The derivative as a rate of change
A tank’s volume is changing at -40 when t = 40. Minus forty what, and what does that number actually promise about the water?
Where you are 13 in Derivatives
When you reach for it
The derivative f'(a) was built as the slope of a tangent line and as an instantaneous velocity, in The derivative at a point. Most functions met outside a calculus course are neither: they give dollars as a function of carpets, liters as a function of minutes, meters of depth as a function of days. Their derivatives are still rates, and the question is what a rate says about the situation.
Reach for this lesson whenever a derivative comes with a story and someone has to say, in a sentence with units, what is happening and what is not. Part 1 (day 1) reads a derivative in words. Part 2 (day 2) covers marginal cost and revenue, rates in science, and rates given in a table or on a graph.
The procedure
Part 1: a derivative in words
Units. The derivative is the limit of a difference quotient,
The top is a change in f, measured in the units of f; the bottom is a change in x, measured in the units of x. Units divide like factors (the same bookkeeping as conversion factors), so every quotient carries the units of f per unit of x, and so does their limit. If V(t) is in liters and t in minutes, V'(t) is in liters per minute. When the input already has compound units, keep them whole: if T(v) is a time in hours and v a flow in liters per minute, T'(v) is in hours per (liter per minute). Simplifying that unit would hide what it measures.
The sentence. A full reading names four things: the input value with its unit, the quantity that is changing, the rate with its units, and the direction. For V'(40) = -40: “Forty minutes after the valve opens, the volume of water in the tank is decreasing at 40 liters per minute.”
Sign. If f'(a) > 0, then f is increasing at that instant; if f'(a) < 0, it is decreasing; if f'(a) = 0, it is momentarily neither. The sign is about the quantity f measures, so the reading depends on what f is. A well’s water measured as depth below the ground has a positive derivative while the water falls.
What it does not say. f'(a) is not the value f(a), and not the amount f changes over the next unit of x. It is a rate at one instant. It does give an estimate: for a small change \Delta x,
This holds because the difference quotient \displaystyle \frac{f(a + \Delta x) - f(a)}{\Delta x} is close to its limit f'(a) when \Delta x is small. The units agree: liters per minute times minutes is liters. Bounding the error belongs to linear approximation and differentials; here it is enough that the estimate is good for small \Delta x and can be poor for large \Delta x (Example 2).
A value and a rate together. f(a) says where the quantity is; f'(a) says how fast and which way it is moving. Together they give
the height of the tangent line at a + \Delta x.
Readings that run backward. Two kinds. The first turns a sentence into notation: “on day 20 the storeroom holds 140 kilograms of wool and is losing 4 kilograms a day” is W(20) = 140 and W'(20) = -4. The second reads the inverse function. If time is a function of volume, t = T(V), then T'(V) is in minutes per liter, and at matching points it is the reciprocal of V'(t): losing 40 liters per minute is the same fact as \displaystyle -\frac{1}{40} minute per liter. That reciprocal law is proved in derivatives of inverse functions; Example 3 confirms it in one case by computing both sides.
Part 2: marginal rates, science, tables and graphs
Marginal cost and revenue. Let C(q) be the cost in dollars of producing q items. Economists call C'(q) the marginal cost at production level q; its units are dollars per item. With \Delta q = 1 the estimate from Part 1 reads
The left side is the exact cost of the (q+1)st item, and it equals the average rate of change of C over [q, q + 1], the slope of a secant. When one item is a small step, the graph is nearly straight over that step and the secant slope is close to the tangent slope. So marginal cost is read as “approximately the cost of one more item”: close to it, not equal.
In the same way, if R(q) is the revenue from selling q items, the marginal revenue R'(q) approximates the revenue from one more sale. Profit is P(q) = R(q) - C(q), and by the difference rule from the basic differentiation rules the marginal profit is P'(q) = R'(q) - C'(q). If R'(q) > C'(q), one more item raises profit; if R'(q) < C'(q), it lowers it.
