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Parametric Motion and Arc Length · Tutorial 751 of 999

Exact Arc Length for Lines and Circles

Find exact lengths for line and circular parametrizations by connecting the arc length integral to constant speed and angular change.

Advanced 9 min read

What You'll Learn

  • Apply the parametric arc length integral to affine line parametrizations.
  • Find the constant speed of a line and use it to calculate distance traveled.
  • Derive the arc length formula for a circle parametrized by its angle.
  • Account for clockwise motion, decreasing angles, and multiple turns.
  • Check exact lengths using endpoint distances or circumference fractions.
  • Identify why endpoint distance alone can give the wrong circular arc length.

Why Lines and Circles Have Exact Arc Lengths

A parametric curve describes a point whose coordinates depend on a parameter. Its arc length on a parameter interval is found by integrating its speed. For many curves, this integral is difficult or does not have an elementary antiderivative. Lines and circles are especially accessible: their parametrizations can have constant speed, so the integral reduces to speed multiplied by elapsed parameter time.

For a circle, the angle gives an additional geometric interpretation. In radians, an angular change of one radian corresponds to an arc length equal to the radius. Thus a circular arc’s length is its radius multiplied by the magnitude of its angular change, provided the angle moves steadily in one direction.

Key idea: A line with constant velocity has length equal to its speed times the parameter interval. A circle with constant angular speed has length equal to its radius times the magnitude of the total angular change.

The Arc Length Formula

Let a curve be parametrized by \(x=x(t)\) and \(y=y(t)\) on an interval \([a,b]\), where the coordinate functions are differentiable. Its speed is the magnitude of its velocity vector:

$$ v(t)=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}. $$

The arc length on the interval is the integral of that speed:

$$ L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt. $$

This formula measures distance traveled along the parametrized curve, including any portion traced more than once. In the examples below, the speed is constant, making the integral especially direct.

Line Segments: Constant Speed

An affine parametrization of a line segment has the form \(x(t)=x_0+pt\), \(y(t)=y_0+qt\) on \([a,b]\), where \(p\) and \(q\) are constants. Its velocity components are \(p\) and \(q\), so its speed is the constant \(\sqrt{p^2+q^2}\). Therefore its arc length is that speed multiplied by the interval’s width.

$$ L=\int_a^b\sqrt{p^2+q^2}\,dt =\sqrt{p^2+q^2}(b-a),\qquad a\le b. $$

Equivalently, because the coordinates change by \(p(b-a)\) and \(q(b-a)\), the distance between the endpoints is \(\sqrt{(p(b-a))^2+(q(b-a))^2}\), which gives the same result. For a straight segment traveled once, arc length and endpoint distance agree.

Worked Example: Length of a Line Segment

Find the length of \(x(t)=1+3t\), \(y(t)=-2+4t\) for \(0\le t\le2\).

Differentiate each coordinate:

$$ \frac{dx}{dt}=3,\qquad \frac{dy}{dt}=4. $$

The speed is constant:

$$ v=\sqrt{3^2+4^2}=\sqrt{9+16}=5. $$

The parameter interval has width \(2-0=2\), so the length is

$$ L=\int_0^2 5\,dt=5(2)=10. $$

Check using the endpoints. At \(t=0\), the point is \((1,-2)\). At \(t=2\), it is \((1+6,-2+8)=(7,6)\). Their distance is \(\sqrt{(7-1)^2+(6-(-2))^2}=\sqrt{36+64}=10\), agreeing with the integral.

Circles: Radius Times Angular Change

A circle of radius \(r>0\) centered at \((h,k)\) can be parametrized using an angle \(\theta\) in radians:

$$ x=h+r\cos\theta,\qquad y=k+r\sin\theta. $$

When the angle itself is the parameter, differentiate with respect to \(\theta\): \(dx/d\theta=-r\sin\theta\) and \(dy/d\theta=r\cos\theta\). The speed with respect to angle is therefore constant:

$$ \sqrt{\left(-r\sin\theta\right)^2+\left(r\cos\theta\right)^2} =r\sqrt{\sin^2\theta+\cos^2\theta} =r. $$

The identity \(\sin^2\theta+\cos^2\theta=1\) explains why the center and the particular angle do not affect the speed: one radian of angular change corresponds to \(r\) units of arc length. If the angle runs from \(\theta_1\) to \(\theta_2\) without reversing direction, integrating this constant speed gives \(r|\theta_2-\theta_1|\). The absolute value ensures the length is nonnegative even when the angle decreases.

