Calculus 2

Work as an integral

Lifting a sack a fixed height takes force times distance. What does it take when the force changes along the way, or when different parts of the load travel different distances?

You need

The definite integral as a limit of Riemann sumsCalculus 1

Where you are 10 in Applications of integrals

Calculus8

Applications of integrals10

The shell method Volumes with known cross sections

Work as an integral

Accumulation and net change3 Area between curves ▾ Show all 10 (5 more)

When you reach for it

In physics, a constant force F pushing an object a distance d along its line of action does work W = Fd. With F in newtons and d in meters the work is in newton-meters, called joules; with F in pounds and d in feet it is in foot-pounds. Lifting something at a steady speed takes a force equal to its weight, so raising a 20 pound sack 4 feet takes 80 foot-pounds. In metric units weight is mass times g = 9.8 meters per second squared, so a 10 kilogram mass weighs 98 newtons.

The formula W = Fd stops working in two situations, and this lesson handles both with a definite integral:

the force changes as the object moves (a spring pulls harder the farther it is stretched, a leaking bucket gets lighter as it rises, a cable being wound up gets shorter);
different parts of the load travel different distances (water at the bottom of a tank has farther to go than water at the top).

The procedure

Day 1: a force that changes along the way

The integral. An object moves along an axis from x = a to x = b, pushed in the direction of motion by a force F(x) that depends on position, with F continuous on [a, b]. Cut [a, b] into n pieces of width \Delta x and pick a point x_i^* in the i-th piece. Over a short piece the force barely changes, so it is close to the constant F(x_i^*), and the work over that piece is close to F(x_i^*)\,\Delta x. Adding the pieces gives

W \approx \sum_{i=1}^{n} F(x_i^*)\,\Delta x,

a Riemann sum for F (The definite integral as a limit of Riemann sums). As \Delta x \to 0 the error in treating the force as constant on each piece goes to zero, and the sums tend to

W = \int_a^b F(x)\,dx.

This is the definition of work for a varying force along a line, and it agrees with W = Fd when F is constant. Read as a picture, work is the area under the graph of force against position (Figure 1).

Springs. Hooke’s law, an experimental law stated here without proof, says that a spring stretched or compressed x units beyond its natural length pushes back with a force of size kx, where the spring constant k depends on the spring. Holding it there takes force kx, so stretching it from x = a to x = b takes

W = \int_a^b kx\,dx = \frac{k}{2}\left(b^2 - a^2\right).

The variable x is the stretch beyond natural length, not the length of the spring.

Cables and chains. A cable of weight \lambda per unit length hanging straight down is wound up to the top. The work can be set up in two ways, and both give the same integral. Force times distance: after y units are wound, the length still hanging is L - y, so the force needed is \lambda(L - y) and W = \int_0^L \lambda(L - y)\,dy. Each piece lifted its own distance: a short piece at depth x weighs \lambda\,\Delta x and rises x units to the top, so W = \int_0^L \lambda x\,dx. The substitution x = L - y turns one integral into the other.

A load that changes. For a bucket that loses water as it rises, write its weight as a function of its height h; if the leak is given per second, the steady lifting speed converts time into height. The force at height h is the weight of the bucket and water plus the weight of whatever rope is still hanging.

Day 2: pumping a liquid out of a tank

When a tank is emptied by pumping its liquid up to an outlet, a single force-times-distance does not apply: water near the top rises a little, water near the bottom rises a lot. So slice the liquid instead of the path. Put a vertical y-axis through the tank and cut the liquid into thin horizontal layers of thickness \Delta y. A layer at height y with cross-sectional area A(y) has volume close to A(y)\,\Delta y. If the liquid weighs \delta per unit volume, the layer weighs close to \delta A(y)\,\Delta y, and all of it rises about the same distance D(y), from its own height to the outlet. Its work is close to \delta A(y) D(y)\,\Delta y, and the layers add to a Riemann sum whose limit is

W = \int_c^d \delta\,A(y)\,D(y)\,dy,

where c and d are the heights of the bottom and the top of the liquid, not of the tank. The two weight densities used here are \delta = 9800 newtons per cubic meter for water in metric units (density 1000 kilograms per cubic meter times g = 9.8) and \delta = 62.4 pounds per cubic foot in US customary units.