Rates in science. Every science names derivatives of its own, all with time as the input: position in meters gives velocity in meters per second; velocity gives acceleration in meters per second per second; temperature gives a rate of cooling in degrees Celsius per minute; volume gives a flow rate in liters per hour; electric charge in coulombs gives current in coulombs per second (amperes). Velocity carries a sign that says direction along the chosen axis; speed is its absolute value. A rate of a rate, such as acceleration, is a Second and higher derivatives.
Tables and graphs. A table may list values of f' next to values of f; each entry is read exactly as in Part 1. On a graph, f'(a) is the slope of the tangent line at a, so two points on that tangent give it, in units of the vertical axis per unit of the horizontal axis. Estimating f' when only values of f are listed is a separate job, done in The derivative as a function.
Worked examples
Day 1: units and sign in three statements
Zarlasht runs a carpet-weaving cooperative; Wahid keeps the village well, which draws on an aquifer. Write a sentence with units for each statement, and say what the sign means. (a) W(d) is the wool in the storeroom, in kilograms, d days after a delivery, and W'(12) = -3.5. (b) D(t) is the depth of the water in the well below ground level, in meters, t days after the dry season begins, and D'(5) = 0.3. (c) P(n) is the price in dollars of a carpet woven at n knots per square centimeter, and P'(30) = 18.
Solution(a) Kilograms per day. Twelve days after the delivery, the wool in the storeroom is decreasing at 3.5 kilograms per day. It does not say 3.5 kilograms are left; that amount is W(12), not given.
(b) The units are meters per day. Five days into the dry season, the depth of the water below ground is increasing at 0.3 meters per day. The sign is positive, but the water is going down: D measures distance below the ground, so a growing D means a falling water surface.
(c) The input has compound units, so the rate is in dollars per (knot per square centimeter). For carpets woven at 30 knots per square centimeter, the price is increasing at 18 dollars for each additional knot per square centimeter of density. A carpet at 31 knots per square centimeter would sell for about 18 dollars more.
Day 1: a value, a rate, and an estimate
Wahid opens the outlet valve of a full storage tank. The tank holds V(t) = 0.25(120 - t)^2 liters t minutes later, for 0 \le t \le 120. Find V(40) and V'(40) and say what they mean together. Estimate V(42) from them and compare with the true value. Then do the same for V(70).
SolutionExpanding by the distributive property, V(t) = 3600 - 60t + 0.25t^2, so by the power rule and the sum rule
Then V(40) = 0.25(80)^2 = 1600 and V'(40) = -60 + 20 = -40. Forty minutes after the valve opens the tank holds 1600 liters, and the volume is decreasing at 40 liters per minute.
For two more minutes, \Delta t = 2, the estimate is
The true value is V(42) = 0.25(78)^2 = 1521 liters, one liter off. For \Delta t = 30 the same reasoning gives 1600 - 1200 = 400 liters, but V(70) = 0.25(50)^2 = 625. The rate -40 holds at t = 40 only; the flow keeps slowing (V'(t) rises toward 0), so assuming 40 liters a minute for half an hour overcounts the loss by 225 liters.
Day 1: reading backward
(a) Write in notation: “Twenty days after the delivery the storeroom holds 140 kilograms of wool and is losing it at 4 kilograms a day.” (b) For Wahid’s tank in Example 2, let T(V) be the time in minutes at which the tank holds V liters. Find T'(1600), give its units, and say what it means.
Solution(a) The value is W(20) = 140. “Losing” means decreasing, so the rate is negative: W'(20) = -4 kilograms per day.
