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