Lesson 2 of 4Applications of Derivatives

Unit 4 · Lesson 2 of 4

Rates of Change in Context

Given a model for a population, a decaying sample, a cooling drink, hours of daylight, a tide or a price index, the derivative gives the instantaneous rate of change. The skill is computing it with the right rule and then saying what it means, with units.

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Key things to remember

1The units of $Q^{'} (t)$ are (units of $Q$ ) per (unit of $t$ ); the sign says increasing or decreasing.
2 $\frac{d}{d t} b^{t} = b^{t} ln b$ and $\frac{d}{d t} e^{k t} = k e^{k t}$ : exponential growth and decay rates are proportional to the amount present.
3For a sinusoidal model $A sin (k (t - c)) + d$ , the rate is $A k cos (k (t - c))$ (radians); it is fastest at the midline and zero at a maximum or minimum.
4Inflation: $P^{'} (t) > 0$ ; deflation: $P^{'} (t) < 0$ ; inflation is fastest where $P^{''} (t) = 0$ and $P^{''}$ changes from positive to negative.
5Relative (percent) rate of change: $\frac{Q ^{'} ( t )}{Q ( t )} \times 100%$ .

Biologists counting butterflies, nurses timing a medical isotope, and economists tracking prices all ask the same question: how fast is this quantity changing RIGHT NOW? A model gives the quantity as a function of time, and the derivative of that function is the instantaneous rate of change. You already have every rule you need from the last unit. What this lesson adds is the habit of reading the answer: its sign (growing or shrinking), its size, and its units.

Reading a derivative in context

Q (t)

is measured in some unit and

t

in another, then

Q^{'} (t)

is measured in (units of

Q

) per (unit of

t

). Its sign says whether

Q

is increasing or decreasing at that instant, and its size says how fast. A statement such as "

P^{'} (5) \approx 505

" is not an answer until it reads "after 5 days the population is growing by about 505 butterflies per day".

VideoPractice turning a derivative value into a sentence with its sign, size and units, the habit this whole lesson is built on.Watch on YouTube

Population growth: the butterfly model

A population of butterflies is modelled by $P (t) = \frac{6000}{1 + 49 ( 0.6 ) ^{t}}$ , where $t$ is the time in days. At $t = 0$ there are $\frac{6000}{50} = 120$ butterflies. As $t$ grows, $(0.6)^{t}$ shrinks toward $0$ , so $P$ levels off near $6000$ , the carrying capacity. To differentiate, write $P (t) = 6000 (1 + 49 (0.6)^{t})^{- 1}$ and use the chain rule, with $\frac{d}{d t} (0.6)^{t} = (0.6)^{t} ln 0.6$ .

P^{'} (t) = - 6000 (1 + 49 (0.6)^{t})^{- 2} \cdot 49 (0.6)^{t} ln 0.6 = \frac{- 294 000 ( 0.6 ) ^{t} ln 0.6}{( 1 + 49 ( 0.6 ) ^{t} ) ^{2}}

ln 0.6 is negative, so the leading minus sign makes P′(t) positive: the population is always growing.

h	Secant slope from t = 5 to t = 5 + h (butterflies per day)
$1$	$578.51$
$0.1$	$512.24$
$0.01$	$505.46$
$0.001$	$504.79$
Derivative $P^{'} (5)$	$504.71$

Numerical check: secant slopes (P(5 + h) − P(5))/h close in on the value from the derivative as h shrinks.

Fig 1Exponential (J-shaped) and logistic (S-shaped) population growth from the same start. The logistic curve levels off at a dashed carrying capacity and grows fastest at half the carrying capacity, the curve's inflection point. The figure uses a carrying capacity of 500; the butterfly model has the same S shape, with carrying capacity 6000.

Decay and cooling

Radioactive decay: a $50$ mg sample of an isotope with a half-life of $8$ days follows $A (t) = 50 (0.5)^{\frac{t}{8}}$ . Then $A^{'} (t) = 50 (0.5)^{\frac{t}{8}} \cdot \frac{ln 0.5}{8}$ . At the start $A^{'} (0) \approx - 4.33$ mg/day; after one half-life $A^{'} (8) \approx - 2.17$ mg/day. The rate halves when the amount halves, because $A^{'} (t)$ is a constant multiple of $A (t)$ .
Newton's law of cooling: a drink at $90° C$ in a $20° C$ room follows $T (t) = 20 + 70 e^{- 0.05 t}$ , with $t$ in minutes. Then $T^{'} (t) = - 3.5 e^{- 0.05 t}$ . At first it cools at $3.5° C/min$ ; after $10$ minutes, $T^{'} (10) \approx - 2.12° C/min$ . The cooling slows as the drink approaches room temperature.

Periodic rates: daylight and tides

Sinusoidal models need the chain rule with radians. At a latitude like Toronto's, the hours of daylight on day $t$ of the year can be modelled by $D (t) = 12.2 + 3.2 sin (\frac{2 π}{365} (t - 80)) .$ It is only a model: it runs from $9.0$ to $15.4$ hours, while Toronto's real day length runs from about 8 h 55 min at the December solstice to about 15 h 26 min at the June solstice. Then $D^{'} (t) = 3.2 \cdot \frac{2 π}{365} cos (\frac{2 π}{365} (t - 80))$ hours per day. The cosine is largest when $t = 80$ (near the March equinox), where the model gives $D^{'} (80) = \frac{6.4 π}{365} \approx 0.0551$ h/day, about $3.3$ minutes of extra daylight each day. Near the solstices the cosine is close to $0$ , so the day length barely changes from one day to the next.

Tides work the same way. If the water depth in a harbour is $h (t) = 3 + 2.5 cos (\frac{2 π t}{12.4})$ metres, $t$ hours after high tide, then $h^{'} (t) = - 2.5 \cdot \frac{2 π}{12.4} sin (\frac{2 π t}{12.4}) .$ The water falls fastest a quarter-period after high tide, at $t = 3.1$ h, where $h^{'} (3.1) = - \frac{5 π}{12.4} \approx - 1.27$ m/h. That is the midpoint between high and low water, not the high or low point itself.

Prices, inflation, volume and flow

If $P (t)$ is a price index, then $P^{'} (t) > 0$ means prices are rising (inflation) and $P^{'} (t) < 0$ means they are falling (deflation). The inflation rate is greatest where $P^{'}$ has a maximum, which (inside the time interval) is where $P^{''} (t) = 0$ and changes from positive to negative: an inflection point of the price graph. Economists usually quote inflation as a percentage, the relative rate $\frac{P ^{'} ( t )}{P ( t )} \times 100%$ . In the same way, if $V (t)$ is the volume of water in a tank, $V^{'} (t)$ is the rate of flow in or out, and if $H (t)$ is a child's height, $H^{'} (t)$ is the growth rate in centimetres per year.

Fig 2A secant line drawn three times with a shrinking interval h of 3, 1.5 and 0.5; its slope falls from 2.5 to 1.75 to 1.25, closing in on the tangent slope of 1. This is how a graphing tool checks an instantaneous rate: zoom in until the secant and the tangent agree.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

A butterfly population is

P (t) = \frac{6000}{1 + 49 ( 0.6 ) ^{t}}

t

in days. Find the rate of growth after

5

days, exactly and to one decimal place.

Example 2

A drink cools according to

T (t) = 20 + 70 e^{- 0.05 t}

(

° C

t

in minutes). How fast is it cooling at

t = 10

, and when is it cooling at exactly

1° C

per minute?

Example 3

Daylight is modelled by

D (t) = 12.2 + 3.2 sin (\frac{2 π}{365} (t - 80))

hours, where

t

is the day of the year. How fast is the day length changing on day

45

(mid-February), in minutes per day?

Example 4

A price index is modelled by

P (t) = - 0.1 t^{3} + 1.5 t^{2} + 2 t + 100,

t

in years. When is the index rising fastest (the rate

P^{'} (t)

greatest), and what is that rate, in index points per year and as a percentage of the index at that time?

Where marks get lost

Treating $0. 6^{t}$ like a power function and writing its derivative as $t (0.6)^{t - 1}$ . The variable is in the exponent, so $\frac{d}{d t} (0.6)^{t} = (0.6)^{t} ln 0.6$ .
Forgetting the inner derivative in a sinusoidal model: $\frac{d}{d t} sin (\frac{2 π t}{12.4})$ is $\frac{2 π}{12.4} cos (\frac{2 π t}{12.4})$ , not just $cos (\frac{2 π t}{12.4})$ .
Evaluating trig models in degree mode. The derivative rules for sine and cosine hold only in radians.
Reporting a rate without units or interpretation, or mixing units: a daylight rate in hours per day must be multiplied by 60 to give minutes per day.
Thinking the fastest change happens at the highest point. At a maximum of a smooth model the rate is $0$ ; for tides and daylight the fastest change is at the midline.

Flashcards 8

Units of

Q^{'} (t)

\frac{d}{d t} b^{t}

Derivative of

A e^{k t}

Inflation vs deflation

+ 4 more

Practice

P (t)

is a town's population and

t

is in years. What are the units of

P^{'} (t)

A sample decays as

A (t) = 50 (0.5)^{\frac{t}{8}}

mg,

t

in days. Find

A^{'} (8)

in mg/day, to 2 decimal places.

mg/day

For the butterfly model

P (t) = \frac{6000}{1 + 49 ( 0.6 ) ^{t}}

, find

P^{'} (10)

to the nearest whole number.

butterflies/day

Harbour depth is

h (t) = 3 + 2.5 cos (\frac{2 π t}{12.4})

t

hours after high tide. Find

h^{'} (2)

in m/h, to 2 decimal places.

m/h

A drink cools as

T (t) = 20 + 70 e^{- 0.05 t}

. At what time, in minutes, is it cooling at

2° C

per minute? Give one decimal place.

min

A tank holds

V (t) = 1000 (1 - \frac{t}{40})^{2}

litres of water as it drains,

0 \leq t \leq 40

minutes. Find

V^{'} (10)

in litres per minute.

L/min

Match each derivative to what it measures.

V^{'} (t)

, where

V

is the water in a tank

P^{'} (t)

, where

P

is a price index

H^{'} (t)

, where

H

is a child's height

T^{'} (t)

, where

T

is a drink's temperature

The graph of a price index rises for 10 years in an S shape: slowly at first, then more and more steeply, then more gently again. At which point on the graph was inflation fastest?

For

D (t) = 12.2 + 3.2 sin (\frac{2 π}{365} (t - 80)),

which is true near the day with the most daylight?

On an S-shaped (logistic) population graph, the point where the population is growing fastest is called the point of ______.

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