Theorem: For a circle of radius \(r\) parametrized by \(x=h+r\cos\theta\), \(y=k+r\sin\theta\), the length traced while the angle changes steadily from \(\theta_1\) to \(\theta_2\) is $$ L=r|\theta_2-\theta_1|. $$ The angles must be measured in radians. If the angle is a constant-rate function \(\theta(t)=\omega t+\phi\) on \(a\le t\le b\), then \(L=r|\omega|(b-a)\).

The constant-rate version follows directly by differentiating with respect to \(t\). In this parametrization, \(dx/dt=-r\omega\sin(\omega t+\phi)\) and \(dy/dt=r\omega\cos(\omega t+\phi)\). Hence the speed is \(r|\omega|\), and its integral over an interval of width \(b-a\) is \(r|\omega|(b-a)\). Since \(\theta(b)-\theta(a)=\omega(b-a)\), this is also \(r|\theta(b)-\theta(a)|\).

Worked Example: A Semicircle from Its Angular Interval

Find the length traced by \(x=2+5\cos\theta\), \(y=-1+5\sin\theta\) as \(\theta\) increases from \(\pi/6\) to \(7\pi/6\).

The radius is \(r=5\). The angular change is

$$ \Delta\theta=\frac{7\pi}{6}-\frac{\pi}{6}=\frac{6\pi}{6}=\pi. $$

Therefore the length is

$$ L=5|\pi|=5\pi. $$

Check by differentiating with respect to \(\theta\): \(dx/d\theta=-5\sin\theta\) and \(dy/d\theta=5\cos\theta\). The speed is \(\sqrt{25\sin^2\theta+25\cos^2\theta}=5\). Integrating over an interval of width \(\pi\) gives \(\int_{\pi/6}^{7\pi/6}5\,d\theta=5\pi\), as required. The angular interval is half a full turn, so the result is also half the circumference \(2\pi(5)\).

Constant Angular Speed in a Time Parameter

The parameter need not be the angle itself. A circle may be described by \(x=h+r\cos(\omega t+\phi)\), \(y=k+r\sin(\omega t+\phi)\), where \(\omega\) and \(\phi\) are constants. The center is \((h,k)\), \(r\) is the radius, and \(\omega\) is the signed angular rate. Its sign gives the direction of angular motion, while its magnitude gives the rate in radians per unit of \(t\).

Worked Example: A Circle Traced with Constant Angular Rate

Find the distance traveled by \(x=1+2\cos(3t-\pi/4)\), \(y=4+2\sin(3t-\pi/4)\) for \(0\le t\le\pi/3\).

The radius is \(2\), and the angular rate is \(\omega=3\). The interval width is \(\pi/3\). Thus the constant speed is

$$ v=r|\omega|=2|3|=6. $$

The length is

$$ L=6\left(\frac{\pi}{3}-0\right)=2\pi. $$

Check using the angular change. At \(t=0\), the angle is \(-\pi/4\). At \(t=\pi/3\), it is \(3(\pi/3)-\pi/4=\pi-\pi/4=3\pi/4\). The change is \(3\pi/4-(-\pi/4)=\pi\), so \(r|\Delta\theta|=2\pi\), in agreement with the speed integral. The angle changes by half a turn, giving half the circumference of a radius-\(2\) circle.

Direction, Repeated Tracing, and Conditions

Length is not signed. A circle traced clockwise has the same length as the same angular change traced counterclockwise. That is why the formula uses an absolute value. For a constant angular rate, the sign of \(\omega\) changes direction but not speed.