Three things change from tank to tank. The shape gives A(y), usually through similar triangles or the equation of a circle. The outlet gives D(y): if it is at height H, then D(y) = H - y, and an outlet above the top simply makes H larger. The amount of liquid gives the limits. Where to put y = 0 is a free choice; the bottom of the tank, the center of a circle and the rim are all common, and the one that makes A(y) simplest is usually best.

Worked examples

Example 1

Day 1: Tupou’s valve spring

Tupou fits a spring into the float valve of the farm cistern. Its natural length is 25 cm, and a force of 40 N holds it stretched to a length of 30 cm. How much work does it take to stretch it from 30 cm to 40 cm? Use SI units.

Solution

Hooke’s law uses the stretch beyond natural length, in meters. At 30 cm the stretch is 0.05 m, so 40 = k(0.05) and k = 800 newtons per meter. Lengths of 30 cm and 40 cm are stretches of 0.05 m and 0.15 m:

\begin{aligned} W &= \int_{0.05}^{0.15} 800x\,dx = 400x^2\Big|_{0.05}^{0.15} \\ &= 400(0.0225 - 0.0025) \\ &= 8 \text{ joules}. \end{aligned}

For comparison, stretching it from its natural length to 30 cm takes \int_0^{0.05} 800x\,dx = 1 joule. The second stretch is twice as long but takes eight times the work, because the force grows along the way. Figure 1 shows the 8 joules as the area of a trapezoid under the force line.

Figure 1Example 1. The work to stretch the spring from x = 0.05 to x = 0.15 is the area under the force graph, a trapezoid of area 8 joules.

Example 2

Day 1: winding a cable out of a shaft

Tahu’s well shaft holds a 30 m cable of mass 2 kg per meter hanging straight down from a winch at ground level. Use SI units. (a) How much work does it take to wind up the whole cable? (b) How much to wind up just the first 10 m? (c) The cable carries a 50 kg pump at its lower end. How much work does it take to wind up the cable and raise the pump to the top?

Solution

Each meter of cable weighs 2 \times 9.8 = 19.6 newtons.

(a) Slice the cable (Figure 2). The piece at depth x, of length \Delta x, weighs 19.6\,\Delta x newtons and is lifted x meters, so

\begin{aligned} W &= \int_0^{30} 19.6x\,dx = 9.8x^2\Big|_0^{30} \\ &= 8820 \text{ joules}. \end{aligned}

Thinking instead of the force: after y meters are wound, 30 - y meters hang, and \int_0^{30} 19.6(30 - y)\,dy = 19.6\,(900 - 450) = 8820 joules, the same.

(b) Here the force view is the natural one, because the question stops partway. For 0 \le y \le 10 the hanging weight is 19.6(30 - y) newtons:

\begin{aligned} W &= \int_0^{10} 19.6(30 - y)\,dy \\ &= 19.6\,(300 - 50) = 4900 \text{ joules}. \end{aligned}

The first third of the cable takes more than half of the total, because it is wound while the most cable is hanging.

(c) The pump weighs 50 \times 9.8 = 490 newtons, a constant force over 30 meters, so it adds 490 \times 30 = 14{,}700 joules. The total is 8820 + 14{,}700 = 23{,}520 joules.

Figure 2Example 2. The piece at depth x has weight 19.6\,\Delta x newtons and is lifted x meters.

Example 3

Day 1: Kanoa’s leaking bucket

Kanoa raises a bucket from a well whose water surface is 40 ft below the top. Bucket and water weigh 35 lb as the bucket leaves the water. He raises it at a steady 2 feet per second while it leaks 0.15 pounds of water per second, and the rope weighs 0.2 pounds per foot. Find the work to raise the bucket to the top, in US customary units.

Solution

Measure h in feet up from the water. At 2 feet per second the bucket reaches height h after \displaystyle \frac{h}{2} seconds and has lost \displaystyle 0.15 \cdot \frac{h}{2} = 0.075h pounds, so it weighs 35 - 0.075h. The rope still hanging is 40 - h feet long and weighs 0.2(40 - h) pounds. Since the lift is steady, the force equals the total weight:

\begin{aligned} F(h) &= 35 - 0.075h + 0.2(40 - h) \\ &= 43 - 0.275h. \end{aligned}

Then

\begin{aligned} W &= \int_0^{40} (43 - 0.275h)\,dh \\ &= 43h - 0.1375h^2\Big|_0^{40} \\ &= 1720 - 220 = 1500 \text{ foot-pounds}. \end{aligned}

The pieces make sense on their own: the bucket alone takes 1400 - 60 = 1340 foot-pounds, and the rope takes \int_0^{40} 0.2(40 - h)\,dh = 160 foot-pounds. The leak rate in pounds per second matters only through the speed; a bucket raised twice as fast would lose half as much water on the way.

