Arc Length

Let's see how to find the length of a curve, using the same process of slicing, approximating, and integrating that we've used to find areas and volumes.

1. Slice

Arc length

Here's a curve that connects two points, AA and BB. We can approximate this curve with a straight line, but it's not a great approximation. If, however, we split the interval in half and use two line segments, the approximation gets better. Of course, if we use more line segments, the approximation is even better, and if we let the number of segments go to infinity, we'll find the exact length of the curve.

To start, find the length of a single segment.

Single segment

This length dd can be found using the distance formula (as the hypotenuse of a right triangle with sides x2−x1x_2-x_1 and y2−y1y_2-y_1): d=(x2−x1)2+(y2−y1)2=(Δx)2+(Δy)2=(Δx)2(1+(Δy)2(Δx)2)=1+(Δy)2(Δx)2 Δx\begin{aligned} d &= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\ &= \sqrt{(\Delta x)^2 + (\Delta y)^2}\ &= \sqrt{(\Delta x)^2\left(1 + \dfrac{(\Delta y)^2}{(\Delta x)^2}\right)}\ &= \sqrt{1 + \dfrac{(\Delta y)^2}{(\Delta x)^2}}\ \Delta x \end{aligned}

2. Approximate

The total length of all these segments is L=∑i=1n1+(Δy)2(Δx)2 ΔxL = \sum_{i=1}^n \sqrt{1 + \dfrac{(\Delta y)^2}{(\Delta x)^2}}\ \Delta x

3. Integrate

As n→∞n \to \infty, the deltas turn into differentials: L=∫ab1+(dydx)2 dx=∫ab1+(f′(x))2 dx\ans{\begin{aligned} L &= \int_a^b \sqrt{1 + \left(\dfrac{dy}{dx}\right)^2}\ dx\ &= \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\ dx \end{aligned}}

Find the length of the curve f(x)=x3/2f(x)=x^{3/2} on the interval [0,4][0,4].

Solution

All we have to do is find f′(x)f'(x) and plug it into the arc length formula: f′(x)=32x1/2L=∫ab1+(f′(x))2 dx=∫041+(32x1/2)2 dx=∫041+94x dx\begin{aligned} f'(x) &= \dfrac{3}{2}x^{1/2}\ L &= \int_a^b \sqrt{1 + \left(f'(x)\right)^2}\ dx\ &= \int_0^4 \sqrt{1 + \left(\dfrac{3}{2}x^{1/2}\right)^2}\ dx\ &= \int_0^4 \sqrt{1 + \dfrac{9}{4}x}\ dx\ \end{aligned}

We can evaluate this integral using u-substitution, where u=1+94x⟶du=94 dx⟶49 du=dxu=1+\dfrac{9}{4}x \longrightarrow du = \dfrac{9}{4}\ dx \longrightarrow \dfrac{4}{9} \ du = dx ∫041+94x dx=∫11049u du=827u3/2∣110≈9.1\begin{aligned} \int_0^4 \sqrt{1+\dfrac{9}{4}x}\ dx &= \int_1^{10} \dfrac{4}{9} \sqrt{u}\ du\ &= \dfrac{8}{27}u^{3/2} \bigg|_1^{10}\ &\approx \ans{9.1} \end{aligned}

  1. Find the length of the curve f(x)=x3f(x)=x^3 on the interval [−2,4].[-2,4].

    2. L=∫−241+(3x2)2 dx=∫−241+9x4 dx≈73.3 (using a computer to do numerical integration)\begin{aligned} L &= \int_{-2}^4 \sqrt{1+(3x^2)^2}\ dx\ &= \int_{-2}^4 \sqrt{1+9x^4}\ dx\ &\approx \ans{73.3} \textrm{ (using a computer to do numerical integration)} \end{aligned}