The angular-change formula also counts distance traveled, not just the size of the shortest arc between the endpoints. If the angle changes by more than \(2\pi\), the curve completes one or more full turns, and those turns contribute to the length. For example, one complete turn has angular change \(2\pi\), so a circle of radius \(r\) has circumference \(2\pi r\).

The constant-rate condition matters. If an angle varies and reverses direction during the interval, its endpoint change alone may undercount the distance traveled. In general, the speed for a circular parametrization with a differentiable angle \(\theta(t)\) is \(r|\theta'(t)|\), so the length is \(\int_a^b r|\theta'(t)|\,dt\). When \(\theta'(t)\) has a constant sign, the angle does not reverse, and this integral equals \(r|\theta(b)-\theta(a)|\). The constant-rate formula is a particularly simple case.

Worked Example: Decreasing Angle and Clockwise Travel

A circle of radius \(3\) is parametrized by \(x=2+3\cos(2-4t)\), \(y=-1+3\sin(2-4t)\) for \(0\le t\le\pi/4\). Find the length traveled.

The angular rate is \(\omega=-4\), so the speed is constant:

$$ v=3|-4|=12. $$

The interval width is \(\pi/4\). Therefore

$$ L=12\left(\frac{\pi}{4}\right)=3\pi. $$

Check from the angle endpoints. The initial angle is \(2\), and the final angle is \(2-4(\pi/4)=2-\pi\). Thus the signed change is \((2-\pi)-2=-\pi\), whose magnitude is \(\pi\). The angular formula gives \(3|\!-\!\pi|=3\pi\), matching the speed calculation. The negative change indicates clockwise motion, not negative length.

A Common Pitfall: Confusing Arc Length with Endpoint Distance

For a line segment traveled once, the endpoint distance equals the arc length. For a circular arc, it usually does not: the straight chord between the endpoints cuts across the circle. For example, after half a turn, the endpoints are opposite ends of a diameter, but the traveled path is a semicircle.

A related error is to use a signed angular change directly as a length. In the preceding example, writing \(L=3(-\pi)=-3\pi\) is incorrect. It confuses clockwise direction with distance. The correction is \(L=3|-\pi|=3\pi\).

1
Identify the curve and interval.
For a line, locate the constant coordinate rates. For a circle, identify its radius and angular expression.
2
Find the speed or angular change.
For a line, compute \(\sqrt{(dx/dt)^2+(dy/dt)^2}\). For a circle with constant angular rate, compute \(r|\omega|\) or \(r|\Delta\theta|\).
3
Calculate and check.
Multiply constant speed by the interval width, or radius by the magnitude of angular change. Check with endpoints for a line or with the circumference fraction for a circle.

Why These Exact Formulas Matter

These cases show how the arc length integral connects calculus to familiar geometry. On a line, constant velocity makes distance a speed-times-time calculation. On a circle, the trigonometric identity makes angular speed translate directly into linear speed, with the radius as the conversion factor. Recognizing these structures avoids unnecessary integration and gives an exact answer, often involving \(\pi\).

Before using a shortcut, check its conditions: the line parametrization must have constant coordinate rates, and the circle formula \(r|\Delta\theta|\) applies directly when the angle travels steadily in one direction. Always use radians for angular change. Finally, distinguish the distance along the curve from the straight-line distance between two points unless the path is a straight segment.

Check Your Understanding

Use the constant-speed and angular-change ideas to answer each question. Give exact lengths.

  1. Find the length of \(x=2-5t\), \(y=1+12t\) for \(0\le t\le1\). Check by finding the distance between its endpoints.
  2. A circle of radius \(4\) is traced as its angle increases from \(\pi/3\) to \(4\pi/3\). What length is traveled?
  3. For a circle of radius \(2\) with angular rate \(\omega=-3\) over \(0\le t\le2\pi/3\), find the length and explain what the negative rate means.
  4. Why does the formula \(r|\theta_2-\theta_1|\) count a full turn when the angular change is \(2\pi\)?
  5. A circle’s angle decreases steadily by \(5\pi/4\) radians. What is the length of the arc traced if its radius is \(6\)?

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