Grade by Grade
A. Rate of Change · Derivatives and Differentiation Rules

Polynomial Models: Rates of Change and Higher Derivatives

A3.3About 70 minutes
Learning goals
  • I can interpret a derivative as a rate of change with units in a real context.
  • I can tell average rates of change (secant slopes) from instantaneous rates (derivatives).
  • I can find when a polynomial model reaches a given rate of change.
  • I can find second and higher derivatives and interpret the second derivative as a rate of change of a rate.

Words to know

marginal cost
The derivative C′(x) of a cost function: approximately the cost of producing one more item.
second derivative
The derivative of the derivative, written f″(x) or d²y/dx².
higher derivatives
Third, fourth and later derivatives: f‴(x), f⁽⁴⁾(x), …
velocity
The rate of change of position with respect to time, s′(t).
acceleration
The rate of change of velocity with respect to time, s″(t).

Derivatives with units

In real problems, the derivative is a rate with units: people per year, dollars per item, centimetres per second. The units of dy/dx are always (units of y) per (unit of x).

This lesson uses polynomial models from population studies, business and geometry. You will compare average and instantaneous rates, interpret derivatives in words, and take derivatives of derivatives (higher derivatives). The unit Applications of Derivatives builds on this with motion problems and optimization.

Average or instantaneous?

  • The average rate of change of f from x = a to x = b is the slope of the secant: [f(b) − f(a)] ÷ (b − a). It describes a whole interval.
  • The instantaneous rate of change at x = a is f′(a), the slope of the tangent. It describes one moment.

Read the question carefully: "How fast was the population growing in 2025?" asks for f′. "What was the average growth rate from 2020 to 2025?" asks for a secant slope.

Worked example

Example 1: A growing town

A town's population is modelled by P(t) = 12 000 + 150t + 6t², where t is the number of years since 2020. (a) Find the average rate of growth from 2020 to 2025. (b) Find the instantaneous rate of growth in 2025. (c) When will the town be growing at 300 people per year?

  1. (a) P(0) = 12 000 and P(5) = 12 000 + 750 + 150 = 12 900. Average rate = (12 900 − 12 000) ÷ 5 = 180 people/year.
  2. (b) P′(t) = 150 + 12t, so P′(5) = 150 + 60 = 210 people/year.
  3. (c) 150 + 12t = 300, so t = 12.5: halfway through 2032.

(a) 180 people/year (b) 210 people/year (c) t = 12.5, mid-2032. The rate keeps increasing because P′ increases with t.

tP24681012 00012 50013 00013 50014 000(5, 12 900)solid: secant from t = 0 to t = 5, slope 180 people/yeardashed: tangent at t = 5, slope 210 people/yearP(t)t (years since 2020)
The population model P(t) = 12 000 + 150t + 6t². The secant gives the average growth rate from 2020 to 2025 (180 people per year); the tangent gives the instantaneous rate in 2025 (210 people per year). The vertical axis starts at 11 800.
Worked example

Example 2: Marginal cost

A small shop in Hamilton makes wooden hockey sticks. The cost of making x sticks in a month is C(x) = 1500 + 12x − 0.004x² dollars, for 0 ≤ x ≤ 1000. (a) Find C′(500) and interpret it. (b) Compare with the actual cost of the 501st stick.

  1. (a) C′(x) = 12 − 0.008x, so C′(500) = 12 − 4 = 8.
  2. C′(500) = 8 dollars per stick: when 500 sticks are being made, each extra stick adds about $8 to the cost. Economists call C′ the marginal cost.
  3. (b) Actual cost of the 501st stick: C(501) − C(500) = 12(1) − 0.004(501² − 500²) = 12 − 0.004(1001) = 12 − 4.004 = 7.996.
  4. The marginal cost $8.00 is an excellent estimate of the true extra cost, $7.996.

C′(500) = $8 per stick, very close to the true cost of the 501st stick, $7.996.

Worked example

Example 3: A growing snowball

A snowball rolling down a hill is a sphere with volume V = (4/3)πr³. (a) Find dV/dr. (b) Evaluate it when r = 10 cm and explain what it means.

  1. (a) dV/dr = (4/3)π(3r²) = 4πr². (Here π is just a constant.)
  2. (b) At r = 10: 4π(100) = 400π ≈ 1256.6 cm³ per cm.
  3. Meaning: when the radius is 10 cm, the volume is increasing by about 1257 cm³ for each additional centimetre of radius (at that instant).

dV/dr = 4πr²; at r = 10 cm it is 400π ≈ 1257 cm³/cm. Notice that 4πr² is the surface area of the sphere: a thin new layer of snow adds volume of about (surface area) × (thickness).

Higher derivatives

The derivative of a polynomial is another polynomial, so you can differentiate again.

  • The second derivative f″(x) (read "f double prime") is the derivative of f′(x). In Leibniz notation it is d²y/dx².
  • The third derivative is f‴(x) or d³y/dx³, and after that we write f⁽⁴⁾(x), f⁽⁵⁾(x), …

The second derivative is the rate of change of the rate of change. If s(t) is a position, s′(t) is the velocity and s″(t) is the acceleration. In Curve Sketching: Connecting f, f′ and f″ you will see that f″ tells you which way a graph bends.

Worked example

Example 4: Derivatives of derivatives

For f(x) = x⁴ − 3x³ + 2x, find f′, f″, f‴, f⁽⁴⁾ and f⁽⁵⁾.

  1. f′(x) = 4x³ − 9x² + 2.
  2. f″(x) = 12x² − 18x.
  3. f‴(x) = 24x − 18.
  4. f⁽⁴⁾(x) = 24.
  5. f⁽⁵⁾(x) = 0, and every higher derivative is 0.

Each derivative lowers the degree by 1, so a degree-4 polynomial has a constant 4th derivative and every derivative from the 5th on is 0.

Worked example

Example 5: A thrown ball

A ball is thrown upward from a height of 1.5 m. Its height after t seconds is s(t) = −4.9t² + 20t + 1.5 metres. Find its velocity and acceleration at t = 1 s and t = 3 s.

  1. v(t) = s′(t) = −9.8t + 20 and a(t) = s″(t) = −9.8.
  2. t = 1: v = 10.2 m/s (moving up); a = −9.8 m/s².
  3. t = 3: v = −29.4 + 20 = −9.4 m/s (moving down); a = −9.8 m/s².
  4. The acceleration is constant: it is the acceleration due to gravity near Earth's surface, about 9.8 m/s² downward.

At 1 s: v = 10.2 m/s, a = −9.8 m/s². At 3 s: v = −9.4 m/s, a = −9.8 m/s².

t1020height s(t) = −4.9t² + 20t + 1.5 (m)t-2020velocity v(t) = s′(t) = −9.8t + 20 (m/s)t1234-10acceleration a(t) = s″(t) = −9.8 (m/s²)t ≈ 2.04 s: top of the flight, v = 0
A ball thrown upward. The height s is a parabola, its derivative v (velocity) is a line, and the second derivative a (acceleration) is the constant −9.8. At the top of the flight (dashed line) the velocity is 0.
Just for fun

The third derivative of position (the rate of change of acceleration) really is called jerk, and engineers who design elevators and subway trains work hard to keep it small so riders don't get, well, jerked around.

Hands-on activity

Marginal cost of a batch of cookies

You need: a grocery receipt or online prices for flour, sugar, butter and chocolate chips, a calculator and paper. (A grown-up should help if you go on to bake: ovens are hot.)

  1. Suppose a bake-sale table pays a fixed $10 for a table fee plus ingredients, and buying in bulk makes each extra dozen slightly cheaper. A model might be C(d) = 10 + 4.5d − 0.05d², where d is the number of dozens (0 ≤ d ≤ 20).
  2. Find C′(d). Then find C′(5) and C′(15) and explain what each means.
  3. Compare C′(5) with the actual extra cost C(6) − C(5).
  4. Using your own receipt, estimate the ingredient cost of one dozen and write your own cost model.

What to look for: C′(d) = 4.5 − 0.1d, so C′(5) = $4.00 per dozen and C′(15) = $3.00 per dozen: each extra dozen costs less as the order grows. C(6) − C(5) = 35.2 − 31.25 = $3.95, close to $4.00.

Your turn

1

If y is measured in litres and x in minutes, what are the units of dy/dx?

2

For P(t) = 12 000 + 150t + 6t², find P′(10) and interpret it.

3

For f(x) = x³ − x, find the average rate of change from x = 0 to x = 2, and f′(2).

4

Find f″(x) for f(x) = 5x³ − 2x² + 7.

5

Find d²y/dx² for y = x⁶.

6

The area of a circle is A = πr². Find dA/dr when r = 3 m and interpret it.

7

A stone dropped from a bridge has height h(t) = 40 − 4.9t² metres. Find its velocity at t = 2 s.

More practice: warm-up

1

What does f″(x) mean in words?

2

Find f″(x) if f(x) = x².

3

What is the 4th derivative of x³?

4

Units of C′(x) if C is in dollars and x is in items?

5

If s′(t) = −9.8t + 20, find s″(t).

More practice: on your own

1

A colony of bacteria has N(t) = 500 + 40t + 3t² cells after t hours. Find the average growth rate over the first 4 hours and the instantaneous rate at t = 4.

2

For C(x) = 1500 + 12x − 0.004x², find the number of sticks at which the marginal cost is $6 per stick.

3

For f(x) = 2x⁵ − x³, find f‴(x) and f‴(1).

4

A cube has volume V = x³ (x in cm). Find dV/dx when x = 5 and explain why it equals half the surface area.

5

The ball in Example 5: at what time is its velocity 0, and how high is it then?

Challenge

1

Find a polynomial f with f″(x) = 6x, f′(0) = 2 and f(0) = −1.

2

Show that y = x³ + 3x satisfies the equation y″ − 6x = 0 for all x.

3

The revenue from selling x tickets to a school play is R(x) = 15x − 0.01x² dollars (0 ≤ x ≤ 750). Find the marginal revenue at x = 300 and x = 750, and explain what the second value suggests.

Check your understanding

Quiz

1.

If V is in cubic metres and t in hours, the units of dV/dt are:

2.

For f(x) = x² + 1, the average rate of change from x = 1 to x = 3 is:

3.

What is f″(x) for f(x) = 4x³ − x?

4.

Which quantity is the second derivative of position with respect to time?

5.

C(x) = 200 + 5x − 0.002x² dollars. What is the marginal cost at x = 1000?

6.

For any polynomial of degree 3, the 4th derivative is:

For parents and teachers

Big idea (A3.3): a derivative is a rate with units. Your teen should be able to say what f′(a) means in a sentence, tell average rates (secant) from instantaneous rates (tangent), and take second and higher derivatives.

How to help: Whenever your teen gets a number such as P′(5) = 210, ask "210 what, per what?" Insist on a full sentence: "In 2025 the population is growing at about 210 people per year."

Common difficulties:

  • Answering an "average rate" question with f′ (or the reverse).
  • Forgetting units or using the units of f instead of the units of f′.
  • Confusing f″(x) with [f′(x)]² (the second derivative is not a square).
  • Mixing up "when" (solve f′(t) = value) with "how fast" (evaluate f′ at a given t).