Difficulty: Intermediate | Prerequisites: Basic integration, u-substitution
Arc length measures how long a curve is between two points using an integral involving the derivative. Work problems use the same slice-and-integrate idea to compute how much energy is needed to move an object (lift a cable, reel in a fish, pump water). Both rely on setting up small pieces, approximating, summing, and taking the limit to get a definite integral.
Arc length
The distance measured along a curve between two points. Not the straight-line distance, but the length you would get by laying a string along the curve and then stretching it out flat.
Arc length element (ds)
The infinitesimal piece of curve length: ds = sqrt(1 + (dy/dx)^2) dx. Think of it as the hypotenuse of a tiny right triangle with legs dx and dy.
Work (in physics / calculus)
The energy required to move an object through a distance against a force. W = Force x distance when force is constant. When force varies, W = integral of F(x) dx. In simple terms, work is "effort times distance," but when the effort changes as you go, you need an integral.
Six-step process (for work problems)
A structured approach used in this course: (1) slice the interval, (2) choose a sample point, (3) write the work for one slice, (4) sum over all slices, (5) take the limit to get a Riemann sum, (6) write the definite integral. Think of it as the formal version of "chop it up, approximate, add, take the limit."
Variable force
A force that changes depending on position. Examples: the weight of a hanging cable (less cable remains as you pull it up) or a spring force (Hooke's law: F = kx).
For a curve y = f(x) on the interval [a, b]:
L = integral from a to b of sqrt(1 + (dy/dx)^2) dx
The key algebraic trick in exam problems is choosing a function y whose derivative squares nicely so that 1 + (dy/dx)^2 becomes a perfect square. This lets you drop the square root and integrate a polynomial.
y = x^4/4 + 1/(8x^2) on [1, 3]
Rewrite as y = (1/4)x^4 + (1/8)x^(-2).
dy/dx = x^3 - (1/4)x^(-3)
(dy/dx)^2 = x^6 - 1/2 + (1/16)x^(-6)
1 + (dy/dx)^2 = x^6 + 1/2 + (1/16)x^(-6) = (x^3 + (1/4)x^(-3))^2
The square root simplifies to x^3 + (1/4)x^(-3) (positive on [1, 3]).
L = integral from 1 to 3 of (x^3 + (1/4)x^(-3)) dx
= [x^4/4 - (1/8)x^(-2)] evaluated from 1 to 3
= (81/4 - 1/72) - (1/4 - 1/8)
= 181/9
y = x^5/5 + 1/(12x^3) on [1, 2]
Rewrite as y = (1/5)x^5 + (1/12)x^(-3).
dy/dx = x^4 - (1/4)x^(-4)
(dy/dx)^2 = x^8 - 1/2 + (1/16)x^(-8)
1 + (dy/dx)^2 = x^8 + 1/2 + (1/16)x^(-8) = (x^4 + (1/4)x^(-4))^2
The square root simplifies to x^4 + (1/4)x^(-4).
L = integral from 1 to 2 of (x^4 + (1/4)x^(-4)) dx
= [x^5/5 - (1/12)x^(-3)] evaluated from 1 to 2
= (32/5 - 1/96) - (1/5 - 1/12)
= 3011/480
This course uses a structured six-step approach for work problems:
Slice the interval [a, b] into n subintervals of equal width delta x.
Choose a sample point x_k* in each subinterval.
Write the work W_k for lifting/moving the k-th slice: W_k = F_k * d_k (force times distance for that slice).
Sum over all slices: W is approximately the sum of all W_k.
Take the limit as n goes to infinity to get a Riemann sum.
Recognise the Riemann sum as a definite integral and evaluate.
A rescue cable attached to a helicopter weighs 2 lb/ft. A 160-lb man grabs the end and is pulled from the ocean. The helicopter is 40 ft above the water. Find the total work.
Two components of work:
Cable: The force to lift the cable varies because less cable hangs as you pull it up. At position x (measured from the bottom), a slice of cable of width dx weighs 2 dx lb, and it must be lifted a distance of x.
Man: The man weighs a constant 160 lb and is lifted a constant 40 ft.
Work for the cable:
W_cable = integral from 0 to 40 of 2x dx = [x^2] from 0 to 40 = 1600 ft-lb
Work for the man:
W_man = 160 * 40 = 6400 ft-lb
Total: W = 1600 + 6400 = 8000 ft-lb
A fisherman reels in an 11-lb fish located 13 ft below. The fishing line weighs 1 oz per foot (= 1/16 lb per foot). Find the total work.
Two components again:
Line: At position x, a slice of line of width dx weighs (1/16) dx lb and is lifted a distance of x.
Fish: The fish weighs 11 lb and is lifted 13 ft.
