Congratulations for escaping from the first ‘layer’ of sections in this website (assuming you didn’t just come straight here – if you did, then welcome!).
In popular media, you often hear the phrase “the ‘something-or-other’ has grown exponentially.” Quite often in such situations, the word ‘exponentially’ is used imprecisely to mean ‘quite a lot’, but in science it has a very specific meaning. In this section, we’ll discuss what that meaning is.
In Section 4 we investigated the differential equation
[latex]\displaystyle mg-kv^2=m\frac{dv}{dt}[/latex]
which it can be useful sometimes to rewrite (by dividing through by [latex]m[/latex] and re-ordering):
[latex]\displaystyle \frac{dv}{dt}=g-(\frac{k}{m})v^2[/latex]
This equation describes the velocity of an object falling through air. You can see that the rate of change of the velocity depends on the velocity (as the velocity increases its rate of change decreases), which we could write in general terms as [latex]\frac{dv}{dt}=f(v)[/latex].
In this section we will ask what happens if the rate of change of a quantity is not just a function of the quantity, but is proportional to it, so that [latex]\frac{dx}{dt} \propto x[/latex]
This is a worthwhile question because there are many situations in science in which it is true. For example:
- Without any limiting factors, Malthusian population growth rate is proportional to the population: [latex]\frac{dp}{dt}=rp[/latex]
- The rate of decay of radioactivity nuclei is proportional to how many there are: [latex]\frac{dN}{dt}=-\lambda N[/latex]
- The rate of flow of charge from a capacitor during its discharge is proportional to how much charge is stored on it: [latex]\frac{dQ}{dt}=-\frac{Q}{RC}[/latex]
- In a chemical reaction that is first order with respect to a reactant [latex]A[/latex], the rate at which that reactant is used up is proportional to its concentration [latex][A][/latex]: [latex]\frac{d[A]}{dt}=-k[A][/latex]
Notice that, in the examples above:
- One has a positive constant of proportionality, which leads to what is called ‘exponential growth’ because the rate of change acts to increase the quantity, which then changes faster
- Three have negative constants of proportionality, which leads to ‘exponential decay’ because the rate of change acts to decrease the quantity, which then decreases further, but more slowly
What kind of a function could describe this behaviour?
If we want to describe quantitatively a situation in which the rate of change of a quantity is proportional to its value, then we need a mathematical function equal to its own derivative, as described in the video below.
So a function of the type ‘a constant [latex]k[/latex] raised to the power [latex]x[/latex]’, (i.e. [latex]k^x[/latex]), seems like a promising function for our purposes. But which value of the constant [latex]k[/latex] does the job we want? Use the slider in the animation below to find the value of [latex]k[/latex] that makes the value of [latex]k^x[/latex] equal to its derivative (the blue line is the ‘target’). The rest of the page would act as a spoiler, so do this before you move on!
Hopefully, you found that the ‘magic number’ that makes just the right amount of growth/decay is 2.72. If you could zoom in infinitely and discriminate between the lines perfectly (!), and if your slider had more decimal places (!!), you would find that the ‘magic number’ is 2.71828… (this number carries on forever and never recurs). It is given the name ‘[latex]e[/latex]’. This means that:
- [latex]e^x[/latex] is the function for exponential growth
- [latex]e^{-x}[/latex] is the function for exponential growth
An explanation of ‘[latex]e[/latex]’ that arrives at it from a different direction (by considering compound interest) is given in the Numberphile YouTube channel. It will also consolidate your understanding of ‘limits’ from Section 1.
Some properties of the functions [latex]f((x)=e^x[/latex] and [latex]f(x)=e^{-x}[/latex] are shown in the video below.
Since we were so sniffy about the ‘popular’ use of the word ‘exponential’ at the top of the page, we had better give a better description… From what we learned in this video, we can say that an exponential change is one in the rate of change of the quantity is proportional to the value of the quantity. An exponential change results in equal fractional changes in equal time intervals.
The exponential function in solutions to differential equations
Now we know that the exponential function will help us in situations in which the rate of change of a quantity is proportional to the quantity, we will show you the solutions to the differential equations we listed at the top of the page (at the moment, we won’t worry about how to get from one to the other).
[table id=1 /]
Spend a moment to see how the solution to these examples relates to the differential equation (but don’t worry – we aren’t expecting you to be able to arrive at the solution yourself!). You should find they all behave identically, mathematically speaking.
More about [latex]e^x[/latex]
It is a curious fact that many functions of [latex]x[/latex] can be expressed as a series consisting of powers of [latex]x[/latex] with appropriate coefficients – that is, in the form:
[latex]\displaystyle f(x)=a_0+a_1x+a_2x^2+…+a_nx^n[/latex]
Strictly speaking, this statement is subject to some admittedly fairly stringent criteria, so it isn’t true for all functions, but they luckily apply in this case – look up ‘Taylor expansions’if you are interested in finding out more about this…
For the example of the function we are interested in, [latex]f(x)=e^x[/latex], the appropriate ‘power series’ is:
[latex]\displaystyle e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+…+\frac{x^n}{n!}+…=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/latex]
Power series like this are an essential tool for scientists, and we explore them in Section 7. We will see how they are used to make approximations in Section 10, and we will use them to arrive at Euler’s equation (the most magical bit of mathematics any scientist could hope for) in Section 12.
