Section 9.4 : Arc Length with Parametric Equations
In the previous two sections we’ve looked at a couple of Calculus I topics in terms of parametric equations. We now need to look at a couple of Calculus II topics in terms of parametric equations.
In this section we will look at the arc length of the parametric curve given by,
\[x = f\left( t \right)\hspace{0.5in}y = g\left( t \right)\hspace{0.5in}\alpha \le t \le \beta \]We will also be assuming that the curve is traced out exactly once as \(t\) increases from \(\alpha\) to \(\beta\). We will also need to assume that the curve is traced out from left to right as \(t\) increases. This is equivalent to saying,
\[\frac{{dx}}{{dt}} \ge 0\hspace{0.5in}{\mbox{for }}\alpha \le t \le \beta \]So, let’s start out the derivation by recalling the arc length formula as we first derived it in the arc length section of the Applications of Integrals chapter.
\[L = \int{{ds}}\]where,
\[\begin{align*}ds & = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx\,\hspace{0.25in}{\mbox{if }}y = f\left( x \right),\,\,a \le x \le b\\ ds & = \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy\,\hspace{0.25in}{\mbox{if }}x = h\left( y \right),\,\,c \le y \le d\end{align*}\]We will use the first \(ds\) above because we have a nice formula for the derivative in terms of the parametric equations (see the Tangents with Parametric Equations section). To use this we’ll also need to know that,
\[dx = f'\left( t \right)\,dt = \frac{{dx}}{{dt}}\,dt\]The arc length formula then becomes,
\[L = \int_{{\,\alpha }}^{{\,\beta }}{{\sqrt {1 + {{\left( {\frac{{\,\,\frac{{dy}}{{dt}}\,\,}}{{\frac{{dx}}{{dt}}}}} \right)}^2}} \,\,\,\,\,\frac{{dx}}{{dt}}dt\,}} = \int_{{\,\alpha }}^{{\,\beta }}{{\sqrt {1 + \frac{{{{\left( {\frac{{dy}}{{dt}}} \right)}^2}}}{{{{\left( {\frac{{dx}}{{dt}}} \right)}^2}}}} \,\,\,\,\frac{{dx}}{{dt}}dt\,}}\]This is a particularly unpleasant formula. However, if we factor out the denominator from the square root we arrive at,
\[L = \int_{{\,\alpha }}^{{\,\beta }}{{\frac{1}{{\left| {\frac{{dx}}{{dt}}} \right|}}\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \,\,\,\,\frac{{dx}}{{dt}}dt\,}}\]Now, making use of our assumption that the curve is being traced out from left to right we can drop the absolute value bars on the derivative which will allow us to cancel the two derivatives that are outside the square root and this gives,
Arc Length for Parametric Equations
Notice that we could have used the second formula for \(ds\) above if we had assumed instead that
\[\frac{{dy}}{{dt}} \ge 0\hspace{0.5in}{\mbox{for }}\alpha \le t \le \beta \]If we had gone this route in the derivation we would have gotten the same formula.
Let’s take a look at an example.
We know that this is a circle of radius 3 centered at the origin from our prior discussion about graphing parametric curves. We also know from this discussion that it will be traced out exactly once in this range.
So, we can use the formula we derived above. We’ll first need the following,
\[\frac{{dx}}{{dt}} = 3\cos \left( t \right)\hspace{0.5in}\hspace{0.25in}\frac{{dy}}{{dt}} = - 3\sin \left( t \right)\]The length is then,
\[\begin{align*}L & = \int_{{\,0}}^{{\,2\pi }}{{\sqrt {9{{\sin }^2}\left( t \right) + 9{{\cos }^2}\left( t \right)} \,\,dt\,}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{3\sqrt {{{\sin }^2}\left( t \right) + {{\cos }^2}\left( t \right)} \,\,dt\,}}\\ & = 3\int_{{\,0}}^{{\,2\pi }}{{\,dt\,}}\\ & = 6\pi \end{align*}\]Since this is a circle we could have just used the fact that the length of the circle is just the circumference of the circle. This is a nice way, in this case, to verify our result.
Let’s take a look at one possible consequence if a curve is traced out more than once and we try to find the length of the curve without taking this into account.
Notice that this is the identical circle that we had in the previous example and so the length is still 6\(p\). However, for the range given we know it will trace out the curve three times instead once as required for the formula. Despite that restriction let’s use the formula anyway and see what happens.
In this case the derivatives are,
\[\frac{{dx}}{{dt}} = 9\cos \left( {3t} \right)\hspace{0.5in}\hspace{0.25in}\frac{{dy}}{{dt}} = - 9\sin \left( {3t} \right)\]and the length formula gives,
\[\begin{align*}L & = \int_{{\,0}}^{{\,2\pi }}{{\sqrt {81{{\sin }^2}\left( {3t} \right) + 81{{\cos }^2}\left( {3t} \right)} \,\,dt\,}}\\ & = \int_{{\,0}}^{{\,2\pi }}{{9\,\,dt\,}}\\ & = 18\pi \end{align*}\]The answer we got form the arc length formula in this example was 3 times the actual length. Recalling that we also determined that this circle would trace out three times in the range given, the answer should make some sense.
If we had wanted to determine the length of the circle for this set of parametric equations we would need to determine a range of \(t\) for which this circle is traced out exactly once. This is, \(0 \le t \le \frac{{2\pi }}{3}\). Using this range of \(t\) we get the following for the length.
\[\begin{align*}L & = \int_{{\,0}}^{{\,\frac{{2\pi }}{3}}}{{\sqrt {81{{\sin }^2}\left( {3t} \right) + 81{{\cos }^2}\left( {3t} \right)} \,\,dt\,}}\\ & = \int_{{\,0}}^{{\,\frac{{2\pi }}{3}}}{{9\,\,dt\,}}\\ & = 6\pi \end{align*}\]which is the correct answer.
Be careful to not make the assumption that this is always what will happen if the curve is traced out more than once. Just because the curve traces out \(n\) times does not mean that the arc length formula will give us \(n\) times the actual length of the curve!
Before moving on to the next section let’s notice that we can put the arc length formula derived in this section into the same form that we had when we first looked at arc length. The only difference is that we will add in a definition for \(ds\) when we have parametric equations.
The arc length formula can be summarized as,
\[L = \int{{ds}}\]where,
\[\begin{align*}ds & = \sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \,dx\,\hspace{0.5in}{\mbox{if }}y = f\left( x \right),\,\,a \le x \le b\\ ds & = \sqrt {1 + {{\left( {\frac{{dx}}{{dy}}} \right)}^2}} \,dy\,\hspace{0.5in}{\mbox{if }}x = h\left( y \right),\,\,c \le y \le d\\ ds & = \sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} \,\,dt\hspace{0.25in}{\mbox{if }}x = f\left( t \right),y = g\left( t \right),\,\,\alpha \le t \le \beta \end{align*}\]