D. Characteristics of Functions · Rates of ChangeRates of Change of Exponential and Logarithmic Functions
D1.4 D1.6 D1.8About 70 minutes
Learning goals- I can estimate instantaneous rates of change of exponential functions and see that the rate is proportional to the value.
- I can estimate instantaneous rates of change of logarithmic functions and describe how they shrink as x grows.
- I can explain why the tangent slopes of inverse functions at reflected points are reciprocals.
- I can interpret rates of change in growth, decay and logarithmic models.
Words to know
proportional
Two quantities are proportional if one is always a constant multiple of the other.
half-life
The time it takes a decaying quantity to fall to half its amount.
inverse functions
Functions that undo each other, such as y = 2ˣ and y = log₂ x. Their graphs are reflections in the line y = x.
the number e
A special base, about 2.718, for which the tangent to y = eˣ at (0, 1) has slope exactly 1. You will use it in MCV4U.
Growth that feeds itself
A rumour, a bank balance, a bacteria culture: the more there is, the faster it grows. In the unit Exponential and Logarithmic Functions you studied these models and their inverses, the logarithms. Now you will measure how fast they change, moment by moment.
You will discover two big patterns:
- For an exponential function, the instantaneous rate of change is proportional to the value of the function.
- For a logarithmic function, the instantaneous rate of change shrinks as x grows, and it never quite reaches 0.
Worked exampleExample 1: Tangent slopes of y = 2ˣ
Use centred intervals with h = 0.001 to estimate the tangent slopes of y = 2ˣ at x = 0, 1, 2 and 3. Compare each slope with the value of the function.
- At x = 0: (2^0.001 − 2^(−0.001)) ÷ 0.002 ≈ 0.6931.
- At x = 1: ≈ 1.3863. At x = 2: ≈ 2.7726. At x = 3: ≈ 5.5452.
- Divide each slope by the function value: 0.6931 ÷ 1, 1.3863 ÷ 2, 2.7726 ÷ 4, 5.5452 ÷ 8. Every quotient is about 0.6931.
Slope of y = 2ˣ at x ≈ 0.693 × 2ˣ. The rate of change is proportional to the value: double the value, double the rate.
Tangents to y = 2ˣ at x = 0, 1 and 2. Each time the value doubles, the slope doubles too, so slope ÷ value stays about 0.693.Why the pattern holds
For f(x) = bˣ, the difference quotient at x = a is
(b^(a + h) − bᵃ) ÷ h = bᵃ × (bʰ − 1) ÷ h.
The factor (bʰ − 1) ÷ h does not depend on a at all. As h shrinks, it approaches a constant that depends only on the base b. Call it k. So the tangent slope at any point is k × bᵃ: a constant times the value.
- For b = 2: k ≈ 0.693.
- For b = 3: k ≈ 1.099.
- For b = 10: k ≈ 2.303.
- For b between 0 and 1 (decay), k is negative: the function decreases, and the size of the rate shrinks with the value.
There is one base for which k is exactly 1. It is about 2.718 and is called e. In MCV4U you will meet it when you study derivatives of exponential functions.
Worked exampleExample 2: A medicine leaving the body (decay)
A 400 mg dose of a medicine is cleared from the body with a half-life of 6 h, modelled by A(t) = 400(0.5)^(t/6). Estimate the rate of change of the amount at t = 0, 6 and 12 h.
- Centred intervals with h = 0.001 give: at t = 0, ≈ −46.21 mg/h; at t = 6, ≈ −23.10 mg/h; at t = 12, ≈ −11.55 mg/h.
- Amounts: A(0) = 400, A(6) = 200, A(12) = 100.
- Rate ÷ amount: −46.21 ÷ 400 ≈ −23.10 ÷ 200 ≈ −11.55 ÷ 100 ≈ −0.1155.
The amount drops at about 46.2, 23.1 and 11.6 mg/h. Each half-life, both the amount and the rate are halved. Decay is fastest at the start.
Worked exampleExample 3: Tangent slopes of y = log x
Estimate the tangent slopes of y = log x (base 10) at x = 1, 2, 5 and 10. Multiply each slope by x. What do you notice?
- Centred intervals, h = 0.001: at x = 1, ≈ 0.4343; at 2, ≈ 0.2171; at 5, ≈ 0.0869; at 10, ≈ 0.0434.
- Slope × x: 0.4343, 0.4343, 0.4343, 0.4343.
- So the slope is about 0.4343 ÷ x.
The slope of y = log x is about 0.4343 ÷ x. It is always positive but gets smaller as x grows: the graph is increasing at a decreasing rate. (0.4343 is 1 ÷ 2.3026, the reciprocal of k for base 10. The next example shows why.)
