D. Characteristics of Functions · Rates of Change

Solving Problems with Rates of Change

D1.1 D1.9About 75 minutes

Learning goals

I can decide whether a problem asks for an average or an instantaneous rate of change.
I can choose a numerical, graphical or algebraic method that fits the information I have.
I can solve multi-step problems, including finding when a rate reaches a given value.
I can check that my answer is reasonable and explain it with units and in context.

Words to know

reasonableness check
Asking whether the sign, size and units of an answer make sense in the situation.

rate of change of volume with respect to radius
How fast a volume grows per centimetre of radius, in cm³/cm.

guess and check (numerical solving)
Trying values, seeing whether the result is too big or too small, and narrowing in on the answer.

Mean Value Theorem (preview)
For a smooth curve, somewhere between a and b the tangent is parallel to the secant over [a, b]. You will study it in university calculus.

Putting it all together

You now have a full toolkit: secant slopes for average rates; tangents, shrinking intervals and the difference quotient for instantaneous rates; and patterns for polynomial, exponential, logarithmic, sinusoidal and rational functions. Real problems rarely say which tool to use. This lesson is about choosing and combining them.

The flowchart below is a good place to start every problem.

Choosing a method. First decide whether the question is about an interval (average) or a single moment (instantaneous). Then use the method that fits the information you have.

A problem-solving routine

Read for key words: "from… to", "over", "during" (average) or "at", "when", "at the moment" (instantaneous).
Identify the variables, their units, and the units of the rate.
Choose a method: formula for averages; difference quotient (polynomials, simple rationals), shrinking h (any equation), centred interval (tables) or a drawn tangent (graphs) for instantaneous rates.
Solve, keeping extra digits until the end.
Check: Is the sign right (growing or shrinking)? Is the size sensible? Does a different method agree?
Communicate: a sentence with the number, the units and what it means.

Worked example

Example 1: A growing balloon

A spherical balloon is being inflated. Its volume is V = (4/3)πr³ (cm³) when its radius is r cm. (a) Find the average rate of change of volume as r goes from 4 cm to 6 cm. (b) Find the instantaneous rate of change of volume with respect to radius when r = 5 cm. (c) Compare with the surface area at r = 5.

(a) V(6) ≈ 904.78 and V(4) ≈ 268.08. Average = (904.78 − 268.08) ÷ 2 ≈ 318.35 cm³/cm.
(b) Centred interval with h = 0.001: (V(5.001) − V(4.999)) ÷ 0.002 ≈ 314.16 cm³/cm.
(c) Surface area 4πr² = 4π(25) ≈ 314.16 cm².
Check: (b) is close to (a), as expected for an interval centred at 5, and positive because the volume grows.

(a) about 318 cm³ per cm; (b) about 314 cm³ per cm; (c) the instantaneous rate equals the surface area. That makes sense: a tiny increase in radius adds a thin shell whose volume is about (surface area) × (thickness).

Worked example

Example 2: When does the rate reach a target?

A town's population is P(t) = 12 000(1.018)^t, t years from now. Planners need a new school when the town is growing by 300 people per year. When will that happen?

From Rates of Change of Exponential and Logarithmic Functions, the growth rate is k × P(t), where k is the slope of 1.018ᵗ at t = 0.
Estimate k: (1.018^0.001 − 1) ÷ 0.001 ≈ 0.017 84.
Set k × P = 300: P = 300 ÷ 0.017 84 ≈ 16 816 people.
Solve 12 000(1.018)^t = 16 816 with logarithms (from Exponential and Logarithmic Functions): 1.018^t ≈ 1.4013, so t = log 1.4013 ÷ log 1.018 ≈ 18.9.
Check with a centred interval at t = 18.9: the rate is about 300 people per year ✓.

In about 19 years (when the population is about 16 800), the town will be growing by 300 people per year.

Worked example

Example 3: Where does the tangent match the secant?

For f(x) = x², find the average rate of change over [1, 3], then find the point in that interval where the instantaneous rate of change equals it.

Average: (9 − 1) ÷ (3 − 1) = 4.
The tangent slope of x² at x = a is 2a (from The Difference Quotient lesson).
Set 2a = 4: a = 2. The point is (2, 4).

At x = 2, the tangent is parallel to the secant. University calculus proves this always happens for smooth functions (the Mean Value Theorem): on a trip with an average speed of 80 km/h, your speedometer read exactly 80 km/h at least once.

The secant over [1, 3] and the tangent at x = 2 on y = x² are parallel: both have slope 4.

Worked example

Example 4: Best box

An open box is made from a 30 cm by 20 cm card by cutting x cm squares from the corners: V(x) = x(30 − 2x)(20 − 2x). At x = 4 cm, should the cut be made larger or smaller to get more volume?

Estimate the rate of change of V at x = 4: (V(4.001) − V(3.999)) ÷ 0.002 ≈ −8 cm³ per cm.
The rate is negative: making the cut bigger would decrease the volume.
Try x = 3.9: rate ≈ 2.5 (positive). So the rate changes sign between 3.9 and 4: the maximum volume is there.

