Arc Length of a Curve: Formula, Derivation & Examples

What Is The Arc Length Of A Curve?

The arc length of a curve is the distance measured along the curve between two points, as if you unbent it into a straight string and laid a ruler beside it. For a smooth function $y = f(x)$ whose derivative is continuous on $[a, b]$, the arc length is:

$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2},dx$$

Sometimes the curve is easier to describe as $x = g(y)$, a function of $y$. Then you integrate with respect to $y$ instead, and the roles of the two variables swap:

$$L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2},dy$$

Both forms say the same thing: the length is the integral of a stretch factor that measures how much longer the true curve is than a flat horizontal (or vertical) step. Where the curve is steep, the stretch factor is large; where the curve is flat, it is close to $1$. The definite integral then adds every stretched step from one end to the other.

How Is The Arc Length Formula Derived?

The whole formula grows out of one picture: a curve, chopped into short segments, each of which is almost a straight line.

Split $[a, b]$ into $n$ small sub-intervals of width $\Delta x$. Over one sub-interval the curve rises by some amount $\Delta y$. The straight segment joining the two endpoints of that little piece is the hypotenuse of a right triangle with legs $\Delta x$ and $\Delta y$, so the distance formula, which is the Pythagoras theorem in disguise, gives its length:

$$\Delta s \approx \sqrt{(\Delta x)^2 + (\Delta y)^2}$$

Now factor $\Delta x$ out from under the root. That turns the raw vertical rise into a slope:

$$\Delta s \approx \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2};\Delta x$$

The total length is the sum of all these little hypotenuses, $L \approx \sum \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2},\Delta x$. As the pieces shrink, $\Delta x \to 0$, the ratio $\frac{\Delta y}{\Delta x}$ becomes the derivative $\frac{dy}{dx}$, and the sum becomes an integral. That limit of a sum is exactly a Riemann sum turning into the definite integral:

$$L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2},dx$$

Geometrically, $\sqrt{1 + (dy/dx)^2},dx$ is the length of the tiny slanted step the curve actually takes while $x$ advances by $dx$. Algebraically, it is the integrand you evaluate. The two readings are the same object seen from two sides.

What Is The Parametric Form Of Arc Length?

When both coordinates are written in terms of a parameter $t$, as $x = x(t)$ and $y = y(t)$, the curve may loop, double back, or run vertically, so a single $y = f(x)$ rule no longer fits. The derivation is the same, except each leg of the little triangle now comes from its own rate of change. Over a small step $\Delta t$ the point moves about $\Delta x \approx \frac{dx}{dt},\Delta t$ across and $\Delta y \approx \frac{dy}{dt},\Delta t$ up. Feeding those into $\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}$ and factoring out $\Delta t$ gives the parametric arc length:

$$L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2},dt$$

For a curve traced by parametric equations, this is the form to reach for. The Cartesian formula is just the special case where the parameter is $x$ itself, so $\frac{dx}{dt} = 1$ and the root collapses back to $\sqrt{1 + (dy/dx)^2}$.

How Do You Calculate Arc Length Of A Curve? Worked Examples

Each example is fully stepped, and every answer is checked by differentiating the chosen expression back to confirm the integrand. The examples are picked so that $1 + (dy/dx)^2$ simplifies cleanly, which is exactly what makes them workable by hand.

Example 1: A curve whose radicand collapses (dy/dx form).

Find the length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 3$.

Differentiate: $\dfrac{dy}{dx} = \dfrac{2}{3}\cdot\dfrac{3}{2}x^{1/2} = x^{1/2}$, so $\left(\dfrac{dy}{dx}\right)^2 = x$. The integrand becomes clean:

$$L = \int_0^3 \sqrt{1 + x},dx$$

Substitute $u = 1 + x$, so $du = dx$; the limits run from $u = 1$ to $u = 4$. This is a short integration by substitution:

$$L = \int_1^4 u^{1/2},du = \left[\frac{2}{3}u^{3/2}\right]_1^4 = \frac{2}{3}\left(4^{3/2} - 1\right) = \frac{2}{3}(8 - 1) = \frac{14}{3}$$

Check: differentiating $\frac{2}{3}(1+x)^{3/2}$ gives $(1+x)^{1/2} = \sqrt{1+x}$, the integrand, so the antiderivative is right.

