What's the difference between arc length and distance traveled?

They use the identical integral _^((dx/dt)^2+(dy/dt)^2)\,dt, because the integrand is the particle's speed. Arc length is the physical length of the path; distance traveled is how far the particle actually moves. If the particle retraces part of the curve, both count that stretch again. Displacement, by contrast, is just the straight-line gap between start and end points.

Do I need to worry about the curve intersecting itself?

For the basic AP arc length formula, no — you integrate speed over the given t-interval regardless of self-intersections. The integral measures the total length swept as t increases, so if the path crosses itself or repeats, that length is included each time. Just integrate over exactly the interval the problem specifies.

When can I use my calculator on these problems?

On the calculator-active sections you may set up the exact integral and then evaluate it numerically, rounding to at least three decimal places. On no-calculator sections the radicand will always simplify — usually through factoring or the Pythagorean identity — so if it won't simplify, recheck your derivatives for algebra errors.

Why isn't there a '+1' under the radical like in the y = f(x) arc length formula?

The +1 appears only when you parametrize a function y=f(x) using x itself as the parameter, which forces dx/dx=1. In general parametric form, both dx/dt and dy/dt are genuine derivatives, so both are squared and added with no stray constant. Using the +1 formula on parametric curves is a common and costly error.

AP-CALCBC-9.3

U9.3 Arc Length of Parametric Curves

Learn to compute parametric arc length with L = ∫√((dx/dt)²+(dy/dt)²) dt. Setup, simplification, and AP exam tips with worked examples.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U9.3 Arc Length of Parametric Curves, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U9.3 Arc Length of Parametric Curves free

What this lesson covers

When a curve is traced by parametric equations

x(t)

and

y(t)

, its total length isn't something you can read off a single function — you have to add up the infinitely many tiny straight-line steps the point takes as

t

increases. That's exactly what the parametric arc length integral does.

In this lesson you'll learn the formula

L=\int_\alpha^\beta \sqrt{(dx/dt)^2+(dy/dt)^2}\,dt

, why it works geometrically, how to set it up cleanly, and where students lose points on the AP exam. Since most of these integrals are messy, you'll also practice recognizing when to reach for your calculator versus when a clean algebraic simplification is hiding in plain sight.

Where the Formula Comes From

Imagine a particle moving along a curve with position

(x(t),y(t))

. Over a tiny time interval

dt

, it moves a small horizontal amount

dx

and a small vertical amount

dy

. By the Pythagorean theorem, the tiny distance traveled is

ds=\sqrt{dx^2+dy^2}

.

Factor

dt

out from under the radical. Since

dx=\frac{dx}{dt}\,dt

and

dy=\frac{dy}{dt}\,dt

, we get

ds=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\;dt.

Adding up (integrating) every tiny piece from

t=\alpha

t=\beta

gives the total arc length

L=\int_\alpha^\beta \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt.

Notice the deep connection to motion: the integrand

\sqrt{(dx/dt)^2+(dy/dt)^2}

is exactly the speed of the particle, and integrating speed over time gives total distance traveled. That is why arc length and the distance-traveled problems in U9.4 share the same integral. The key distinction: arc length measures the physical length of the path, so if the particle backtracks or traces a segment twice, the integral counts that length again.

Setting Up the Integral Correctly

A reliable four-step process keeps you from making setup errors.

Step	Action
1	Differentiate to find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
2	Square each derivative and add them
3	Take the square root of the sum
4	Integrate over the given $t$ -interval $[\alpha,\beta]$

The most common mistake is confusing this with the single-variable arc length formula

\int\sqrt{1+(dy/dx)^2}\,dx

. For parametric curves you do NOT have a lone

1

under the radical — both derivatives are squared. The

1

only appears in the

y=f(x)

case because there

dx/dx=1

.

Another frequent error: dropping the square root or trying to integrate

\sqrt{a^2+b^2}

a+b

. The square root of a sum is never the sum of the square roots.

Always confirm the interval is in terms of

t

, not

x

y

. If a problem gives

x

-bounds, convert them to the corresponding

t

-values first.