If the previous line of algebra held no fears for you, and if you can immediately see WHY this power series would have the properties of [latex]e^x[/latex], then you may want to skip this next video. Otherwise, here’s a video to explain it …
The following animation shows the relative importance of the different terms in the expansion. Use the slider to ‘truncate’ the power series at different places:
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Introducing natural logs
If you start with a number (call it ‘[latex]p[/latex]’, say) and raise ‘[latex]e[/latex]’ to the power of it, you will get a new number (let’s call it ‘[latex]q[/latex]’). Mathematically, that previous sentence can be written very concisely:
[latex]\displaystyle q=e^p[/latex]
Natural logs (mathematical symbol ‘[latex]\ln[/latex]’, or less commonly ‘[latex]\log_e[/latex]’) are the inverse operation. That is, if ‘raising [latex]e[/latex] to the power of…’ takes you from [latex]p[/latex] to [latex]q[/latex], then taking natural logs takes you from [latex]q[/latex] back to [latex]p[/latex]. That is:
[latex]\displaystyle q=e^p \;\Leftrightarrow\; p=\ln q[/latex]
where the ‘double arrow’ means that each statement implies the other (in contrast, [latex]A ; \Rightarrow ;B[/latex] would mean that [latex]A[/latex] implies [latex]B[/latex] but not that [latex]B[/latex] implies [latex]A[/latex]).
The other most common logarithms are ‘logs to the base 10’, for which [latex]\log_{10} x[/latex] represents the number you need to raise 10 to, in order to provide [latex]x[/latex]. So [latex]\log_{10} 1000[/latex] is 3, because you need to raise 10 to the power of 3 to get to 1000. Again, ‘raising 10 to the power of…’ and ‘taking logs to the base 10’ are inverse operations, so that:
[latex]y=10^x \; \Leftrightarrow \; x=\log_{10} y[/latex]
Natural logs have ‘[latex]e[/latex]’ as their base. You can form logs with any base, but in science ‘[latex]e[/latex]’ and 10 are the two most commonly used. For this reason, they are the ones that your calculator has keys for.
Why have we introduced natural logs? Well, good reasons include:
- We have introduced the exponential function, so it seems only fair to introduce its inverse (which was the approach we took with differentiation and integration)
- When you differentiate [latex]\ln x[/latex], you get [latex]\frac{1}{x}[/latex]. That is: [latex]\frac{x}{dx}(\ln x)=\frac{1}{x}[/latex], and this is going to be a really important result for the rest of the page. This also means that integrating [latex]\frac{1}{x}[/latex] gives [latex]\ln x[/latex], so that [latex]\int \frac{1}{x} \: dx=\ln|x|+c[/latex]
Just take a minute to take in the second bullet point – it is extremely important, and not that obvious at first glance (to many people, anyway). If you are happy to take it at face value, you can skip this next video, but if you want to know why it is true, press play…
Using natural logs
Natural logs can be used to give information about the elapsed time in exponential processes. For example, consider a sample of a radioactive substance containing 1020 nuclei, whose decay constant is 0.1 s-1 (that means that the probability of any given nucleus decaying is 0.1 per second). Let’s say we want to know how long it would take before only 1 % of the nuclei remain (such a question would be relevant – although it is certainly simplified – in the case of a nuclear accident, and estimating how long it would take for contamination to drop to safe levels).
Well, we would use the radioactive decay law: [latex]N=N_0e^{- \lambda t}[/latex], with the following values:
- [latex]N[/latex] is the number of nuclei at time [latex]t[/latex], which is 1018, because this is 1 % of the initial value
- [latex]N_0[/latex] is the initial number of nuclei: 1020
- [latex]\lambda[/latex] is the decay constant: 0.1
- [latex]t[/latex] is the quantity that we are trying to find
Substituting these values into the decay equation gives:
[latex]\displaystyle 10^18=10^20e^{-0.1t}[/latex]
giving (by dividing both sides by 1020):
[latex]\displaystyle 10^{-2}=e^{-0.1t}[/latex]
Now, this equation contains one unknown, [latex]t[/latex], and numbers whose values are known. However, with the usual laws of multiplication/division and addition/subtraction, it is not possible to manipulate this algebraically to make [latex]t[/latex] the subject of the equation (because it is in the exponent, and those operations will not move it from there). However, we CAN move [latex]t[/latex] from the exponent by using natural logs – we just have to remember to apply natural logs to the other side of the equation aswell:
[latex]\displaystyle \ln (10^{-2})=-0.1t[/latex]
And now all we need to do is use a calculator to evaluate [latex]\ln (10^{-2})[/latex] and divide by [latex]-0.1[/latex], which gives:
[latex]\displaystyle t=46 s[/latex]
In other words, in 46 s time, there will be 1 % of the radioactive nuclei remaining.
If you need to, take a moment here to consolidate your understanding of the use of natural logs in solving the equation for [latex]t[/latex].
At some point we’ll do a blog post in which we take a semi-humorous look at radioactivity in the movies. We’ll post a link here when we’ve done it…
Natural logs will also be used in a different way in the next video, in which we derive the radioactive decay law from its governing differential equation.
So where should you go next? Well, now you understand exponentials, a whole load of science is unlocked for you…
Link to Section 06 – Complex numbers
This is the start of a journey that will allow you to use numbers that are not ‘real’ to make predictions about objects that are! That journey will continue from Section 12 to Section 14.
Link to Home Page
And you can always return here…
Finally, we couldn’t resist showing you this from @benorlin, aka ‘Maths with bad diagrams’, a bit harshly named if you ask us…