Final answer: $L = \dfrac{14}{3} \approx 4.6667$ units.

Example 2: A perfect-square radicand (dy/dx form).

Find the length of $y = \frac{1}{3}\left(x^2 + 2\right)^{3/2}$ from $x = 0$ to $x = 3$.

Differentiate with the chain rule: $\dfrac{dy}{dx} = \dfrac{1}{3}\cdot\dfrac{3}{2}\left(x^2 + 2\right)^{1/2}(2x) = x\left(x^2 + 2\right)^{1/2}$. Square it:

$$\left(\frac{dy}{dx}\right)^2 = x^2\left(x^2 + 2\right) = x^4 + 2x^2$$

Add $1$, and the radicand is a perfect square:

$$1 + \left(\frac{dy}{dx}\right)^2 = x^4 + 2x^2 + 1 = \left(x^2 + 1\right)^2$$

The root of a perfect square is clean, $\sqrt{(x^2+1)^2} = x^2 + 1$ (positive on this interval), so the integral is an ordinary polynomial:

$$L = \int_0^3 \left(x^2 + 1\right)dx = \left[\frac{x^3}{3} + x\right]_0^3 = \left(9 + 3\right) - 0 = 12$$

Check: differentiating $\frac{x^3}{3} + x$ returns $x^2 + 1$, the integrand.

Final answer: $L = 12$ units exactly.

Example 3: A curve in parametric form (the cycloid).

A wheel of radius $1$ rolls along a line, and a point on its rim traces a cycloid $x = t - \sin t$, $y = 1 - \cos t$. Find the length of one full arch, from $t = 0$ to $t = 2\pi$.

Differentiate each coordinate: $\dfrac{dx}{dt} = 1 - \cos t$ and $\dfrac{dy}{dt} = \sin t$. Add the squares:

$$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = (1 - \cos t)^2 + \sin^2 t = 1 - 2\cos t + \cos^2 t + \sin^2 t = 2 - 2\cos t$$

Using the identity $1 - \cos t = 2\sin^2\left(\frac{t}{2}\right)$, the radicand becomes $4\sin^2\left(\frac{t}{2}\right)$, another perfect square. On $[0, 2\pi]$ the value $\sin\left(\frac{t}{2}\right)$ is not negative, so the root is $2\sin\left(\frac{t}{2}\right)$:

$$L = \int_0^{2\pi} 2\sin\left(\frac{t}{2}\right)dt = \left[-4\cos\left(\frac{t}{2}\right)\right]_0^{2\pi} = -4\cos\pi + 4\cos 0 = 4 + 4 = 8$$

Check: differentiating $-4\cos\left(\frac{t}{2}\right)$ gives $-4\cdot\left(-\sin\frac{t}{2}\right)\cdot\frac{1}{2} = 2\sin\frac{t}{2}$, the integrand.

Final answer: one arch of the cycloid has length $L = 8$ units, exactly eight times the wheel's radius.

What Are The Formulas For Arc Length In Every Form?

The four common ways a curve gets described each have their own version of the same integral. This table gathers them in one place.

Table: The arc-length integral in each way a curve can be described.

Every row is a distance-formula sum in disguise. Choose the row that matches how the curve is written, and the rest is finding the derivative and evaluating the integral. Arc length is one of the core applications of integration, alongside area and volume.

What If The Arc Length Integral Cannot Be Solved By Hand?

Here is the honest truth textbooks often soften: most arc-length integrals have no elementary antiderivative. The examples above were chosen so that $1 + (dy/dx)^2$ collapses to a perfect square or a simple substitution. Change the function slightly and that luck vanishes.

Take the plain cubic $y = x^3$ on $[0, 1]$. Then $\frac{dy}{dx} = 3x^2$ and the length integral is:

$$L = \int_0^1 \sqrt{1 + 9x^4},dx$$

There is no elementary function whose derivative is $\sqrt{1 + 9x^4}$, so no amount of substitution will finish it by hand. You estimate it numerically instead. Using Simpson's rule with four sub-intervals gives $L \approx 1.5482$; refining to eight sub-intervals gives $L \approx 1.5478$, and finer still keeps converging near $1.5476$. A course on numerical integration makes this exact, and the trapezoidal rule is the simpler cousin of Simpson's rule for the same job.

