Speed and Arc Length of a Parametrized Curve

From Motion to Distance Along a Curve

A parametrized curve describes a point’s position as a parameter varies. When the parameter represents time, the velocity vector tells us both the direction and rate of motion. Its magnitude gives the instantaneous speed, a scalar that measures how quickly the point moves without recording its direction.

To find how far the point travels along its path, we accumulate that speed over the parameter interval. The result is the curve’s arc length. This is generally different from the straight-line distance between the starting and ending positions: a moving point can follow a curved route while its endpoints remain comparatively close.

Central formula: For a continuously differentiable curve $\mathbf{r}(t)$ on $a\le t\le b$, its speed is $\|\mathbf{r}'(t)\|$, and its arc length over that interval is the integral of this speed.

Velocity and Speed

Write a curve in three-dimensional space as $\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle$, where each coordinate is a differentiable function of the parameter $t$. Its velocity vector is the derivative of its position vector:

\mathbf{v}(t)=\mathbf{r}'(t) =\left\langle x'(t),y'(t),z'(t)\right\rangle.

The norm of a vector is its magnitude. In three dimensions, the magnitude of $\langle u_1,u_2,u_3\rangle$ is $\sqrt{u_1^2+u_2^2+u_3^2}$. Thus the speed of the curve is the magnitude of its velocity.

Definition: The speed of a differentiable parametrized curve $\mathbf{r}(t)$ is the nonnegative scalar function $$ v(t)=\|\mathbf{r}'(t)\| =\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}. $$ For a plane curve $\mathbf{r}(t)=\langle x(t),y(t)\rangle$, omit the $z$-component.

Velocity and speed are related but are not interchangeable. Velocity is a vector, so it includes direction; speed is a scalar, so it records magnitude only. Speed is never negative, even if a coordinate is decreasing or the point reverses direction.

Arc Length as Accumulated Speed

Suppose the parameter increases from $a$ to $b$. At each parameter value, the curve has an instantaneous speed. Integrating that rate over the interval accumulates the distance traveled along the path. If the parameter is time, this is the usual relationship between speed, elapsed time, and distance; the same calculation applies when $t$ is simply a curve parameter.

Arc-Length Formula: If $\mathbf{r}(t)$ is continuously differentiable on $[a,b]$, then its arc length on that interval is $$ L=\int_a^b \|\mathbf{r}'(t)\|\,dt =\int_a^b \sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}\,dt. $$ For a piecewise continuously differentiable curve with finitely many pieces, compute the integral on each piece and add the results.

A short justification comes from the meaning of speed. The instantaneous rate at which distance along a moving path accumulates is the speed $\|\mathbf{r}'(t)\|$. If $s(t)$ denotes the distance traveled from the starting parameter $a$ up to $t$, then $s'(t)=\|\mathbf{r}'(t)\|$. By the Fundamental Theorem of Calculus, the total distance from $a$ to $b$ is therefore the integral of this rate:

L=s(b)-s(a) =\int_a^b s'(t)\,dt =\int_a^b \|\mathbf{r}'(t)\|\,dt.

This reasoning assumes the curve is smooth enough for its velocity and speed to be defined continuously on the interval, or piecewise continuously on finitely many pieces. A zero velocity at an isolated parameter value does not prevent the integral from giving the length. The speed is zero at that instant, but the curve may still move before and after it.

Worked Examples

Worked Example: Length of a Helical Path

Find the speed and arc length of $\mathbf{r}(t)=\langle 3\cos t,3\sin t,4t\rangle$ for $0\le t\le \frac{\pi}{2}$.

Differentiate each coordinate to find the velocity:

\mathbf{r}'(t)=\langle -3\sin t,3\cos t,4\rangle.

Take its magnitude and simplify using $\sin^2t+\cos^2t=1$:

\begin{aligned} v(t)&=\sqrt{(-3\sin t)^2+(3\cos t)^2+4^2}\\ &=\sqrt{9\sin^2t+9\cos^2t+16}\\ &=\sqrt{9(\sin^2t+\cos^2t)+16}\\ &=\sqrt{25}=5. \end{aligned}

The speed is constant, so integrate it over the given interval:

L=\int_0^{\pi/2}5\,dt =5\left[t\right]_0^{\pi/2} =\frac{5\pi}{2}.

Check: Differentiating the accumulated-length function $s(t)=5t$ gives $s'(t)=5$, the computed speed. The length is positive, as expected. If $t$ is measured in seconds and the coordinates in metres, the speed is $5$ metres per second and the length is $\frac{5\pi}{2}$ metres.

Worked Example: A Curve with Variable Speed

Find the speed and arc length of $\mathbf{r}(t)=\left\langle \frac{t^2}{2},\frac{t^3}{3}\right\rangle$ for $0\le t\le 1$.

Differentiate the two coordinates:

\mathbf{r}'(t)=\langle t,t^2\rangle.

Then calculate the speed. Since $t\ge 0$ on this interval, $\sqrt{t^2}=t$:

v(t)=\sqrt{t^2+t^4} =\sqrt{t^2(1+t^2)} =t\sqrt{1+t^2}.