(b) Solve V = 0.25(120 - t)^2 for t. On 0 \le t \le 120 the quantity 120 - t is not negative, so taking square roots gives \displaystyle \sqrt{V} = 0.5(120 - t), and
By the power rule, \displaystyle T'(V) = -2 \cdot \tfrac{1}{2}V^{-1/2} = -\dfrac{1}{\sqrt{V}}, so T'(1600) = -\dfrac{1}{40} = -0.025 minute per liter. When the tank holds 1600 liters, each liter less is reached about 0.025 minute later; the sign is negative because time grows as volume shrinks. This is the reciprocal of V'(40) = -40 at the matching point: the same fact, read backward.
Day 2: marginal cost and the next carpet
The cost of weaving q carpets in a month at Zarlasht’s cooperative is C(q) = 2400 + 180q - 1.5q^2 + 0.02q^3 dollars. Find C'(20), say what it means, and compare it with the exact cost of the 21st carpet. What does the 2400 contribute to the marginal cost?
SolutionTerm by term, C'(q) = 180 - 3q + 0.06q^2 dollars per carpet, so
At 20 carpets a month, cost is rising at 144 dollars per carpet, so the 21st carpet should cost about 144 dollars. Exactly,
so the 21st carpet costs 143.72 dollars, 28 cents from the marginal cost: over one carpet the graph of C is so nearly straight that the secant over [20, 21] and the tangent at 20 almost agree.
The 2400 dollars is a fixed cost, paid even when no carpet is made (rent and looms, say). Its derivative is 0: fixed costs do not change the cost of one more carpet.
Day 2: marginal revenue and marginal profit
A trader buys the cooperative’s carpets at 400 - 2q dollars each when the cooperative offers q carpets in a month, so the revenue is R(q) = q(400 - 2q) = 400q - 2q^2 dollars. With the cost from Example 4, find the marginal revenue and marginal profit at q = 30 and at q = 60, and say whether weaving one more carpet pays at each level.
SolutionR'(q) = 400 - 4q dollars per carpet. At q = 30, R'(30) = 280 and C'(30) = 180 - 90 + 54 = 144, so the marginal profit is
A 31st carpet adds about 280 dollars of revenue and 144 of cost, so it raises profit by about 136 dollars. Notice that R'(30) = 280 is less than the price 400 - 60 = 340: the extra carpet sells for about 340 dollars, but offering it lowers the price of all 30 others by 2 dollars each, about 60 dollars in all.
At q = 60, R'(60) = 160 and C'(60) = 180 - 180 + 216 = 216, so P'(60) = -56. A 61st carpet costs more to make than it brings in and lowers profit by about 56 dollars. Somewhere between 30 and 60 the marginal profit passes through 0; finding where profit is largest is optimization.
Day 2: velocity and a cooling dye bath
(a) Mursal lowers a bucket into the well and draws it back up. Its height above the water is h(t) = 0.5t^2 - 6t + 20 meters t seconds after she starts, for 0 \le t \le 10. Find its velocity and acceleration, and describe the motion at t = 2, t = 6 and t = 10. (b) At the cooperative, wool is dyed in a vat whose temperature is H(t) = 95 - 4t + 0.05t^2 degrees Celsius t minutes after the heat is turned off, for 0 \le t \le 40. Find and interpret H'(10).
Solution(a) Velocity is the derivative of position: v(t) = h'(t) = t - 6 meters per second, positive meaning up. At t = 2, v(2) = -4: the bucket is h(2) = 10 meters above the water, moving down at a speed of 4 meters per second. At t = 6, v(6) = 0: the bucket is at its lowest, h(6) = 2 meters above the water, and is momentarily at rest as it turns around. At t = 10, v(10) = 4: it is back at h(10) = 10 meters and rising at 4 meters per second. Acceleration is the derivative of velocity, a(t) = v'(t) = 1 meter per second per second: the upward velocity grows by 1 meter per second every second, throughout. Reading stops and turns from the sign of velocity is the subject of Position, velocity and acceleration.
