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Polar Calculus and Applications · Tutorial 995 of 998

Arc Length of an Archimedean Spiral

Learn to derive and evaluate the arc length of \(r=a+b\theta\), including the role of the radial rate \(b\) and the sign of the substitution.

Advanced 9 min read

What You'll Learn

  • Derive the speed of an Archimedean spiral from the polar arc-length formula.
  • Set up the arc-length integral on a specified finite angular interval.
  • Use \(u=a+b\theta\) to reduce the integral to a standard form.
  • Evaluate the standard antiderivative and check it by differentiation.
  • Handle positive, negative, and zero values of the radial rate.
  • Recognize and correct a common error involving the derivative of \(r\).

Why the Archimedean Spiral Has a Manageable Arc-Length Integral

The Archimedean spiral \(r=a+b\theta\) changes its radius at a constant rate as the angle changes. Its shape can wind around the pole, but its arc length over a finite angular interval can be found by reducing the polar arc-length integral to an ordinary integral of the form \(\int\sqrt{u^2+c^2}\,du\).

The key is to keep track of the rate \(b\). It affects both the speed along the curve and the change of variable used to evaluate the integral. In particular, replacing \(dr/d\theta\) by \(1\) is only valid when \(b=1\).

From the Polar Formula to the Spiral Integral

For a polar curve \(r=r(\theta)\) on an interval \([\theta_0,\theta_1]\), where \(\theta_0\leq\theta_1\), the arc length is

$$ L=\int_{\theta_0}^{\theta_1}\sqrt{r(\theta)^2+\left(\frac{dr}{d\theta}\right)^2}\,d\theta. $$

This formula follows by writing the curve in Cartesian coordinates as \(x=r\cos\theta\) and \(y=r\sin\theta\). Differentiating gives \(x'=r'\cos\theta-r\sin\theta\) and \(y'=r'\sin\theta+r\cos\theta\). Squaring and adding cancels the cross terms:

$$ (x')^2+(y')^2 =(r')^2(\cos^2\theta+\sin^2\theta) +r^2(\sin^2\theta+\cos^2\theta) =(r')^2+r^2. $$

The speed is therefore \(\sqrt{(x')^2+(y')^2}=\sqrt{r^2+(r')^2}\), which gives the polar arc-length formula.

For the Archimedean spiral \(r=a+b\theta\), the derivative is \(dr/d\theta=b\). Thus, when \(b\neq0\), its length over a finite interval is

$$ L=\int_{\theta_0}^{\theta_1} \sqrt{(a+b\theta)^2+b^2}\,d\theta. $$
Definition: On a finite angular interval \([\theta_0,\theta_1]\), the arc length of \(r=a+b\theta\) is the integral of its polar speed. For \(b\neq0\), that speed is \(\sqrt{(a+b\theta)^2+b^2}\). The formula applies for real constants \(a,b\); \(b\) may be positive or negative.

The integral is finite on every finite interval because its integrand is continuous there. If \(b=0\), then \(r=a\) is constant and the speed is \(|a|\), so the length is \(|a|(\theta_1-\theta_0)\). The rest of the evaluation focuses on the nonconstant case \(b\neq0\).

Substitution and the Sign of \(b\)

Set \(u=a+b\theta\). Then \(du=b\,d\theta\), and the endpoint values become \(u_0=a+b\theta_0\) and \(u_1=a+b\theta_1\). The integral becomes

$$ L=\frac{1}{b}\int_{u_0}^{u_1}\sqrt{u^2+b^2}\,du. $$

This expression also works when \(b<0\). In that case \(u_1<u_0\), so the integral with respect to \(u\) is negative, as is \(1/b\); their product is positive. Equivalently, the positive length can be written using an antiderivative \(F\) as \(\bigl(F(u_1)-F(u_0)\bigr)/b\).

Important: Do not replace \(1/b\) by \(1/|b|\) while leaving the endpoints in their original order. The signed factor \(1/b\) and the transformed endpoint order work together. Alternatively, reorder the endpoints and use a positive factor consistently.

