Arc Length of a Space Curve Calculator

Published on: July 21, 2026
Final Answer: Free • Full Steps: Pro

This Arc Length Calculator finds the length of a space curve \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\). It differentiates each component, forms the speed \(\lVert \mathbf{r}'(t) \rVert\), and integrates it over the interval - step by step.

Step-by-step method

  1. Set up the components of r(t) and the interval.
  2. Differentiate each component to get r'(t).
  3. Form the speed |r'(t)| = sqrt(x'^2 + y'^2 + z'^2).
  4. Integrate the speed over the interval.
  5. Evaluate to get the arc length.

Formula:

\(L = \int_{a}^{b} \sqrt{\left(x'\right)^2 + \left(y'\right)^2 + \left(z'\right)^2}\,dt\)

Example 1:

\(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle,\; t \in [0,2 \pi]\)

Step 1 - Set up the components of r(t) and the interval.

In this problem: The curve is \(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle\) for \(t \in [0, 2 \pi]\).

\(\mathbf{r}(t) = \langle \cos{\left(t \right)},\sin{\left(t \right)},t \rangle,\; t \in [0,2 \pi]\)

Step 2 - Differentiate each component to get r'(t).

In this problem: Differentiating each component: \(\mathbf{r}'(t) = \langle - \sin{\left(t \right)},\cos{\left(t \right)},1 \rangle\).

\(\mathbf{r}'(t) = \langle - \sin{\left(t \right)},\cos{\left(t \right)},1 \rangle\)

Step 3 - Form the speed |r'(t)| = sqrt(x'^2 + y'^2 + z'^2).

In this problem: The speed is \(\lVert \mathbf{r}'(t) \rVert = \sqrt{2}\).

\(\lVert \mathbf{r}'(t) \rVert = \sqrt{2}\)

Step 4 - Integrate the speed over the interval.

In this problem: Integrate the speed over the interval.

\(L = \int_{0}^{2 \pi} \sqrt{2}\,dt\)

Step 5 - Evaluate to get the arc length.

In this problem: This evaluates to \(L = 2 \sqrt{2} \pi \approx 8.89\).

\(L = 2 \sqrt{2} \pi \approx 8.89\)

Final answer:

\(L = 2 \sqrt{2} \pi \approx 8.89\)

Example 2:

\(\mathbf{r}(t) = \langle t,t,t \rangle,\; t \in [0,1]\)

Step 1 - Set up the components of r(t) and the interval.

In this problem: The curve is \(\mathbf{r}(t) = \langle t,t,t \rangle\) for \(t \in [0, 1]\).

\(\mathbf{r}(t) = \langle t,t,t \rangle,\; t \in [0,1]\)

Step 2 - Differentiate each component to get r'(t).

In this problem: Differentiating each component: \(\mathbf{r}'(t) = \langle 1,1,1 \rangle\).

\(\mathbf{r}'(t) = \langle 1,1,1 \rangle\)

Step 3 - Form the speed |r'(t)| = sqrt(x'^2 + y'^2 + z'^2).

In this problem: The speed is \(\lVert \mathbf{r}'(t) \rVert = \sqrt{3}\).

\(\lVert \mathbf{r}'(t) \rVert = \sqrt{3}\)

Step 4 - Integrate the speed over the interval.

In this problem: Integrate the speed over the interval.

\(L = \int_{0}^{1} \sqrt{3}\,dt\)

Step 5 - Evaluate to get the arc length.

In this problem: This evaluates to \(L = \sqrt{3} \approx 1.73\).

\(L = \sqrt{3} \approx 1.73\)

Final answer:

\(L = \sqrt{3} \approx 1.73\)