Rate of change

Differential calculus has many applications, and they all follow from the meaning we give to the function $f$ . For example, if $f(x)$ is a rule that tells us where a car is at any given time $x$ , then the slope of the secant takes the meaning of average speed, and the slope of the tangent, that is $f'(x)$ , takes the meaning of instantaneous speed. In the following, we will discuss this in more detail.

Speed is an example of a rate. Generally speaking, a rate is a ratio between two numbers with different units. We speak of a rate of change (or, to be more precise, of an average rate of change) if these these two numbers represent differences or changes. Here we are only interested in rates of change.

Speed, for example, is a rate of change, because it is distance travelled in metres (change in position) over a period of time in seconds (change in time). Typically, in the context of rates we use the word per rather than "over", that is, we say speed is the change in position per change in time.

Other typical rates of change are

acceleration: change of speed per change in time
radioactive decay: change in particles due to radioactive decay per change in time
power: amount of energy transferred per change in time
rate of return: the ratio of money gained or lost on an investment relative to the amount of money invested
inflation: increase in the general price level of goods and services in an economy over a period of time.

So clearly, rates are important ... but what does this have to do with differential calculus? To see this connection, we return again to the defining relation of the derivative:

f'(x) \approx \frac{f(x+h)-f(x)}{h}

where the right side is the difference quotient. So far we interpreted the left side as "slope of tangent" and the right side as "slope of secant". We also know that $h$ can be any number, but should be close to zero for the approximation to be good. And to get the exact value of $f'(x)$ we let $h$ move towards $0$ , and can then replace the $\approx$ sign with the $=$ sign.

Observe that the difference quotient is a rate of change, but without units. The numerator is a difference ( $f(x+h)-f(x)$ ), and the denominator as well, as $h$ describes the difference between $x$ and $x+h$ .

But depending on the context, the units arise naturally, and the difference quotient becomes a proper rate of change. Here is an example:

Example 1: Speed

Assume that $f(x)$ describes the position of a car at time $x$ along a straight road. For example, for

f(x)=\sqrt{x}

the car is at time $x=0$ at the starting position $0$ , at time $x=4s$ ( $s$ is seconds) the car is at position $2m$ ( $m$ is metres), and so on.

The ratio

\frac{f(x+h)-f(x)}{h}

is then the change in position (that is, the distanced travelled) per change in time, which is the definition of speed. The units are $m/s$ .

Let us calculate the speed of the car between the times $x=4s$ and $x=25s$ (so $h$ is $21$ ): it is

v = \frac{f(25)-f(4)}{21}=\frac{5-2}{21}=\frac{1}{7} m/s

If we let $h$ move towards $0$ , the difference quotient becomes the derivative $f'(x)$ , and we call this the instantaneous rate of change. In the case of speed, for example, $f'(x)$ is the instantaneous speed, that is, the speed at time $x$ . Let us revisit the example from above.

Example 2: Instantaneous speed

The function $f(x)=\sqrt{x}$ still describes the position of the car at time $x$ along a straight road.

Because of

f'(x)=\frac{1}{2}x^{-1/2}=\frac{1}{2\sqrt{x}}

the instantaneous speed of the car at time $x=4s$ is

f'(4)=\frac{1}{2\sqrt{4}}=\frac{1}{4} m/s

Note that instantaneous rate of change is widely used in science, but is actually a subtle concept. For example, what does it mean that a car has an instantaneous speed of $0.25m/s$ at time $4s$ ? (see example above). Clearly, for $h\rightarrow 0$ the period of time from $4s$ to $4+h$ seconds becomes a point in time (a single instant). And clearly no car can move in a single instant. The whole concept of speed does not work any more, right?

Yes, but we know from differential calculus that if we let $h$ approach $0$ , the speed of the car in the period of time between $4s$ and $4+h$ seconds approaches a single value ( $f'(4)=\frac{1}{4} m/s$ ). This seems to justify the name "instantaneous speed", but it is an abstract entity, which can be approximated and calculated if we know $f$ , but never directly measured. In the end, we use it because it turns out that instantaneous rates of change are super useful!

Exercise 1

Q1

The position of a car at every time $t$ (seconds) can be described with the following function: $s(t)=5\sqrt{t^5}$ .

What is the average speed between the time $t=0s$ and $t=9s$ ?
Determine the instantaneous speed at time $t=9s$
Find the instantaneous acceleration at time $t=9$ (Hint: the instantaneous acceleration is the instantaneous change of instantaneous speed)
When does the car have an instantaneous acceleration of $75m/s^2$ ?

Q2

The radioactive substance Carbon-14 has a a half-life of $5730$ years. Recall that radioactive decay is an exponential decay process.

Starting with $10$ grams, find the formula that let's you calculate the weight after $t$ years.
Determine the instantaneous rate of decay at time $t=10'000$ years and at $t=20'000$ years.

Solution

A1

$v_{av} = \frac{s(9)-s(0)}{9}=\frac{5\sqrt{9^5}-5\sqrt{0^5}}{9}=\underline{135 m/s}$ (because $h=9$ ).
With $s(t)=5\sqrt{t^5}=5 \cdot t^{2.5}$ we get for the instantaneous speed $v(t)=s'(t)=5\cdot 2.5\cdot t^{1.5}=12.5\cdot t^{1.5}$ . Thus $v(9)=12.5\cdot 9^{1.5}=\underline{337.5 m/s}$ .
Instant acceleration is instantaneous change of speed, which is $a(t)=v'(t)=12.5\cdot 1.5 \cdot t^{0.5}=18.75\sqrt{t}$ . Thus, $a(9)=18.75\sqrt{9}=\underline{56.25 m/s^2}$ .
Find $t$ with $a(t)=75m/s^2$ , that is, solve the equation

18.75\sqrt{t}=75

It follows $t=\underline{16 s}$

A2

Let N(t) be that amount of Carbon-14 left after $t$ years.

\begin{array}{ccccccccccc} 10 & \xrightarrow{\cdot 0.5}& 5 & \xrightarrow{\cdot 0.5} & 2.5 & \xrightarrow{\cdot 0.5} & ... &\xrightarrow{\cdot 0.5} & N(t)\\ 0 & \rightarrow & 5730 & \rightarrow & 11460 & \rightarrow & ... & \rightarrow & t\\ \end{array}

Thus, we have
$N(t)=10 \cdot 0.5^{(t-0)/5730}=\underline{10\cdot 0.5^{t/5730}}$
Instantaneous rate of decay is the derivative of
$N(t)=10\cdot 0.5^{t/5730}$
This is a chain of functions, $N(t)=u(v(t))$ with $u(v)=10\cdot 0.5^{v}$ and $v(t)=t/5730$ , we get
$N'(t)=10 \cdot \ln(0.5)\cdot 0.5^{t/5730} \cdot \frac{1}{5730}$
Thus, the instantaneous decay per year at $t=10\,000$ years is $N'(10000)=\underline{-0.000360839}$ , and at $t=20\,000$ years it is $N'(10000)=\underline{-0.000107635}$ .