Inverses have reciprocal slopes
In Exponential and Logarithmic Functions you learned that y = bˣ and y = log_b x are inverses: each graph is the reflection of the other in the line y = x.
Reflecting in y = x swaps x and y, so it swaps rise and run. A tangent with slope m (rise m for run 1) becomes a tangent with rise 1 for run m, which has slope 1 ÷ m.
So if y = bˣ has slope m at (p, q), then y = log_b x has slope 1 ÷ m at (q, p).
Worked exampleExample 4: Typing practice (a logarithmic model)
A student's typing speed is modelled by W(t) = 15 + 20 log(t + 1) words per minute after t weeks of practice. Estimate the rate of improvement at t = 1 and t = 9 weeks.
- W(1) = 15 + 20 log 2 ≈ 21.0 and W(9) = 15 + 20 log 10 = 35.
- Centred intervals with h = 0.001: at t = 1, ≈ 4.34 words per minute per week; at t = 9, ≈ 0.87 words per minute per week.
After 1 week she improves by about 4.3 wpm per week; after 9 weeks, only about 0.9 wpm per week. Logarithmic models describe skills that improve quickly at first, then level off (but never completely stop improving).
Just for funFold a 0.1 mm sheet of paper in half 42 times and, in theory, it would be about 440 000 km thick, farther than the Moon (about 384 400 km away). In practice, people have only managed about a dozen folds. Exponential growth runs out of paper fast.
Hands-on activityFolding frenzy
You need: a sheet of printer paper, a ruler, paper and a pencil.
- Count the layers before folding (1). Fold the sheet in half and count again (2). Keep folding and counting until you cannot fold any more.
- Make a table: number of folds n and number of layers L.
- Compute the average rate of change of L for each step (n to n + 1). What do you notice?
- Divide each rate by the number of layers at the start of that step.
- A stack of 500 printer sheets is about 5 cm thick, so one sheet is about 0.1 mm. Estimate the thickness after each fold.
What to look for: the layers are 1, 2, 4, 8, 16, …, so the rate (layers per fold) is also 1, 2, 4, 8, …: it equals the number of layers at the start of each step. Rate ÷ value = 1 every time. This is the discrete version of "the rate is proportional to the value". Most people manage about 6 or 7 folds with printer paper.
Your turn
1Estimate the tangent slope of y = 2ˣ at x = 4 using a centred interval with h = 0.001. Check with slope ≈ 0.693 × 2ˣ.
≈ 11.09. Check: 0.6931 × 16 ≈ 11.09 ✓.
2Use slope ≈ 0.693 × 2ˣ to estimate the tangent slopes of y = 2ˣ at x = 5 and at x = −1.
x = 5: 0.693 × 32 ≈ 22.18. x = −1: 0.693 × 0.5 ≈ 0.347.
3Estimate the tangent slope of y = 3ˣ at x = 2.
Centred, h = 0.001: ≈ 9.89. (Check: 1.0986 × 9 ≈ 9.89.)
4Use slope ≈ 0.4343 ÷ x to estimate the tangent slope of y = log x at x = 4.
5For the medicine A(t) = 400(0.5)^(t/6), estimate the rate of change at t = 18 h.
A(18) = 50, and rate ≈ −0.1155 × 50 ≈ −5.78 mg/h (or compute a centred difference).
6In a sentence, explain why an exponential growth model grows faster and faster.
Its instantaneous rate is proportional to its value: as the quantity gets bigger, it grows faster, which makes it bigger still.
7The tangent slope of y = 2ˣ at (3, 8) is about 5.545. What is the tangent slope of y = log₂ x at (8, 3)?
The functions are inverses, so the slopes at reflected points are reciprocals: 1 ÷ 5.545 ≈ 0.180.
More practice: warm-up
1Is y = 2ˣ increasing at an increasing rate or at a decreasing rate?
2Is y = log x (for x > 0) increasing at an increasing rate or at a decreasing rate?
3Can the tangent slope of y = 2ˣ ever be 0? Explain.
No. The slope is about 0.693 × 2ˣ, and 2ˣ is always positive.
4Is the tangent slope of y = (0.5)ˣ positive or negative?
Negative (it is a decay function).
5The tangent slope of y = 2ˣ at x = 0 is about 0.693. What is it at x = 1?
About 1.386 (twice as much, because 2¹ = 2 × 2⁰).
More practice: on your own
1A town's population is P(t) = 12 000(1.018)^t, t years from now. Estimate the growth rate now and in 10 years.
