Deriving the Polar Arc-Length Formula

Why Polar Arc Length Has Two Contributions

A polar curve changes position in two ways as its angle varies: its distance from the pole can change, and its direction can change. Both changes contribute to the distance traveled along the curve. The polar arc-length formula combines them in a single expression.

The derivation begins with ordinary Cartesian distance. A small change in a point’s coordinates produces a small displacement whose length is determined by the Pythagorean theorem. Writing the Cartesian coordinates in terms of the polar radius and angle will show exactly how radial and angular changes contribute.

Definition: A polar curve described by $r=f(\theta)$ on an interval $[\alpha,\beta]$ is the parametrized curve with Cartesian coordinates $x(\theta)=r(\theta)\cos\theta$ and $y(\theta)=r(\theta)\sin\theta$, where $r(\theta)=f(\theta)$. Its arc length is the distance traveled as $\theta$ runs through the specified interval, counting any retracing of the curve.

Deriving the Differential

Suppose $r$ is differentiable as a function of $\theta$. The Cartesian coordinate functions are products of $r(\theta)$ with trigonometric functions. Differentiating each coordinate gives

\frac{dx}{d\theta} = \frac{dr}{d\theta}\cos\theta-r\sin\theta, \qquad \frac{dy}{d\theta} = \frac{dr}{d\theta}\sin\theta+r\cos\theta.

For a small parameter change $d\theta$, the corresponding Cartesian changes are $dx$ and $dy$. The Euclidean distance differential satisfies $ds^2=dx^2+dy^2$. Substituting the coordinate differentials yields

\begin{aligned} ds^2 &= \left(\frac{dr}{d\theta}\cos\theta-r\sin\theta\right)^2d\theta^2 + \left(\frac{dr}{d\theta}\sin\theta+r\cos\theta\right)^2d\theta^2\\ &= \left[ \left(\frac{dr}{d\theta}\right)^2\cos^2\theta -2r\frac{dr}{d\theta}\sin\theta\cos\theta +r^2\sin^2\theta\right.\\ &\qquad\left.+ \left(\frac{dr}{d\theta}\right)^2\sin^2\theta +2r\frac{dr}{d\theta}\sin\theta\cos\theta +r^2\cos^2\theta \right]d\theta^2\\ &= \left[ \left(\frac{dr}{d\theta}\right)^2(\cos^2\theta+\sin^2\theta) +r^2(\sin^2\theta+\cos^2\theta) \right]d\theta^2\\ &= \left[r^2+\left(\frac{dr}{d\theta}\right)^2\right]d\theta^2. \end{aligned}

The two mixed terms cancel, and each pair of squared trigonometric functions sums to $1$. Taking the nonnegative square root gives the arc-length differential. When $\theta$ increases, $d\theta$ is positive; for an interval traversed in the decreasing direction, the differential uses $|d\theta|$.

Theorem: If $r=f(\theta)$ is continuously differentiable on $[\alpha,\beta]$, where $\alpha\leq\beta$, then the length traced as $\theta$ increases from $\alpha$ to $\beta$ is $$ L=\int_\alpha^\beta \sqrt{r(\theta)^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. $$ Equivalently, the differential is $ds=\sqrt{r^2+(dr/d\theta)^2}\,d\theta$ for increasing $\theta$. For a decreasing parameter, use the absolute value of $d\theta$, or reverse the limits to write the length as a nonnegative integral.

Why the Integral Gives Length

The Cartesian parametrization is $ (x(\theta),y(\theta))$. For a continuously differentiable parametrized curve, its speed is the magnitude of its velocity vector, $\sqrt{(dx/d\theta)^2+(dy/d\theta)^2}$. The arc length over an increasing parameter interval is the integral of that speed. This result reflects how length is built from short displacements: on a sufficiently small parameter interval, the curve’s displacement is approximated by its tangent vector times the parameter change, so its chord length is approximated by speed times that change. Summing these contributions and taking the limiting integral gives the length.

Here the derivation above calculates that speed directly:

\sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2} = \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}.

Thus the polar formula is not a separate rule to memorize without explanation. It is the standard Cartesian speed formula expressed using polar coordinates.

Worked Examples

Worked Example: A Constant Radius on a Quarter Turn

Find the length traced by $r=3$ as $\theta$ runs from $0$ to $\pi/2$.

First differentiate the radius:

r=3, \qquad \frac{dr}{d\theta}=0.

Substitute these values and the chosen limits into the formula:

\begin{aligned} L &=\int_0^{\pi/2}\sqrt{3^2+0^2}\,d\theta\\ &=\int_0^{\pi/2}3\,d\theta\\ &=3\left[\theta\right]_0^{\pi/2}\\ &=3\left(\frac{\pi}{2}-0\right)\\ &=\frac{3\pi}{2}. \end{aligned}

Check using Cartesian velocity. The coordinates are $x=3\cos\theta$ and $y=3\sin\theta$, so $dx/d\theta=-3\sin\theta$ and $dy/d\theta=3\cos\theta$. The speed is

\sqrt{(-3\sin\theta)^2+(3\cos\theta)^2} = \sqrt{9(\sin^2\theta+\cos^2\theta)} =3.