The Standard Antiderivative

For a fixed nonzero real number \(c\), an antiderivative of \(\sqrt{u^2+c^2}\) is

$$ F_c(u)=\frac12\left( u\sqrt{u^2+c^2} +c^2\ln\left(u+\sqrt{u^2+c^2}\right) \right). $$

The logarithm is defined for every real \(u\) because \(\sqrt{u^2+c^2}>|u|\) when \(c\neq0\), and hence \(u+\sqrt{u^2+c^2}>0\). Here is a derivation, so the logarithmic term and its coefficient are accounted for rather than memorized without justification.

1
Integrate by parts.
Write \(J=\int\sqrt{u^2+c^2}\,du\). Using \(u\) as one factor and \(\sqrt{u^2+c^2}\) as the other gives \(J=u\sqrt{u^2+c^2}-\int \frac{u^2}{\sqrt{u^2+c^2}}\,du\).
2
Separate the remaining integral.
Since \(u^2=(u^2+c^2)-c^2\), the integral on the right is \(J-c^2K\), where \(K=\int \frac{1}{\sqrt{u^2+c^2}}\,du\). Therefore \(2J=u\sqrt{u^2+c^2}+c^2K\).
3
Evaluate \(K\).
Differentiating the logarithm gives \(\frac{d}{du}\ln(u+\sqrt{u^2+c^2})=\frac{1}{\sqrt{u^2+c^2}}\). Thus \(K=\ln(u+\sqrt{u^2+c^2})\), up to a constant.
4
Combine the terms.
Substituting this expression for \(K\) into \(2J\) and dividing by \(2\) gives the stated formula for \(F_c\).

For the spiral, take \(c=b\); the antiderivative depends on \(b^2\), so its sign does not change when \(b\) is negative. The arc-length formula is therefore

$$ L= \frac{F_b(a+b\theta_1)-F_b(a+b\theta_0)}{b}, \qquad F_b(u)=\frac12\left( u\sqrt{u^2+b^2} +b^2\ln\left(u+\sqrt{u^2+b^2}\right) \right), \quad b\neq0. $$

As a check on the derivation, differentiating \(F_b\) gives \(\sqrt{u^2+b^2}\). Consequently, differentiating \(F_b(a+b\theta)/b\) with respect to \(\theta\) gives \(\sqrt{(a+b\theta)^2+b^2}\), the original speed.

Worked Examples

Worked Example: A Spiral Starting at the Pole

Find the length of \(r=\theta\) for \(0\leq\theta\leq1\). Here \(a=0\), \(b=1\), and \(dr/d\theta=1\). The speed and its integral are

$$ \sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} =\sqrt{\theta^2+1}, \qquad L=\int_0^1\sqrt{\theta^2+1}\,d\theta. $$

Use \(F_1(u)=\frac12\left(u\sqrt{u^2+1}+\ln(u+\sqrt{u^2+1})\right)\). At the upper endpoint,

$$ F_1(1)=\frac12\left(\sqrt2+\ln(1+\sqrt2)\right). $$

At the lower endpoint, \(F_1(0)=\frac12\ln(1)=0\). Hence

$$ L=F_1(1)-F_1(0) =\frac12\left(\sqrt2+\ln(1+\sqrt2)\right) \approx 1.1478. $$

Check: differentiating \(F_1(\theta)\) gives \(\sqrt{\theta^2+1}\), the integrand. Evaluating the antiderivative at \(1\) and \(0\) therefore verifies the stated length.

Worked Example: A Spiral with Radial Rate Two

Find the length of \(r=1+2\theta\) for \(0\leq\theta\leq1\). Its derivative is \(2\), so the speed is not \(\sqrt{r^2+1}\); it is

$$ \sqrt{(1+2\theta)^2+4}. $$

Set \(u=1+2\theta\), so \(du=2\,d\theta\). As \(\theta\) goes from \(0\) to \(1\), \(u\) goes from \(1\) to \(3\). Thus

$$ L=\frac12\int_1^3\sqrt{u^2+4}\,du =\frac{F_2(3)-F_2(1)}{2}. $$

Using \(F_2(u)=\frac12\left(u\sqrt{u^2+4}+4\ln(u+\sqrt{u^2+4})\right)\), the endpoint values are

$$ F_2(3)=\frac12\left(3\sqrt{13}+4\ln(3+\sqrt{13})\right), \qquad F_2(1)=\frac12\left(\sqrt5+4\ln(1+\sqrt5)\right). $$

