Mathematics in Medicine: Exercise 2

by Batool Akmal

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    00:01 Have a look at our second example.

    00:03 We mentioned lots of times through this course that differentiation helps you find the growth and decay of different objects.

    00:10 So be it a bacteria or chemical or a tumor, you can model accurately the growth and the decay of those objects.

    00:20 So we’ve been given the growth of a certain bacteria, and we’ve been told that the growth rate is dN by dt equals to 2N over T.

    00:29 So if I just write that here, we have dN by dt equals to 2N over t.

    00:37 And it says T is the time in minutes, but the question is asking us to find N in terms of T.

    00:43 Now, let’s just understand this for a minute.

    00:46 Now that you’ve done a calculus course, you can actually analyze what this is saying.

    00:50 So if you look at this differential equation, it says dN by dt is giving you some type of a value.

    00:58 Now, if you look at this closely, dN by dt is giving you the rate of change.

    01:03 So you’re saying that the growth rate, N, is changing over time, and it’s changing with that function.

    01:11 So you can say that it’s almost proportional to N because you could see here, if I just use this, you can say that dN by dt is proportional to this N value here.

    01:24 Which really is saying that the growth rate is growing.

    01:27 So the rate of growth is increasing as N is increasing.

    01:31 So you can do some analysis on this, just looking at it.

    01:36 But the question is asking us to find N in terms of t.

    01:40 Let me just explain what they’re wanting.

    01:42 We know the differential equation so it’s telling you the rate of change of N with t.

    01:47 So we know what the rate of change is.

    01:52 But the question is asking you to find the original equation.

    01:55 So where does this rate of change actually come from? It’s quite common in real life for us to start with differential equations, to start with rate of change and then go back to finding the original equation for ourselves or for more calculations.

    02:09 And this is exactly that kind of a question.

    02:12 It’s asking you to solve a differential equation.

    02:16 Now, hopefully will you remember that the opposite of differentiation or in order to solve differentiation or differential equations, we need to do the opposite of differentiation.

    02:26 And if you think really hard, you’ll remember that the opposite method of differentiation is integration.

    02:33 We’ll have to use a slightly different technique on this.

    02:35 And this technique is known as separation of variables, which you can later develop by just studying integration.

    02:42 But it’s fairly straightforward, so I’ll just show that to you here.

    02:46 We have dN by dt if I just rewrite it so we can see as 2N over t.

    02:53 We have a slight problem here because you have dN on this side, but you have the N function on this side.

    03:00 And then you have the dt on this side, but you have the t function on the other side.

    03:05 This problem can easily be resolved.

    03:07 All you have to do is swap places.

    03:09 Take all your t's to one side and take all your N's to one side.

    03:13 And the process is known as separation of variables.

    03:16 You’ve got 2 variables, you can separate them out on each side of the equation.

    03:22 Let me show you what I mean.

    03:23 So I’ll bring that 2 – Not 2, I’ll leave the 2 there.

    03:27 I’ll bring the N down so I’ll rewrite this as dN over N.

    03:31 It was multiplying here, so you can divide on this side.

    03:33 And on the other side, I still have 2t left, and you can take the dt to the other side.

    03:41 So the dt is dividing here, so when you move it to the other side, it’s multiplying.

    03:46 You can see that what you’ve done here is created two integrals for yourselves.

    03:50 So if I put this little integration signs on both sides, you’ll be able to see that you have an integration question here and an integration question here.

    03:59 Each with their own respective variables.

    04:03 So if I rewrite this in a way that we can see this, we have 1 over N dN equals to 2 over t dt.

    04:14 And all we have to do now is integrate them separately.

    04:18 So you do one integral and you do the other integral with respect to their variables.

    04:24 Think back at what this is now.

    04:25 Straightforward answer.

    04:27 1 over N dN.

    04:29 Can you remember what result this gave you? If I remind you that when you have ln of x, and you differentiate it, you get 1 over x.

    04:41 So think of what’s going to happen when you have to integrate 1 over X, that gives you ln of x.

    04:47 So 1 over x integrates to give you ln of x.

    04:51 And so all you have to do is change the variable.

    04:53 Rather than X, let’s use N.

    04:55 1 over N dN integrates to ln of N.

    05:02 On the other side, you can take the 2 out of the integral and write it as T dt.

    05:08 And you’ll see that the other side also gives you an ln function.

    05:13 So we have ln of N equals to 2 ln of t.

    05:21 And don’t forget to add a plus C there.

    05:27 The question said express N in terms of t, so we don’t quite have N by itself here.

    05:32 So what we can do is solve this equation.

    05:35 In order to get rid of ln’s, you can E both sides, so exponentials are the opposite of ln’s.

    05:41 So we can do e to the ln of N equals to e to 2ln of t, plus e to the power of C.

    05:55 So I’ve exponentiated every term here.

    05:59 You could now rewrite this because e and ln cancel out.

    06:03 You could write this side as N, which is good news.

    06:07 This side doesn’t quite cancel out yet because you have a 2 at front.

    06:12 Let me just show that to you on the side.

    06:13 So we’ve got e 2ln of t.

    06:17 Now, if it was e and ln next to each other, you could cancel it out, but you have this 2.

    06:23 Think of the rules of logs that we spoke about through this course and this is the third rule of log, which allows you to take the power up.

    06:30 So you could rewrite this as e to the ln of t squared.

    06:36 You can take the 2 up and at which case, the e in ln will just be the inverse of each other so they just disappear.

    06:42 So this leaves you with t squared plus e to C.

    06:47 e to the C is just any number, it’s just a constant, so you could just write it as N equals to t squared plus let’s just call it C.

    06:56 And so we found our equation.

    06:58 So the equation that this differential must have come from would look like this, we have N equals to t squared plus C.

    07:06 C, remember we mentioned before that you can find out when you know some initial conditions.

    07:11 So if I was told what the time was when N is a certain number, I can substitute it into this equation and calculate my C.

    07:20 But if you’re not given a C or a constant, you just leave it there.

    07:23 It could be zero or it could be another number, but we don’t want to risk just removing it and assuming it’s zero.

    About the Lecture

    The lecture Mathematics in Medicine: Exercise 2 by Batool Akmal is from the course Calculation Methods: Exercises.

    Included Quiz Questions

    1. N(t) = ce^(rt)
    2. N(t) = e^(rt) + c
    3. N = e^(-rt) + c
    4. N = ce^(-rt)
    5. N = t² + c
    1. ct³
    2. t³ + c
    3. ct²
    4. t² + c
    5. ce^(3t)
    1. N = ct²
    2. N = t²
    3. N = t
    4. N = t³
    5. N = ceᵗ
    1. N(t) = ce^(2r√t)
    2. N(t) = ce^(-2r√t)
    3. N(t) = ct^r
    4. N(t) = ce^(r√t)
    5. N(t) = 2ce^(-r√t)

    Author of lecture Mathematics in Medicine: Exercise 2

     Batool Akmal

    Batool Akmal

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