A. Rate of Change · Derivatives and Differentiation Rules

Solving Rate-of-Change Problems with the Derivative Rules

A3.3 A3.5About 75 minutes

Learning goals

I can model a real situation with a polynomial, rational or radical function and differentiate it.
I can decide whether a problem asks for a rate at a given time, a time for a given rate, or an average rate.
I can combine the product, quotient and chain rules to solve multi-step problems.
I can communicate my answers in sentences with correct units and check that they make sense.

Words to know

mathematical model
A function that describes a real situation, with a stated domain where it makes sense.

rate of change
How fast one quantity changes compared with another; the derivative.

horizontal asymptote
A horizontal line that a graph approaches as x (or t) grows without bound.

closest approach
The point where a moving object is nearest to a fixed point; the distance has derivative 0 there.

Putting the whole unit together

This lesson is about solving problems: reading a situation, building or using a model, choosing the right derivative rules, and explaining what the answer means. The models involve polynomial, rational and radical functions, and most need more than one rule.

Think of it as practice for the unit Applications of Derivatives, where you will also optimize quantities and relate rates that change together.

Problem-solving steps

Read and sketch. What quantity changes? With respect to what? Draw a diagram if there is any geometry.
Model. Write the function (or use the one given). Note its domain.
Decide what is asked. A rate at a given time (substitute into the derivative)? A time when the rate has a given value (solve an equation)? An average rate (secant)?
Differentiate using the rules, outside in.
Answer in a sentence with units, and check that it makes sense (sign, size).

Worked example

Example 1: Stocking a lake

A lake is stocked with trout. The model P(t) = 5000(t + 1)/(t + 5) gives the number of trout t years after stocking. (a) How many trout were stocked? (b) Find the average growth rate over the first 5 years. (c) Find the growth rate at t = 0 and t = 5. (d) What happens to the growth rate in the long run?

(a) P(0) = 5000(1)/5 = 1000 trout.
(b) P(5) = 5000(6)/10 = 3000. Average rate = (3000 − 1000)/5 = 400 trout per year.
(c) Quotient rule: P′(t) = 5000[(t + 5) − (t + 1)]/(t + 5)² = 20 000/(t + 5)².
P′(0) = 20 000/25 = 800 trout/year; P′(5) = 20 000/100 = 200 trout/year.
(d) As t grows, (t + 5)² grows without bound, so P′(t) approaches 0. The population levels off toward 5000 (the horizontal asymptote, as in MHF4U).

(a) 1000 (b) 400 trout/year (c) 800 and 200 trout/year (d) the growth rate approaches 0 as the population nears 5000.

Worked example

Example 2: Draining a rain barrel

The volume of water left in a draining rain barrel is V(t) = 200(1 − t/20)² litres, 0 ≤ t ≤ 20, where t is in minutes. (a) Find V′(t). (b) How fast is the water draining at t = 0 and t = 10? (c) Explain why the rate slows down.

(a) Chain rule: V′(t) = 200 · 2(1 − t/20) · (−1/20) = −20(1 − t/20) = t − 20.
(b) V′(0) = −20 L/min; V′(10) = −10 L/min. The water drains at 20 L/min at first and 10 L/min after 10 minutes.
(c) Water drains faster when it is deeper, because the pressure at the spout is greater. As the barrel empties, the drain rate drops, reaching 0 at t = 20.

(a) V′(t) = t − 20 L/min (b) draining at 20 L/min and 10 L/min (c) less water above the spout means slower flow.

A boat sails a straight course at 20 km/h. At t = 1.5 h it passes point P, the closest point to the lighthouse, 10 km away. Its distance along the course from P is 20t − 30 km (negative before it reaches P), so the Pythagorean theorem gives d(t). (Sketch not to scale.)

Worked example

Example 3: Passing a lighthouse

A boat sails a straight course at 20 km/h. Its closest point to a lighthouse is 10 km away, and it reaches that point at t = 1.5 h. The distance from the boat to the lighthouse is d(t) = √(100 + (20t − 30)²) km. Find d′(t), and find how fast the distance is changing at t = 1, t = 1.5 and t = 2.

Write d(t) = [100 + (20t − 30)²]^(1/2).
Chain rule (twice): d′(t) = (1/2)[100 + (20t − 30)²]^(−1/2) · 2(20t − 30) · 20 = 20(20t − 30)/√(100 + (20t − 30)²).
t = 1: 20(−10)/√200 = −200/14.142 ≈ −14.1 km/h.
t = 1.5: numerator 0, so d′ = 0.
t = 2: 20(10)/√200 ≈ 14.1 km/h.

d′(t) = 20(20t − 30)/√(100 + (20t − 30)²). The distance is decreasing at about 14.1 km/h at t = 1, not changing at t = 1.5 (closest approach) and increasing at about 14.1 km/h at t = 2. These are less than 20 km/h because the boat is not heading straight at or away from the lighthouse.

The distance from the boat to the lighthouse. The derivative is negative while the boat approaches, 0 at the closest point (t = 1.5 h) and positive as it moves away.

