Parametric Arc Lengths Parametric Arc Lengths

Introduction

Finding the arc lengths of parametric curves is similar to the process for explicit functions.

For parametric functions defined by

x(t)
and
y(t)
, the arc length
L
from
t=a
to
t=b
is given by:
\begin{align*} L &= \sum^n_{i=0}{\sqrt{\Delta x^2 + \Delta y^2}} \\ &= \int_{a}^{b} \sqrt{dx^2 + dy^2} \\ &= \int_{a}^{b} \sqrt{dx^2 + dy^2} \cdot \frac{dt}{dt} \\ &= \int_{a}^{b} \sqrt{\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}} \, dt \\ &= \int_{a}^{b} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \\ \end{align*}
Same as our first arc length formula, this derives from euclidean distance.

By summing infinitely many infinitely small lines approximating the function, we can find the length of the function.

Example 1: Straight Line Verification

Find the length of the parametric curve from

t=0
to
t=1
:
\begin{align*} x(t) &= 3t \\ y(t) &= 4t \end{align*}
Solution:
\frac{dx}{dt} = 3,\quad \frac{dy}{dt} = 4
\begin{align*} L &= \int_{0}^{1} \sqrt{3^2 + 4^2}\,dt \\ &= \int_{0}^{1} 5\,dt \\ &= 5(1) - 5(0) \\ &= 5 \end{align*}
Note that this parametric equation is a straight line, so the result is verifiable using the distance between the two points.
\sqrt{(3(1) - 3(0))^2 + (4(1) - 4(0))^2} = \sqrt{3^2 + 4^2} = 5

Example 2: Circular Arc

Calculate the length of a quarter circle defined by:

\begin{align*} x(t) &= 2\cos(t) \\ y(t) &= 2\sin(t) \\ &0 \leq t \leq \frac{\pi}{2} \end{align*}
Solution:
\frac{dx}{dt} = -2\sin(t),\quad \frac{dy}{dt} = 2\cos(t)
\begin{align*} L &= \int_{0}^{\pi/2} \sqrt{(-2\sin(t))^2 + (2\cos(t))^2}\,dt \\ &= \int_{0}^{\pi/2} \sqrt{4\sin^2(t) + 4\cos^2(t)}\,dt \\ &= \int_{0}^{\pi/2} 2\sqrt{(\sin^2(t) + \cos^2(t))}\,dt \\ &= \int_{0}^{\pi/2} 2\,dt \\ &= 2\left(\frac{\pi}{2}\right) \\ &= \pi \end{align*}
Since these functions represent
1/4
of a circle of radius 2, we can verify the arc length using the formula for circumference:
\frac{1}{4} \times 2\pi r = \frac{1}{4} \times 4\pi = \pi