B. Derivatives and Their Applications · Applications of Derivatives: Motion, Rates and Optimization

Instantaneous Rates in Models: Growth, Decay and Tides

B2.3About 70 minutes

Learning goals

I can find instantaneous rates of change for exponential, sinusoidal, polynomial and rational models using the derivative rules.
I can interpret the rate with units and a sentence.
I can show that exponential growth and decay rates are proportional to the amount present.
I can find when a sinusoidal quantity changes fastest.

Words to know

exponential growth or decay
A model A(t) = A₀bᵗ or A₀e^(kt); its rate of change is proportional to the amount present.

half-life
The time for a decaying quantity to fall to half its amount.

instantaneous rate of change
The derivative at one moment: how fast the quantity is changing at that instant.

sinusoidal model
A model such as c + a sin(bt) or c + a cos(bt), used for repeating patterns like tides and daylight.

radian mode
The calculator setting needed for derivatives of sine and cosine.

How fast, right now?

In the last lesson you read rates from graphs and simple polynomials. Now you will use every derivative rule from this course: power, product and chain rules, and the derivatives of eˣ, aˣ, sin x and cos x from Derivatives of Exponential and Sinusoidal Functions. With them you can find the instantaneous rate of change for the models scientists actually use: exponential growth and decay, periodic models such as tides, and rational models.

The method is always the same:

Write (or read) the model y = f(t).
Differentiate it with the rules. (Do not worry about making the derivative look pretty. The goal is a correct derivative you can evaluate, not a perfectly simplified one.)
Substitute the time you care about.
Interpret the number with units and a sentence.

Worked example

Example 1: A butterfly population

The population of a colony of butterflies in a conservatory is modelled by P(t) = 250e^(0.1t), where t is in days. Find the growth rate after 5 days, and check it with a secant.

Chain rule: P′(t) = 250e^(0.1t) × 0.1 = 25e^(0.1t).
P′(5) = 25e^(0.5) ≈ 25(1.6487) ≈ 41.2.
Check with a short secant: [P(5.01) − P(5)] ÷ 0.01 ≈ (412.59 − 412.18) ÷ 0.01 ≈ 41.2. ✓
Notice P′(t) = 0.1 P(t): the growth rate is always 10% of the current population. That is the signature of exponential growth.

After 5 days the population (about 412) is growing at about 41 butterflies per day.

The butterfly model P(t) = 250e^(0.1t). The dashed tangent at t = 5 has slope P′(5) ≈ 41.2. The secant to t = 10 (dotted) is steeper, because the curve bends upward: short secants give much better estimates.

Worked example

Example 2: Radioactive decay

Iodine-131, used in some medical treatments, has a half-life of about 8 days. A sample starts with 100 mg, so the amount left is A(t) = 100(0.5)^(t/8) mg after t days. (a) Find the rate of decay at t = 8. (b) Show that A′(t) is proportional to A(t).

Use d/dt aᵘ = aᵘ ln a × u′. Here u = t/8, so u′ = 1/8.
A′(t) = 100(0.5)^(t/8) × ln 0.5 × (1/8).
(a) At t = 8: A′(8) = 100(0.5) × (−0.6931) ÷ 8 ≈ −4.33 mg/day.
(b) A′(t) = [ln 0.5 ÷ 8] × 100(0.5)^(t/8) = −(ln 2 ÷ 8) A(t) ≈ −0.0866 A(t).

At 8 days, 50 mg are left and the sample is losing about 4.33 mg per day. The decay rate is always about 8.66% of the amount present, which is why the half-life never changes.

Worked example

Example 3: Tides

In a harbour, the water depth is modelled by d(t) = 5 + 3 cos(2πt/12.4) metres, where t is the number of hours after high tide (a tide cycle of about 12.4 h). (a) How fast is the depth changing 2 h after high tide? (b) When is the water falling fastest, and how fast?

Chain rule: d′(t) = −3 sin(2πt/12.4) × (2π/12.4) = −(6π/12.4) sin(2πt/12.4) ≈ −1.520 sin(2πt/12.4).
(a) d′(2) ≈ −1.520 sin(1.0134) ≈ −1.520(0.8486) ≈ −1.29 m/h. (Radian mode!)
(b) The size of d′ is largest when sin(2πt/12.4) = ±1. It equals +1 (so d′ is most negative) when 2πt/12.4 = π/2, so t = 12.4/4 = 3.1 h.
Then d′(3.1) ≈ −1.52 m/h. At that moment the depth is d(3.1) = 5 + 3 cos(π/2) = 5 m, the middle level.

(a) Falling at about 1.29 m/h. (b) Falling fastest at 3.1 h after high tide, at about 1.52 m/h, when the water passes its mean level.

The tide model d(t) = 5 + 3 cos(2πt/12.4). The water changes fastest as it passes the middle level (5 m) and not at all at high and low tide, where the tangent is horizontal.

Worked example

Example 4: A rational model

After a medication is taken, its concentration in the blood is modelled by C(t) = 5t ÷ (t² + 4) mg/L, t hours after the dose. Find the rate of change at t = 1 h and t = 3 h.

