Practical Applications of Rates of Change
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.
So 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.
The rate of change over time (or over an interval of space) tells us so much more than a single instant in time. When analysing real life phenomena, it's the rate at which things are changing that tells us so much.
If we take the example of technology and computing speed we can see why the rate of change is more useful to us than examining a single point in time.
Right now, as you wait for the next model of your favourite mobile phone to be released, the progress of technological advancement may seem slow to you! But looking back at the rate of change of the technological progress of phones in the last $20$20 years would certainly indicate the exact opposite.
Have a look at the graph below, created by the somewhat controversial Ray Kurtzweil. You can see that the rate of change of computing power appears to be exponential and you can see when he predicts computing power will be faster than all the human brains on the planet!
Instantaneous Rate of Change vs Average 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. Put more simply, we are simply calculating the rise over the run between two points on a curve.
The instantaneous rate of change is the exact gradient at a given point on the curve. We know that to calculate an exact gradient of a curve, we first differentiate to find the gradient function and we then use this gradient function to calculate the value of the gradient.
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.
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 units.
Do: $\frac{P\left(2\right)-P\left(0\right)}{2-0}=\frac{390-\left(-150\right)}{2}=270$P(2)−P(0)2−0=390−(−150)2=270
Therefore the average profit in this interval is $\$270$$270
(b) Calculate $P'(2)$P′(2).
Think: This notation tells us we want the instantaneous rate of change of profit at the point where $200$200 items are produced. To do this we will need the gradient function.
Do:
$P'\left(x\right)=18x^2-244x+490$P′(x)=18x2−244x+490
$P'\left(2\right)=74$P′(2)=74
Therefore the rate of change of profit at that point is $\$74$$74.
Worked Examples
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
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.