When It Simplifies vs. When to Use a Calculator

Most parametric arc length integrals do not have elementary antiderivatives, so on the calculator-active portion of the AP exam you set up the integral exactly, then evaluate it numerically. Writing the correct definite integral earns the setup points even if you never simplify.

But examiners love integrands engineered to collapse. Watch for these patterns:

Pattern	Simplification
Perfect square trinomial under the root	$\sqrt{(f+g)^2}=\|f+g\|$
Pythagorean identity $\sin^2+\cos^2$	Combines to a constant
$\sqrt{a^2\cos^2 t + a^2\sin^2 t}$	Equals $\|a\|$

For example, a cycloid or an involute problem often produces

\sqrt{2-2\cos t}

, which becomes

2\left|\sin\frac{t}{2}\right|

using a half-angle identity.

On the no-calculator section, the integral is guaranteed to simplify — so if it doesn't, recheck your derivatives. On the calculator section, don't waste time forcing an antiderivative; set up the integral, then use numerical integration and round to three decimal places as the exam requires.

How the AP Exam Tests This

Parametric arc length appears both as a standalone multiple-choice item and as part of a larger FRQ that blends U9.1 differentiation, U9.4 motion, and arc length. In an FRQ, a single part might ask for the length of the path traced on a given interval, worth roughly two points: one for the correct integral setup with limits, one for the numerical answer.

Graders want the integral written with the square root, the squared derivatives inside, and correct

t

-limits before any numerical value. A bare answer with no visible integrand risks losing the setup point. Show the integral, then the number.

A subtle trap: 'distance traveled' versus 'displacement' versus 'arc length.' Arc length and total distance traveled use the same speed integral. Displacement is

\sqrt{(x(\beta)-x(\alpha))^2+(y(\beta)-y(\alpha))^2}

— a straight-line measurement, not an integral. Read the verb in the prompt carefully.

Units and rounding also matter: if the problem is calculator-active, present at least three digits after the decimal, and never round intermediate values in a way that changes the final digits.

Key terms

Arc length.: The total length of a curve measured along the path itself, computed by integrating $ds$ over the parameter interval.
Parameter interval.: The range $[\alpha,\beta]$ of the parameter $t$ over which the curve is traced and the arc length is integrated.
Differential of arc length ( $ds$ ).: The infinitesimal path length $ds=\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$ for a parametric curve.
Speed.: The magnitude of the velocity vector, $\sqrt{(dx/dt)^2+(dy/dt)^2}$ ; integrating it over time gives distance traveled, identical to arc length.
Displacement.: The straight-line distance between start and end points, $\sqrt{(\Delta x)^2+(\Delta y)^2}$ ; not an integral and generally smaller than arc length.
Pythagorean identity.: The relation $\sin^2\theta+\cos^2\theta=1$ , frequently used to collapse the radicand in trigonometric parametric curves.

Worked example

A curve is defined by

x(t)=t^2

and

y(t)=\frac{2}{3}t^3

for

0\le t\le 2

. Find the exact arc length of the curve.

Start by differentiating. We have

\frac{dx}{dt}=2t

and

\frac{dy}{dt}=2t^2

.

Square and add:

\left(2t\right)^2+\left(2t^2\right)^2=4t^2+4t^4

.

Factor inside the radical:

4t^2+4t^4=4t^2(1+t^2)

. Take the square root:

\sqrt{4t^2(1+t^2)}=2|t|\sqrt{1+t^2}

. Since

0\le t\le 2

, we have

|t|=t

, so the integrand is

2t\sqrt{1+t^2}

.

Set up the integral:

L=\int_0^2 2t\sqrt{1+t^2}\,dt.

Use the substitution

u=1+t^2

, so

du=2t\,dt

. When

t=0

u=1

; when

t=2

u=5

. The integral becomes

L=\int_1^5 \sqrt{u}\,du=\left[\frac{2}{3}u^{3/2}\right]_1^5=\frac{2}{3}\left(5^{3/2}-1\right).

Since

5^{3/2}=5\sqrt{5}

, the exact arc length is

L=\frac{2}{3}\left(5\sqrt{5}-1\right)\approx 6.786.

