Arc Length For Cartesian Parametric Curves

From Parametric Motion to Length

A parametrization describes a point in the plane by giving its coordinates as functions of a parameter: \(x=x(t)\) and \(y=y(t)\). As \(t\) increases over an interval, the point moves along a curve. The length of the curve over that interval depends on how far the point travels, not merely on the distance between its starting and ending positions.

The earlier material on speed from parametric derivatives gives the speed as the magnitude of the velocity vector. For a curve in the plane, that magnitude is \(\sqrt{(x'(t))^2+(y'(t))^2}\). Integrating speed over the parameter interval gives the total distance traveled along the curve.

Arc-Length Formula: Suppose \(x(t)\) and \(y(t)\) have continuous first derivatives on \([a,b]\), where \(a\le b\). The length of the parametrized curve traced as \(t\) runs from \(a\) to \(b\) is

L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\,dt.

The integral counts every portion of the path as it is traversed, including any portion that is traced more than once.

The quantity under the integral is the speed, which is never negative. This matters when simplifying the square root: for example, \(\sqrt{t^2}=|t|\), not \(t\) for all real \(t\). The parameter interval is part of the problem; changing it can change which portions of the curve are included and how many times they are traversed.

Why the Formula Gives Length

Length can be defined by approximating a curve with straight line segments. Choose parameter values \(a=t_0<t_1<\cdots<t_n=b\), and join the successive points \((x(t_i),y(t_i))\) by chords. The sum of the chord lengths is an approximation to the length of the curve.

For a small parameter increment \(\Delta t=t_{i+1}-t_i>0\), differentiability gives the coordinate increments, to first order, as \(x'(t_i)\Delta t\) and \(y'(t_i)\Delta t\). The corresponding chord length is therefore approximately

\sqrt{\bigl(x'(t_i)\Delta t\bigr)^2+ \bigl(y'(t_i)\Delta t\bigr)^2} = \sqrt{(x'(t_i))^2+(y'(t_i))^2}\,\Delta t.

Adding these approximations over a partition gives a Riemann sum for the integral of the speed. Because the coordinate derivatives are continuous, their first-order approximations have errors that become uniformly small as the largest partition interval shrinks. The total chord-length sums thus approach the integral of the speed. This justifies the arc-length formula.

Key takeaway: Differentiate both coordinate functions, form the nonnegative speed \(\sqrt{(x')^2+(y')^2}\), simplify it, and integrate over the full parameter interval.

A Circle: Simplify Before Integrating

Worked Example: A Quarter-Circle

Find the length traced by \(x(t)=3\cos t\), \(y(t)=3\sin t\) for \(0\le t\le \frac{\pi}{2}\).

Differentiate each coordinate:

x'(t)=-3\sin t, \qquad y'(t)=3\cos t.

Substitute into the speed formula and simplify before integrating:

\begin{aligned} \sqrt{(x'(t))^2+(y'(t))^2} &=\sqrt{9\sin^2 t+9\cos^2 t}\\ &=\sqrt{9(\sin^2 t+\cos^2 t)}\\ &=3. \end{aligned}

Therefore, the length is

\begin{aligned} L &=\int_0^{\pi/2}3\,dt\\ &=3\left[t\right]_0^{\pi/2}\\ &=\frac{3\pi}{2}. \end{aligned}

As a check, differentiating the antiderivative \(3t\) gives \(3\), the speed in the integrand. The result is also the length of a quarter-circle of radius \(3\), namely one quarter of \(2\pi\cdot 3\), which is \(\frac{3\pi}{2}\).

A Cycloid Arch and the Parameter Interval

Worked Example: One Cycloid Arch

Find the length of the curve \(x(t)=t-\sin t\), \(y(t)=1-\cos t\) for \(0\le t\le 2\pi\).

The coordinate derivatives are

x'(t)=1-\cos t, \qquad y'(t)=\sin t.

Now simplify the speed:

\begin{aligned} \sqrt{(x'(t))^2+(y'(t))^2} &=\sqrt{(1-\cos t)^2+\sin^2 t}\\ &=\sqrt{1-2\cos t+\cos^2 t+\sin^2 t}\\ &=\sqrt{2-2\cos t}\\ &=\sqrt{4\sin^2(t/2)}\\ &=2|\sin(t/2)|. \end{aligned}

On \(0\le t\le2\pi\), we have \(0\le t/2\le\pi\), so \(\sin(t/2)\ge0\). Thus the speed on this interval is \(2\sin(t/2)\), and

\begin{aligned} L &=\int_0^{2\pi}2\sin(t/2)\,dt\\ &=\left[-4\cos(t/2)\right]_0^{2\pi}\\ &=-4\cos\pi+4\cos 0\\ &=4+4\\ &=8. \end{aligned}

To check, the derivative of \(-4\cos(t/2)\) is \(2\sin(t/2)\), as required. On the longer interval \([0,4\pi]\), the point traces two consecutive cycloid arches, so the length includes both arches. Indeed, \(x(t+2\pi)=x(t)+2\pi\) and \(y(t+2\pi)=y(t)\): the second arch is shifted horizontally by \(2\pi\), not a repeat of the same arch. Each arch has length \(8\), so the length on \([0,4\pi]\) is \(16\).

