Arc Length of a Parametric Curve

In addition to finding the area under a parametric curve, we sometimes need to find the arc length of a parametric curve. In the case of a line segment, arc length is the same as the distance between the endpoints. If a particle travels from point A to point B along a curve, then the distance that particle travels is the arc length. To develop a formula for arc length, we start with an approximation by line segments as shown in the following graph.

Figure 3. Approximation of a curve by line segments.

Given a plane curve defined by the functions [latex]x=x\left(t\right),y=y\left(t\right),a\le t\le b[/latex], we start by partitioning the interval [latex]\left[a,b\right][/latex] into n equal subintervals: [latex]{t}_{0}=a<{t}_{1}<{t}_{2}<\cdots <{t}_{n}=b[/latex]. The width of each subinterval is given by [latex]\Delta t=\frac{\left(b-a\right)}{n}[/latex]. We can calculate the length of each line segment:

Then add these up. We let s denote the exact arc length and [latex]{s}_{n}[/latex] denote the approximation by n line segments:

If we assume that [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] are differentiable functions of t, then the Mean Value Theorem (Introduction to the Applications of Derivatives) applies, so in each subinterval [latex]\left[{t}_{k - 1},{t}_{k}\right][/latex] there exist [latex]\hat{t}_{k}[/latex] and [latex]\tilde{t}_{k}[/latex] such that

Therefore [latex]s[/latex] becomes

This is a Riemann sum that approximates the arc length over a partition of the interval [latex]\left[a,b\right][/latex]. If we further assume that the derivatives are continuous and let the number of points in the partition increase without bound, the approximation approaches the exact arc length. This gives

When taking the limit, the values of [latex]\hat{t}_{k}[/latex] and [latex]\tilde{t}_{k}[/latex] are both contained within the same ever-shrinking interval of width [latex]\Delta t[/latex], so they must converge to the same value.

We can summarize this method in the following theorem.

At this point a side derivation leads to a previous formula for arc length. In particular, suppose the parameter can be eliminated, leading to a function [latex]y=F\left(x\right)[/latex]. Then [latex]y\left(t\right)=F\left(x\left(t\right)\right)[/latex] and the Chain Rule gives [latex]{y}^{\prime }\left(t\right)={F}^{\prime }\left(x\left(t\right)\right){x}^{\prime }\left(t\right)[/latex]. Substituting this into the theorem gives

Here we have assumed that [latex]{x}^{\prime }\left(t\right)>0[/latex], which is a reasonable assumption. The Chain Rule gives [latex]dx={x}^{\prime }\left(t\right)dt[/latex], and letting [latex]a=x\left({t}_{1}\right)[/latex] and [latex]b=x\left({t}_{2}\right)[/latex] we obtain the formula

which is the formula for arc length obtained in the Introduction to the Applications of Integration.

Example: Finding the Arc Length of a Parametric Curve

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.

Figure 4. The arc length of the semicircle is equal to its radius times [latex]\pi[/latex].

Watch the following video to see the worked solution to Example: Finding the Arc Length of a Parametric Curve.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).

We now return to the problem posed at the beginning of the section about a baseball leaving a pitcher’s hand. Ignoring the effect of air resistance (unless it is a curve ball!), the ball travels a parabolic path. Assuming the pitcher’s hand is at the origin and the ball travels left to right in the direction of the positive x-axis, the parametric equations for this curve can be written as

where t represents time. We first calculate the distance the ball travels as a function of time. This distance is represented by the arc length. We can modify the arc length formula slightly. First rewrite the functions [latex]x\left(t\right)[/latex] and [latex]y\left(t\right)[/latex] using v as an independent variable, so as to eliminate any confusion with the parameter t:

Then we write the arc length formula as follows:

The variable v acts as a dummy variable that disappears after integration, leaving the arc length as a function of time t. To integrate this expression we can use a formula from Appendix A,

We set [latex]a=140[/latex] and [latex]u=-32v+2[/latex]. This gives [latex]du=-32dv[/latex], so [latex]dv=-\frac{1}{32}du[/latex]. Therefore

and

This function represents the distance traveled by the ball as a function of time. To calculate the speed, take the derivative of this function with respect to t. While this may seem like a daunting task, it is possible to obtain the answer directly from the Fundamental Theorem of Calculus:

Therefore

One third of a second after the ball leaves the pitcher’s hand, the distance it travels is equal to

This value is just over three quarters of the way to home plate. The speed of the ball is

This speed translates to approximately 95 mph—a major-league fastball.

Surface Area Generated by a Parametric Curve

Recall the problem of finding the surface area of a volume of revolution. In Curve Length and Surface Area, we derived a formula for finding the surface area of a volume generated by a function [latex]y=f\left(x\right)[/latex] from [latex]x=a[/latex] to [latex]x=b[/latex], revolved around the x-axis:

We now consider a volume of revolution generated by revolving a parametrically defined curve [latex]x=x\left(t\right),y=y\left(t\right),a\le t\le b[/latex] around the x-axis as shown in the following figure.

Figure 5. A surface of revolution generated by a parametrically defined curve.

The analogous formula for a parametrically defined curve is

provided that [latex]y\left(t\right)[/latex] is not negative on [latex]\left[a,b\right][/latex].

Example: Finding Surface Area

Find the surface area of a sphere of radius [latex]r[/latex] centered at the origin.

Show Solution

We start with the curve defined by the equations

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

 

This generates an upper semicircle of radius r centered at the origin as shown in the following graph.

Figure 6. A semicircle generated by parametric equations.

When this curve is revolved around the [latex]x[/latex]-axis, it generates a sphere of radius [latex]r[/latex]. To calculate the surface area of the sphere, we use the above equation:

[latex]\begin{array}{cc}\hfill S&=2\pi {\displaystyle\int_{a}^{b}} y\left(t\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime}\left(t\right)\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}} r\sin{t}\sqrt{{\left(-r\sin{t}\right)}^{2}+{\left(r\cos{t}\right)}^{2}}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}{\sin}^{2}t+{r}^{2}{\cos}^{2}t}dt\hfill \\ &=2\pi {\displaystyle\int }_{0}^{\pi }r\sin{t}\sqrt{{r}^{2}\left({\sin}^{2}t+{\cos}^{2}t\right)}dt\hfill \\ &=2\pi {\displaystyle\int_{0}^{\pi}{r}^{2}\sin{t}dt}\hfill \\ &=2\pi{r}^{2}\left(-\cos{t}|_{0}^{\pi}\right)\hfill \\ & =2\pi {r}^{2}\left(-\cos\pi +\cos0\right)\hfill \\ & =4\pi {r}^{2}.\hfill \end{array}[/latex]

 

This is, in fact, the formula for the surface area of a sphere.

Watch the following video to see the worked solution to Example: Finding Surface Area.

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “7.2 Calculus of Parametric Curves” here (opens in new window).