Arc Length of a Parametric Curve

# Arc Length of a Parametric Curve

Recall that if we had an equation of a continuous curve on the interval $[a, b]$, then we could calculate the length of the arc using the following formula:

(1)
\begin{align} L = \int_a^b \sqrt{1 + \left ( \frac{dy}{dx} \right)^2 } \: dx \end{align}

Now suppose that we have a curve defined by the parametric equations $x = f(t)$ and $y = g(t)$. We can come up with a formula to calculate the length of the arc on the interval $[\alpha, \beta]$ where $\alpha ≤ t ≤ \beta$, $f'$ and $g'$ are continuous on this interval, and $a = f(\alpha)$, $b = f(\beta)$ OR $a = f(\beta)$, $b = f(\alpha)$. First, we will make the substitution that $dx = \frac{dx}{dt} dt$ and let's assume that $\frac{dx}{dt} > 0$:

(2)
\begin{align} L = \int_a^b \sqrt{1 + \left ( \frac{dy}{dx} \right)^2 } \: dx = \int_{\alpha}^{\beta} \sqrt{1 + \left ( \frac{dy}{dx} \right)^2 } \cdot \frac{dx}{dt} \: dt \\ L = \int_{\alpha}^{\beta} \sqrt{1 + \left ( \frac{dy}{dx} \right)^2 } \cdot \sqrt{ \left( \frac{dx}{dt} \right) ^2 } \: dt \\ L = \int_{\alpha}^{\beta} \sqrt{ \left ( 1 + \left ( \frac{dy}{dx} \right)^2 \right) \left( \frac{dx}{dt} \right) ^2} \: dt \\ L = \int_{\alpha}^{\beta} \sqrt{ \left ( 1 + \left ( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \right)^2 \right) \left( \frac{dx}{dt} \right) ^2} \: dt \\ L = \int_{\alpha}^{\beta} \sqrt{ \left ( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right )^2 } \: dt \end{align}

## Example 1

Determine the length of the arc given by the parametric equations $x = \cos t$ and $y = \sin t$ on the interval $0 ≤ t ≤ 2\pi$.

We note that these parametric equations yield a circle with radius $1$ since $x^2 + y^2 = \cos ^2 t + \sin ^2 t = 1$. Hence we should expect that the circumference of this circle to be $2\pi r = 2\pi$. Now let's verify it using the formula we just obtained.

We acknowledge that $\frac{dx}{dt} = -\sin t$, and $\frac{dy}{dt} = \cos t$. Hence it follows that:

(3)
\begin{align} L = \int_0^{2 \pi} \sqrt{(-\sin t)^2 + (\cos t)^2} \: dt \\ L = \int_0^{2 \pi} \sqrt{\sin ^2 t + \cos^2 t} \: dt \\ L = \int_0^{2 \pi} \sqrt{1} \: dt \\ L = \int_0^{2 \pi} 1\: dt \\ L = t \bigg |_0^{2\pi} \\ L = 2\pi \end{align}

## Example 2

Find the length of the curve given by the parametric equations $x = 1 + 3t^2$ and $y = 4 + 2t^3$ for $0 ≤ t ≤ 1$.

We first note that this length is traced out only once on our interval $0 ≤ t ≤ 1$. We also note that $\frac{dx}{dt} = 6t$, and $\frac{dy}{dt} = 6t^2$. Now we just need to substitute this information into our formula as follows:

(4)
\begin{align} L = \int_0^1 \sqrt{(6t)^2 + (6t^2)^2} \: dt \\ L = \int_0^1 \sqrt{36t^2 + 36t^4} \: dt \\ L = \int_0^1 \sqrt{36t^2(1 + t^2)} \: dt \\ L = \int_0^1 6t \sqrt{1 + t^2} \: dt \end{align}

Now let's make a substitution. Let $u = 1 + t^2$, and thus $du = 2t \: dt$. We also need to change our limits of integration. We get that they become $1$ and $2$ using our substitution equation. Hence we get:

(5)
\begin{align} L = \int_1^2 3 \sqrt{u} \: du \\ L = 3 \int_1^2 u^{1/2} \: du \\ L = 3 \left ( \frac{2u^{3/2}}{3} \right ) \bigg |_1^2 \\ L = 3 ( \frac{2 \sqrt{8}}{3} - \frac{1}{3} ) \\ L = 2 \sqrt{8} - 1 \\ L = 4\sqrt{2} - 1 \approx 3.656 \end{align}

The following image shows the length that we have just calculated: