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1

Find the average rate of change or average velocity

f

1
$[a, b]$
2
$f(a)$ $Use the original function, not its derivative.$
3
$\dfrac{f(b) - f(a)}{b - a}$
4
$f$ $50(e^{0.6} - e^{0.2})$

Remember: Change in output over change in input.

See it in a problem3 worked examples

s(t) = t^2 - 4t + 7

1
$[a, b]$ $[a, b] = [1, 4]$
2
$f(a)$ $s(1) = 1 - 4 + 7 = 4$
3
$\dfrac{f(b) - f(a)}{b - a}$ $\dfrac{7 - 4}{4 - 1} = \dfrac{3}{3} = 1$
4
$f$ $1$

1

Common mistakes3

Computing $f'$ at all. The average rate needs only two values of $f$ — no derivative.
Averaging two velocities. Average velocity is (change in position) ÷ (change in time), not $\dfrac{v(a) + v(b)}{2}$ .
Subtracting in different orders: $\dfrac{f(b) - f(a)}{b - a}$ , later minus earlier on top and bottom.

2

Find the instantaneous rate of change or velocity

t = 4

1
$Differentiate the quantity with respect to its variable$
2
$Plug in the given moment$
3
$Attach units: units of the quantity per unit of input$
4
$Interpret the sign in words$ $Positive: increasing (rising, moving forward). Negative: decreasing (falling, moving backward, draining).$

Remember: "How fast" means derivative.

See it in a problem3 worked examples

h(t) = 30t - 4.9t^2

1
$Differentiate the quantity with respect to its variable$ $h'(t) = 30 - 9.8t$
2
$Plug in the given moment$ $h'(4) = 30 - 39.2 = -9.2$
3
$Attach units: units of the quantity per unit of input$ $-9.2$
4
$Interpret the sign in words$ $9.2$

h'(4) = -9.2

Common mistakes3

Plugging the moment into $f$ instead of $f'$ . $h(4)$ is a height; $h'(4)$ is a velocity.
Dropping a negative sign. $-9.2$ m/s means falling at $9.2$ m/s; the sign is the direction.
Wrong units. A derivative of liters with respect to minutes is in L/min, not L.

3

Solve an exponential growth or decay problem

e.g. “ grows at a rate proportional to its size ”

1
$Q(t) = Q_0e^{kt}$ $\dfrac{dQ}{dt} = kQ$
2
$Q_0 = Q(0)$ $1000$
3
$k$ $e^{kT} = \text{ratio}$
4
$t$

k

See it in a problem3 worked examples

500

1
$Q(t) = Q_0e^{kt}$ $Q(t) = Q_0e^{kt}$
2
$Q_0 = Q(0)$ $Q_0 = 500$
3
$k$ $1500 = 500e^{2k} \Rightarrow e^{2k} = 3 \Rightarrow k = \dfrac{\log 3}{2}$
4
$t$ $Q(6) = 500e^{3\log 3} = 500 \cdot 27 = 13\,500$

13\,500

Common mistakes3

A positive $k$ for decay. Decay has $k < 0$ ; check that $Q$ shrinks.
Using a base-10 log (the calculator's "log" key on most models). Here $\log$ is the natural log, the inverse of $e^x$ .
Forgetting to divide by $Q_0$ before taking the log: from $1500 = 500e^{2k}$ , $e^{2k} = 3$ .

4

Solve a Newton's law of cooling problem

e.g. “ An object cools (or warms) at a rate proportional to the difference between its temperature and the surroundings ”

1
$T(t) - T_a = (T_0 - T_a)e^{kt}$ $T_a$
2
$T_a$
3
$k$ $T'(0)$
4
$t$ $\dfrac{dT}{dt} = k(T - T_a)$

Remember: The difference from the room decays.

See it in a problem2 worked examples

90^\circ

1
$T(t) - T_a = (T_0 - T_a)e^{kt}$ $T(t) - T_a = (T_0 - T_a)e^{kt}$
2
$T_a$ $T - 20 = 70e^{kt}$
3
$k$ $76 - 20 = 70e^{5k} \Rightarrow e^{5k} = 0.8 \Rightarrow k = \dfrac{\log 0.8}{5} \approx -0.0446$
4
$t$ $48 - 20 = 70e^{kt} \Rightarrow e^{kt} = 0.4 \Rightarrow t = \dfrac{5\log 0.4}{\log 0.8} \approx 20.5$

t = \dfrac{5\log 0.4}{\log 0.8} \approx 20.5

Common mistakes3

Writing $T = T_0e^{kt}$ . It is the difference $T - T_a$ that is exponential.
Mixing up $T_a$ and $T_0$ : $T_a$ is the room, $T_0$ the object at $t = 0$ .
For warming, $T_0 - T_a$ is negative. Keep the sign; $k$ is still negative.

Rates of Change, Velocity & Growth

Find the average rate of change or average velocity

Find the instantaneous rate of change or velocity

Solve an exponential growth or decay problem

Solve a Newton's law of cooling problem

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Find the average rate of change or average velocity

Find the instantaneous rate of change or velocity

Solve an exponential growth or decay problem

Solve a Newton's law of cooling problem