Finding Arc Lengths of Curves Given by Parametric Equations: Study Guide (Calculus BC 9.3)

BC-only material. Parametric arc length is part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

You need four earlier ideas. If any is shaky, revisit it from the Calculus BC roadmap first.

Prerequisite	Topic	What you use it for here
Parametric derivatives dx/dt and dy/dt	9.1	They go inside the arc length integral
Arc length of y = f(x)	8.13	The same idea, now with a parameter
Riemann sums and the definite integral	6.2 to 6.4	Adding up many tiny lengths
Substitution	6.9	Some exact arc lengths need it

Notation on this page: ∫ from a to b of f(t) dt is the definite integral with lower limit a and upper limit b. x′(t) means dx/dt and y′(t) means dy/dt.

The idea: Pythagoras on tiny pieces

A parametric curve is drawn by the point (x(t), y(t)) as t runs from a to b. You want the length of the path the point draws.

Cut the interval a ≤ t ≤ b into many short pieces of width Δt. Over one piece, the point moves a horizontal distance Δx and a vertical distance Δy. If Δt is small, the curve is almost straight there, so its length Δs is close to the length of the straight chord:

Δs ≈ √((Δx)² + (Δy)²)

Now bring in derivatives. For a small Δt, Δx ≈ x′(t) Δt and Δy ≈ y′(t) Δt. Substitute:

Δs ≈ √((x′(t) Δt)² + (y′(t) Δt)²) = √((x′(t))² + (y′(t))²) · Δt

Adding all the pieces gives a Riemann sum. As Δt → 0 the sum becomes a definite integral:

L = ∫ from a to b of √((dx/dt)² + (dy/dt)²) dt

The integrand √((dx/dt)² + (dy/dt)²) is “length per unit of t”. In motion problems (Topic 9.6) it is the speed, and the same integral gives the distance travelled.

When does the integral give the length?

Two conditions matter.

1. The curve is smooth on [a, b]. dx/dt and dy/dt are continuous. This makes the integrand continuous, so the integral exists.
2. The curve is traced exactly once as t goes from a to b. The integral adds up every bit of path the point covers. If the point goes over part of the curve twice, that part is counted twice.

Example of condition 2. The circle x = 2 cos t, y = 2 sin t has radius 2. Its integrand is √(4 sin² t + 4 cos² t) = 2.

• For 0 ≤ t ≤ 2π: ∫ from 0 to 2π of 2 dt = 4π. That is the circumference 2π · 2. ✓
• For 0 ≤ t ≤ 4π: the integral is 8π, but the circle is still only 4π long. The point went round twice.

How to check “traced once”. The quickest proof is a component that is always increasing or always decreasing on the interval. That proof is enough, but it is not the only way. A curve can turn back and still be traced once, as long as the point never returns to a place it has already been. If neither component is monotonic, look at where the point is at different values of t, or sketch the curve with the values of t marked, before you call the integral a length.

So read the question carefully. “Find the length of the curve” needs a parameter interval that traces it once. “Find the distance the point travels” (Topic 9.6) counts every lap.

Link to Topic 8.13. If a curve is y = f(x), use x as the parameter: x = t, y = f(t). Then dx/dt = 1 and dy/dt = f′(t), and the formula becomes ∫ √(1 + (f′(x))²) dx. The parametric formula is the more general one.

Worked example 1: an exact length by hand

Question. Find the exact length of the curve x = ½t², y = ⅓(2t + 1)^(3/2) for 0 ≤ t ≤ 4. No calculator.

1. Check the curve is traced once. dx/dt = t, which is positive for t > 0. So x keeps increasing and the point never comes back over its path.
2. Differentiate.
   • dx/dt = t
   • dy/dt = ⅓ · (3/2)(2t + 1)^(1/2) · 2 = (2t + 1)^(1/2) (chain rule: the factor 2 from the inside)
3. Square and add. (dx/dt)² + (dy/dt)² = t² + (2t + 1) = t² + 2t + 1 = (t + 1)²
4. Take the square root. √((t + 1)²) = |t + 1| = t + 1, because t + 1 > 0 on [0, 4].
5. Integrate. L = ∫ from 0 to 4 of (t + 1) dt = [½t² + t] from 0 to 4 = (8 + 4) − 0 = 12

Answer. The curve is 12 units long.

