Source: PS3 worksheet
Tags: arc length, catenary, cosh, sinh, midpoint rule, numerical approximation, cable length, Calc 2
Difficulty: Intermediate to Advanced | Prerequisites: Calc I derivatives, integration, hyperbolic functions (cosh, sinh), numerical integration methods (midpoint rule)
Arc length is the measurement of how long a curve is. It appears everywhere in engineering: the length of a cable strung between poles, the amount of material needed to form a corrugated sheet, the distance along a road or track. The arc length formula involves an integral of √(1 + (dy/dx)²), which is often difficult to evaluate exactly, hence the midpoint rule in Q5. The catenary (Q6) is the specific curve a hanging cable forms under its own weight, and it has the pleasant property that its arc length integral simplifies cleanly via hyperbolic identities.
Arc length = ∫√(1 + (y')²) dx. For a sinusoidal roof sheet, you approximate this integral numerically. For a catenary y = C + A cosh(x/A), the integral simplifies to 2A sinh(B/A) because the hyperbolic identity 1 + sinh² = cosh² collapses the square root.
Arc length
The total length measured along a curve between two points. For y = f(x) from x = a to x = b, the formula is L = ∫ₐᵇ √(1 + (f'(x))²) dx.
Catenary
The curve formed by a uniform, flexible cable hanging under its own weight between two supports. Its equation is y = C + A cosh(x/A), where A is related to the cable's weight per unit length and the horizontal tension.
cosh (hyperbolic cosine)
cosh(u) = (eᵘ + e⁻ᵘ)/2. Its derivative is sinh(u). The identity cosh²(u) = 1 + sinh²(u) is key for arc length calculations.
sinh (hyperbolic sine)
sinh(u) = (eᵘ - e⁻ᵘ)/2. Its derivative is cosh(u). Think of it as the odd counterpart to cosh.
Midpoint rule
A numerical integration method: divide [a, b] into n equal subintervals, evaluate the function at each subinterval's midpoint, and multiply by the subinterval width. More accurate than left or right Riemann sums for the same n.
The problem
A roofing sheet is moulded into the shape y = sin(πx/9). The finished sheet is 22.5 inches wide, meaning x runs from 0 to 22.5. You need the arc length of the curve over this interval, which tells you how wide the flat sheet must be before moulding.
Setting up the arc length integral
First, find dy/dx:
y = sin(πx/9)
dy/dx = (π/9) cos(πx/9)
The arc length formula gives:
L = \int_0^{22.5} \sqrt{1 + \left(\frac{\pi}{9}\cos\frac{\pi x}{9}\right)^2}\, dx = \int_0^{22.5} \sqrt{1 + \frac{\pi^2}{81}\cos^2\frac{\pi x}{9}}\, dxThis integral has no closed-form antiderivative (it is an elliptic integral), so you approximate numerically.
Applying the midpoint rule with n = 5
The interval [0, 22.5] is divided into 5 subintervals, each of width Δx = 22.5/5 = 4.5.
The midpoints are: x₁ = 2.25, x₂ = 6.75, x₃ = 11.25, x₄ = 15.75, x₅ = 20.25.
Let f(x) = √(1 + (π²/81) cos²(πx/9)). Evaluate at each midpoint:
f(2.25) = √(1 + (π²/81) cos²(π(2.25)/9)) = √(1 + (π²/81) cos²(π/4))
cos(π/4) = √2/2, so cos²(π/4) = 1/2
f(2.25) = √(1 + π²/162) ≈ √(1.0610) ≈ 1.0297
f(6.75) = √(1 + (π²/81) cos²(3π/4))
cos²(3π/4) = 1/2
f(6.75) ≈ 1.0297
f(11.25) = √(1 + (π²/81) cos²(5π/4))
cos²(5π/4) = 1/2
f(11.25) ≈ 1.0297
f(15.75) = √(1 + (π²/81) cos²(7π/4))
cos²(7π/4) = 1/2
f(15.75) ≈ 1.0297
f(20.25) = √(1 + (π²/81) cos²(9π/4))
cos²(9π/4) = 1/2
f(20.25) ≈ 1.0297
Result
L ≈ Δx × [f(2.25) + f(6.75) + f(11.25) + f(15.75) + f(20.25)]
L ≈ 4.5 × 5 × 1.0297 ≈ 4.5 × 5.1484 ≈ 23.1680 inches
So the flat sheet should be approximately 23.1680 inches wide (to four decimal places, as requested).
The curve
The catenary is y = C + A cosh(x/A), with poles at x = -B and x = +B.
Finding dy/dx
dy/dx = A × (1/A) × sinh(x/A) = sinh(x/A)
Setting up the arc length integral
L = \int_{-B}^{B} \sqrt{1 + \sinh^2\left(\frac{x}{A}\right)}\, dxSimplifying with the hyperbolic identity
The key identity is: 1 + sinh²(u) = cosh²(u). Therefore:
\sqrt{1 + \sinh^2\left(\frac{x}{A}\right)} = \cosh\left(\frac{x}{A}\right)The integral becomes:
L = \int_{-B}^{B} \cosh\left(\frac{x}{A}\right) dxEvaluating
The antiderivative of cosh(x/A) is A sinh(x/A):
L = \left[A\sinh\left(\frac{x}{A}\right)\right]_{-B}^{B} = A\sinh\left(\frac{B}{A}\right) - A\sinh\left(\frac{-B}{A}\right)Since sinh is an odd function, sinh(-B/A) = -sinh(B/A):
L = A\sinh\left(\frac{B}{A}\right) + A\sinh\left(\frac{B}{A}\right) = 2A\sinh\left(\frac{B}{A}\right)Result
The cable length is L = 2A sinh(B/A). Notice that C does not appear in the answer: the cable's length depends only on the span (2B) and the parameter A, not on the height C above the ground.
