against the original field.

Include the speed for the line integral.
Compute \(\|\mathbf{r}'(t)\|\) and multiply it by the restricted field before integrating over the parameter interval.

The distinction between the last two steps is useful. The restriction \(f(\mathbf{r}(t))\) is determined solely by the field and the parametrization. The factor \(\|\mathbf{r}'(t)\|\) enters because the scalar line integral is accumulated with respect to arc length.

Common Mistake: Dropping the Parameter

Suppose \(f(x,y)=x^2+y\) and \(\mathbf{r}(t)=(t,2t)\), for \(0\le t\le1\). A common incorrect substitution is to write \(f(\mathbf{r}(t))=t^2+2\), as though \(y\) were the constant \(2\). But the second coordinate of the curve is \(y(t)=2t\), not \(2\). The wrong calculation would yield

\sqrt{5}\int_0^1(t^2+2)\,dt =\sqrt{5}\left(\frac13+2\right) =\frac{7\sqrt{5}}{3}.

This is not the scalar line integral for the stated field and curve. Correct substitution gives

f(\mathbf{r}(t)) =f(t,2t) =t^2+2t.

Since \(\mathbf{r}'(t)=(1,2)\), its speed is \(\sqrt{5}\), and the correct integral is

\sqrt{5}\int_0^1(t^2+2t)\,dt =\sqrt{5}\left(\frac13+1\right) =\frac{4\sqrt{5}}{3}.

Checking the substitution coordinate by coordinate catches the error: \(x^2\) becomes \(t^2\), and \(y\) becomes \(2t\). The incorrect answer fails this direct comparison with the field.

Why the Restricted Function Matters

Thinking of \(f(\mathbf{r}(t))\) as a separate one-variable function makes the calculation easier to organize. It reveals which values the field takes along the particular curve, exposes simplifications that may be hidden in the spatial formula, and makes clear how the parameter bounds determine the interval of integration.

It also separates two questions that can otherwise be confused: what value does the field have at the point on the curve, and how much arc length corresponds to a small change in parameter? The first is answered by composition; the second by the speed factor. Keeping those roles distinct is a reliable way to build the correct scalar line-integral integrand.

Check Your Understanding

For each question, write the substitution explicitly before simplifying.

If \(f(x,y)=x^2+y\) and \(\mathbf{r}(t)=(t,3t)\), what is \(f(\mathbf{r}(t))\)?
If \(f(x,y,z)=e^z+xy\) and \(\mathbf{r}(t)=(t,1,t^2)\), what is the restricted function?
For \(f(x,y)=\cos x+y\) and \(\mathbf{r}(t)=(2t,\sin t)\), write the scalar line-integral integrand, including the speed.
In the definition \(g(t)=f(\mathbf{r}(t))\), what does \(g(t)\) represent geometrically?
Why is substituting \(y=2\) incorrect for the curve \(\mathbf{r}(t)=(t,2t)\)?