Difficulty: Intermediate | Prerequisites: Differentiation (power rule, chain rule), definite integrals, square roots in integrands.
TL;DR: Arc length measures how long a curve is between two points. The formula involves integrating √(1 + (dy/dx)²). The hardest part is usually simplifying the expression under the square root so the integral is doable by hand.
Arc length is another application of the "slice and sum" philosophy that runs through Calculus II. Instead of summing areas (for area problems) or volumes (for solids of revolution), you are summing tiny straight-line segments along a curve to approximate its total length. This is the mathematical version of laying a piece of string along a curve and then measuring the string. Engineers use arc length when calculating the length of cable needed for a suspension bridge, the distance a particle travels along a path, or the perimeter of a cam profile. The prerequisite is comfort with differentiation and with setting up definite integrals.
Arc length
The total distance along a curve between two points. Not the straight-line distance between the endpoints, but the actual path length.
Arc length element (ds)
The infinitesimal piece of length along the curve. In terms of x: ds = √(1 + (dy/dx)²) dx. In terms of y: ds = √(1 + (dx/dy)²) dy. Think of it as the hypotenuse of a tiny right triangle with legs dx and dy.
Smooth curve
A curve whose derivative is continuous on the interval. The arc length formula requires this; if the curve has a cusp or corner, you must split the integral there.
The arc length formula (in x)
L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dxThe arc length formula (in y)
When the curve is more naturally x = g(y):
L = \int_c^d \sqrt{1 + \left(\frac{dx}{dy}\right)^2}\, dyThe key skill: simplifying under the radical
Most arc length problems are designed so that (dy/dx)² simplifies neatly. Often the expression 1 + (dy/dx)² turns into a perfect square, making the square root collapse. If it does not simplify, the integral is likely not solvable by hand (and would not appear on this exam).
Example approach (Problem 8): x = (2/3)(y - 1)^(3/2), from y = 1 to y = 4. (Answer: 14/3)
Differentiate: dx/dy = (y - 1)^(1/2).
Compute (dx/dy)² = y - 1.
The integrand becomes √(1 + y - 1) = √y.
L = ∫ from 1 to 4 of √y dy = (2/3)y^(3/2) evaluated from 1 to 4 = (2/3)(8 - 1) = 14/3.
Notice how the expression under the radical simplified to a perfect form. This is by design.
Example approach (Problem 9): y = (1/6)x³ + (1/2)x^(-1), from x = 1 to x = 2. (Answer: 17/12)
Differentiate: dy/dx = (1/2)x² - (1/2)x^(-2).
Compute (dy/dx)² = (1/4)x⁴ - (1/2) + (1/4)x^(-4).
Then 1 + (dy/dx)² = 1 + (1/4)x⁴ - (1/2) + (1/4)x^(-4) = (1/4)x⁴ + (1/2) + (1/4)x^(-4) = [(1/2)x² + (1/2)x^(-2)]².
The square root collapses: √(...) = (1/2)x² + (1/2)x^(-2).
L = ∫ from 1 to 2 of [(1/2)x² + (1/2)x^(-2)] dx = [(1/6)x³ - (1/2)x^(-1)] from 1 to 2 = (8/6 - 1/4) - (1/6 - 1/2) = 17/12.
The pattern: when the function contains both a power of x and its reciprocal, the cross term in (dy/dx)² often produces -1/2, which cancels the +1 in the formula, leaving a perfect square.
Forgetting to square the derivative before adding 1. The formula is √(1 + (dy/dx)²), not √(1 + dy/dx). The squaring is essential.
Confusing arc length with displacement. Arc length is the total distance along the curve, not the net change in position. A curve that doubles back on itself has a longer arc length than the straight-line distance between its endpoints.
Not recognising the perfect-square pattern. Many students expand (dy/dx)² correctly but then fail to see that 1 + (dy/dx)² is itself a perfect square. Look for the pattern a² + 2ab + b² = (a + b)².
⚠️ Arc length problems on this exam are designed to simplify. If your expression under the radical does not collapse, recheck your derivative.
⚠️ Problem 8 uses x as a function of y. Be ready to use the dy version of the formula without hesitation.
⚠️ The algebra of completing the square under the radical is where marks are lost. Show every step.
⚠️ Only two problems on the review cover arc length (8 and 9), but the section is explicitly listed on the exam syllabus. Expect at least one arc length question.
Fill in the blank: The arc length formula in terms of x is L = ∫ √(1 + ______) dx. (dy/dx)².
True or false: If y = x², the arc length from x = 0 to x = 1 is simply ∫ from 0 to 1 of x² dx. False. You need √(1 + 4x²) dx.
True or false: Arc length problems on exams are typically chosen so the radical simplifies to a nice form. True.
Fill in the blank: When the curve is given as x = g(y), the integrand becomes √(1 + (______)²) dy. dx/dy.
Q: (Problem 8) Determine the length of x = (2/3)(y - 1)^(3/2) between y = 1 and y = 4. (Answer: 14/3)
A: dx/dy = (y - 1)^(1/2). Then 1 + (dx/dy)² = 1 + (y - 1) = y. So L = ∫ from 1 to 4 of √y dy = (2/3)y^(3/2) from 1 to 4 = (2/3)(8) - (2/3)(1) = 14/3.
Q: (Problem 9) Find the arc length of y = (1/6)x³ + (1/2)x^(-1) from x = 1 to x = 2. (Answer: 17/12)
A: dy/dx = (1/2)x² - (1/2)x^(-2). Square it: (1/4)x⁴ - (1/2) + (1/4)x^(-4). Add 1: (1/4)x⁴ + (1/2) + (1/4)x^(-4) = [(1/2)x² + (1/2)x^(-2)]². Take the root: (1/2)x² + (1/2)x^(-2). Integrate from 1 to 2: [(1/6)x³ - (1/2)x^(-1)] from 1 to 2 = (4/3 - 1/4) - (1/6 - 1/2) = 13/12 + 4/12 = 17/12.
Q: Write the arc length integral for y = sin(x) from x = 0 to x = π. (Do not evaluate.)
A: L = ∫ from 0 to π of √(1 + cos²(x)) dx. This integral does not have a closed-form antiderivative, which is why exam problems are chosen to simplify.
Arc length is closely related to surface area of revolution (often covered nearby in the textbook), which uses the same ds element multiplied by 2πr. The idea of summing infinitesimal lengths along a curve extends to line integrals in multivariable calculus and vector calculus, where you integrate a function along a path.
The Pythagorean theorem underlies the whole derivation: each tiny segment of arc is approximated as the hypotenuse √(dx² + dy²).
Tags: arc length, arc length formula, curve length, ds, length of a curve, Pythagorean theorem, Calculus II, Calc 2, NYU, section 8.2, arc length integral, smooth curve