Work for the line:
W_line = integral from 0 to 13 of (x/16) dx = [x^2/32] from 0 to 13 = 169/32 approx 5.3 ft-lb
Work for the fish:
W_fish = 11 * 13 = 143 ft-lb
Total: W = 5.3 + 143 = 148.3 ft-lb (rounded to the nearest tenth)
Arc length
L = integral from a to b of sqrt(1 + (dy/dx)^2) dx
Equivalently, if x = g(y): L = integral from c to d of sqrt(1 + (dx/dy)^2) dy
Work (constant force)
W = F * d
Work (variable force)
W = integral from a to b of F(x) dx
Work for lifting a rope/cable/chain of weight w lb/ft
W = integral from 0 to L of w * x dx, where L is the length of the rope
Students often forget the "+1" inside the square root for arc length. The formula is sqrt(1 + (dy/dx)^2), not sqrt((dy/dx)^2). That "+1" accounts for the horizontal component of the curve.
In work problems, students sometimes set up the integral for the variable part (cable/rope) but forget to add the constant part (the object at the end). The total work is the integral for the rope plus the constant force times distance for the object.
A common error is confusing which direction x is measured from. In cable-lifting problems, x is typically measured from the bottom (the end being lifted), so the distance each slice travels equals x. Measuring from the top gives a different (but equivalent) integral with adjusted limits.
When the arc length integrand does not simplify to a perfect square, students panic. On this exam, the functions are always designed so that 1 + (dy/dx)^2 factors neatly. If it does not, check your derivative.
Arc length problems on this exam always feature functions designed so that 1 + (dy/dx)^2 becomes a perfect square. Recognising the pattern (terms of the form x^n and x^(-n)) is the entire difficulty.
Work problems require the six-step process to be shown explicitly for full credit. Skipping to the integral without showing the Riemann sum setup loses marks.
Cable/rope/chain problems are the most common type of work problem on this exam. Know the template: variable-weight component plus constant-weight object.
Arc length answers are often ugly fractions. Do not assume you have made an error just because the answer is something like 181/9 or 3011/480.
Fill in the blank: The arc length formula is L = integral of sqrt(1 + ____) dx. (Answer: (dy/dx)^2)
True or false: In a work problem, if the force is constant, you still need an integral. (False. Constant force means W = F * d, no integral needed.)
Fill in the blank: A cable weighing 3 lb/ft and 20 ft long requires work W = integral from 0 to 20 of ____ dx to be fully lifted. (Answer: 3x)
True or false: If 1 + (dy/dx)^2 does not simplify to a perfect square, you should try a trig substitution. (True in general, but on this exam the function is chosen so that it does simplify.)
Fill in the blank: The total work to lift a 50-lb bucket on a 30-ft rope weighing 2 lb/ft is ____ + ____. (Answer: integral from 0 to 30 of 2x dx + 50 * 30)
Q: Find the arc length of y = x^3/3 + 1/(4x) on [1, 2].
A: dy/dx = x^2 - 1/(4x^2). Then (dy/dx)^2 = x^4 - 1/2 + 1/(16x^4). So 1 + (dy/dx)^2 = (x^2 + 1/(4x^2))^2. The integral becomes integral from 1 to 2 of (x^2 + 1/(4x^2)) dx = [x^3/3 - 1/(4x)] from 1 to 2 = (8/3 - 1/8) - (1/3 - 1/4) = 59/24.
Q: A 10-ft chain weighing 5 lb/ft hangs from a ledge. How much work is needed to pull the entire chain to the top?
A: W = integral from 0 to 10 of 5x dx = 5[x^2/2] from 0 to 10 = 250 ft-lb.
Q: In Problem 8A, why is the work for the man simply 160 * 40 rather than an integral?
A: The man's weight is constant (160 lb) and he travels a fixed distance (40 ft). When force is constant, W = F * d. Only the cable requires an integral because its effective weight changes as it is pulled up.
Q: Why does the arc length formula contain sqrt(1 + (dy/dx)^2) rather than just dy/dx?
A: Arc length is the total distance along the curve. Each infinitesimal piece has both a horizontal component (dx) and a vertical component (dy). By the Pythagorean theorem, ds = sqrt(dx^2 + dy^2) = sqrt(1 + (dy/dx)^2) dx. The "+1" captures the horizontal contribution.
Arc length and work both use the same core idea as volumes of revolution: slice a continuous quantity into small pieces, approximate each piece, sum, and take the limit. Surface area of revolution (covered later in the course) extends the arc length formula by rotating the curve, combining arc length with volumes of revolution. Work problems connect to centre of mass and moments, which also slice a physical quantity and integrate.
Related Terms / Search Tags
arc length formula, arc length integral, work integral, variable force, six-step process, Riemann sum, cable work, rope lifting, fishing line work, rescue cable, calculus applications, M 408D, 2414 exam, UT Austin calculus, integral applications, sqrt(1 + dy/dx squared)