What are wrong limits?

Integrate over the parameter interval in t, not over an x-interval; substitute the t-values given.

College Board AP 2022 (BC, style)-style practice: Section I (multiple choice, calculator). A curve is given by x = t^2, y = t^3 for 0 t 1. Its length is (A) _0^1sqrt(4t^2 + 9t^4)\,dt (B) _0^1sqrt(2t + 3t^2)\,dt (C) _0^1(4t^2 + 9t^4)\,dt (D) _0^1sqrt(t^4 + t^6)\,dt

The correct answer is (A). (dx)/(dt) = 2t, (dy)/(dt) = 3t^2, so ((dx)/(dt))^2 + ((dy)/(dt))^2 = 4t^2 + 9t^4. Length = _0^1sqrt(4t^2 + 9t^4)\,dt.

College Board AP 2024 (BC, style)-style practice: Section II (free response). A particle moves along x = 3t^2, y = 2t^3 for 0 t 2. (a) Set up the integral for the distance traveled. (b) The integrand simplifies to 6tsqrt(1 + t^2); evaluate the integral exactly.

A 4-point parametric arc-length problem. (a) (2 points) (dx)/(dt) = 6t, (dy)/(dt) = 6t^2, so distance = _0^2sqrt(36t^2 + 36t^4)\,dt = _0^2 6tsqrt(1 + t^2)\,dt. (b) (2 points) Let u = 1 + t^2, du = 2t\,dt: _0^2 6tsqrt(1+t^2)\,dt = 3_1^5sqrt(u)\,du = 3·(2)/(3)[u^3/2]_1^5 = 2(5^3/2 - 1) = 2(55 - 1).

United StatesCalculusSyllabus dot point

How do you find the length of a parametric curve over an interval of the parameter?

Topic 9.3 Finding Arc Lengths of Curves Given by Parametric Equations: compute the length of a parametric curve using the integral of the square root of (dx/dt)^2 + (dy/dt)^2 (BC).

A focused answer to AP Calculus BC Topic 9.3, computing the arc length of a parametric curve with the integral of the square root of (dx/dt)^2 + (dy/dt)^2 over the parameter interval, with worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
The formula and its meaning as speed
Why it is the same idea as Topic 8.13
A worked exact arc length
Distance traveled versus displacement
When to use the calculator

What this topic is asking

The College Board (Topic 9.3, BC only) gives the arc length of a curve described parametrically by $x = x(t)$ , $y = y(t)$ . This is the parametric version of Topic 8.13, and it is also the distance traveled by a particle whose position is $(x(t), y(t))$ , which makes it one of the most heavily tested ideas in Unit 9.

The formula and its meaning as speed

Why it is the same idea as Topic 8.13

The planar arc-length formula $L = \int\sqrt{1 + (dy/dx)^2}\,dx$ of Topic 8.13 is the special case of this one where the parameter is $x$ . If $t = x$ , then $\frac{dx}{dt} = 1$ and $\frac{dy}{dt} = \frac{dy}{dx}$ , so $\sqrt{(dx/dt)^2 + (dy/dt)^2} = \sqrt{1 + (dy/dx)^2}$ . The parametric form is more general because it treats $x$ and $y$ symmetrically, which is why it can handle loops, vertical pieces, and retraced paths that a single function $y = f(x)$ cannot. Seeing the two formulas as the same Pythagorean idea, $ds = \sqrt{(dx)^2 + (dy)^2}$ , expressed with different parameters, makes the unit hang together.

A worked exact arc length

A clean parametric length

A curve is given by $x = t^2$ , $y = \frac{2}{3}t^3$ for $0 \le t \le \sqrt3$ . Find its length.

step 1 Differentiate

$\frac{dx}{dt} = 2t$ , $\frac{dy}{dt} = 2t^2$ .

step 2 Form the integrand

$\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 4t^2 + 4t^4 = 4t^2(1 + t^2)$ , so $\sqrt{\cdots} = 2t\sqrt{1 + t^2}$ (for $t \ge 0$ ).

step 3 Set up and substitute

L = \int_0^{\sqrt3} 2t\sqrt{1 + t^2}\,dt.

Let

u = 1 + t^2

du = 2t\,dt

; limits

t = 0\Rightarrow u = 1

t = \sqrt3\Rightarrow u = 4

step 4 Integrate and evaluate

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

Distance traveled versus displacement

A vital distinction on the exam is distance traveled versus displacement. The arc-length integral $\int_a^b\sqrt{(dx/dt)^2 + (dy/dt)^2}\,dt$ gives the total distance the particle travels, always positive, accounting for every wiggle and reversal. The displacement is the straight-line change in position, $\left(x(b) - x(a),\, y(b) - y(a)\right)$ , which can be smaller (or even zero if the particle returns to its start). When a question asks "how far does the particle travel," it wants the arc-length integral; when it asks "what is the net change in position" or "where does it end up," it wants displacement. Reading which is requested prevents a costly mismatch.

When to use the calculator

As with Topic 8.13, the integrand $\sqrt{(dx/dt)^2 + (dy/dt)^2}$ is usually not integrable by hand, so parametric arc-length problems most often appear on the calculator-active sections. The expected work is the correct integral expression with correct limits, after which the numerical value comes from the calculator. The exact-by-hand cases, like the worked example, are engineered so the expression under the root factors into a perfect square or yields to substitution. A quick test: form the integrand, and if it does not simplify to something with an elementary antiderivative, set up and evaluate numerically rather than struggling for an exact form.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2022 (BC, style)1 marksSection I (multiple choice, calculator). A curve is given by

x = t^2

y = t^3

for

0 \le t \le 1

. Its length is (A)

\int_0^1\sqrt{4t^2 + 9t^4}\,dt

(B)

\int_0^1\sqrt{2t + 3t^2}\,dt

(C)

\int_0^1(4t^2 + 9t^4)\,dt

(D)

\int_0^1\sqrt{t^4 + t^6}\,dt

Show worked answer →

The correct answer is (A).

$\frac{dx}{dt} = 2t$ , $\frac{dy}{dt} = 3t^2$ , so $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 = 4t^2 + 9t^4$ . Length $= \int_0^1\sqrt{4t^2 + 9t^4}\,dt$ .

AP 2024 (BC, style)4 marksSection II (free response). A particle moves along

x = 3t^2

y = 2t^3

for

0 \le t \le 2

. (a) Set up the integral for the distance traveled. (b) The integrand simplifies to

6t\sqrt{1 + t^2}

; evaluate the integral exactly.

Show worked answer →

A 4-point parametric arc-length problem.

(a) (2 points) $\frac{dx}{dt} = 6t$ , $\frac{dy}{dt} = 6t^2$ , so distance $= \int_0^2\sqrt{36t^2 + 36t^4}\,dt = \int_0^2 6t\sqrt{1 + t^2}\,dt$ .
(b) (2 points) Let $u = 1 + t^2$ , $du = 2t\,dt$ : $\int_0^2 6t\sqrt{1+t^2}\,dt = 3\int_1^5\sqrt{u}\,du = 3\cdot\frac{2}{3}\left[u^{3/2}\right]_1^5 = 2(5^{3/2} - 1) = 2(5\sqrt5 - 1)$ .

Related dot points

Sources & how we know this

AP Calculus AB and BC Course and Exam Description — College Board (2020)