Arc Length of Parametrized Plane Curves

From Parametric Equations to Arc Length

A plane curve may be described by giving its two coordinates as functions of a parameter $t$. This is useful when the curve is not naturally described as a single graph $y=f(x)$, or when a parameter traces a curve whose $x$-coordinates repeat. The parameter records where the moving point is; its derivatives describe how quickly each coordinate changes.

To find the distance traveled along the curve, combine the rates of change of both coordinates. The result is an integral of the curve’s speed. In particular, the integrand is not usually the sum of the absolute coordinate rates: it is their Euclidean magnitude.

Definition: Let $x(t)$ and $y(t)$ be continuously differentiable on $[a,b]$, and let the parametrized plane curve be $\mathbf r(t)=(x(t),y(t))$. Its arc length from $t=a$ to $t=b$ is $$ L=\int_a^b \sqrt{(x'(t))^2+(y'(t))^2}\,dt. $$ The integrand is the speed $\lVert\mathbf r'(t)\rVert$. If the coordinates are only piecewise continuously differentiable, the integral is evaluated piece by piece and the lengths are added.

The interval matters: the formula measures the distance traced as $t$ runs from $a$ to $b$. If the parametrization retraces part of the curve, that part is counted each time. A constant portion contributes zero length because both coordinate derivatives vanish there.

Why the Formula Uses Speed

For a small parameter change $\Delta t$, the change in position is approximately $\mathbf r'(t)\Delta t$. Its length is therefore approximately $\lVert\mathbf r'(t)\rVert\Delta t$, by the distance formula in the plane. Adding these short distances along the interval leads to the integral of speed. This is the same arc-length principle used for vector-valued curves; in two dimensions, the vector derivative is $(x'(t),y'(t))$, so its magnitude is exactly the square root in the definition.

\lVert\mathbf r'(t)\rVert =\sqrt{\bigl(x'(t)\bigr)^2+\bigl(y'(t)\bigr)^2}.

Thus the computation has two main steps: differentiate each coordinate, then combine those derivatives with the Pythagorean theorem before integrating. The sum of the coordinate rates would measure motion along a path made of horizontal and vertical segments, not the length of the given curve.

Worked Example: A Quarter Circle

Worked Example: Finding the Length of a Quarter Circle

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

Differentiate each coordinate:

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

The speed is

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

Therefore the length is

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

Check: the speed calculated from the coordinate derivatives is $3$ throughout the interval, so the length is the constant speed multiplied by the parameter interval length, $3(\frac{\pi}{2}-0)=\frac{3\pi}{2}$, as found.

Ellipses and Their Arc-Length Integrals

An ellipse centered at the origin with semiaxes $a$ and $b$ has a standard parametrization $x(t)=a\cos t$, $y(t)=b\sin t$. For $a\ge b>0$, one traversal of the ellipse is obtained as $t$ runs from $0$ to $2\pi$. Differentiating and applying the arc-length formula gives

L=\int_0^{2\pi}\sqrt{a^2\sin^2t+b^2\cos^2t}\,dt.

The four quarter-arcs have equal length, so symmetry also gives $L=4\int_0^{\pi/2}\sqrt{a^2\sin^2t+b^2\cos^2t}\,dt$. Unlike the circle case, the square root generally does not simplify to a constant. The resulting integral is an example of an elliptic integral; an exact answer can be expressed using a standard integral defined for this purpose.

Definition: For $0\le k<1$, the complete elliptic integral of the second kind is $$ E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2u}\,du. $$ For an ellipse with semiaxes $a\ge b>0$, set $e=\sqrt{1-b^2/a^2}$. Its perimeter is $4aE(e)$.

Worked Example: The Perimeter of an Ellipse

Worked Example: An Ellipse with Semiaxes 5 and 3

Find an exact expression for the perimeter of the ellipse $x(t)=5\cos t$, $y(t)=3\sin t$, for $0\le t\le2\pi$.

First find the coordinate derivatives and speed:

x'(t)=-5\sin t,\qquad y'(t)=3\cos t,

\lVert\mathbf r'(t)\rVert =\sqrt{25\sin^2t+9\cos^2t}.

Using the four equal quarter-arcs, the perimeter is

L=4\int_0^{\pi/2}\sqrt{25\sin^2t+9\cos^2t}\,dt.

Factor out $25$ from inside the square root and rewrite the remaining expression:

25\sin^2t+9\cos^2t =25\left(\sin^2t+\frac{9}{25}\cos^2t\right) =25\left(1-\frac{16}{25}\cos^2t\right).