Write C(t) = 5t(t² + 4)⁻¹ and use the product and chain rules. (The quotient rule from Derivatives and Differentiation Rules gives the same result; either method works.)
C′(t) = 5(t² + 4)⁻¹ + 5t × (−1)(t² + 4)⁻² × 2t = 5/(t² + 4) − 10t²/(t² + 4)².
C′(1) = 5/5 − 10/25 = 1 − 0.4 = 0.6 mg/L per hour.
C′(3) = 5/13 − 90/169 = 65/169 − 90/169 = −25/169 ≈ −0.148 mg/L per hour.

At 1 h the concentration is rising at 0.6 mg/L per hour; at 3 h it is falling at about 0.15 mg/L per hour. Somewhere between, it peaked (see the Challenge questions).

Radian mode, and units

Every sinusoidal derivative in this course assumes radians. Before you evaluate sin or cos, check your calculator. And always attach units: amount per unit of time (mg/day, m/h, butterflies per day).

Just for fun

A radioactive sample is the most patient thing in the universe. Every half-life it loses half of what it has, and it never complains that the amount it loses keeps getting smaller. (Its derivative does all the worrying: A′ = −kA.)

Hands-on activity

Cooling curve in the kitchen

You need: a mug, warm water from the tap (not boiling; ask an adult if you use a kettle), a kitchen or aquarium thermometer, a phone stopwatch, and paper.

Fill the mug with the warmest tap water. Put the thermometer in and wait 30 s.
Record the temperature every 2 minutes for 20 minutes. Also record the room temperature.
Make a table of t and T. Estimate the rate of cooling in each interval with ΔT/Δt.
Next to each rate, write the difference (T − room temperature) at the start of that interval. Divide the rate by that difference.

What to look for: the cooling rate is largest at the start and shrinks over time. The ratio rate ÷ (T − room temperature) should be roughly constant (and negative). That is Newton's law of cooling, T(t) = T_room + Ce^(−kt), whose derivative is T′(t) = −k(T − T_room): the same kind of proportional rate as the decay in Example 2.

Your turn

A bacteria population is P(t) = 800e^(0.05t), t in hours. Find the growth rate at t = 10.

A sample of carbon-14 (half-life about 5730 years) has A(t) = 200(0.5)^(t/5730) mg. Find the rate of decay at t = 1000 years.

A Ferris wheel rider's height is h(t) = 12 + 10 sin(πt/30) metres, t in seconds. Find the rate of change of height at t = 10 s.

An insect population is N(t) = 50t² − t³ for 0 ≤ t ≤ 30 days. Find the growth rate on day 20.

A substance dissolves so that the amount dissolved is D(t) = 10t ÷ (t + 1) grams after t minutes. Find D′(4).

A cup of tea cools so that T(t) = 20 + 70e^(−0.1t) °C after t minutes. How fast is it cooling at t = 5?

For the harbour tide d(t) = 5 + 3 cos(2πt/12.4), at what time after high tide is the water rising fastest?

More practice: warm-up

Differentiate e^(3t).

Differentiate 5(2)ᵗ.

Differentiate sin(2t).

Differentiate 100e^(−0.2t).

f(t) = 4t³. Find f′(2).

More practice: on your own

For the butterflies P(t) = 250e^(0.1t), when will the population be growing at 100 butterflies per day?

For iodine-131, A(t) = 100(0.5)^(t/8) mg. When is the sample decaying at 1 mg/day?

A town's population is P(t) = 20 000 ÷ (1 + 4e^(−0.3t)), t in years. Find the growth rate at t = 0.

For the harbour, find the rate of change of depth 9 h after high tide.

A simple model for the hours of daylight in an Ontario city is D(t) = 12 + 3 sin(2π(t − 80)/365), t = day of the year. How fast is the length of day changing at t = 80 (near the spring equinox)? Give your answer in minutes per day.

Challenge

For a decay model A(t) = A₀e^(−kt) with half-life h, show that A′(t) = −kA(t) and that k = (ln 2)/h.

A tide is modelled by d(t) = c + a cos(2πt/T) with a > 0. Show that the greatest rate of change of depth is 2πa/T, and find it for a 12.4 h cycle with a 6 m tidal range.

For C(t) = 5t ÷ (t² + 4) from Example 4, when does the concentration stop increasing, and what is the peak concentration?

Check your understanding

Quiz

P(t) = 300e^(0.2t). What is P′(t)?

A(t) = 100(0.5)^(t/10). Which expression is A′(t)?

For any exponential decay model, the rate of decay is…

Depth d(t) = 4 + 2 sin(πt/6) m, t in hours. What is d′(0)?

For a sinusoidal tide model, when is the water level changing fastest?

A population is P(t) = 1000 + 50t − t² (t in years). How fast is it changing at t = 30?

For parents and teachers

Big idea (B2.3): with the derivative rules, you can find the instantaneous rate of change of any model a scientist might use: population growth, radioactive decay, tides, and drug concentration. The curriculum stresses applying the rules correctly, not simplifying long expressions.

How to help: ask your teen to say the answer as a sentence with units ("after 5 days the colony is growing by about 41 butterflies a day"). Check that the calculator is in radian mode for any sine or cosine.

Common difficulties:

Using the power rule on aᵗ (the exponent is the variable, so you need aᵗ ln a).
Forgetting the chain-rule factor (the 0.1 in e^(0.1t), the 2π/12.4 in a tide model).
Degree mode on the calculator.
Mixing up the amount and the rate (P(5) versus P′(5)).