Example 4

Day 2: a cone-shaped cistern, point down

A cistern is a cone with its point down, 4 m deep with a top radius of 2 m, and it is full of water. Find the work to pump all the water out over the rim, in SI units.

Solution

Put y = 0 at the point, so the rim is at y = 4 (Figure 3). By similar triangles the radius at height y is to y as 2 is to 4, so the radius is \displaystyle \frac{y}{2} and the layer at height y has area \displaystyle A(y) = \pi\left(\frac{y}{2}\right)^2 = \frac{\pi y^2}{4}. It rises D(y) = 4 - y meters. With \delta = 9800:

\begin{aligned} W &= \int_0^4 9800 \cdot \frac{\pi y^2}{4}\,(4 - y)\,dy \\ &= 2450\pi \int_0^4 \left(4y^2 - y^3\right)dy \\ &= 2450\pi\left(\frac{256}{3} - 64\right) \\ &= \frac{156{,}800\pi}{3}, \end{aligned}

about 164{,}200 joules.

A check on size: the water’s volume is \displaystyle \frac{1}{3}\pi(2)^2(4) = \frac{16\pi}{3} cubic meters and its weight \displaystyle \frac{156{,}800\pi}{3} newtons, so the answer says the water rises 1 meter on average. Most of a point-down cone’s water is near the top, so a short average lift is reasonable.

Figure 3Example 4. The layer at height y has radius \displaystyle \frac{y}{2} and is lifted 4 - y meters to the rim.

Example 5

Day 2: a cylinder on its side, half full

The farm’s spare water tank is a cylinder lying on its side, with radius 3 ft and length 10 ft, filled with water to the center line. Water is pumped out through an outlet at the very top of the tank. Find the work in US customary units.

Solution

Put y = 0 on the center line of the circular end, so the circle is x^2 + y^2 = 9, the water fills -3 \le y \le 0, and the outlet is at y = 3 (Figure 4). The layer at height y is a thin rectangle 10 feet long and \displaystyle 2\sqrt{9 - y^2} feet wide, and it rises 3 - y feet:

\begin{aligned} W &= \int_{-3}^{0} 62.4 \cdot 10 \cdot 2\sqrt{9 - y^2}\,(3 - y)\,dy \\ &= 3744\int_{-3}^{0} \sqrt{9 - y^2}\,dy \\ &\quad - 1248\int_{-3}^{0} y\sqrt{9 - y^2}\,dy. \end{aligned}

The first integral is the area of a quarter of a circle of radius 3 (Area of a circle), \displaystyle \frac{9\pi}{4}. In the second, u = 9 - y^2 gives du = -2y\,dy, and u runs from 0 to 9:

\begin{aligned} -\int_{-3}^{0} y\sqrt{9 - y^2}\,dy &= \frac{1}{2}\int_0^9 \sqrt{u}\,du \\ &= \frac{1}{2}\cdot\frac{2}{3}\cdot 27 = 9. \end{aligned}

So \displaystyle W = 3744\cdot\frac{9\pi}{4} + 1248\cdot 9, which is 8424\pi + 11{,}232 \approx 37{,}697 foot-pounds.

Figure 4Example 5. With the center at y = 0, the layer at height y is \displaystyle 2\sqrt{9 - y^2} feet wide and is lifted 3 - y feet to the outlet at the top.

Example 6

Day 2: a hemisphere and an outlet above the rim

Kanoa’s farm has a hemispherical cistern of radius 2 m, flat side up, full of water. The pump sends the water to a trough 1.5 m above the rim. Find the work in SI units, and compare it with the work to pump the water just over the rim.