The most famous case is the ellipse. The perimeter of an ellipse cannot be written with elementary functions at all; it defines its own family of elliptic integrals, named precisely because this length resisted every closed-form attempt. When a curve resists integration, that is not a failure of method. It is a reminder that a simple-looking shape can hide an integral no formula can name, which is why numerical tools sit right beside the pencil-and-paper formula.

Why Does The Arc Length Formula Work?

The formula can feel like a trick with square roots until you see the geometry underneath it.

• A curve is a sum of tiny straight lines. Zoom in far enough on any smooth curve and each piece looks straight. The length of a straight piece is a hypotenuse, and adding hypotenuses is all the integral does.

• The integrand is a stretch factor. The term $\sqrt{1 + (dy/dx)^2}$ answers one question: while $x$ moves forward by a tiny bit, how much longer is the slanted path the curve takes than that flat horizontal bit? A steeper slope means a longer slanted step, so a larger integrand.

• It never dips below the straight-line distance. Because $\sqrt{1 + (dy/dx)^2} \ge 1$, the arc length is always at least $b - a$, matching the fact that the shortest path between two points is the straight one. A curve can only be longer than the chord, never shorter.

Read together, these say the formula is not an arbitrary definition. It is the only sum consistent with measuring length by the distance formula, taken to the limit of infinitely fine pieces.

Who Discovered How To Measure Arc Length?

For most of mathematical history, finding the exact length of a curved line was thought impossible, a problem called rectification, meaning "making straight." Many believed no curved arc could ever equal a whole-number length. The breakthrough came from a young Englishman.

Two more figures completed the picture:

• Hendrik van Heuraet (c. 1634–1660, Dutch Republic) published a general method for rectifying curves in 1659, turning Neile's single result into a repeatable procedure, one of the direct ancestors of the modern arc-length integral.

• Christopher Wren (1632–1723, England), better known as the architect of St Paul's Cathedral, found the exact length of one arch of the cycloid, the very curve worked in Example 3 above.

Where Is Arc Length Of A Curve Used In The Real World?

Any time a real quantity follows a curved path, its true length is an arc-length integral.

• Engineering and construction: the amount of cable in a hanging power line, the steel in a suspension-bridge deck, or the track along a curved railway is the arc length of the curve the structure follows.

• Roller coasters and roads: designers compute the length of a track or a highway curve to estimate material, travel time, and the forces a rider feels along the bend.

• Navigation and mapping: the distance along a winding river, coastline, or GPS route is the arc length of the path traced on the map, not the straight-line distance between endpoints.

• Manufacturing and 3D printing: a cutting tool or a print head following a curved profile travels an arc-length distance, which sets the time and the wear on the machine.

• Physics: the distance a projectile or an orbiting body covers along its trajectory is the arc length of that trajectory, distinct from its straight-line displacement.

One integral converts "how curvy is this path" into "how long is it," which is why the same formula shows up in a bridge cable and a spacecraft's orbit alike.

What Are The Most Common Mistakes With Arc Length?

These four errors account for most lost marks on arc-length problems, and each matches a question real students ask on r/calculus, r/learnmath, and AP review guides.

Forgetting to square the derivative.

Where it slips in:

A student writes the integrand as $\sqrt{1 + \frac{dy}{dx}}$ instead of $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$, dropping the square on the slope.

Don't do this:

Do not put the bare derivative under the root. The derivation squared a leg of a right triangle, so the square is not optional.

The correct way:

Compute $\frac{dy}{dx}$, then square it before adding $1$. For $y = \frac{2}{3}x^{3/2}$, the slope is $x^{1/2}$ and its square is $x$, giving $\sqrt{1 + x}$, never $\sqrt{1 + x^{1/2}}$.

Using the wrong variable for the limits.

Where it slips in:

On a parametric problem a student sets up the integral in $t$ but plugs in $x$-values for the limits, or mixes an $x$-integral with $y$-bounds.