Make the cut slightly smaller. The maximum volume (about 1056 cm³) happens at a cut of about 3.9 cm, where the rate of change is 0. This is a preview of optimization in MCV4U.

Just for fun

The balloon in Example 1 shows a neat fact: the rate of change of a sphere's volume is its surface area, and the rate of change of a circle's area is its circumference (2πr). Math is basically one big set of nesting dolls.

Hands-on activity

The melting ice cube

You need: an ice cube from the freezer, a plate, a ruler, a timer, paper and a pencil. (Wipe up any water on the floor right away.)

Put the ice cube on the plate at room temperature. Measure one side length (to the nearest millimetre) at t = 0 and then every 5 minutes for about 40 minutes.
Make a table of side length s (cm) against time t (min).
Calculate the average rate of change over the whole time, and over each 5-minute interval.
Estimate the instantaneous rate of change at t = 15 min with a centred interval [10, 20].
Plot the data, draw a smooth curve, draw a tangent at t = 15 and find its slope. Compare with step 4.
Write a sentence interpreting your rate at t = 15, with units.

What to look for: all rates are negative (the cube shrinks). Typical values are a few hundredths of a centimetre per minute, depending on the room. Your centred estimate and your tangent slope should roughly agree; differences come from measuring an irregular, slippery cube.

Your turn

For the balloon V = (4/3)πr³, estimate the rate of change of volume with respect to radius when r = 10 cm.

For a circle, A = πr². Find the instantaneous rate of change of area with respect to radius at r = 3 cm. What familiar quantity is it?

For P(t) = 12 000(1.018)^t, estimate the growth rate now (t = 0).

For f(x) = x² on [0, 4], find the average rate of change and the point where the tangent is parallel to the secant.

For the ball h(t) = −4.9t² + 20t + 1.5 with velocity v(t) = 20 − 9.8t, when is the ball falling at 10 m/s? Check that it has not landed yet.

For the Ferris wheel h(t) = 21 − 20 cos(πt/30), compare the average vertical rate over the first 20 s with the instantaneous rate at t = 20 s.

For the car V(t) = 28 000(0.85)^t, the rate of change is about −0.1625 × V(t). When is the car losing value at $2000 per year?

More practice: warm-up

Average or instantaneous: "How fast is the tank filling at 3 p.m.?"

You have water-level data every 10 minutes. Which interval gives the best estimate of the rate at t = 20 min?

What are the units of the rate of change of volume (cm³) with respect to radius (cm)?

A tangent passes through (0, 5) and (4, 13). What is its slope?

For y = x² on [2, 4], at which x is the tangent parallel to the secant?

More practice: on your own

Tea cools as T(t) = 20 + 70(0.9)^t (°C, t in minutes). The cooling rate is about −0.1054 × (T − 20). When is the tea cooling at 2 °C per minute?

For C(t) = 5t/(t² + 1) mg/L, is the concentration rising or falling at t = 1.5 h, and how fast?

For the box V(x) = x(30 − 2x)(20 − 2x), estimate the rate of change at x = 2 and at x = 6. Interpret both.

On a graph of temperature against time, a tangent at noon passes through (10:00, −6 °C) and (13:00, 3 °C). How fast is the temperature rising at noon?

A square has side s cm and area A = s². Find the rate of change of area at s = 5 cm and explain it with a picture in words.

Challenge

Use the difference quotient to show that the rate of change of V = (4/3)πr³ with respect to r at r = a is 4πa², the surface area.

For the box V(x) = x(30 − 2x)(20 − 2x), use centred estimates and guess-and-check to find, to two decimal places, the cut x where the rate of change is 0. What is the maximum volume?

For the car V(t) = 28 000(0.85)^t, find the average rate of change over the first 6 years, then find when the instantaneous rate equals it.

Check your understanding

Quiz

"How fast is the balloon's volume changing when its radius is 5 cm?" What kind of rate does this ask for?

For a sphere, V = (4/3)πr³. What is the rate of change of volume with respect to radius at r = 5 cm?

A table gives data every 10 minutes. What is the best estimate of the instantaneous rate at t = 20 min?

For f(x) = x², the average rate of change over [1, 3] is 4. At which x is the tangent parallel to that secant?

For the box V(x) = x(30 − 2x)(20 − 2x), the rate of change at x = 4 is about −8 cm³ per cm. What should you do to increase the volume?

Which is the best quick check that an answer to a rate problem is reasonable?

For parents and teachers

Big ideas (D1.1, D1.9): solving real problems means first deciding between average and instantaneous rates, then choosing a method that fits the information (equation, table or graph), and finally checking and explaining the answer with units.

How to help: Ask your teen to talk through a problem out loud before calculating: "Is this over an interval or at a moment? What do you know? What method will you use?" Then ask: "Does the sign make sense? Are the units right?"

Common difficulties:

Answering an "at" question with an average over a long interval.
Losing track of what the rate means (for example, mixing up the population and its growth rate).
Rounding k or intermediate values too early in multi-step problems.
Forgetting to check that the answer fits the situation (for example, a ball that has already landed).

Looking ahead: the box problem (Example 4) is an optimization problem. In MCV4U you will solve problems like it exactly, using derivatives.