9.3 Finding Arc Lengths of Curves Given by Parametric Equations

The arc length of a curve given by parametric equations $x(t)$ and $y(t)$ on $[a, b]$ is $S = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt$. You find both derivatives, plug them under the square root, and evaluate the definite integral. For AP Calculus BC, remember that the bounds are parameter values, not x- or y-values.

Why This Matters for the AP Calculus Exam

Parametric arc length builds directly on the Pythagorean idea behind every arc length formula: tiny pieces of a curve act like hypotenuses of small triangles. On the AP Calculus BC exam, this topic shows up in multiple-choice and free-response questions where you set up or evaluate an arc length integral, often connected to planar motion (the same integral gives total distance traveled when $x(t)$ and $y(t)$ describe a moving particle).

The exam rewards clean setup. Many questions just ask you to write the correct integral, and on calculator-active sections you can evaluate it numerically. Choosing the right procedure and writing precise notation matters for clear work, since errors in algebra, trig, or derivatives cause most lost points here.

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Key Takeaways

• The parametric arc length formula is $S = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\, dt$, where $a$ and $b$ are values of the parameter $t$.
• Take $\frac{dx}{dt}$ and $\frac{dy}{dt}$ separately, then square and add them inside the radical.
• The bounds are values of $t$, not $x$ or $y$. Use the parameter interval given in the problem.
• The expression $\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$ is the speed of a particle, so this same integral gives total distance traveled.
• Watch for trig identities like $\sin^2 t + \cos^2 t = 1$, which often simplify the integrand.
• On calculator-active questions, you can set up the integral and evaluate it numerically.

Reviewing Arc Length

Arc length is the distance along a curve between two points. Picture a paperclip: if you mark two points, bend it, then straighten it out and measure with a ruler, you get the arc length. Calculus does the same thing by adding up many tiny straight pieces.

Where the Formula Comes From (Cartesian)

Each small piece of the curve is the hypotenuse of a right triangle with legs $dx$ and $dy$. By the Pythagorean Theorem, $c = \sqrt{a^2 + b^2}$, so each tiny length $ds$ satisfies:

Factoring out $dx$ gives the familiar single-variable arc length formula:

This version works for curves written as $y = f(x)$. Parametric curves need a slightly adjusted version, since both $x$ and $y$ depend on $t$.

Arc Length of Parametric Curves

Connecting Parametric and Cartesian Forms

Before adapting the formula, it helps to see how parametric and Cartesian forms describe the same curve. A circle of radius $r$ can be written parametrically as $x(t) = r\cos(t)$ and $y(t) = r\sin(t)$.

Building the Parametric Formula

Start from $ds = \sqrt{dx^2 + dy^2}$. When the curve is parametric, both $x$ and $y$ change with $t$, so replace $dx$ with $\frac{dx}{dt}\,dt$ and $dy$ with $\frac{dy}{dt}\,dt$:

Adding these up over the parameter interval gives the parametric arc length formula:

It looks like the Cartesian version, but now you account for how both $x$ and $y$ change as $t$ moves.

Worked Examples

Example 1

Find the arc length over $[0, \pi]$ for the parametric curve $x(t) = \sin(t)$ and $y(t) = \cos(t)$.

Start by finding both derivatives:

Substitute into the formula with $a = 0$ and $b = \pi$:

Since $\cos^2(t) + \sin^2(t) = 1$, the integrand simplifies to $\sqrt{1} = 1$:

The arc length is $\pi$. That makes sense: this is half of a unit circle, which has circumference $2\pi$.

Example 2

Find the arc length over $[0, \pi]$ for the parametric curve $x(t) = 2$ and $y(t) = t^2$.

Take the derivatives:

Substitute with $a = 0$ and $b = \pi$:

Since $t \geq 0$ on this interval, $\sqrt{4t^2} = 2t$:

The arc length is $\pi^2$.

How to Use This on the AP Calculus Exam

Problem Solving

• Read off $x(t)$ and $y(t)$, then compute $\frac{dx}{dt}$ and $\frac{dy}{dt}$ before anything else.
• Square each derivative, add them, and place the sum under the radical.
• Confirm your bounds are $t$-values from the parameter interval.
• Check whether a trig identity collapses the integrand. Many AP problems are designed so $\sin^2 t + \cos^2 t = 1$ appears.

Free Response

• If the integral is messy, write the correct setup first. A correct unevaluated integral shows you chose the right procedure.
• On calculator-active questions, evaluate numerically and round as the problem directs.
• When the curve represents motion, state that this integral equals the total distance the particle travels, since the integrand is the speed.

Common Trap

• Pulling a constant out of the square root incorrectly, or forgetting to square the derivatives, are frequent errors. Keep each step written out.

Common Misconceptions

• Using $x$ or $y$ bounds instead of $t$ bounds. The limits of integration are values of the parameter $t$. Do not substitute endpoints in terms of $x$ or $y$.
• Adding the derivatives before squaring. You need $\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2$, not $\left(\frac{dx}{dt} + \frac{dy}{dt}\right)^2$. Square first, then add.
• Dropping a derivative when one is zero. If $\frac{dx}{dt} = 0$, you still keep $\frac{dy}{dt}$ under the radical. A zero term just means that coordinate is not changing at that moment.
• Forgetting that $\sqrt{4t^2} = 2|t|$. On an interval where $t$ could be negative, the absolute value matters. Check the sign of $t$ before simplifying.
• Confusing arc length with displacement. Arc length (and total distance traveled) uses speed under the radical and is always nonnegative. Displacement is a separate net-change idea.

Related AP Calculus Guides

• 9.1 Defining and Differentiating Parametric Equations
• 9.4 Defining and Differentiating Vector-Valued Functions
• Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
• 9.2 Second Derivatives of Parametric Equations
• 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve