24.5 Rates of Change
SL 5.1
- the difference between a constant, an average and an instantaneous rate of change
- that the derivative \(\frac{dy}{dx}\) is the instantaneous rate of change of \(y\) with respect to \(x\)
- how to interpret derivatives in context (for example speed, marginal cost, rate of flow)
Average rate of change
A rate of change tells us how one quantity changes compared with another. Speed (metres per second) and heart rate (beats per minute) are examples.
For a function \(y = f(x)\), the average rate of change of \(y\) with respect to \(x\) between the points \((x_1, y_1)\) and \((x_2, y_2)\) is
\[ \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{change in } y}{\text{change in } x} . \]
This is the gradient of the straight line (the chord) joining the two points.
- For a linear function \(y = mx + c\), the gradient is always \(m\). We say the rate of change is constant. For example, if an electricity bill is \(C = 0.3n + 12\) dollars for \(n\) units, then \(C\) increases at a constant rate of \(0.3\) dollars per unit.
- For a non-linear function, the gradient changes from point to point. The gradient of a chord gives only the average rate of change over that interval.
For example, if a car has travelled \(s = 80\) m after \(5\) s and \(s = 200\) m after \(8\) s, its average speed over this interval is \(\frac{200 - 80}{8 - 5} = 40\) m s\(^{-1}\).
Instantaneous rate of change
Take two points \(P(x_1, y_1)\) and \(Q(x_2, y_2)\) on a curve. As \(Q\) moves towards \(P\) (so \(x_2 - x_1 \to 0\)), the chord \(PQ\) gets closer and closer to the tangent at \(P\).
Instantaneous rate of change Not in the formula booklet
The instantaneous rate of change of \(y = f(x)\) with respect to \(x\) at \(x = x_1\) is the limit of the average rate of change. It is the gradient of the tangent:
\[ \left.\frac{dy}{dx}\right|_{x = x_1} = f'(x_1) . \]
- \(x_1\): the value of \(x\) where we find the rate of change
- \(f'(x_1)\): the gradient of the tangent at \(x = x_1\)
The limit idea is studied more carefully in Unit 25.
Example 1 The distance–time graph shows a cyclist’s first \(55\) seconds. It is made of two line segments \(P\) and \(Q\), and a curve \(R\) with equation \(s = 30 + \frac{1}{25}(t - 30)^2\) for \(30 \le t \le 55\), where \(s\) is in metres and \(t\) in seconds.
(a)
(i) Find the gradients of \(P\) and \(Q\).
(ii) What type of rate of change do they show?
(iii) Describe the motion for \(0 \le t \le 30\).
(b) Find the gradients of the chords \(AB\), \(AC\) and \(AD\).
(c)
(i) Explain how to find the instantaneous rate of change at \(A\).
(ii) Which chord in (b) gives the best estimate? Find the exact value.
(iii) Is the speed constant, increasing or decreasing for \(30 < t < 55\)?
(a)
(i) \(P\): \(\frac{30 - 0}{15 - 0} = 2\). \(\quad\) \(Q\): \(\frac{30 - 30}{30 - 15} = 0\).
(ii) Straight lines have constant rates of change.
(iii) For \(0 \le t \le 15\) the cyclist moves at a constant speed of \(2\) m s\(^{-1}\). For \(15 \le t \le 30\) she is not moving (speed \(0\)).
(b) \(AB\): \(\frac{39 - 34}{45 - 40} = 1\). \(\quad\) \(AC\): \(\frac{46 - 34}{50 - 40} = 1.2\). \(\quad\) \(AD\): \(\frac{55 - 34}{55 - 40} = 1.4\). These are average rates of change (average speeds).
(c)
(i) Find the gradient of the tangent at \(A\): take chords \(AX\) with \(X\) closer and closer to \(A\), or differentiate.
(ii) \(AB\) is the best, because \(B\) is closest to \(A\). Exactly: \(\frac{ds}{dt} = \frac{2}{25}(t - 30)\), so at \(t = 40\) the speed is \(\frac{2}{25} \times 10 = 0.8\) m s\(^{-1}\).
(iii) Increasing: the curve gets steeper (it is concave up), and \(\frac{ds}{dt} = \frac{2}{25}(t - 30)\) increases with \(t\).
Interpreting derivatives in context
If \(y\) depends on \(x\), then \(\frac{dy}{dx}\) has units “units of \(y\) per unit of \(x\)”. Always give the units, and say in words what the number means.
Example 2 A patch of algae on a lake grows so that its area after \(t\) days is \(A = 9t\) km\(^2\).
(a) The patch is modelled as a circle. Write the radius \(r\) in terms of \(t\).
(b) Find \(\frac{dr}{dt}\).
(c) How fast is the radius increasing after \(4\) days?
(a) \(\pi r^2 = 9t\), so \(r = \sqrt{\frac{9t}{\pi}} = \frac{3}{\sqrt{\pi}}\, t^{\frac{1}{2}}\).
