Arc Length of a Curve

# 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)
\begin{align} L = \int_a^b \sqrt{1 + \left ( \frac{dy}{dx} \right )^2} \: dx \quad \mathbf{or} \quad L = \int_a^b \sqrt{1 + [f'(x)]^2} \: dx \end{align}

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)
\begin{align} L = \int_c^d \sqrt{1 + \left ( \frac{dx}{dy} \right )^2} \: dy \quad \mathbf{or} \quad L = \int_c^d \sqrt{1 + [g'(y)]^2} \: dy \end{align}

## 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)
\begin{align} L = \int_a^b \sqrt{1 + (0)^2} \: dx \\ L = \int_a^b \sqrt{1} \: dx \\ L = \int_a^b 1 \: dx \\ L = x \: \bigg | _a^b \\ L = b - a \end{align}

## 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)
\begin{align} L = \int_a^b \sqrt{1 + 1} \: dx \\ L = \int_a^b \sqrt{2} \: dx \\ L = \sqrt{2} \cdot x \: \bigg | _a^b \\ L = \sqrt{2} ( b - a ) \end{align}

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)
\begin{align} L = \sqrt {2}(b - a) \end{align}

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)
\begin{align} L = \int_0^3 \sqrt {1 + ( \frac{3}{5} x^{1/2})^2} \: dx \\ L = \int_0^3 \sqrt {1 + \frac{9}{25} x} \: dx \\ L = \int_{1}^{2.08} \sqrt { u } \cdot \frac{25}{9} \: du \\ L = \frac{25}{9} \int_{1}^{2.08} \sqrt { u } \: du \\ L = \frac{25}{9} \left ( \frac{2}{3} x^{3/2} \right ) \bigg | _{1}^{2.08} \\ L = \frac{25}{9} \cdot \frac{2}{3} x^{3/2} \bigg |_{1}^{2.08} \\ L = \frac{50}{27} x^{3/2} \bigg |_{1}^{2.08} \\ L \approx \frac{50}{27} (3 - 1) \\ L \approx \frac{100}{27} \end{align}

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.