Now integrate. The substitution $u=1+t^2$ gives $du=2t\,dt$, so $t\,dt=\frac12\,du$. When $t=0$, $u=1$; when $t=1$, $u=2$.

\begin{aligned} L&=\int_0^1 t\sqrt{1+t^2}\,dt\\ &=\frac12\int_1^2 u^{1/2}\,du\\ &=\frac12\left[\frac{2}{3}u^{3/2}\right]_1^2\\ &=\frac13\left(2^{3/2}-1\right) =\frac{2\sqrt{2}-1}{3}. \end{aligned}

Check: An antiderivative of the speed is $F(t)=\frac13(1+t^2)^{3/2}$. Differentiating gives $F'(t)=t\sqrt{1+t^2}$, exactly the speed. Its endpoint difference is $F(1)-F(0)=\frac{2\sqrt{2}-1}{3}$, as calculated.

Worked Example: A Curve That Momentarily Stops

Find the speed and arc length of $\mathbf{r}(t)=\left\langle \frac{t^2-1}{2},\frac{t^2+1}{2}\right\rangle$ for $-1\le t\le 2$.

Differentiate the coordinates and take the magnitude:

\mathbf{r}'(t)=\langle t,t\rangle, \qquad v(t)=\sqrt{t^2+t^2}=\sqrt{2}\,|t|.

The absolute value matters because $t$ is negative on part of the interval, but speed cannot be negative. Split the integral at $t=0$, where the formula for $|t|$ changes:

\begin{aligned} L&=\int_{-1}^{0}\sqrt{2}(-t)\,dt +\int_{0}^{2}\sqrt{2}t\,dt\\ &=\sqrt{2}\left[-\frac{t^2}{2}\right]_{-1}^{0} +\sqrt{2}\left[\frac{t^2}{2}\right]_{0}^{2}\\ &=\frac{\sqrt{2}}{2}+2\sqrt{2} =\frac{5\sqrt{2}}{2}. \end{aligned}

At $t=0$, the velocity vector is $\langle 0,0\rangle$, so the point momentarily stops. The zero speed at one parameter value does not invalidate the arc-length calculation.

Check: On $[-1,0]$, an antiderivative of the speed is $-\frac{\sqrt{2}}{2}t^2$, whose derivative is $-\sqrt{2}t=\sqrt{2}|t|$. On $[0,2]$, an antiderivative is $\frac{\sqrt{2}}{2}t^2$, whose derivative is $\sqrt{2}t=\sqrt{2}|t|$. The two pieces give the stated total length.

A Reliable Calculation Process

For a parametrized curve, keep the vector differentiation separate from the scalar integration. This helps prevent sign errors and makes it clear which quantity is being accumulated.

Differentiate the position.
Find $\mathbf{r}'(t)$ by differentiating every coordinate with respect to the same parameter.

Compute the speed.
Take the magnitude of the velocity vector. Simplify the expression, taking care that a square root such as $\sqrt{t^2}$ equals $|t|$, not always $t$.

Integrate over the interval.
Evaluate the definite integral of the nonnegative speed. Split the interval if the speed formula changes, for example because of an absolute value or a piecewise definition.

Check the result.
Differentiate the antiderivative to recover the speed, check the endpoint evaluation, and confirm that the length has appropriate units and is nonnegative.

Common Pitfall: Using Displacement Instead of Arc Length

Arc length measures the distance traveled along the entire path. The magnitude of the difference between the endpoint positions measures only the straight-line displacement. These agree for a straight segment traversed without reversal, but they need not agree for a curved path.

For example, consider one full revolution of the unit circle, $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$, for $0\le t\le 2\pi$. A mistaken calculation uses only the endpoint positions, which are both $\langle 1,0\rangle$:

\|\mathbf{r}(2\pi)-\mathbf{r}(0)\| =\|\langle 1,0\rangle-\langle 1,0\rangle\| =0.

This is the displacement, not the length of the path. The correct calculation differentiates and integrates the speed:

\mathbf{r}'(t)=\langle-\sin t,\cos t\rangle, \qquad v(t)=\sqrt{\sin^2t+\cos^2t}=1, \qquad L=\int_0^{2\pi}1\,dt=2\pi.

The curve returns to its starting point, so its net displacement is zero, but it travels a full circumference. The distinction between displacement and distance traveled is essential whenever a curve bends, loops, or reverses direction.

Why the Formula Is Useful

The speed-and-integral method converts a geometric question into a calculation involving derivatives and a definite integral. It applies to curves in two or three dimensions and works whether speed is constant or varies along the path. It also makes units transparent: if the coordinates are lengths and $t$ is time, speed has units of length per time, and integrating with respect to time produces length.

Some speed functions do not have elementary antiderivatives. The arc-length formula still identifies the correct quantity; evaluating such integrals may require methods beyond the examples here. In every case, the integrand must be the magnitude of velocity, not a signed coordinate derivative or the magnitude of endpoint displacement.

Check Your Understanding

For each question, distinguish the vector velocity from the scalar speed and use the parameter interval as given.

For $\mathbf{r}(t)=\langle 2t,3t\rangle$ on $0\le t\le 4$, find the velocity, speed, and arc length.
Why is speed computed using the magnitude of velocity rather than adding the coordinate derivatives?
For $\mathbf{r}(t)=\langle\cos t,\sin t\rangle$ on $0\le t\le \pi$, find the speed and arc length.
If $\mathbf{r}'(t)=\langle t,-t\rangle$ on $[-2,0]$, write the speed using an absolute value and state whether the curve has positive speed throughout the interval.
Explain why a curve can have zero displacement between its endpoints but positive arc length.