Grade by Grade
A. Rate of Change · Derivatives of Exponential and Sinusoidal Functions

Solving Problems with Exponential Rates of Change

A3.5 A2.5About 75 minutes
Learning goals
  • I can build and differentiate exponential models of growth, decay and cooling.
  • I can find the rate of change at a given time and interpret it with correct units and sign.
  • I can find the time when a rate reaches a given value by solving an exponential equation with ln.
  • I can explain why, in an exponential model, the rate is proportional to the amount (or to the difference from a limiting value).

Words to know

exponential growth model
P(t) = P₀ e^(kt) with k > 0, or P₀ bᵗ with b > 1.
exponential decay model
A(t) = A₀ e^(−kt) with k > 0, or A₀ bᵗ with 0 < b < 1.
Newton's law of cooling
An object's temperature T approaches the room temperature Tₛ, with T′ proportional to T − Tₛ. Solutions look like T = Tₛ + C e^(−kt).
instantaneous rate of change
The derivative at a single moment, with units of output per unit of input.
half-life
The time for a decaying quantity to fall to half its amount.

Rates in the real world

A town's population, a radioactive tracer in a hospital, a cup of tea on the counter: each changes exponentially, and the important questions are about rates. How fast is it changing now? When will it change at a certain speed? Is the rate growing or shrinking?

This lesson puts together everything from the exponential half of the unit: d/dx eˣ = eˣ, d/dx aˣ = aˣ ln a, and the chain and product rules. Use the four-step roadmap below for every problem.

1. Modelwrite the function2. Differentiatechain/product rules3. Substituteor solve for t4. Interpretsign and unitsA rate always comes with units: (units of the quantity) per (unit of time).
A four-step plan for rate problems. Step 4 matters most: a negative rate means the quantity is decreasing, and every answer needs units.
Worked example

Example 1: A growing town

A town's population is modelled by P(t) = 25 000e^(0.018t), where t is years after 2026. (a) How fast is it growing in 2036? (b) In what year will it be growing by 1000 people per year?

  1. Differentiate: P′(t) = 25 000 × 0.018e^(0.018t) = 450e^(0.018t).
  2. (a) 2036 means t = 10: P′(10) = 450e^(0.18) ≈ 538.7 people per year.
  3. (b) Solve 450e^(0.018t) = 1000: e^(0.018t) = 2.2222, so 0.018t = ln 2.2222 ≈ 0.7985 and t ≈ 44.4.
  4. t ≈ 44.4 years after 2026 is during 2070.

(a) about 539 people per year in 2036; (b) about 44 years later, during 2070. Notice P′ = 0.018P: the town grows by 1.8% of its population each year, so a bigger town adds more people.

Worked example

Example 2: A medical tracer

Iodine-131, used in some medical treatments, has a half-life of about 8 days. A sample of 200 mg decays as A(t) = 200(0.5)^(t/8), t in days. Find A′(0) and A′(8). How do they compare?

  1. A′(t) = 200(0.5)^(t/8) × ln 0.5 × (1/8) = 25 ln 0.5 × (0.5)^(t/8).
  2. A′(0) = 25 ln 0.5 ≈ −17.33 mg per day.
  3. A′(8) = 25 ln 0.5 × 0.5 ≈ −8.66 mg per day.

At first the sample loses about 17.3 mg per day; after one half-life the rate has also halved, to about 8.7 mg per day. The rate is always proportional to the amount left.

Worked example

Example 3: A cooling cup of tea

A cup of tea cools according to T(t) = 22 + 68e^(−0.04t) °C, t minutes after it is poured into a mug in a 22°C room. (a) How fast is it cooling at the start and after 10 minutes? (b) When is it cooling at exactly 1°C per minute, and how hot is it then?

  1. T′(t) = 68 × (−0.04)e^(−0.04t) = −2.72e^(−0.04t).
  2. (a) T′(0) = −2.72°C per minute. T′(10) = −2.72e^(−0.4) ≈ −1.82°C per minute.
  3. (b) Solve −2.72e^(−0.04t) = −1: e^(−0.04t) = 1/2.72, so −0.04t = ln(1/2.72) and t = ln 2.72 ÷ 0.04 ≈ 25.0 minutes.
  4. T(25.0) = 22 + 68 × (1/2.72) = 22 + 25 = 47°C.

(a) 2.72°C/min at first, 1.82°C/min after 10 minutes; (b) after about 25 minutes, when the tea is 47°C.

2040608020406080100t (min)T (°C)slope −2.72 at t = 0slope −1.00 at t ≈ 25room temperature 22°CT(t) = 22 + 68e^(−0.04t)The tea cools fastestat the start, when itis hottest comparedwith the room.T′ = −0.04(T − 22)
The tea's temperature falls towards 22°C. The tangent at t = 0 is steep (−2.72°C per minute); by t ≈ 25 minutes the tangent slope has dropped to −1°C per minute.

Why exponential models fit these situations

In each example the derivative turned out to be a multiple of something you can see:

  • Population: P′ = 0.018P. More people means more births, so faster growth.
  • Tracer: A′ = (ln 0.5 ÷ 8)A. More atoms means more decays per day.
  • Tea: T′(t) = −2.72e^(−0.04t) = −0.04(T − 22). The bigger the gap between the tea and the room, the faster it cools.

Whenever a quantity changes at a rate proportional to its size (or to its distance from a fixed level), an exponential function is the right model. That is the idea of expectation A2.5 at work, and it is how differential equations begin in first-year university calculus, physics, biology and economics.