(b) H'(t) = -4 + 0.1t, so H'(10) = -3 degrees Celsius per minute. Ten minutes after the heat is turned off, the dye bath is cooling at 3 degrees Celsius per minute; its temperature then is H(10) = 95 - 40 + 5 = 60 degrees Celsius.
Day 2: rates given in a table
A pump fills a storage tank from Wahid’s well while villagers draw from it. The table gives the volume V(t) in liters and its rate V'(t) in liters per hour, t hours after 6 a.m.
| t | V(t) | V'(t) |
|---|---|---|
| 0 | 1800 | 500 |
| 2 | 2600 | 300 |
| 4 | 3000 | 0 |
| 6 | 2800 | -250 |
| 8 | 2100 | -400 |
(a) Interpret V(4) and V'(4) together. (b) At which listed time is the tank filling fastest, and at which is it emptying fastest? (c) Estimate V(6.5). (d) Find the average rate of change of V over [2, 6], and explain why it is not one of the listed rates.
Solution(a) At 10 a.m. the tank holds 3000 liters, and the volume is momentarily neither increasing nor decreasing: inflow and outflow balance at that instant.
(b) Filling fastest at t = 0, where V'(0) = 500 liters per hour is the largest positive rate listed. Emptying fastest at t = 8: V'(8) = -400 is the most negative rate, so the volume is falling at 400 liters per hour. “Fastest emptying” compares sizes of negative rates, not the rates themselves.
(c) From t = 6, with \Delta t = 0.5,
(d) The average rate is
in liters per hour.
It is the net change over four hours spread evenly; the tank rose to 3000 liters and fell back, and 50 describes no listed instant. An average rate belongs to an interval, a derivative to an instant.
Boundaries
- The rate holds at one input. f'(a) describes f at a only. Using it over a stretch of inputs assumes the rate stays about the same, which is fair for small \Delta x and can fail badly for large \Delta x (Example 2).
- Marginal cost approximates the next item only when one item is a small step. If q counts hundreds of carpets, C'(q) approximates the cost of one more hundred, and a graph that bends sharply within a single step makes the secant and tangent disagree. A cost defined only at whole numbers has a derivative only after it is modeled by a smooth function.
- The sign describes the quantity measured. A positive derivative of depth below ground means falling water. Read the sign against the definition of f, never against the words “up” and “down” alone.
- The derivative must exist, and the model must hold. At a corner there is no single rate to read, and a formula valid on 0 \le t \le 120 says nothing about the tank outside that domain.
Where this goes next
Linear approximation makes f(a) + f'(a)\,\Delta x precise and bounds its error. Related rates connect two rates through an equation, such as the falling depth and shrinking volume of one tank. Optimization finds where marginal profit is zero, and the net change theorem totals a rate in liters per hour back into liters.
Exercises
Day 1: units, sign, value and rate
For Exercises 1 to 4, give the units of the derivative and write a sentence that says what the statement means.
- A(t) is the area of carpet the cooperative has finished, in square meters, t days into the month; A'(10) = 1.5.
- D(t) is the depth of Wahid’s well water below ground, in meters, t days after the rains stop; D'(30) = 0.12. Is the water rising or falling?
- G(w) is the weight in kilograms of a carpet of fixed length and width w meters; G'(2) = 9.5.
- T(v) is the number of hours a tank takes to empty when the outflow is v liters per minute; T'(50) = -0.4. Also give the change in minutes.
- Write in notation, with d(t) the depth of the water below ground in meters t hours after the pump is switched off: “Three hours after the pump is switched off, the water is 14 meters below ground and rising at 0.6 meters per hour.”
- Write in notation, with S(q) the revenue in dollars from q carpets: “When 25 carpets have been sold, revenue is 9{,}500 dollars and is rising at 210 dollars per carpet.”
- A second tank drains so that it holds V(t) = 0.4(90 - t)^2 liters after t minutes, for 0 \le t \le 90. Find V(30) and V'(30) and interpret them together.
