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Arc Length and Curvature · Tutorial 181 of 999

Arc Length of Polar Curves

Derive and apply the polar arc-length formula, with careful attention to the curve and angular interval being measured.

Intermediate 9 min read

What You'll Learn

  • Derive the polar arc-length formula from the Cartesian parametrization of a polar curve
  • Identify the differentiability and interval conditions needed to apply the formula
  • Calculate lengths of circular arcs using constant-radius and nonconstant polar descriptions
  • Set up the exact length integral for one petal of a rose curve
  • Evaluate the length of a spiral on a specified angular interval
  • Avoid omitting the contribution from angular change or counting the wrong interval

Why Polar Arc Length Has Two Contributions

A polar curve describes points by their distance \(r\) from the origin and their direction \(\theta\). As \(\theta\) changes, the point can move because its distance from the origin changes, because its direction changes, or because both change at once. Arc length must account for both effects.

For example, a circle centered at the origin has constant radius, but it still has positive length: its point moves as its direction changes. A spiral can change both radius and direction. The polar arc-length formula combines these contributions into one speed and then accumulates that speed over the angular interval.

From a Polar Equation to a Parametrized Curve

Let \(r=f(\theta)\) describe a polar curve for \(\alpha\leq\theta\leq\beta\). The corresponding Cartesian parametrization is

$$ \mathbf{r}(\theta) = \bigl(f(\theta)\cos\theta,\ f(\theta)\sin\theta\bigr). $$

Here the bold symbol denotes the position vector; the scalar polar radius is \(r=f(\theta)\). If \(f\) is continuously differentiable on the interval, the parametrized arc length is the integral of its speed. Differentiating the two coordinates gives

$$ \mathbf{r}'(\theta) = \bigl(f'(\theta)\cos\theta-f(\theta)\sin\theta,\ f'(\theta)\sin\theta+f(\theta)\cos\theta\bigr). $$

The speed is the magnitude of this derivative. Squaring its components and adding them produces terms involving \(f'(\theta)^2\), \(f(\theta)^2\), and cross terms. The cross terms cancel, since one is \(-2f'f\sin\theta\cos\theta\) and the other is \(2f'f\sin\theta\cos\theta\). The remaining trigonometric factors satisfy \(\sin^2\theta+\cos^2\theta=1\).

$$ \begin{aligned} \|\mathbf{r}'(\theta)\|^2 &= \bigl(f'\cos\theta-f\sin\theta\bigr)^2 +\bigl(f'\sin\theta+f\cos\theta\bigr)^2\\ &=f'(\theta)^2\bigl(\cos^2\theta+\sin^2\theta\bigr) +f(\theta)^2\bigl(\sin^2\theta+\cos^2\theta\bigr)\\ &=f'(\theta)^2+f(\theta)^2. \end{aligned} $$
Polar arc-length formula: If \(r=f(\theta)\) is continuously differentiable on \([\alpha,\beta]\), then the length traced as \(\theta\) runs from \(\alpha\) to \(\beta\), with \(\alpha\leq\beta\), is $$ L=\int_\alpha^\beta \sqrt{f(\theta)^2+\bigl(f'(\theta)\bigr)^2}\,d\theta. $$

The formula adds the squared radial rate \(f'(\theta)^2\) to the squared contribution from turning, \(f(\theta)^2\), before taking the square root. It applies whether \(f\) is positive, zero, or negative. A negative polar radius represents a point in the opposite direction; it does not invalidate the Cartesian parametrization or the formula.

Why the Formula Works

The cancellation above is a direct proof from the Cartesian parametrization. There is also a useful geometric interpretation. The unit vectors in the radial and angular directions are perpendicular. A small change in the curve has a radial component \(f'(\theta)\,d\theta\) and a perpendicular turning component \(f(\theta)\,d\theta\). By the Pythagorean theorem, its length to first order is determined by the square root of the sum of their squares.

This interpretation explains two features of the formula. When the radius is constant, the radial contribution is zero, but the turning contribution remains. When the angle is fixed and the radius changes, the radial contribution measures that change. In general, neither contribution can be discarded.