Checks.

• Straight-line distance: the curve runs from (0, ⅓) to (8, 9). The chord length is √(8² + (26/3)²) = √(64 + 75.11…) ≈ 11.795. The curve must be at least this long, and 12 is just a little more, which fits a curve that bends only gently. ✓
• Calculator (where allowed): a numerical integral of √(t² + 2t + 1) from 0 to 4 gives 12.000. ✓

Why this worked. The expression under the root became a perfect square. That is the only reason an exact answer was possible. Questions that want an exact length are built this way, so after squaring and adding, look for a perfect square.

Worked example 2: a calculator length in context

Context. A fictional survey drone flies over a field. Its position, in metres, is x(t) = 8t and y(t) = 20 + 6 sin(πt/5), where t is the time in seconds, 0 ≤ t ≤ 10. Find the length of the drone’s path. Then compare it with the straight-line distance between its start and end.

1. Check the path is traced once. dx/dt = 8 > 0, so the drone always moves in the positive x direction. No part of the path is repeated.
2. Differentiate.
   • dx/dt = 8
   • dy/dt = 6 · (π/5) cos(πt/5) = (6π/5) cos(πt/5)
3. Set up the integral. L = ∫ from 0 to 10 of √(64 + (6π/5)² cos²(πt/5)) dt
4. Can you integrate by hand? No. 64 + (a constant)·cos²(…) is not a perfect square. Use the calculator’s numerical integration (radian mode).
5. Evaluate. L ≈ 84.272 m.

Checks.

• Straight-line distance: at t = 0 the drone is at (0, 20); at t = 10 it is at (80, 20 + 6 sin 2π) = (80, 20). The chord is 80 m. The path is a wave, so it must be longer than 80 m. ✓
• Upper bound: the integrand is at most √(64 + (6π/5)²) ≈ 8.844 m per second, so in 10 seconds the path is at most about 88.4 m. 84.272 m lies between 80 m and 88.4 m. ✓
• Units: the integrand is in metres per second; multiplied by seconds it gives metres. ✓

Interpretation. The drone’s path over the 10 seconds is about 84.3 m long, about 4.3 m more than the straight-line distance between its start and end points, because of the up-and-down wave.

A strategy for any arc length question

1. Find dx/dt and dy/dt (watch the chain rule).
2. Check the parameter interval traces the curve once. A component that is always increasing or always decreasing is the quickest proof.
3. Write L = ∫ from a to b of √((dx/dt)² + (dy/dt)²) dt with the correct limits. The limits are values of t, not of x or y.
4. Expand (dx/dt)² + (dy/dt)². If it is a perfect square, integrate by hand. If not, and a calculator is allowed, evaluate numerically.
5. Check: is the answer at least the straight-line distance between the end points? Are the units right?

Common misconceptions

• Forgetting the square root, or squaring after adding: ∫ ((dx/dt)² + (dy/dt)²) dt and ∫ (dx/dt + dy/dt)² dt are not lengths. In Worked example 2 the first one gives about 711, which is far too big.
• Adding the derivatives without squaring. ∫ (dx/dt + dy/dt) dt = [x + y], which only measures the net change in x + y. For the drone it gives 80, the chord length, by coincidence, not the arc length.
• Using x-values or y-values as limits. The integral is in t, so the limits are the starting and finishing values of t.
• Losing the chain-rule factor inside a derivative, such as writing dy/dt = 6 cos(πt/5) instead of (6π/5) cos(πt/5).
• Ignoring retracing. A circle traced twice gives double its circumference. Check the parameter interval.
• Mixing up √(a² + b²) with a + b. √(t² + 2t + 1) simplifies only because it is a perfect square. √(64 + 14.2 cos² u) does not split into 8 + 3.77 cos u.
• Calculator in degree mode for trigonometric components. Calculus uses radians.

Where this leads

The integrand √((dx/dt)² + (dy/dt)²) comes back in Topic 9.6 as the speed of a particle, and this integral becomes the total distance it travels. Next, the Topic 9.4 study guide writes the same curves as vector-valued functions; the arc length integrand is then the size of the derivative vector. If you need to review parametric derivatives first, go back to the Topic 9.2 study guide.

Try the practice questions now, then use the revision notes and the checklist to consolidate.