(b) \(\frac{dr}{dt} = \frac{3}{\sqrt{\pi}} \times \frac{1}{2}t^{-\frac{1}{2}} = \frac{3}{2\sqrt{\pi t}}\) km per day.
(c) \(t = 4\): \(\frac{dr}{dt} = \frac{3}{2\sqrt{4\pi}} = \frac{3}{4\sqrt{\pi}} \approx 0.423\) km per day.
In economics, the derivative of the cost function \(C(x)\) is called the marginal cost \(C'(x)\). It estimates the cost of making one more item.
Example 3 The cost of making \(x\) chairs per week is \(C(x) = 2500 + 8x + 0.02x^2\) dollars.
(a) Find the marginal cost function \(C'(x)\).
(b) Find \(C'(150)\). What does it estimate?
(c) Find \(C(151) - C(150)\) and compare it with your answer to (b).
(a) \(C'(x) = 8 + 0.04x\).
(b) \(C'(150) = 8 + 6 = 14\). It estimates that the \(151\)st chair costs about \(14\) dollars to make (when \(150\) chairs are already being made).
(c) \(C(151) - C(150) = 8(1) + 0.02(151^2 - 150^2) = 8 + 0.02 \times 301 = 14.02\) dollars. This is the exact cost of the \(151\)st chair. It is very close to \(C'(150)\), because the graph of \(C\) is almost straight over one unit.
Example 4 A pool holds \(30\,000\) litres of water. When the drain is opened, all the water runs out in \(60\) minutes. The volume left after \(t\) minutes is
\[ V = 30\,000\left(1 - \frac{t}{60}\right)^2 \text{ litres}, \qquad 0 \le t \le 60 . \]
(a) Find \(\frac{dV}{dt}\), and sketch the graph of \(\frac{dV}{dt}\) against \(t\).
(b) At what time is the water flowing out fastest?
(c) Show that \(\frac{d^2V}{dt^2}\) is constant and positive. Interpret this.
(a) Chain rule:
\[ \frac{dV}{dt} = 30\,000 \times 2\left(1 - \frac{t}{60}\right) \times \left(-\frac{1}{60}\right) = -1000\left(1 - \frac{t}{60}\right) = -1000 + \frac{50}{3}t . \]
The graph is a straight line from \((0, -1000)\) to \((60, 0)\).
(b) The outflow rate is \(-\frac{dV}{dt}\). It is largest when \(t = 0\) (at the start): \(1000\) litres per minute.
(c) \(\frac{d^2V}{dt^2} = \frac{50}{3} > 0\) for all \(t\). The rate \(\frac{dV}{dt}\) increases (from \(-1000\) towards \(0\)) at a constant rate, so the water flows out more and more slowly, and the outflow rate falls by \(\frac{50}{3}\) litres per minute every minute.
HL students: continue with 24.6 Related Rates.
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| Average rate used as the instantaneous rate | Instantaneous rate \(= f'(x_1)\) | The chord gradient is only an approximation. |
| Answer without units or meaning | “decreasing at \(8\) km h\(^{-1}\)” | Rates in context need units and a direction. |
Exercises
1 For \(f(x) = x^2 + 3x\), find
(a) the average rate of change of \(f\) between \(x = 1\) and \(x = 4\),
(b) the instantaneous rate of change at \(x = 1\).
(a) \(f(1) = 4\), \(f(4) = 28\): \(\frac{28 - 4}{4 - 1} = 8\).
(b) \(f'(x) = 2x + 3\), \(f'(1) = 5\).
2 A cup of tea cools so that its temperature after \(t\) minutes is \(T = 20 + 80\mathrm{e}^{-0.1t}\) °C. Find the rate of change of temperature
(a) at \(t = 0\),
(b) at \(t = 10\) (exact value). Explain the sign.
\(\frac{dT}{dt} = 80 \times (-0.1)\mathrm{e}^{-0.1t} = -8\mathrm{e}^{-0.1t}\).
(a) \(-8\) °C per minute.
(b) \(-8\mathrm{e}^{-1} = -\frac{8}{\mathrm{e}}\) °C per minute (\(\approx -2.94\)).
The sign is negative because the tea is cooling (the temperature decreases).
3 A shop’s daily profit from selling \(x\) lamps is \(P(x) = 40x - \frac{1}{2}x^2 - 200\) dollars.
(a) Find the marginal profit \(P'(x)\).
(b) Find \(P'(30)\) and explain what it means.
(c) Find the value of \(x\) for which the marginal profit is zero. What does this tell the shop?
(a) \(P'(x) = 40 - x\) dollars per lamp.
(b) \(P'(30) = 10\). When \(30\) lamps are sold, selling one more lamp adds about \(10\) dollars to the profit.
(c) \(40 - x = 0\) gives \(x = 40\). For \(x < 40\) the profit increases, and for \(x > 40\) it decreases (\(P'(x) < 0\)). So selling about \(40\) lamps a day gives the largest profit.