Worked Example: An Arc of a Circle Centered at the Origin

Worked Example: An Arc of a Circle Centered at the Origin

Find the length of the arc \(r=3\) for \(\pi/6\leq\theta\leq\pi/2\). Suppose the radius is measured in centimeters.

Here \(f(\theta)=3\), so \(f'(\theta)=0\). The speed in the polar formula is therefore

$$ \sqrt{f(\theta)^2+\bigl(f'(\theta)\bigr)^2} = \sqrt{3^2+0^2} =3. $$

Integrating over the specified angular interval gives

$$ L=\int_{\pi/6}^{\pi/2}3\,d\theta =3\left(\frac{\pi}{2}-\frac{\pi}{6}\right) =3\left(\frac{\pi}{3}\right) =\pi\text{ cm}. $$

Check: The angular change is \(\pi/3\) radians, so a circle of radius \(3\) cm has arc length \(3(\pi/3)=\pi\) cm for this interval. This agrees with the integral.

Angles in the formula are measured in radians. The differential \(d\theta\) and the derivative \(f'(\theta)\) are understood with respect to the radian-valued angle; replacing radians with degrees without adjusting the calculation gives an incorrect length.

Worked Example: A Circle with a Nonconstant Polar Radius

Worked Example: A Circle with a Nonconstant Polar Radius

Find the length traced by \(r=4\cos\theta\) as \(\theta\) runs from \(0\) to \(\pi\).

Although the radius depends on \(\theta\), the curve is a circle. First, \(f'(\theta)=-4\sin\theta\), so the speed is

$$ \begin{aligned} \sqrt{f(\theta)^2+\bigl(f'(\theta)\bigr)^2} &=\sqrt{16\cos^2\theta+16\sin^2\theta}\\ &=\sqrt{16}=4. \end{aligned} $$

Thus the length over the full specified interval is

$$ L=\int_0^\pi 4\,d\theta=4(\pi-0)=4\pi. $$

Check: In Cartesian coordinates, \(x=4\cos^2\theta\) and \(y=4\sin\theta\cos\theta\). Therefore \(x-2=2\cos(2\theta)\) and \(y=2\sin(2\theta)\), so

$$ (x-2)^2+y^2 =4\cos^2(2\theta)+4\sin^2(2\theta) =4. $$

This is a circle of radius \(2\), and \(2\theta\) runs from \(0\) to \(2\pi\), tracing it once. Its circumference is \(2\pi(2)=4\pi\), as the integral gives. This example also shows why differentiating \(r\) matters: a changing radius can combine with angular motion to produce a constant speed.

Worked Example: One Petal of a Rose Curve

Worked Example: One Petal of a Rose Curve

Find the length of the petal traced by \(r=2\cos(3\theta)\) on \(-\pi/6\leq\theta\leq\pi/6\).

On this interval, \(3\theta\) runs from \(-\pi/2\) to \(\pi/2\), so \(r\) is nonnegative. The radius is zero at both endpoints and reaches \(2\) at \(\theta=0\); this interval traces the petal pointing along the positive horizontal axis. The derivative is \(f'(\theta)=-6\sin(3\theta)\). Substitution into the formula gives

$$ \begin{aligned} \sqrt{f(\theta)^2+\bigl(f'(\theta)\bigr)^2} &=\sqrt{4\cos^2(3\theta)+36\sin^2(3\theta)}\\ &=2\sqrt{1+8\sin^2(3\theta)}. \end{aligned} $$

Consequently, the exact length is

$$ \begin{aligned} L &=\int_{-\pi/6}^{\pi/6}2\sqrt{1+8\sin^2(3\theta)}\,d\theta\\ &=\frac{2}{3}\int_{-\pi/2}^{\pi/2}\sqrt{1+8\sin^2 u}\,du\\ &=\frac{4}{3}\int_0^{\pi/2}\sqrt{1+8\sin^2 u}\,du. \end{aligned} $$

In the second line, \(u=3\theta\), so \(d\theta=du/3\). In the last line, the integrand is even, so the integral over the symmetric interval is twice the integral over its positive half. The displayed definite integral is an exact specification of the petal’s length; it does not simplify to an elementary antiderivative.

