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Mathematics in Medicine: Exercise 1

by Batool Akmal
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    00:01 Have a look at out first example.

    00:03 We’ve been given an equation or a function for drug sensitivity and we’re calling that R.

    00:10 M here is the dosage of the drug in milligrams.

    00:13 So we’re looking at a drug sensitivity of a certain drug and it’s given in terms of this function.

    00:19 The question is asking us to find the rate of change of R with respect to M.

    00:24 Now, really think about this, we’ve mentioned this so many times through this course.

    00:29 Rate of change with respect to something.

    00:32 I’ll take you back to when we looked at the change in Y over change in X.

    00:36 How does Y change with respect to X? And what does that give you? It gives you the gradient and it gives you dy/dx.

    00:44 So with that thinking, you can conclude that this question is actually asking us to differentiate R with respect to M.

    00:54 Let’s try that out on this function So given function R of M.

    00:59 So that’s just saying that R is a function of M.

    01:02 We have 2M square root of 10 plus 0.5M.

    01:10 Okay, we’ve discussed that the question is asking us for the rate of change of R with respect to M.

    01:16 So how does a drug sensitivity change with respect to the dosage? Does it increase? Does it decrease? And it’s been given to us in terms of this function.

    01:27 Let’s have a look at what we need to do.

    01:29 So if it’s asking us for the rate of change of R with respect to M, we need to do dR/dM.

    01:35 So that’s change in R over change in M.

    01:39 And that is differentiation.

    01:41 So we are differentiating this function here.

    01:46 Hopefully, lots and lots of different techniques are going through your minds right now.

    01:50 You’ve got two functions in terms of M.

    01:54 You’ve got 2M that’s multiplying with something else.

    01:57 You’re starting to think how do you differentiate two functions that multiply together.

    02:03 And I hope you’ve concluded that we need to use the product rule.

    02:06 Let me remind you of the product rule here on this side.

    02:10 dy/dx or a product rule in terms of Y and X is vdu dx plus udv dx.

    02:18 So in very simple terms, leave the first function as it is, differentiate the second, leave the second function as it is, differentiate the first.

    02:26 We’ve done this using this formula and we’ve done this faster when we were doing implicit differentiation.

    02:32 To make this a little bit easier, let me just write this as 2M and rather than the square root, I’m going to write that as a power.

    02:39 So I have 10 plus 0.5 M to the power of a half.

    02:44 And here, you can see two clear functions.

    02:47 We’ve got this function and we’ve got this function.

    02:50 You can call one of them U and one of them V and we can bring them together.

    02:54 Or alternatively, we can try and use the faster method where we leave one term as it is, differentiate the second, leave the second as it is, differentiate the first.

    03:04 So remember that we’re now differentiating with respect to M, which makes complete sense because this equation is in terms of M.

    03:13 I’m going to start with leaving my 2M as it is, so that’s my first term.

    03:17 So 2M stays and I now have to differentiate this.

    03:21 Look at this closely and decide what kind of function this is firstly and what method of differentiation you’d need.

    03:29 You can see here that you have something to the power of a half and then you have something inside of that function.

    03:35 This is a function of a function and then you have to think back as to what method you need to use to differentiate a function of a function.

    03:44 We’ll have to use the chain rule method to differentiate the function of a function.

    03:49 So the chain rule stated differentiate the outside function first as a whole and then multiply it with the differential of the inside.

    03:58 So let’s try to – I’ll put brackets around this because this seems a bit more complicated.

    04:02 Bring the power down.

    04:03 So I’ll bring the half down, leave the inside function as it is, and decrease the power by one.

    04:12 Don’t forget now to multiply it with the differential of the inside.

    04:17 So the inside says that you have 10 plus a half M, differential of that is just going to be a half or 0.5, however you prefer to write it.

    04:26 So the differential of the inside, I’m writing as half because I already have a half there, it’s easier for me to combine it.

    04:33 That’s the first half of the product rule then.

    04:35 The second half, remember that the first bit I left this term as it is and I differentiated the second.

    04:42 I’m now going to differentiate this term and leave the second term as it is.

    04:46 So the differential of 2M with respect to the M is just 2, and the second term can stay as it is, we don’t have to fiddle around with it.

    04:54 That can just stay as it was.

    04:57 And so we’ve used the product rule to find dR/dM or the rate of change of R with respect to M.

    05:05 It obviously isn’t very tidy, so we just need to do some algebra to tidy this up.

    05:10 I can multiply those two numbers together.

    05:12 So half and a half.

    05:13 So I’ve got 2M and then I’ve got a quarter on the inside and then I have 10 plus 0.5M to the minus half.

    05:22 You can take this down if you want or you can leave it for a minute and then we’ll take it back in the next step.

    05:28 Here, you can rewrite this as 2 and you can change this to square root of 10 plus 0.5M.

    05:36 These two numbers will cancel, so the 2 and the 4 cancels to give you 2, giving you M over 2.

    05:42 And then like I said, you can bring this down.

    05:45 So because this is to the power of minus half, you can write it as positive half , as 10 plus 0.5M plus 2 square root of 10, plus 0.5M.

    05:57 And what you’ve just found here is dR by dM or another way of writing this is just R differentiate it with respect to M.

    06:07 So if I’m using the similar notation to what they’ve used in the question, they gave us R as a function of M.

    06:13 If I differentiate it, you can write it as R’M or dR by dM and we’ve differentiated it.

    06:19 So really, you found a function which represents the gradient of how drug sensitivity is changing with respect to dosage.

    06:31 If we now look at answering the second part of the question, so it says find the rate of change of R with respect to M, which we found here as an expression.

    06:40 It then says find R’ of 50.

    06:43 So it’s saying when you put 50 mg into this equation, what is the drug sensitivity just as a number? So we’ll do that here on the side, it’s just a simple matter of substituting numbers.

    06:56 We have R’ of 50.

    06:58 And all we’re doing is changing our M, which was here, to 50.

    07:03 So we’re using exactly the same equation.

    07:05 So we’re using this equation and replacing all our Ms here with 50.

    07:10 So I’ll end up with 50 over 2, root of 10 plus a half times 50.

    07:20 And then I have 2 root of 10 plus a half times 50.

    07:28 If we continue just simplifying this out, the 2 and the 50 cancels out to give you 25.

    07:33 A half times 50 gives you 25 plus 10, so I have root 35 and then I also have 2 root of 35 here.

    07:42 You can take 35 as a common denominator.

    07:45 So you get 25 plus 2 times 35.

    07:50 So remember when you multiply this – if I do that on the side – root 35 multiplied by root 35 is just going to give you 35.

    07:59 Thirty-five, that should read.

    08:01 You can try that out on your calculators because we said we’re allowing us calculators here so that it gives you 35.

    08:07 But those are just rules of surds here.

    08:10 This gives you 25 plus 70 over 35, which then gives you 95 over 35.

    08:20 So 95 over 35 gives you an answer of 3 and we’ve got a nice whole number and so we’re saying, if you want it to, that R’ of 50 gives you an answer of 3.

    08:32 And then you can do all types of mathematical analysis on those results and you can try and change the dosage and see how drug sensitivity varies.


    About the Lecture

    The lecture Mathematics in Medicine: Exercise 1 by Batool Akmal is from the course Calculation Methods: Exercises.


    Author of lecture Mathematics in Medicine: Exercise 1

     Batool Akmal

    Batool Akmal


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