Absolute Values and a Path That Turns Around

Worked Example: A Parametrization with Zero Speed

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

Differentiate:

x'(t)=6t, \qquad y'(t)=8t.

The speed is

\begin{aligned} \sqrt{(6t)^2+(8t)^2} &=\sqrt{36t^2+64t^2}\\ &=\sqrt{100t^2}\\ &=10|t|. \end{aligned}

The absolute value changes form at \(t=0\). Thus

\begin{aligned} L &=\int_{-1}^{0}(-10t)\,dt+\int_0^2 10t\,dt\\ &=\left[-5t^2\right]_{-1}^{0}+\left[5t^2\right]_0^2\\ &=5+20\\ &=25. \end{aligned}

For a check, the point moves from \((3,4)\) to \((0,0)\) as \(t\) goes from \(-1\) to \(0\), a distance of \(5\). It then moves from \((0,0)\) to \((12,16)\), a distance of \(20\). The total is \(5+20=25\), matching the integral. The point travels along the same ray on both parts of the interval, but it first moves toward the origin and then away from it. Both portions count toward the path length.

A Common Mistake: Dropping the Absolute Value

In the preceding example, a tempting but incorrect simplification is \(\sqrt{100t^2}=10t\). That identity fails for negative \(t\), because a square root is nonnegative. If used over the whole interval, it would give

\int_{-1}^{2}10t\,dt =\left[5t^2\right]_{-1}^{2} =20-5 =15.

This is wrong: the integrand \(10t\) is negative on \([-1,0)\), even though speed cannot be negative. The correct simplification is \(\sqrt{100t^2}=10|t|\). Splitting at \(t=0\), where the sign changes, gives the correct length \(25\). In general, simplify square roots of squared expressions using absolute values, then remove the absolute value only on intervals where the sign is known.

Choosing and Checking an Integration Setup

Before integrating, confirm that the parameter bounds describe the intended part of the curve. A parametrization may trace one arch, several consecutive pieces, or a path that turns back over part of itself. The formula measures the distance traveled during the stated parameter interval, so repeated traversal is counted again.

A reliable calculation follows these steps:

Differentiate the coordinates.
Find \(x'(t)\) and \(y'(t)\) separately.

Form the speed.
Use \(\sqrt{(x'(t))^2+(y'(t))^2}\); do not omit either coordinate derivative.

Simplify carefully.
Use trigonometric identities and keep absolute values until the sign is determined on the interval.

Integrate over the given bounds.
Split the integral if an absolute value changes form or if the parametrization has separate smooth pieces.

Check the result.
Differentiate the antiderivative and, when possible, compare the result with the geometry or a direct distance calculation.

If \(x'(t)\) and \(y'(t)\) are continuous on the interval, the formula applies even at parameter values where both derivatives vanish. Such a value gives speed zero at that instant; it does not invalidate the length integral. If the coordinate functions are only piecewise continuously differentiable, calculate on each smooth subinterval and add the resulting lengths.

Arc length converts a parametrized description into a geometric quantity. It is useful when a curve is easier to describe through motion than through an equation \(y=f(x)\), and it keeps track of the full route rather than only the endpoints. The main calculation is often not the integration itself, but reducing the speed expression correctly before integrating.

Check Your Understanding

For each question, pay particular attention to the parameter interval and the sign of the speed expression.

State the arc-length formula for a plane curve parametrized by \(x(t)\) and \(y(t)\) on \([a,b]\).
For \(x(t)=4\cos t\), \(y(t)=4\sin t\) on \(0\le t\le\pi\), what is the simplified speed?
Why is \(\sqrt{t^2}\) equal to \(|t|\), rather than \(t\) for every real \(t\)?
For \(x(t)=t-\sin t\), \(y(t)=1-\cos t\), does \([2\pi,4\pi]\) trace the same arch as \([0,2\pi]\), or a horizontally shifted arch?
What does the arc-length integral count if a parametrized point travels along a segment and then retraces it?