Don't do this:

Do not let the limits disagree with the variable of integration. A $dt$ integral needs $t$-limits; a $dx$ integral needs $x$-limits.

The correct way:

Match them deliberately. For the cycloid arch integrated in $t$, the limits are $t = 0$ and $t = 2\pi$, the parameter values at the ends, not the $x$-coordinates $0$ and $2\pi$.

Assuming every integral has a closed form.

Where it slips in:

A student expects a clean exact answer and keeps searching for a substitution on an integral like $\int_0^1 \sqrt{1 + 9x^4},dx$ that has none.

Don't do this:

Do not force an elementary antiderivative onto a radicand that has none. Most arc-length integrands do not collapse.

The correct way:

Check whether $1 + (dy/dx)^2$ simplifies. If it does not, switch to a numerical method such as Simpson's rule and report an approximate length to a stated number of decimal places.

Losing the $+1$ under the root.

Where it slips in:

A student copies the parametric habit into the Cartesian form and writes $\sqrt{\left(\frac{dy}{dx}\right)^2}$, which just returns the slope and measures nothing.

Don't do this:

Do not drop the $1$. Without it, the "length" would be zero on any horizontal stretch, which is wrong.

The correct way:

Keep the constant term: the Cartesian integrand is always $\sqrt{1 + (dy/dx)^2}$. The $1$ is the horizontal leg $\Delta x$ that every step of the curve still takes.

Practice Problems On Arc Length Of A Curve

Work each one, then check against the verified answer.

1. Find the length of $y = \frac{2}{3}x^{3/2}$ from $x = 0$ to $x = 8$.
   (Answer: $\int_0^8 \sqrt{1+x},dx = \left[\frac{2}{3}(1+x)^{3/2}\right]_0^8 = \frac{2}{3}(27 - 1) = \frac{52}{3} \approx 17.3333$.)

2. Find the length of $y = \frac{1}{3}(x^2 + 2)^{3/2}$ from $x = 0$ to $x = 1$.
   (Answer: integrand $x^2 + 1$, so $\int_0^1 (x^2+1),dx = \frac{1}{3} + 1 = \frac{4}{3} \approx 1.3333$.)

3. A curve is given by $x = g(y) = \frac{2}{3}y^{3/2}$ for $y = 0$ to $y = 3$. Find its length.
   (Answer: $\frac{dx}{dy} = y^{1/2}$, so $\int_0^3 \sqrt{1+y},dy = \frac{2}{3}(8 - 1) = \frac{14}{3} \approx 4.6667$.)

4. Find the length of the parametric curve $x = t^2$, $y = t^3$ from $t = 0$ to $t = 1$.
   (Answer: $\int_0^1 \sqrt{4t^2 + 9t^4},dt = \int_0^1 t\sqrt{4 + 9t^2},dt = \frac{1}{27}\left(13^{3/2} - 8\right) \approx 1.4397$.)

5. Set up (do not evaluate) the length of $y = e^x$ from $x = 0$ to $x = 1$, and say why it needs a numerical method.
   (Answer: $\int_0^1 \sqrt{1 + e^{2x}},dx$; the radicand has no elementary antiderivative, so use Simpson's rule; the value is about $2.0035$.)

6. Find the length of one arch of the cycloid $x = t - \sin t$, $y = 1 - \cos t$ from $t = 0$ to $t = 2\pi$.
   (Answer: $\int_0^{2\pi} 2\sin\frac{t}{2},dt = 8$.)

Where Should You Go Next After Arc Length Of A Curve?

Arc length is one branch of using integrals to measure geometry, and several natural doors open from here.

1. Applications of integration. See how the same integrate-the-tiny-pieces idea gives area, volume, and average value, not just length.

2. Parametric equations. Master the curves that loop and double back, where the parametric arc-length form is the only one that works.

3. Numerical integration. Learn Simpson's rule and the trapezoidal rule properly, the tools that finish every arc-length integral no antiderivative can.

4. If your child is meeting arc length for the first time, a live Bhanzu trainer teaches it from the triangle picture up in the Bhanzu math classes.