Before launching into the many applications of differentiation, let's recap a few key ideas around rates of change.
What does rate of change mean?
When talking about the rate of change, we're referring to the rate at which the dependent variable, often denoted by $y$y, increases or decreases as the independent variable, often denoted by $x$x, increases.
For example, we can talk about the rate at which the volume of a cylinder increases as the radius increases or we can talk about the rate at which the price of a commodity decreases as time increases.
Average rate of change vs instantaneous rate of change
The average rate of change is the change in the $y$y values divided by the change in the $x$x values over a given interval for $x$x. This is equivalent to calculating the gradient of the secant between two points on a curve.
The instantaneous rate of change is given by the gradient of the tangent at a given point on the curve. We initially estimated this but we can now use differentiation to calculate this value exactly. We first differentiate to find the gradient function, then we substitute a particular value into this gradient function to calculate the instantaneous rate of change at that value.
The average rate of change of a function, $f\left(x\right)$f(x), from $x=a$x=a to $x=b$x=b, is given by:
$\text{Average rate of change}=\frac{f\left(b\right)-f\left(a\right)}{b-a}$Average rate of change=f(b)−f(a)b−a
If $A$A is the point on $f\left(x\right)$f(x) at $x=a$x=a and $B$B is the point of $f\left(x\right)$f(x) at $x=b$x=b, then the average rate of change is the gradient of the secant through points $A$A and $B$B.
The instantaneous rate of change of a function, $y=f\left(x\right)$y=f(x), at $x=a$x=a, is given by:
The derivative evaluated at $x=a$x=a, denoted $f'\left(a\right)$f′(a)
This is the gradient of the tangent to the curve at $x=a$x=a.
When answering questions about practical applications, it's very important to ensure you read carefully to ascertain whether you're being asked for the average rate of change or the instantaneous rate of change. Also remember to include units for the rate of change. Units for a rate of change are the dependent variable units over the independent variable units.
Worked example
Example 1
The profit function, in dollars, for producing $x$x hundred items is given by $P\left(x\right)=6x^3-122x^2+490x-150$P(x)=6x3−122x2+490x−150.
(a) Calculate the average profit for producing and selling the first $200$200 items.
Think: As we've been asked for the average profit, we need to calculate the change in profit over the change production for the first $200$200 units. That is, the rate of change between $x=0$x=0 and $x=2$x=2. Note $x$x is in units of $100$100 items.
Do:
| Average rate of change | $=$= | $\frac{P\left(2\right)-P\left(0\right)}{2-0}$P(2)−P(0)2−0 |
| $=$= | $\frac{390-\left(-150\right)}{2}$390−(−150)2 | |
| $=$= | $270$270 |
Therefore, the average profit in this interval is $\$270$$270 per $100$100 items.
(b) Calculate the instantaneous rate of change of profit at the point where $200$200 items are produced.
Think: The instantaneous rate of change of profit at the point where $200$200 items are produced is equivalent to asking the gradient of the tangent at $x=2$x=2 or $P'\left(2\right)$P′(2).
Do:
| $P'\left(x\right)$P′(x) | $=$= | $18x^2-244x+490$18x2−244x+490 |
Differentiate $P\left(x\right)$P(x) to find the gradient function |
| $\therefore P'\left(2\right)$∴P′(2) | $=$= | $18\left(2\right)^2-244\left(2\right)+490$18(2)2−244(2)+490 |
Substitute the value $x=2$x=2 into the derivative |
| $=$= | $74$74 |
Therefore, the rate of change of profit at that point is $\$74$$74 per $100$100 items.
Practice questions
Question 1
The electrical resistance, $R$R, of a component at temperature, $t$t, is given by $R=9+\frac{t}{17}+\frac{t^2}{108}$R=9+t17+t2108.
Find $\frac{dR}{dt}$dRdt, the instantaneous rate of increase of resistance with respect to temperature.
Question 2
Researchers have created a model to project the country’s population for the next $10$10 years, so that the population $t$t years from now is given by the function $P\left(t\right)=\frac{57460e^{\frac{t}{7}}}{t+13}$P(t)=57460et7t+13, where $P$P is the population (in thousands).
What is the current population of the country (in thousands)? Write your answer as an equation with $P\left(0\right)$P(0) as the subject.
According to the model, what is the current rate of growth of the population (in thousands) to the nearest whole number?
At what rate will the population (in thousands) be growing $7$7 months from now, to the nearest whole number?
At what rate will the population (in thousands) be growing $10$10 years from now, to the nearest whole number?
Question 3
As the sand in a hourglass is poured, the radius, $r$r, of the cone formed by the sand expands according to the rule $r=\frac{3t}{5}$r=3t5, where $t$t is the time in seconds.
Given that the sand falls such that the height of the cone is the same as the radius at all times, determine an equation for the volume, $V$V, of the cone of sand with respect to time, $t$t.
Determine an equation for $\frac{dV}{dt}$dVdt, the rate of change of the volume of the cone of sand with respect to time.
Hence calculate the instantaneous rate of change of the volume when $t=4$t=4.
Give an exact answer.
Rates of change and exponential functions
Exponential growth or decay occurs when the rate of change of a variable is proportional to the value of the variable itself. We saw many examples of this when reviewing applications of exponential functions, such as:
- compound interest on financial deposits
- radioactive decay
- the cooling of a body that is hot compared with its environment
- the growth of certain microbiological cultures
In all living things, there is a process where new cells are constantly being made to create growth and to replace old cells. One common way these new cells are made is by cell division:
- On day 0, we start with $1$1 cell.