Consequently, the quarter-perimeter integral is $5\int_0^{\pi/2}\sqrt{1-\frac{16}{25}\cos^2t}\,dt$. Use $u=\frac{\pi}{2}-t$, so $\cos t=\sin u$; the endpoints still run from $0$ to $\frac{\pi}{2}$ after reversing their order. Thus

L=20\int_0^{\pi/2}\sqrt{1-\frac{16}{25}\sin^2u}\,du =20E\left(\frac45\right).

This is an exact expression under the definition of $E$ above. The check is that $e=\sqrt{1-\frac{3^2}{5^2}}=\sqrt{\frac{16}{25}}=\frac45$, and the perimeter formula $4aE(e)$ gives $4(5)E(\frac45)=20E(\frac45)$, matching the integral calculation.

For a circle, $a=b$, so $e=0$ and $E(0)=\int_0^{\pi/2}1\,du=\frac{\pi}{2}$. The formula then gives $4aE(0)=2\pi a$, the usual circumference. This also explains why the circle’s speed simplifies to a constant, while a noncircular ellipse usually retains a nonconstant square root.

Worked Example: Different Trigonometric Coordinate Functions

Worked Example: A Curve with a Trigonometric and a Linear-Trigonometric Coordinate

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

Differentiate the coordinates:

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

Substitute into the speed formula and simplify using $\sin^2t+\cos^2t=1$:

\begin{aligned} \lVert\mathbf r'(t)\rVert &=\sqrt{(1+\cos t)^2+(-\sin t)^2}\\ &=\sqrt{1+2\cos t+\cos^2t+\sin^2t}\\ &=\sqrt{2+2\cos t}\\ &=\sqrt{4\cos^2(t/2)} =2\left|\cos(t/2)\right|. \end{aligned}

Because $0\le t/2\le\pi/2$, $\cos(t/2)\ge0$, so the absolute value can be removed on this interval. Hence

L=\int_0^\pi 2\cos(t/2)\,dt =4\left[\sin(t/2)\right]_0^\pi =4.

Check: an antiderivative of $2\cos(t/2)$ is $4\sin(t/2)$, since its derivative is $4\cdot\frac12\cos(t/2)=2\cos(t/2)$. Evaluating at $t=\pi$ and $t=0$ gives $4-0=4$.

Conditions and a Common Mistake

The formula applies directly when the coordinate functions have continuous derivatives on the parameter interval. If a coordinate derivative has a finite number of discontinuities but the curve is piecewise continuously differentiable, split the interval at those points and add the resulting lengths. In either case, make sure the endpoints match the portion of the curve requested. A full ellipse requires a full parameter interval; a quarter-ellipse does not.

A frequent error is to add the absolute values of the coordinate derivatives. Consider again the quarter circle $x(t)=3\cos t$, $y(t)=3\sin t$, $0\le t\le\frac{\pi}{2}$. The incorrect calculation would be

\int_0^{\pi/2}\bigl(|x'(t)|+|y'(t)|\bigr)\,dt =\int_0^{\pi/2}3(\sin t+\cos t)\,dt =6.

This is the sum of the horizontal and vertical distances, not the distance along the circular arc. The correct integrand is $\sqrt{(x')^2+(y')^2}=3$, so the correct length is $\int_0^{\pi/2}3\,dt=\frac{3\pi}{2}$, not $6$. The correction is to combine the coordinate rates using the Pythagorean theorem before integrating.

Arc-length calculations are useful whenever a curve is specified parametrically rather than as a graph. They also show why a curve’s shape alone does not determine the integrand: the coordinate derivatives determine the speed at each parameter value. Simplifying that speed carefully—using identities, factoring, and checking the sign before removing absolute values—is often the central work of the problem.

Check Your Understanding

For each question, identify the coordinate derivatives before deciding how to integrate.

Write the arc-length integral for $x(t)=4\cos t$, $y(t)=2\sin t$ on $0\le t\le\frac{\pi}{2}$.
What smoothness condition allows the arc-length formula to be applied directly on a closed interval?
For an ellipse with semiaxes $a\ge b>0$, what is the eccentricity parameter used in the perimeter expression $4aE(e)$?
Why is $\int_a^b(|x'(t)|+|y'(t)|)\,dt$ generally not the arc length?
For $x(t)=t+\sin t$, $y(t)=\cos t$ on $0\le t\le\pi$, why may $|\cos(t/2)|$ be replaced by $\cos(t/2)$?