Centred, h = 0.001: now ≈ 214 people/year; at t = 10 (P ≈ 14 344) ≈ 256 people/year.
2For the typing model W(t) = 15 + 20 log(t + 1), show that the rate at t = 9 is one fifth of the rate at t = 1. Why does that make sense?
Rates: about 4.34 and 0.87, and 0.87 ≈ 4.34 ÷ 5 ✓. The slope of log(t + 1) is about 0.4343 ÷ (t + 1), which is 0.4343 ÷ 2 at t = 1 and 0.4343 ÷ 10 at t = 9: one fifth as large.
3For the car V(t) = 28 000(0.85)^t, the tangent slope at t = 3 is about −2794.59 and V(3) = 17 195.50. Find the slope ÷ value. Why is it −0.1625 and not −0.15?
−2794.59 ÷ 17 195.50 ≈ −0.1625. At each instant the car is losing value at a rate of about 16.25% of its current value per year. Because the value keeps shrinking during the year, those moment-by-moment losses add up to only 15% over a whole year.
4Find the average rate of change of y = log x over [1, 10] and over [10, 100]. Compare.
[1, 10]: (1 − 0) ÷ 9 ≈ 0.111. [10, 100]: (2 − 1) ÷ 90 ≈ 0.0111. The second is one tenth of the first: the graph gets flatter as x grows.
5A cup of tea cools as T(t) = 20 + 70(0.9)^t. Is the rate of cooling proportional to T? To what is it proportional?
The slope of 70(0.9)^t is k × 70(0.9)^t with k ≈ −0.105, so the rate is proportional to T − 20, the difference between the tea and the room, not to T itself. (This is Newton's law of cooling.)
Challenge
1Show that the difference quotient of f(x) = bˣ at x = a equals bᵃ times the difference quotient at x = 0. What does this prove about tangent slopes of exponential functions?
(b^(a + h) − bᵃ) ÷ h = bᵃ(bʰ − 1) ÷ h, and (bʰ − 1) ÷ h = (b^(0 + h) − b⁰) ÷ h is the difference quotient at 0. As h → 0, slope at a = bᵃ × (slope at 0): the tangent slope is proportional to the function's value.
2Find, to two decimal places, the base b for which the tangent to y = bˣ at (0, 1) has slope 1. Use (b^0.001 − 1) ÷ 0.001 to test values.
b = 2.7 gives ≈ 0.9937 (too small); b = 2.72 gives ≈ 1.0011 (too big); b = 2.718 gives ≈ 1.0004. So b ≈ 2.72. (The exact value is e ≈ 2.71828.)
3Sound level in decibels is L = 10 log(I ÷ I₀), where I is intensity. Show that doubling the intensity always raises the level by about 3.01 dB. What does this tell you about the average rate of change of L as I grows?
L(2I) − L(I) = 10 log(2I ÷ I₀) − 10 log(I ÷ I₀) = 10 log 2 ≈ 3.01 dB, for every I. The interval [I, 2I] gets longer as I grows, but the rise stays 3.01, so the average rate 3.01 ÷ I gets smaller: the graph flattens, as every log graph does.
Quiz
1. The tangent slope of y = 2ˣ at x = 0 is about 0.693. About what is the tangent slope at x = 3?
2. Which statement describes the instantaneous rate of change of an exponential function y = a·bˣ?
3. As x increases, the tangent slopes of y = log x:
4. The tangent to y = 2ˣ at (1, 2) has slope about 1.386. What is the slope of the tangent to y = log₂ x at (2, 1)?
5. A medicine decays as A(t) = 400(0.5)^(t/6). How does its rate of change behave over time?
6. For which base b does the tangent to y = bˣ at (0, 1) have slope exactly 1?
Big ideas (D1.4, D1.6, D1.8): an exponential function's instantaneous rate of change is proportional to its value (slope = k × value). A logarithmic function's rate shrinks as x grows (for log x, slope ≈ 0.4343 ÷ x). Because they are inverses, their slopes at reflected points are reciprocals.
How to help: Use money: "If our savings earn interest, does the balance grow by the same number of dollars each year?" Then use skills: "When you started guitar (or skating, or a language), did you improve faster at the start or now?" These are exponential and logarithmic rates in real life.
Common difficulties:
- Thinking the slope of 2ˣ at x = 3 is 3 × 0.693 (it is 2³ × 0.693).
- Calculator brackets: 400 × 0.5^(t ÷ 6) needs brackets around t ÷ 6.
- Using log (base 10) when the question uses log₂ (use the change of base formula from Exponential and Logarithmic Functions).
- Expecting a log graph to level off at a maximum. It keeps rising forever, just more and more slowly.