Arc Length of a Parametric Curve

Besides finding areas, we often need to calculate the arc length of parametric curves. Arc length measures the actual distance along a curve—if a particle travels from point A to point B along a curve, the arc length tells us exactly how far that particle traveled. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

For a plane curve defined by [latex]x = x(t)[/latex], [latex]y = y(t)[/latex] where [latex]a \le t \le b[/latex], we’ll approximate the curve using line segments.

Start by partitioning [latex][a,b][/latex] into [latex]n[/latex] equal subintervals: [latex]t_0 = a < t_1 < t_2 < \cdots < t_n = b[/latex], where each subinterval has width [latex]\Delta t = \frac{b-a}{n}[/latex].

Using the distance formula, we can calculate the length of each line segment:

The approximate arc length [latex]s_n[/latex] is the sum of all these segments:

If [latex]x(t)[/latex] and [latex]y(t)[/latex] are differentiable, the Mean Value Theorem tells us that in each subinterval [latex][t_{k-1}, t_k][/latex], there exist points [latex]\hat{t}_k[/latex] and [latex]\tilde{t}_k[/latex] where:

Substituting these expressions into our sum and factoring out [latex]\Delta t[/latex]:

This sum is a Riemann sum approximating the arc length over the partition of [latex][a,b][/latex]. As [latex]n \to \infty[/latex] and assuming the derivatives are continuous, we get:

As the partition gets finer, [latex]\hat{t}_k[/latex] and [latex]\tilde{t}_k[/latex] both lie in the same shrinking interval of width [latex]\Delta t[/latex], so they converge to the same value.

We can summarize this method in the following theorem.

Find the arc length of the semicircle defined by the equations

[latex]x\left(t\right)=3\cos{t},y\left(t\right)=3\sin{t},0\le t\le \pi[/latex].

 

Show Solution

The values [latex]t=0[/latex] to [latex]t=\pi[/latex] trace out the red curve in Figure 9. To determine its length, use the theorem:

[latex]\begin{array}{cc}\hfill s& ={\displaystyle\int }_{{t}_{1}}^{{t}_{2}}\sqrt{{\left(\frac{dx}{dt}\right)}^{2}+{\left(\frac{dy}{dt}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{{\left(-3\sin{t}\right)}^{2}+{\left(3\cos{t}\right)}^{2}}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{9{\sin}^{2}t+9{\cos}^{2}t}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }\sqrt{9\left({\sin}^{2}t+{\cos}^{2}t\right)}dt\hfill \\ & ={\displaystyle\int }_{0}^{\pi }3dt={3t|}_{0}^{\pi }=3\pi .\hfill \end{array}[/latex]

 

Note that the formula for the arc length of a semicircle is [latex]\pi r[/latex] and the radius of this circle is 3. This is a great example of using calculus to derive a known formula of a geometric quantity.