Arc Length of a Space Curve Calculator

Published on: June 28, 2026

This Arc Length of a Space Curve Calculator helps you find the length of a vector-valued curve in three-dimensional space. First differentiate the vector function, then find the magnitude of the derivative and use it in the arc length formula. If limits are given, the calculator sets up and evaluates the definite integral for the total length. It is a simple way to check answers, understand the method clearly, and practise vector calculus step by step.

Step-by-step method

  1. Identify the vector-valued function r(t) and the interval if it is given.
  2. Differentiate the full vector to get r′(t), then solve each component.
  3. Find the magnitude ||r′(t)||.
  4. Set up the arc length formula L = ∫ ||r′(t)|| dt.
  5. If an interval is given, evaluate the definite integral.

Formulas:

Derivative of the curve

r′( t ) = ⟨ x′( t ), y′( t ), z′( t ) ⟩

Speed formula

2D: ||r′( t )|| = (x′(t))2 + (y′(t))2
3D: ||r′( t )|| = (x′(t))2 + (y′(t))2 + (z′(t))2

Arc length formula

L = ba||r′( t )|| dt

Example 1: r(t)=<t,cos(t),sin(t)>; [0,2*pi]

Step 1A - Identify the vector-valued function.

In this problem: Read the given vector function.

r( t ) = ⟨ t, cos(t), sin(t) ⟩

Step 1B - Identify the interval.

In this problem: Read the interval for the arc length.

Interval = [ 0, 2π ]

Step 2A - Differentiate the full vector function.

In this problem: Show the derivative of each component before simplifying.

r′( t ) =
d
dt
(t),
d
dt
(cos(t)),
d
dt
(sin(t)) ⟩

Step 2B - Differentiate component 1.

In this problem: Differentiate with respect to t.

d
dt
(t) = 1

Step 2C - Differentiate component 2.

In this problem: Differentiate with respect to t.

d
dt
(cos(t)) = −sin(t)

Step 2D - Differentiate component 3.

In this problem: Differentiate with respect to t.

d
dt
(sin(t)) = cos(t)

Step 2E - Combine into the derivative vector.

In this problem: Put the differentiated components into ⟨ … ⟩.

r′( t ) = ⟨ 1, −sin(t), cos(t) ⟩

Step 3A - Use the magnitude formula.

In this problem: Square each velocity component, then add and take the square root.

||r′( t )|| = 12 + −sin(t)2 + cos(t)2

Step 3B - Simplify the magnitude.

In this problem: Simplify the square root.

||r′( t )|| = 2 ≈ 1.41

Step 4A - Set up the arc length integral.

In this problem: Substitute the interval into the formula.

L = 02 dt

Step 4B - Evaluate the integral.

In this problem: Find the arc length on the given interval.

L = 2π2 ≈ 8.89

Final answer: Arc length computed

Example 2: r(t)=<3*cos(t),3*sin(t),4*t>; [0,1]

Step 1A - Identify the vector-valued function.

In this problem: Read the given vector function.

r( t ) = ⟨ 3·cos(t), 3·sin(t), 4t ⟩

Step 1B - Identify the interval.

In this problem: Read the interval for the arc length.

Interval = [ 0, 1 ]

Step 2A - Differentiate the full vector function.

In this problem: Show the derivative of each component before simplifying.

r′( t ) =
d
dt
(3·cos(t)),
d
dt
(3·sin(t)),
d
dt
(4t) ⟩

Step 2B - Differentiate component 1.

In this problem: Differentiate with respect to t.

d
dt
(3·cos(t)) = −3·sin(t)

Step 2C - Differentiate component 2.

In this problem: Differentiate with respect to t.

d
dt
(3·sin(t)) = 3·cos(t)

Step 2D - Differentiate component 3.

In this problem: Differentiate with respect to t.

d
dt
(4t) = 4

Step 2E - Combine into the derivative vector.

In this problem: Put the differentiated components into ⟨ … ⟩.

r′( t ) = ⟨ −3·sin(t), 3·cos(t), 4 ⟩

Step 3A - Use the magnitude formula.

In this problem: Square each velocity component, then add and take the square root.

||r′( t )|| = −3·sin(t)2 + 3·cos(t)2 + 42

Step 3B - Simplify the magnitude.

In this problem: Simplify the square root.

||r′( t )|| = 5

Step 4A - Set up the arc length integral.

In this problem: Substitute the interval into the formula.

L = 105 dt

Step 4B - Evaluate the integral.

In this problem: Find the arc length on the given interval.

L = 5 ≈ 5.00

Final answer: Arc length computed