Notice how the algebraic factoring exposed a clean

u

-substitution — a signal that this was a no-calculator style problem.

Practice questions

A particle moves along the curve given by

x(t)=3\cos t

and

y(t)=3\sin t

for

0\le t\le \pi

. What is the length of the path traced?

$3\pi$
$6\pi$
$3$
$9\pi$

Answer: $3\pi$

Differentiate:

dx/dt=-3\sin t

and

dy/dt=3\cos t

. Then

(dx/dt)^2+(dy/dt)^2=9\sin^2 t+9\cos^2 t=9

, so the integrand is

\sqrt{9}=3

. The arc length is

\int_0^\pi 3\,dt=3\pi

. This makes sense geometrically: the curve is a circle of radius 3, and

t

from

0

\pi

traces the top half, a semicircle of length

\frac{1}{2}(2\pi\cdot 3)=3\pi

Set up, but do not evaluate, an integral for the arc length of the curve

x(t)=e^t\cos t

y(t)=e^t\sin t

from

t=0

t=\ln 2

. Then simplify the radicand as far as possible.

Answer: $L=\int_0^{\ln 2}\sqrt{2}\,e^t\,dt$

Differentiate using the product rule:

dx/dt=e^t\cos t-e^t\sin t

and

dy/dt=e^t\sin t+e^t\cos t

. Squaring each and adding, the cross terms cancel and the

\sin^2+\cos^2

terms combine, giving

(dx/dt)^2+(dy/dt)^2=e^{2t}(2\cos^2 t+2\sin^2 t)=2e^{2t}

. Taking the square root yields

\sqrt{2}\,e^t

, so

L=\int_0^{\ln 2}\sqrt{2}\,e^t\,dt

. (For interest, this evaluates to

\sqrt{2}\,(2-1)=\sqrt{2}

For

x(t)=t-\sin t

and

y(t)=1-\cos t

0\le t\le 2\pi

(one arch of a cycloid), which integral gives its arc length?

$\int_0^{2\pi} 2\left|\sin\tfrac{t}{2}\right|\,dt$
$\int_0^{2\pi}\sqrt{2-2\cos t}\,dt$
Both of the above are equal
$\int_0^{2\pi}(2-2\cos t)\,dt$

Answer: Both of the above are equal

Here

dx/dt=1-\cos t

and

dy/dt=\sin t

. Then

(1-\cos t)^2+\sin^2 t=1-2\cos t+\cos^2 t+\sin^2 t=2-2\cos t

, so the radicand is

\sqrt{2-2\cos t}

. Applying the half-angle identity

1-\cos t=2\sin^2\frac{t}{2}

gives

\sqrt{4\sin^2\frac{t}{2}}=2\left|\sin\frac{t}{2}\right|

. Both integrals represent the same length; the last choice is wrong because it omits the square root.

FAQ

What's the difference between arc length and distance traveled?: They use the identical integral $\int_\alpha^\beta\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$ , because the integrand is the particle's speed. Arc length is the physical length of the path; distance traveled is how far the particle actually moves. If the particle retraces part of the curve, both count that stretch again. Displacement, by contrast, is just the straight-line gap between start and end points.
Do I need to worry about the curve intersecting itself?: For the basic AP arc length formula, no — you integrate speed over the given $t$ -interval regardless of self-intersections. The integral measures the total length swept as $t$ increases, so if the path crosses itself or repeats, that length is included each time. Just integrate over exactly the interval the problem specifies.
When can I use my calculator on these problems?: On the calculator-active sections you may set up the exact integral and then evaluate it numerically, rounding to at least three decimal places. On no-calculator sections the radicand will always simplify — usually through factoring or the Pythagorean identity — so if it won't simplify, recheck your derivatives for algebra errors.
Why isn't there a '+1' under the radical like in the y = f(x) arc length formula?: The $+1$ appears only when you parametrize a function $y=f(x)$ using $x$ itself as the parameter, which forces $dx/dx=1$ . In general parametric form, both $dx/dt$ and $dy/dt$ are genuine derivatives, so both are squared and added with no stray constant. Using the $+1$ formula on parametric curves is a common and costly error.

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