Subtract and divide by \(2\):

$$ L=\frac14\left( 3\sqrt{13}-\sqrt5 +4\ln\left(\frac{3+\sqrt{13}}{1+\sqrt5}\right) \right) \approx 2.8588. $$

Check: if \(G(\theta)=F_2(1+2\theta)/2\), then the chain rule gives \(G'(\theta)=F_2'(1+2\theta)=\sqrt{(1+2\theta)^2+4}\). Also, \(G(1)-G(0)\) is exactly the endpoint expression above.

Worked Example: A Negative Radial Rate

Find the length of \(r=3-\theta\) for \(0\leq\theta\leq2\). Here \(b=-1\), so \(dr/d\theta=-1\). The speed is

$$ \sqrt{(3-\theta)^2+(-1)^2} =\sqrt{(3-\theta)^2+1}. $$

Set \(u=3-\theta\), for which \(du=-d\theta\). The endpoints transform from \(\theta=0,2\) to \(u=3,1\). Keeping both the negative differential and the transformed endpoints gives

$$ L=-\int_3^1\sqrt{u^2+1}\,du =\int_1^3\sqrt{u^2+1}\,du =F_1(3)-F_1(1). $$

Substitute the endpoints:

$$ L=\frac12\left( 3\sqrt{10}-\sqrt2+ \ln\left(\frac{3+\sqrt{10}}{1+\sqrt2}\right) \right) \approx 4.5050. $$

Check: the function \(-F_1(3-\theta)\) has derivative \(\sqrt{(3-\theta)^2+1}\). Its value at \(2\) minus its value at \(0\) is \(F_1(3)-F_1(1)\), so the endpoint order and sign give a positive length.

A Common Error: Losing the Radial Rate

A frequent mistake is to use \(1\) in the speed formula in place of \((dr/d\theta)^2\), even when the radius does not have derivative \(1\). For example, for \(r=1+2\theta\), an incorrect setup is

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

This calculation uses the wrong derivative: \(dr/d\theta=2\), not \(1\). The correct setup is

$$ L=\int_0^1\sqrt{(1+2\theta)^2+2^2}\,d\theta =\int_0^1\sqrt{(1+2\theta)^2+4}\,d\theta. $$

The correction affects both parts of the substitution: the constant inside the square root is \(b^2\), and \(d\theta=du/b\). Keeping these linked prevents a wrong speed or a missing factor in the final integral.

Why the Method Is Useful

The substitution works because the spiral's radius is a linear function of the angle. It turns the changing radius \(a+b\theta\) into the integration variable \(u\), while the constant radial rate contributes the fixed term \(b^2\) under the square root. The geometry is thus reduced to one standard integral, regardless of the starting radius or whether the spiral moves outward or inward as \(\theta\) increases.

The formula also offers a reliable way to check a calculation: differentiate the proposed antiderivative with the chain rule and compare the result with the polar speed. When \(b<0\), check the transformed endpoints as well as the sign of \(1/b\). These checks address the two common sources of errors: omitting the radial rate and reversing the substitution inconsistently.

Check Your Understanding

Use the polar speed and the substitution carefully in each question.

  1. For \(r=4+3\theta\), what is \(dr/d\theta\), and what is the speed as a function of \(\theta\)?
  2. For \(r=2-\theta\) on \(0\leq\theta\leq1\), what are the transformed \(u\)-endpoints under \(u=2-\theta\)?
  3. Why does the expression \(\bigl(F_b(u_1)-F_b(u_0)\bigr)/b\) remain positive when \(b<0\) and \(\theta_1>\theta_0\)?
  4. What is the arc length of the constant-radius curve \(r=5\) on \(0\leq\theta\leq\pi\)?
  5. Differentiate \(F_b(a+b\theta)/b\) to verify that it gives the spiral's speed when \(b\neq0\).