Given information
Poles are 80 feet apart, so 2B = 80, meaning B = 40.
Cable length L = 81 feet.
Lowest point of the cable is 25 feet above the ground.
Step 1: Find A
From Q6a, L = 2A sinh(B/A), so:
81 = 2A sinh(40/A)
This equation cannot be solved algebraically. You need to solve it numerically (Desmos, a graphing calculator, or Newton's method). Plot y = 2x sinh(40/x) and find where it equals 81.
Numerically, A ≈ 79.8316 (you should verify this with your own tool).
Step 2: Find C
The lowest point of the catenary y = C + A cosh(x/A) occurs at x = 0 (the midpoint between the poles), where cosh(0) = 1:
y_min = C + A
We need y_min = 25:
25 = C + A, so C = 25 - A ≈ 25 - 79.8316 ≈ -54.83
Step 3: Find the pole height
The poles are at x = ±40. The height of the cable at the poles is:
y(40) = C + A cosh(40/A)
y(40) ≈ -54.83 + 79.8316 × cosh(40/79.8316)
cosh(40/79.8316) = cosh(0.5010) ≈ 1.1272
y(40) ≈ -54.83 + 79.8316 × 1.1272 ≈ -54.83 + 89.98 ≈ 35.15
Result
The poles must be approximately 35.15 feet tall. (Exact value depends on precision of the numerical solve for A; verify with Desmos to two decimal places as the problem requests.)
Arc length (general)
L = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\, dxCatenary equation
y = C + A\cosh\left(\frac{x}{A}\right)Catenary arc length
L = 2A\sinh\left(\frac{B}{A}\right)Key hyperbolic identity
1 + \sinh^2(u) = \cosh^2(u)Midpoint rule
\int_a^b f(x)\, dx \approx \Delta x \sum_{i=1}^{n} f(\bar{x}_i)where Δx = (b - a)/n and each x̄ᵢ is the midpoint of the i-th subinterval.
The catenary curve describes the shape of power lines, suspension bridge cables (approximately, when the cable's own weight dominates), and chains. Engineers use the catenary equation to determine how much cable to order and how tall the support structures need to be.
The sinusoidal arc length problem models real manufacturing: knowing how much flat material becomes a corrugated sheet of a given width. This calculation determines material costs.
Students frequently confuse the catenary with a parabola. A hanging cable is a catenary (cosh), not a parabola. A parabola results only when the load is distributed uniformly along the horizontal (like the deck weight of a suspension bridge), not along the curve itself.
Forgetting the identity 1 + sinh² = cosh² and leaving the integral in unsimplified form. This identity is the whole reason the catenary arc length works out neatly.
In the midpoint rule, students sometimes use the endpoints instead of the midpoints of each subinterval. The midpoint is (xᵢ₋₁ + xᵢ)/2, not xᵢ.
In the numerical catenary problem, students sometimes forget that C can be negative. The constant C shifts the curve vertically, and when A is large and the minimum height is modest, C will be negative.
⚠️ The arc length formula is heavily tested. Memorise L = ∫√(1 + (y')²) dx and practise setting it up from various functions.
⚠️ The hyperbolic identity 1 + sinh² = cosh² is the single most useful identity for catenary problems. It appears in exams regularly.
⚠️ Midpoint rule calculations are often tested as standalone problems. Show all five (or n) function evaluations for full marks.
True or false: The arc length formula involves √(1 + (dy/dx)²).
True.
Fill in the blank: For a catenary y = C + A cosh(x/A), the derivative dy/dx = ______.
sinh(x/A).
True or false: The midpoint rule uses the function value at the left endpoint of each subinterval.
False. It uses the midpoint.
Fill in the blank: 1 + sinh²(u) = ______.
cosh²(u).
True or false: A hanging cable forms a parabola.
False. It forms a catenary (cosh curve).
Q: Write the arc length integral for y = sin(πx/9) from x = 0 to x = 22.5.
A: L = ∫₀²²·⁵ √(1 + (π/9)² cos²(πx/9)) dx.
Q: Why can this integral not be evaluated in closed form?
A: It is an elliptic integral. There is no elementary antiderivative for √(1 + k² cos²(u)) in general.
Q: Derive the arc length of the catenary y = C + A cosh(x/A) from x = -B to x = B.
A: dy/dx = sinh(x/A). Then L = ∫ √(1 + sinh²(x/A)) dx = ∫ cosh(x/A) dx = [A sinh(x/A)] from -B to B = 2A sinh(B/A).
Q: If two poles are 80 feet apart and the cable is 81 feet long, what equation determines the parameter A?
A: 81 = 2A sinh(40/A). This must be solved numerically.
Q: The lowest point of the catenary is at x = 0. Why?
A: cosh(x/A) has its minimum at x = 0, where cosh(0) = 1. Since A > 0, the catenary's minimum height is C + A, occurring at x = 0.
Arc length connects directly to surface area of revolution (which multiplies the arc length element by a circumference factor). The midpoint rule connects to the broader family of numerical integration methods (trapezoidal rule, Simpson's rule) covered later in Calc II. The catenary also appears in differential equations courses as a classic example solved via the Euler-Lagrange equation.
arc length formula, arc length integral, catenary curve, catenary equation, cosh, sinh, hyperbolic cosine, hyperbolic sine, hanging cable, power line sag, midpoint rule, numerical integration, elliptic integral, sinusoidal arc length, corrugated metal sheet, Calc 2 NYU, Calculus II for Engineers