# Arc Length of a Curve

Suppose that we have a function $y = f(x)$. To calculate the arc length on the interval $[a, b]$ where $a ≤ b$, (the length travelling across the function from $x = a$ to $x = b$), we can apply the following formula:

(1)Both formulas above are identical with the only difference being the notation used for the derivative of $y$ with respect to $x$.

Additionally, if we have a curve $x = g(y)$, calculating the arc length of the curve from $y = c$ to $y = d$ where $c ≤ d$ is as follows:

(2)## Example 1

**Calculate the length of the function $y = 1$ on the interval $[a, b]$.**

We know that $y = 1$ is a straight line, so the length of the line on the interval $[a, b]$ where $a ≤ b$ should be $b - a$. We will verify this with the first formula. Note that $\frac{dy}{dx} = 0$.

(3)## Example 2

**Calculate the length of the function $y = x$ on the interval $[a, b]$.**

Applying the first formula once again acknowledging that $\frac{dy}{dx} = 1$ we obtain:

(4)Note that $(b - a)$ represents the length of the horizontal line. Nevertheless, geometrically, our answer should make sense. We know that $f(x) = x$ is a diagonal line with slope $1$. A triangle formed with its hypotenuse on the line can easily be calculated as shown in the following diagram:

We know that the interior angle will be $45^{\circ}$ since the slope is $1$. Hence, we have an angle and the length of one side of the triangle is $(b - a)$, so using elementary trigonometry, we get that $\cos (45) = \frac{b - a}{L}$, or rather $L = \frac{b - a}{\cos (45) }$. We also know that $\cos (45) = \frac{1}{\sqrt{2}}$, hence:

(5)Which is the exact same value for $L$ as we calculated using integration.

## Example 3

**Calculate the arc length of the function $y = \frac{2}{5} x ^{3 / 2}$ on the interval $[0, 3]$.**

Let's first acknowledge that $\frac{dy}{dx} = \frac{2}{5} \cdot \frac{3}{2} x^{1/2} = \frac{3}{5} x^{1/2}$. Now we can apply our first formula and simplify. Note that we make the substitution that $u = 1 + \frac{9}{25} x$, and thus $du = \frac{9}{25} \: dx$. With our substitution, we must also change the limits of integration. When $x = 0$ we have that $u = 1$, and when $x = 3$ we have that $u = \frac{52}{25} = 2.08$.

(6)Hence the length of the curve from $x = 0$ to $x = 3$ is approximately $\frac{100}{27}$. The graph below illustrates the length we have calculated.