Worked example

Example 4: Walking away from a lamp

The brightness of a lamp at distance d metres is I = 900/d² lux. A student walks away so that d = 2 + 1.5t (t in seconds). How fast is the brightness changing at t = 2 s?

I = 900d⁻², so dI/dd = −1800d⁻³. And dd/dt = 1.5.
Chain rule: dI/dt = (dI/dd)(dd/dt) = −1800d⁻³ · 1.5 = −2700/d³.
At t = 2: d = 5, so dI/dt = −2700/125 = −21.6.

The brightness is decreasing at 21.6 lux per second.

Just for fun

A trout stocked in a lake has no idea that its population is modelled by a rational function with a horizontal asymptote. It is mostly interested in the derivative of the number of mayflies.

Hands-on activity

Draining-bottle experiment

You need: an empty 2 L plastic bottle, a pushpin or nail to make a hole (a grown-up should make the hole), a measuring cup marked in millilitres, a marker, a timer and a sink or bathtub.

Have a grown-up make a small hole about 2 cm above the bottom of the bottle. Cover it with your finger or a piece of tape.
Fill the bottle to a mark well above the hole. Mark the water level every 2 cm with the marker before you start.
Uncover the hole and start the timer. Record the time as the water passes each mark.
Make a table of time and height. Estimate the rate of change of height (cm per second) at the start and near the end, using differences between nearby rows.
Compare with the barrel model in Example 2: does your bottle drain more slowly as it empties?

What to look for: the height falls faster at the beginning and more slowly near the hole, so the time between marks grows. A model of the form h(t) = a(1 − t/T)² usually fits well. Exact numbers depend on the hole size.

Your turn

For P(t) = 5000(t + 1)/(t + 5), find P′(15) and interpret it.

For V(t) = 200(1 − t/20)², when is the water draining at 5 L/min?

A balloon's radius is r(t) = 3 + 0.5t cm. Find how fast its volume is changing at t = 2 s. (V = (4/3)πr³)

A cost model is C(x) = 400 + 30√x dollars for x items. Find the marginal cost at x = 100.

For d(t) in Example 3, find d′(0.5).

The temperature of a cup of tea is T(t) = 20 + 70/(1 + 0.1t) °C (t in minutes). How fast is it cooling at t = 10?

Explain why the boat in Example 3 has d′(1.5) = 0 even though it is still moving.

More practice: warm-up

What is the first step in a rate problem?

Units of dI/dt if I is in lux and t in seconds?

A negative derivative for a volume means …

To find "when the rate equals 5", you …

To find "the rate at t = 5", you …

More practice: on your own

The height of a growing maple seedling is h(t) = 60t/(t + 4) cm, t in months. Find h′(t), the growth rate at t = 2, and the height it approaches.

A ladder-free problem: a kite is 30 m high and its horizontal distance from you is x(t) = 10 + 2t m. The string length is L = √(900 + x²). How fast is the string being let out at t = 5?

A company's profit is P(x) = x√(100 − x) thousand dollars for x thousand items (0 ≤ x ≤ 100). Find P′(x) and the rate at x = 36.

For the lamp in Example 4, at what time is the brightness falling at 2700/1000 = 2.7 lux/s?

Snow on a roof melts so that its depth is D(t) = (24 − t)²/24 cm, 0 ≤ t ≤ 24 (t in hours). How fast is it melting at t = 12?

Challenge

In Example 3, find the time when the distance is decreasing at 16 km/h.

A cone-shaped paper cup has height 12 cm and top radius 4 cm. Water fills it to depth h with volume V = (1/27)πh³ (from similar triangles, r = h/3). If h = 2√t cm after t seconds, how fast is the volume increasing at t = 4?

Show that for P(t) = 5000(t + 1)/(t + 5), the growth rate is always positive but always decreasing, and explain what this means.

Check your understanding

Quiz

For P(t) = 5000(t + 1)/(t + 5), what is P′(t)?

V(t) = 200(1 − t/20)². What is V′(10)?

In the lighthouse problem, what does d′(t) = 0 mean?

I = 900/d² and d = 2 + 1.5t. Which expression is dI/dt?

A cost model C(x) = 400 + 30√x. What is C′(25)?

A question asks "At what time is the population growing at 100 per year?" What should you do?

Why is the boat's distance from the lighthouse changing at less than 20 km/h at t = 2?

For parents and teachers

Big idea (A3.3, A3.5): the derivative rules solve real problems: how fast a population grows, how fast a barrel drains, how fast a distance changes. Your teen should model the situation, decide whether the question asks for a rate at a time or a time for a rate, differentiate, and answer in a sentence with units.

How to help: Ask your teen to read a problem aloud and tell you, before any algebra, "What is changing? With respect to what? What units will the answer have?" Then ask whether the sign of their answer makes sense (a draining barrel should have a negative rate).

Common difficulties:

Using the function instead of its derivative (P(5) instead of P′(5)).
Missing an inner derivative in a chain (for example the 1.5 in dd/dt).
Answers without units or without a sentence.
Forgetting the domain of the model (the barrel model only works until it is empty at t = 20).