Before launching into applications of rates of change, let's recap a few key ideas.
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
Recall from earlier chapters the difference between these two.
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 and we then use this gradient function to calculate the instantaneous rate of change at a particular value.
The average rate of change of a function, $f(x)$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 $f(x)$f(x) at $x=a$x=a and $B$B is the point of $f(x)$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, $y=f(x)$y=f(x), at $x=a$x=a, is given by:
The derivative evaluated at $x=a$x=a, denoted $f'(a)$f′(a) or 
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
The profit function, in dollars, for producing $x$x hundred items is given by $P(x)=6x^3-122x^2+490x-150$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\text{ per 100 items}$$270 per 100 items
(b) Calculate the instantaneous rate of change of profit at the point where $200$200 items are produced.
Think: The instantaneous rate of change of profit at the point where $200$200 items are produced is equivalent to asking the gradient of the tangent at $x=2$x=2 or $P'(2)$P′(2).
Do:
| $P'(x)$P′(x) | $=$= | $18x^2-244x+490$18x2−244x+490 | Differentiate $P(x)$P(x) to find the gradient function |
| $\therefore P'\left(2\right)$∴P′(2) | $=$= | $18\left(2\right)^2-244\left(2\right)+490$18(2)2−244(2)+490 | Substitute the value $x=2$x=2 into the derivative |
| $=$= | $74$74 |
Therefore, the rate of change of profit at that point is $\$74\text{ per 100 items}$$74 per 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
The effectiveness of a supplement is given by $\frac{dE}{dg}$dEdg, where $E$E is the body’s reaction to the supplement and $g$g is the number of grams of the supplement administered to the subject.
For a particular supplement, $E=\frac{g^2\left(400-g\right)}{2}$E=g2(400−g)2.
Determine the equation for the effectiveness of this particular supplement $\frac{dE}{dg}$dEdg.
Determine the effectiveness of the drug when $g=30$g=30.
After how many units of the supplement does the effectiveness of the drug start to decrease?
Give your answer to one decimal place.
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.