Solution

Here the rim is the natural reference, so measure y as depth below the rim, 0 \le y \le 2. The hemisphere’s cross-section at depth y is a circle of radius \displaystyle \sqrt{4 - y^2}, by the Pythagorean theorem on a radius of the sphere, so A(y) = \pi(4 - y^2). A layer at depth y rises y meters to the rim and 1.5 more to the trough, D(y) = y + 1.5:

\begin{aligned} W &= 9800\pi\int_0^2 \left(4 - y^2\right)(y + 1.5)\,dy \\ &= 9800\pi\int_0^2 \left(4y - y^3 + 6 - 1.5y^2\right)dy \\ &= 9800\pi\,(8 - 4 + 12 - 4) = 117{,}600\pi, \end{aligned}

about 369{,}451 joules. Over the rim alone, D(y) = y and 9800\pi\int_0^2 \left(4y - y^3\right)dy = 39{,}200\pi joules. The difference, 78{,}400\pi, is exactly the water’s whole weight, \displaystyle 9800 \cdot \frac{2}{3}\pi(2)^3 = \frac{156{,}800\pi}{3} newtons, times 1.5 m: once the water is at the rim, every layer travels the same extra distance, which is a constant force times a distance (Volume of solids gives the hemisphere’s volume).

Example 7

Day 2: stopping partway

The cistern of Example 4 is pumped over the rim only until the water is 2 m deep. Find the work, and compare it with the work to empty the cistern.

Solution

Only the water between y = 2 and y = 4 leaves, so only those layers count, and each still rises 4 - y:

\begin{aligned} W &= 2450\pi\int_2^4 \left(4y^2 - y^3\right)dy \\ &= 2450\pi\left(\frac{224}{3} - 60\right) \\ &= \frac{107{,}800\pi}{3}, \end{aligned}

about 112{,}888 joules. The full emptying took \displaystyle 2450\pi\cdot\frac{64}{3}, so this is \displaystyle \frac{44}{64} = \frac{11}{16} of it. By volume, the water removed is \displaystyle \frac{4^3 - 2^3}{4^3} = \frac{7}{8} of the whole. The last eighth of the water takes \displaystyle \frac{5}{16} of the work, because it sits deepest.

Boundaries

W = \int_a^b F(x)\,dx is for a force along the line of motion, continuous on [a, b] (or continuous on pieces, with the integral split, as with a bucket that runs dry). A force at a slant does work only through its part along the path, found by Vector and scalar projection; a curved path needs a line integral.
Work against gravity here means lifting at a steady speed, so the force equals the weight. Speeding up, friction and the losses of a real pump are left out, which makes the answer the least work that can do the job.
Hooke’s law holds only within the spring’s elastic range, and x is measured from the natural length, in the same length unit as k. Centimeters must become meters before the integral when k is in newtons per meter.
Weight as mass times g = 9.8 assumes the object stays near the earth’s surface. Over distances where gravity weakens, the force is a different function of position and the integral uses that function.
In a tank, the limits run over the liquid, not the tank, and D(y) is measured to the outlet. The layer method needs each layer to be at one height, so the layers must be horizontal.
Mixing units breaks every example: one system per problem, with \delta = 9800 newtons per cubic meter or 62.4 pounds per cubic foot to match.

Where this goes next

Work is one of a family of totals built the same way: a quantity known for a constant rate (force times distance, weight times height) becomes an integral when the rate varies, exactly as distance comes from velocity in The net change theorem in context. The layers of a tank are the slices of the The disk method and the The washer method with a weight and a distance attached; the The shell method supplies another way to cut. The same layering, with pressure at depth in place of lifting distance, gives the force of water on a wall. Exercise 22 shows that pumping work equals weight times the distance the liquid’s center of mass rises, the first appearance of moments, which return in the study of centers of mass and later in line integrals of a force along a curved path.

Exercises

Day 1: varying forces and springs

A force F(x) = 3x^2 + 2 newtons moves an object along an axis from x = 0 to x = 4 meters. Find the work.
A force F(x) = \dfrac{10}{(1 + x)^2} pounds moves an object from x = 0 to x = 4 feet. Find the work.
A spring has natural length 20 cm, and a force of 30 N holds it at a length of 26 cm. Find the work to stretch it from 26 cm to 32 cm.
It takes 9 J of work to stretch a spring from its natural length of 30 cm to a length of 50 cm. Find the spring constant, the force that holds it at 50 cm, and the work to stretch it from 50 cm to 60 cm.
A force of 12 lb holds a spring 3 inches beyond its natural length. Find the work, in foot-pounds, to stretch it 6 inches beyond its natural length.