Worked example

Example 4: Air pressure and altitude

Air pressure is roughly P(h) = 101.3e^(−0.12h) kilopascals (kPa) at an altitude of h kilometres. Find the rate of change of pressure at h = 2 km and explain its meaning for a hiker.

  1. P′(h) = 101.3 × (−0.12)e^(−0.12h) ≈ −12.16e^(−0.12h).
  2. P′(2) ≈ −12.16e^(−0.24) ≈ −9.56 kPa per km.
  3. P(2) ≈ 79.7 kPa.

At 2 km the pressure is about 79.7 kPa and is falling by about 9.6 kPa for each extra kilometre of climb (about 0.96 kPa per 100 m). That is one reason ears 'pop' on mountain roads.

Hands-on activity

Measure a cooling curve

You need: a kitchen thermometer (a digital cooking thermometer works well), a mug, hot water from a kettle or tap, a timer, graph paper and a calculator. Safety: a grown-up should pour the hot water; do not use boiling water, keep the mug on a stable table, and don't hold the thermometer in the steam.

  1. Note the room temperature Tₛ (leave the thermometer on the counter for a few minutes first).
  2. Fill the mug with hot water. Record the temperature every 2 minutes for 30 minutes.
  3. Plot temperature against time. Draw tangents by eye at t = 2 and t = 20 and estimate their slopes.
  4. Model: T(t) = Tₛ + (T₀ − Tₛ)e^(−kt). Estimate k from your data at t = 20: k = −ln[(T(20) − Tₛ)/(T₀ − Tₛ)] ÷ 20.
  5. Find T′(t) from your model and compare T′(2) and T′(20) with your hand-drawn slopes.

What to look for: the water cools quickly at first and then more and more slowly. With a typical ceramic mug, k is often somewhere around 0.02 to 0.05 per minute, depending on the mug and room. Your hand-drawn slopes and the model's T′ values should agree in sign (both negative) and roughly in size; the model fits best if you measure from the same spot in the water each time.

Just for fun

Isaac Newton is said to have published his law of cooling anonymously, in 1701. Even the tea was more famous than the author of that paper for a while.

Your turn

1

A bacteria culture has N(t) = 400e^(0.25t) cells after t hours. Find N′(4).

2

For N(t) = 400e^(0.25t), when is the growth rate 1000 cells per hour?

3

$5000 is invested at 6% per year compounded continuously: A(t) = 5000e^(0.06t). How fast is it growing after 5 years?

4

A medication is eliminated as C(t) = 80e^(−0.2t) mg/L, t in hours. Find C′(5) and interpret it.

5

For T(t) = 22 + 68e^(−0.04t), show that T′ = −0.04(T − 22).

6

A population doubles every 15 years: P(t) = 1200(2)^(t/15). Find P′(15).

More practice: warm-up

1

What are the units of P′(t) if P is a number of people and t is in years?

2

If A′(t) < 0, is the quantity growing or decaying?

3

Differentiate P(t) = 100e^(0.05t).

4

For P(t) = 100e^(0.05t), what is P′(t) ÷ P(t)?

5

Solve e^(0.1t) = 3 for t.

More practice: on your own

1

A radioactive sample has A(t) = 50e^(−0.0866t) grams after t days. Find the half-life and the rate of decay at that time.

2

A hot pie is taken out of the oven and cools as T(t) = 20 + 160e^(−0.05t) °C. When is it cooling at 2°C per minute?

3

The value of a car is V(t) = 32 000(0.85)ᵗ dollars after t years. Find V′(3) and interpret it.

4

For P(t) = 25 000e^(0.018t), find the doubling time and the growth rate at that time.

5

A lake trout population, after a restocking program, is N(t) = 3000 − 2500e^(−0.3t) fish after t years. Find N′(2) and explain why the rate shrinks over time.

Challenge

1

Two investments: A(t) = 1000e^(0.08t) and B(t) = 2000e^(0.05t). When are they growing at the same rate? (Give t to one decimal place.)

2

Show that for any model A(t) = A₀e^(−kt), the rate of decay halves over one half-life.

3

A logistic model for an app's users is U(t) = 10 000 ÷ (1 + 99e^(−0.5t)), t in weeks. Find U′(t) and U′(0), and find U′ when U = 5000 (which happens at t = 2 ln 99 ≈ 9.19).

Check your understanding

Quiz

1.

P(t) = 600e^(0.03t). Which is P′(t)?

2.

A(t) = 100(0.5)^(t/10). What is A′(0)?

3.

For T(t) = 20 + 60e^(−0.1t), when is T′(t) = −3?

4.

In an exponential growth model, the rate of growth is:

5.

A town's population P is in people and t is in years. A value P′(5) = −120 means:

6.

For a cooling object T = Tₛ + Ce^(−kt), the object cools fastest:

7.

Which equation gives the time when 450e^(0.018t) = 900?

For parents and teachers

Big idea (A3.5, A2.5): exponential models describe situations where the rate is proportional to the amount (or to the gap from a limiting value). Students differentiate the model, substitute or solve with ln, and interpret the result with units and a sign.

How to help: for each answer, ask "What are the units?" and "Is it going up or down?" A number like −1.82 only makes sense as "cooling at 1.82°C per minute". The cooling-mug activity is an easy kitchen experiment you can do together; supervise the hot water.

Common difficulties:

  • Forgetting the chain-rule constant (writing d/dt e^(0.018t) = e^(0.018t)).
  • Confusing the amount P(t) with the rate P′(t).
  • Solving e^(kt) = c by dividing by e instead of taking ln.
  • Dropping the negative sign of a decay rate, or reporting it without saying 'decreasing'.