- On day 1, this cell splits into two and forms $2$2 cells.
- On day 2, each of the two cells splits and forms $4$4 cells, and so on.
We have generated the number pattern that is produced by doubling: $1,2,4,8,16,\ldots$1,2,4,8,16,…
Exponential growth exhibited in a colony of E. coli bacteria. Source: Wikimedia commons
What we immediately notice is how quickly the number of cells increases. On top of this, the more cells there are the more rapidly the number of cells increases. Another way to describe this is to say that the quantity is increasing at a rate that is proportional to the amount present. Any time a quantity grows in this way, it is increasing at an exponential rate. This kind of increase is called exponential growth.
Recall, the general equation for an exponential model in base $e$e, for a variable $A$A in terms of time, $t$t, and with initial value $A_0$A0 is:
$A\left(t\right)=A_0e^{kt}$A(t)=A0ekt
Let's differentiate the model in terms of $t$t:
| $A'\left(t\right)$A′(t) | $=$= | $kA_0e^{kt}$kA0ekt |
By the chain rule |
| $=$= | $kA\left(t\right)$kA(t) |
|
Notice that differentiating the function, $A\left(t\right)$A(t), to find the function, $A'\left(t\right)$A′(t), giving the instantaneous rate is the same as multiplying the original function by the constant $k$k. Which is the mathematical way of expressing that "the quantity is increasing at a rate that is proportional to the amount present".
The general form of an exponential function for a variable $A$A in terms of time, $t$t, and with initial value $A_0$A0 is:
$A\left(t\right)=A_0e^{kt}$A(t)=A0ekt
Its rate of change $A'\left(t\right)$A′(t) is proportional to $A\left(t\right)$A(t), with the constant of proportionality being $k$k:
$A'\left(t\right)=kA\left(t\right)$A′(t)=kA(t)
The value of $k$k is positive for exponential growth, and negative for exponential decay. This value describes the continuous rate of change expressed as a percentage of the current value.
Worked example
Example 2
A population of a colony of ants in a nest is given by $P\left(t\right)=2000e^{0.025t}$P(t)=2000e0.025t, where $t$t is time in months after first monitoring.
(a) State the initial population.
Think: The initial population is given by $t=0$t=0. This will return the coefficient of the term since $e^0=1$e0=1.
Do: The initial population is $2000$2000 ants.
(b) Find the average rate of change between $t=0$t=0 and $t=1$t=1.
Think: This can be calculated using the formula for average rate of change.
Do:
| Average rate of change | $=$= | $\frac{A\left(1\right)-A\left(0\right)}{1-0}$A(1)−A(0)1−0 |
| $\approx$≈ | $\frac{2050.63-2000}{1}$2050.63−20001 | |
| $=$= | $50.63$50.63 ants/month |
(c) What is the continuous growth rate used by the model?
Think: Identify the value of $k$k in the model.
Do: The continuous rate of growth is $2.5%$2.5% per month.
(d) What percentage is the population growing per month?
Think: Part c) gave us the continuous rate, as this rate is continually compounding the population will see a growth of a bit more than $2.5%$2.5% per month. We can observe this by calculating the multiplier $e^{0.025}$e0.025 and interpreting it.
Do:
| Percentage growth rate | $=$= | $\left(e^{0.025}-1\right)\times100%$(e0.025−1)×100% |
| $\approx$≈ | $\left(1.0253-1\right)\times100%$(1.0253−1)×100% | |
| $=$= | $2.53%$2.53% per month |
Reflect: This is the average rate of change as a percentage. We could double check this by finding the percentage of the original population that average rate of change found in part b) represents.
(e) What is the instantaneous rate of change at $t=2$t=2?
Think: We could use the chain rule to find the derivative and then use this to find the instantaneous rate of change at $t=2$t=2. This will be the same as multiplying the population at $t=2$t=2 by the constant $k$k.
Do:
| $A'\left(t\right)$A′(t) | $=$= | $kA\left(t\right)$kA(t) |
| $\therefore A'\left(2\right)$∴A′(2) | $=$= | $0.025A\left(2\right)$0.025A(2) |
| $=$= | $0.025\times2000e^{0.025\times2}$0.025×2000e0.025×2 | |
| $\approx$≈ | $52.56$52.56 ants/month |
Practice questions
Question 4
A radioactive isotope decays continuously and can be modelled by $A=A_0e^{kt}$A=A0ekt, where $A$A is the kilograms of isotope remaining after $t$t years.
$10$10 kilograms is produced in an industrial process and the isotope decays at a continuous rate of $30%$30% per year.
State the value of $A_0$A0.
State the value of $k$k.
Hence state the equation for $A$A, the amount remaining after $t$t years.
Determine an equation for the instantaneous rate of change of $A$A after $t$t years.
Hence determine the value of $m$m if $\frac{dA}{dt}=mA$dAdt=mA.
How much of the isotope remains after $4$4 years?
Give your answer to five decimal places.
Question 5
Under certain climatic conditions the number $P$P of blue-green algae satisfies the equation $P=Be^{0.0008t}$P=Be0.0008t, where $t$t is measured in days from when measurement began, and $B$B is constant.
Show that the number of algae increases at a rate proportional to the number present.
At what rate is the number of algae increasing when there are $300000$300000 algae present?
Solve for $t$t, the number of days it takes the initial number of algae to double.
Round your answer to two decimal places.