Day 1: cables and buckets

A chain 20 ft long weighing 3 lb per foot hangs from a winch at the top of a well. Find the work to wind it all up.
For the chain of Exercise 6, find the work to wind up only the top 10 ft, and say what fraction of the total that is.
A 50 m cable with mass 1.5 kg per meter hangs down a shaft with an 80 kg load at its end. Find the work to raise the load to the top, winding up the cable.
A bucket of water has mass 12 kg as it leaves the water and is raised 15 m. It leaks steadily, so that its mass falls in proportion to height and is 9 kg at the top. Ignoring the rope, find the work in joules.
Tupou raises a bucket that weighs 25 lb as it leaves the water, 30 ft below the top of the well. She lifts it at 1.5 feet per second, it leaks 0.3 pounds per second, and the rope weighs 0.5 pounds per foot. Find the work.
A 3 lb bucket holds 10 lb of water at the bottom of a 30 ft well and leaks 0.5 pounds per foot of height. The rope’s weight is negligible. Find the work to raise it to the top. (When does it run dry?)

Day 2: pumping

A tank is a cone with its point down, 9 ft deep with a top radius of 3 ft, full of water. Find the work to pump the water out over the rim.
The cone of Exercise 12 is turned point up and filled, and the water is pumped out through an outlet at the point. Find the work, and compare it with Exercise 12.
A vertical cylindrical cistern has radius 1.5 m and depth 3 m and is full of water. Find the work to pump the water out over the top.
The cistern of Exercise 14 holds water 2 m deep, and the water is pumped to an outlet 1 m above the top. Find the work.
A cylindrical tank lying on its side has radius 2 ft and length 8 ft and is full of water. Find the work to pump it out through an outlet at the top.
Tahu’s watering trough is a box 8 ft long, 3 ft wide and 2 ft deep, full of water. Find the work to pump the water over the rim.
A trough 10 ft long has ends that are triangles, point down, 2 ft across the top and 2 ft deep. It is full of water. Find the work to pump the water over the rim.
A spherical tank of radius 1 m is full of water. Find the work to pump the water out through an outlet at the top of the tank.
In the cistern of Exercise 14, how deep is the water when the pump has done half the work of emptying it over the top?

Reasoning

A cable of length L and weight \lambda per unit length hangs straight down. Show that the force view and the slicing view both give \displaystyle W = \frac{\lambda L^2}{2}, and explain why this equals the cable’s weight times the distance its midpoint rises.
A tank holds liquid of weight density \delta from y = c to y = d with cross-sectional area A(y), and it is pumped to an outlet at height H \ge d. Let V = \int_c^d A(y)\,dy and \displaystyle \bar{y} = \frac{1}{V}\int_c^d y\,A(y)\,dy, the height of the liquid’s center of mass. Show that W = \delta V\,(H - \bar{y}), and check it against Exercise 14.
Explain why stretching a spring from 2 cm to 4 cm beyond its natural length takes three times the work of stretching it from 0 to 2 cm, whatever the spring constant.