- For the tank in Exercise 7, estimate V(33) from V(30) and V'(30), and compare with the true value.
- For the tank in Exercise 7, let T(V) be the time in minutes at which it holds V liters. Find T'(1440) two ways: from a formula for T, and as a reciprocal. Give its units.
- In Exercise 1, suppose also A(10) = 22. Estimate A(12) and A(9).
Day 2: marginal rates, science, tables and graphs
- In a second year the cost of q carpets is C(q) = 1500 + 95q - 0.4q^2 + 0.004q^3 dollars. Find C'(50) and compare it with the exact cost of the 51st carpet.
- That year’s revenue is R(q) = 320q - 1.6q^2 dollars. Find R'(40) and R'(120), and say what each means for the cooperative.
- With C and R from Exercises 11 and 12, find the marginal profit at q = 50. Should the cooperative weave a 51st carpet?
- Find the average cost per carpet \displaystyle \frac{C(50)}{50} and compare it with C'(50). Which one answers “what will one more carpet cost?”
- A dye vat cools so that its temperature is H(t) = 90 - 3t + 0.03t^2 degrees Celsius after t minutes, for 0 \le t \le 50. Find H'(10) and H'(40) with units. When is it cooling faster?
- Mursal hauls a full bucket up from water 40 meters below the rim. Its height above the water is h(t) = 0.3t^2 - 0.01t^3 meters after t seconds, for 0 \le t \le 20. Find its velocity and acceleration at t = 5, with units, and its height and velocity at t = 20. What happens at t = 20?
- In another month, C(10) = 3900, C(20) = 5200, C(30) = 6300 and C(40) = 7600 dollars, with C'(10) = 140, C'(20) = 120, C'(30) = 115 and C'(40) = 145 dollars per carpet. Estimate the cost of the 31st carpet and the cost C(32). At which of these levels is one more carpet cheapest?
- On the graph of a revenue function G(q) in dollars, the tangent line at q = 25 passes through (25, 9000) and (35, 11{,}000). Find G'(25) with units and interpret it.
Reasoning
- For the tank in Exercise 7, Wahid reasons: “It holds 1440 liters and loses 48 liters a minute, so it will be empty in 30 more minutes.” When does it really empty, and what is wrong with the reasoning?
- Starting from the difference quotient, explain why f'(a) has the units of f per unit of x, and why f'(a)\,\Delta x has the units of f.
- Zarlasht reads C'(50) = 85 from Exercise 11 as “each of our 50 carpets cost 85 dollars”; Mursal reads it as “the 50 carpets cost 85 dollars in all.” Explain why both are wrong and give the correct reading.
- Explain why C(q + 1) - C(q) is close to C'(q) for the cost in Example 4, and describe a situation in which the marginal cost would not be a good estimate of the cost of the next unit.
- The ground at Wahid’s well is 412 meters above sea level, so the water’s elevation is E(t) = 412 - D(t) meters, where D(t) is its depth below ground. Show that E'(t) = -D'(t), and explain what this says about the sign of D'(t) when the water is falling.
Answers
- Square meters per day. On day 10 the finished area is increasing at 1.5 square meters per day.
- Meters per day; falling. On day 30 after the rains stop, the depth below ground is increasing at 0.12 meters per day, so the water surface is dropping.
- Kilograms per meter. For a carpet 2 meters wide, weight increases at 9.5 kilograms per meter of extra width; 10 more centimeters adds about 0.95 kilograms.
- Hours per (liter per minute). At an outflow of 50 liters per minute, raising the outflow by 1 liter per minute shortens the emptying time by about 0.4 hours, which is 24 minutes.
- d(3) = 14 and d'(3) = -0.6. Rising water means a shrinking depth below ground, so the rate is negative.
- S(25) = 9500 and S'(25) = 210.
- V(30) = 1440 liters and V'(30) = -48 liters per minute. V(t) = 3240 - 72t + 0.4t^2, so V'(t) = -72 + 0.8t. After 30 minutes the tank holds 1440 liters and is losing 48 liters per minute.