Integrating this speed over the same interval also gives $3\pi/2$, as required.

Worked Example: An Exponentially Changing Radius

Find the length traced by $r=e^\theta$ for $0\leq\theta\leq\ln 2$.

Differentiate the radius:

\frac{dr}{d\theta}=e^\theta.

The radius and its derivative are equal, so the expression under the square root simplifies:

\begin{aligned} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} &=\sqrt{(e^\theta)^2+(e^\theta)^2}\\ &=\sqrt{2e^{2\theta}}\\ &=\sqrt{2}\,e^\theta. \end{aligned}

Since $e^\theta$ is positive, the last equality uses the positive square root. Now integrate over the stated interval:

\begin{aligned} L &=\int_0^{\ln 2}\sqrt{2}\,e^\theta\,d\theta\\ &=\sqrt{2}\left[e^\theta\right]_0^{\ln 2}\\ &=\sqrt{2}\left(e^{\ln 2}-e^0\right)\\ &=\sqrt{2}(2-1)\\ &=\sqrt{2}. \end{aligned}

For a check, differentiating the Cartesian coordinates gives the speed $\sqrt{2}e^\theta$, the same integrand as the polar calculation. Its integral over $0\leq\theta\leq\ln 2$ is $\sqrt{2}(2-1)=\sqrt{2}$. The length is positive, as it must be.

Worked Example: A Radius That Reaches the Pole

Find the length traced by $r=2\sin\theta$ for $0\leq\theta\leq\pi/2$. The radius is zero at the first endpoint, but that does not prevent use of the formula.

Differentiate and substitute:

\frac{dr}{d\theta}=2\cos\theta, \qquad \begin{aligned} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} &=\sqrt{4\sin^2\theta+4\cos^2\theta}\\ &=\sqrt{4}\\ &=2. \end{aligned}

The length is therefore

\begin{aligned} L &=\int_0^{\pi/2}2\,d\theta\\ &=2\left[\theta\right]_0^{\pi/2}\\ &=2\left(\frac{\pi}{2}-0\right)\\ &=\pi. \end{aligned}

Check in Cartesian coordinates: $x=2\sin\theta\cos\theta=\sin 2\theta$ and $y=2\sin^2\theta=1-\cos 2\theta$. Thus $dx/d\theta=2\cos 2\theta$ and $dy/d\theta=2\sin 2\theta$. The speed is $\sqrt{4\cos^2 2\theta+4\sin^2 2\theta}=2$. Its integral over the interval is $\pi$, matching the result.

Using the Interval Correctly

The integral measures the curve as it is traced over the chosen parameter interval, not merely the length of the geometric set of points. If the parametrization retraces some part of the curve, that part contributes each time it is traversed. Therefore, choose limits that describe the intended traversal. A length calculation can be numerically correct for its interval while still answering a different question if the interval traces too much or too little of the curve.

The regularity condition in the theorem is convenient: continuous differentiability ensures the derivatives and speed are well behaved throughout the interval. A curve may still have a meaningful length when a derivative fails at isolated points, but such cases require treating the intervals on either side appropriately. Do not silently apply the stated theorem across a point where its assumptions fail.

The radius $r$ may be zero or negative; neither sign changes the formula because the radial contribution appears as $r^2$. At a pole crossing, the formula remains applicable whenever the polar radius function is continuously differentiable on the interval. Also, angle measurements in calculus formulas are in radians: the derivative of $\sin\theta$ used above, for example, is $\cos\theta$ under the radian convention.

A Common Error

A frequent mistake is to count only the change in radius and write $ds=|dr|$. That is correct only in special situations where the curve moves purely radially. For example, with $r=3$ on $0\leq\theta\leq\pi/2$, this incorrect calculation gives

\int_0^{\pi/2}\left|\frac{dr}{d\theta}\right|\,d\theta = \int_0^{\pi/2}0\,d\theta =0.

But the point moves around a circle even though its radius does not change. The corrected calculation includes the angular contribution:

\int_0^{\pi/2} \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta = \int_0^{\pi/2}\sqrt{3^2+0^2}\,d\theta = \frac{3\pi}{2}.

The error comes from treating a change in distance from the pole as the whole displacement. The Cartesian derivation shows why that is insufficient: radial and angular changes jointly determine the speed.

Check Your Understanding

Use the differential and integral formulas derived above to answer these questions.

Starting from $x=r\cos\theta$ and $y=r\sin\theta$, what happens to the mixed terms when $dx^2+dy^2$ is expanded?
For $r=4$ on $0\leq\theta\leq\pi/3$, set up and evaluate the arc-length integral.
For $r=3\theta$ on $1\leq\theta\leq2$, write the arc-length integral, including the derivative of $r$. You do not need to evaluate it.
Why can a radius that equals zero at an endpoint still be used in the formula?
If an interval makes a parametrized curve trace the same segment twice, how does the arc-length integral count that segment?