Check: Replacing \(u\) by \(3\theta\) in the second line restores the original limits and gives the first line. The factor \(4/3\) in the final line comes from both the substitution factor \(2/3\) and the symmetry factor \(2\). Thus the interval, substitution, and symmetry have all been accounted for.

Worked Example: A Spiral on a Specified Interval

Worked Example: A Spiral on a Specified Interval

Find the length of \(r=\theta\) for \(0\leq\theta\leq1\).

Here \(f(\theta)=\theta\) and \(f'(\theta)=1\). The speed is \(\sqrt{1+\theta^2}\), so

$$ L=\int_0^1\sqrt{1+\theta^2}\,d\theta. $$

An antiderivative is \(\frac12\bigl(\theta\sqrt{1+\theta^2}+\ln(\theta+\sqrt{1+\theta^2})\bigr)\). Evaluating at the endpoints yields

$$ \begin{aligned} L &=\frac12\left(\sqrt{2}+\ln(1+\sqrt{2})\right) -\frac12\left(0+\ln 1\right)\\ &=\frac12\left(\sqrt{2}+\ln(1+\sqrt{2})\right). \end{aligned} $$

Check: If \(F(\theta)=\frac12\bigl(\theta\sqrt{1+\theta^2}+\ln(\theta+\sqrt{1+\theta^2})\bigr)\), differentiation gives

$$ \begin{aligned} F'(\theta) &=\frac12\left( \sqrt{1+\theta^2} +\frac{\theta^2}{\sqrt{1+\theta^2}} +\frac{1}{\sqrt{1+\theta^2}}\right)\\ &=\frac12\left(\sqrt{1+\theta^2} +\frac{\theta^2+1}{\sqrt{1+\theta^2}}\right) =\sqrt{1+\theta^2}. \end{aligned} $$

This is exactly the integrand, confirming the evaluation.

Common Pitfall: Leaving Out the Turning Contribution

A frequent error is to use \(\int_\alpha^\beta |f'(\theta)|\,d\theta\) as the polar arc length. That expression measures only accumulated change in radius, not the distance traveled as the curve turns.

Correction: The speed is \(\sqrt{f(\theta)^2+(f'(\theta))^2}\), not \(|f'(\theta)|\). For \(r=3\), the incorrect expression gives \(\int_{\pi/6}^{\pi/2}0\,d\theta=0\), even though the point travels along a circular arc. The correct formula gives \(\int_{\pi/6}^{\pi/2}3\,d\theta=\pi\).

Another important check is whether the chosen interval traces the desired part of the curve. A polar equation can retrace points on different angular intervals, and a negative radius can represent points opposite the direction indicated by \(\theta\). The formula measures the path traced by the parametrization over the stated interval, including any retracing. Identify the interval first, then apply the integral to those bounds.

Summary

For a continuously differentiable polar curve \(r=f(\theta)\) on \([\alpha,\beta]\), convert to Cartesian coordinates or use the resulting polar speed directly:

$$ L=\int_\alpha^\beta\sqrt{f(\theta)^2+\bigl(f'(\theta)\bigr)^2}\,d\theta. $$

The term \(f'(\theta)^2\) accounts for radial change, while \(f(\theta)^2\) accounts for angular motion. The specified interval determines which portion of the parametrized curve is measured. Verify the function, its derivative, the bounds, and whether the path is traced once or more than once before interpreting the result.

Check Your Understanding

Use the polar arc-length formula and pay attention to what each interval traces.

  1. For \(r=5\), what is the length traced on \(0\leq\theta\leq\pi/4\)?
  2. For \(r=2\sin\theta\), calculate the speed and identify the circle described by the curve.
  3. For \(r=\theta^2\) on \(0\leq\theta\leq1\), write the exact arc-length integral.
  4. Why does \(\int_\alpha^\beta |f'(\theta)|\,d\theta\) fail to give the length of a circle of constant positive radius?
  5. For \(r=2\cos(3\theta)\), why does the interval \([-\pi/6,\pi/6]\) describe a single petal?

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