- About 1296 liters; the true value is 1299.6 liters. 1440 + (-48)(3) = 1296, and V(33) = 0.4(57)^2 = 1299.6.
- \displaystyle T'(1440) = -\frac{1}{48} \approx -0.0208 minute per liter. From V = 0.4(90 - t)^2, \displaystyle T(V) = 90 - \sqrt{2.5V}, so \displaystyle T'(V) = -\frac{\sqrt{2.5}}{2\sqrt{V}}, and at V = 1440, \displaystyle \sqrt{2.5 \cdot 1440} = 60 gives \displaystyle -\frac{2.5}{2 \cdot 60} = -\frac{1}{48}. As a reciprocal, \displaystyle \frac{1}{V'(30)} = -\frac{1}{48}.
- A(12) \approx 25 and A(9) \approx 20.5 square meters. 22 + 1.5(2) and 22 + 1.5(-1).
- C'(50) = 85 dollars per carpet; the 51st carpet costs exactly 85.204 dollars. C'(q) = 95 - 0.8q + 0.012q^2; C(50) = 5750 and C(51) = 5835.204.
- R'(40) = 192 and R'(120) = -64 dollars per carpet. R'(q) = 320 - 3.2q. At 40 carpets one more sale adds about 192 dollars; at 120 it lowers revenue by about 64 dollars.
- P'(50) = 75 dollars per carpet; yes. R'(50) = 320 - 160 = 160 and C'(50) = 85, so the 51st carpet raises profit by about 75 dollars.
- 115 dollars per carpet, against C'(50) = 85. \displaystyle \frac{5750}{50} = 115. The average spreads all costs, fixed ones included, over carpets already made; C'(50) estimates the next one.
- H'(10) = -2.4 and H'(40) = -0.6 degrees Celsius per minute; faster at t = 10. H'(t) = -3 + 0.06t. The size of the rate is larger at t = 10.
- v(5) = 2.25 meters per second and a(5) = 0.3 meters per second per second; h(20) = 40 meters and v(20) = 0. v(t) = 0.6t - 0.03t^2 and a(t) = 0.6 - 0.06t. At t = 20 the bucket reaches the rim and is momentarily at rest.
- About 115 dollars; C(32) \approx 6530 dollars; at q = 30. C(31) - C(30) \approx C'(30) = 115, and 6300 + 115(2) = 6530. The smallest listed marginal cost is 115.
- G'(25) = 200 dollars per carpet. \displaystyle \frac{11000 - 9000}{35 - 25} = 200. At 25 carpets, revenue is rising at 200 dollars per carpet, so a 26th carpet adds about 200 dollars.
- It empties at t = 90, which is 60 more minutes. V(90) = 0. The rate -48 holds only at t = 30; V'(t) = -72 + 0.8t rises toward 0, so the flow slows.
- The quotient \displaystyle \frac{f(a + h) - f(a)}{h} divides a change in f by a change in x, so it has units of f per unit of x, and a limit of such quotients keeps them. Multiplying by \Delta x, in units of x, cancels the unit of x as a conversion factor does, leaving units of f.
- C'(50) = 85 is a rate at q = 50, not a cost of carpets already made; those cost C(50) = 5750 dollars, 115 each on average. At 50 carpets a month cost is rising at 85 dollars per carpet, so a 51st costs about 85 dollars.
- C(q + 1) - C(q) is the secant slope over [q, q + 1]; one carpet is a small step on a gently bending graph, so it nearly matches the tangent slope at q. The estimate fails if one unit is a large batch (a hundred carpets), or if cost jumps, say when the next carpet needs a new loom.
- By the difference and constant rules, E'(t) = 0 - D'(t) = -D'(t). Falling water means E is decreasing, so E'(t) < 0 and therefore D'(t) > 0.