Answers

72 joules: \int_0^4 \left(3x^2 + 2\right)dx = 64 + 8.
8 foot-pounds: \displaystyle 10\left[-\frac{1}{1 + x}\right]_0^4 = 10\left(1 - \frac{1}{5}\right).
2.7 joules: \displaystyle k = \frac{30}{0.06} = 500, and \int_{0.06}^{0.12} 500x\,dx = 250(0.0144 - 0.0036).
k = 450 newtons per meter, 90 N, 11.25 joules: \displaystyle \frac{k}{2}(0.2)^2 = 9 gives k = 450; the force at stretch 0.2 m is 450(0.2); and 225\left(0.3^2 - 0.2^2\right) = 225(0.05).
6 foot-pounds: 3 inches is 0.25 ft, so k = 48 pounds per foot, and \int_0^{0.5} 48x\,dx = 24(0.25).
600 foot-pounds: \displaystyle \int_0^{20} 3x\,dx = \frac{3}{2}(400).
450 foot-pounds, \displaystyle \frac{3}{4} of the total: \int_0^{10} 3(20 - y)\,dy = 3(200 - 50).
57{,}575 joules: the cable weighs 14.7 N per meter, \int_0^{50} 14.7x\,dx = 18{,}375, and the load adds 80(9.8)(50) = 39{,}200.
1543.5 joules: the mass at height h is 12 - 0.2h, and \int_0^{15} 9.8(12 - 0.2h)\,dh = 9.8(180 - 22.5).
885 foot-pounds: the bucket loses 0.2 pounds per foot, so F(h) = 25 - 0.2h + 0.5(30 - h) = 40 - 0.7h, and \int_0^{30} (40 - 0.7h)\,dh = 1200 - 315.
190 foot-pounds: the water is gone at h = 20. The force is 13 - 0.5h for 0 \le h \le 20 and 3 after, so W = (260 - 100) + 3(10).
3790.8\pi \approx 11{,}909 foot-pounds: the radius at height y is \displaystyle \frac{y}{3}, and \displaystyle 62.4\pi\int_0^9 \frac{y^2}{9}(9 - y)\,dy is \displaystyle \frac{62.4\pi}{9}\left(2187 - \frac{6561}{4}\right).
11{,}372.4\pi \approx 35{,}727 foot-pounds, three times Exercise 12: the radius at height y above the base is \displaystyle \frac{9 - y}{3}, and \displaystyle \frac{62.4\pi}{9}\int_0^9 (9 - y)^3\,dy is \displaystyle \frac{62.4\pi}{9}\cdot\frac{6561}{4}. Point up, most of the water is near the bottom and far from the outlet.
99{,}225\pi \approx 311{,}725 joules: A(y) = 2.25\pi, and 9800(2.25\pi)\int_0^3 (3 - y)\,dy = 22{,}050\pi(4.5).
132{,}300\pi \approx 415{,}633 joules: the outlet is at y = 4, and 22{,}050\pi\int_0^2 (4 - y)\,dy = 22{,}050\pi(6).
3993.6\pi \approx 12{,}546 foot-pounds: \displaystyle 62.4(8)\int_{-2}^{2} 2\sqrt{4 - y^2}\,(2 - y)\,dy. The term with y is odd and gives 0, and \displaystyle \int_{-2}^{2} 2\sqrt{4 - y^2}\,dy is the circle’s area 4\pi, so W = 499.2(2)(4\pi).
2995.2 foot-pounds: 62.4(24)\int_0^2 (2 - y)\,dy = 1497.6(2).
832 foot-pounds: the width at height y is y, so \displaystyle 62.4(10)\int_0^2 y(2 - y)\,dy = 624\left(4 - \frac{8}{3}\right).
\displaystyle \frac{39{,}200\pi}{3} \approx 41{,}050 joules: with y = 0 at the center, A(y) = \pi(1 - y^2) and D(y) = 1 - y, and 9800\pi\int_{-1}^{1} \left(1 - y - y^2 + y^3\right)dy is \displaystyle 9800\pi\left(2 - \frac{2}{3}\right).
\displaystyle 3 - \frac{3\sqrt{2}}{2} \approx 0.879 m: lowering the level to d takes \displaystyle 22{,}050\pi\cdot\frac{(3 - d)^2}{2}, half of 22{,}050\pi(4.5) when (3 - d)^2 = 4.5.
Force view: \displaystyle \int_0^L \lambda(L - y)\,dy = \lambda\left(L^2 - \frac{L^2}{2}\right). Slicing view: \displaystyle \int_0^L \lambda x\,dx = \frac{\lambda L^2}{2}. The cable weighs \lambda L and its midpoint, at depth \displaystyle \frac{L}{2}, rises to the top, and \displaystyle \lambda L\cdot\frac{L}{2} = \frac{\lambda L^2}{2}.
By linearity, W = \int_c^d \delta A(y)(H - y)\,dy splits as \delta H\int_c^d A(y)\,dy - \delta\int_c^d y\,A(y)\,dy, which is \delta HV - \delta V\bar{y} = \delta V(H - \bar{y}). For Exercise 14, V = 6.75\pi and \bar{y} = 1.5, so W = 9800(6.75\pi)(1.5) = 99{,}225\pi.
With x in cm, the work from 0 to 2 is \displaystyle \frac{k}{2}(4) and from 2 to 4 is \displaystyle \frac{k}{2}(16 - 4) = \frac{k}{2}(12), three times as much for every k. The force over the second stretch runs from 2k to 4k, averaging three times the first stretch’s average of k.