# Mathematics in Medicine: Exercise 3

by Batool Akmal

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00:01 So let’s look at our final example.

00:03 It’s talking about pressure that’s being exerted on the walls of an artery for a specific patient.

00:09 So we’re talking about blood pressure here.

00:12 And t is the number of seconds since the beginning of the cardiac cycle.

00:18 Now, if you look at what the question is asking, it say’s find t when the pressure is highest.

00:22 So it’s asking us for the time when the pressure for this function is at its maximum.

00:28 Now, it’s really amazing that you were able to do this now because you don’t need any more information.

00:33 You can just use this function and your own knowledge as to when you get maximums and when you get minimums and actually work out and predict what the time should be when the pressure is highest.

00:46 Let’s have a look the function and see what we can do to answer this question.

00:50 I have a function of t, which equals to 90 plus 15 sine 2.5 pi t.

01:01 So this is just a function, it could be anything.

01:03 So any general function, this isn’t a rule, but that’s what the pressure is.

01:07 It could be anything, but we’re just going to learn what to do to deal with it.

01:13 It’s says find t when the pressure is highest, so when this value or this pressure is maximum.

01:21 There must be a lot of theory going through your minds right now.

01:24 How can you calculate the maximum or the minimum of a function? Think about it.

01:30 We’ve done this before when we looked at stationary points.

01:34 We spoke about maximum and minimum points.

01:37 We spoke about how to find maximum and minimum points and we also spoke about how to prove that they are maximums or minimums.

01:45 So if we remind ourselves about stationary points, remember that if you have any type of a function, so any type of graph, these here are your stationary points.

01:56 And we said that these points occur when the gradient is 0.

02:00 So if this is Y-X graph, we can say that it occurs when dy/dx equals to 0.

02:06 That’s how you can find a stationary point.

02:09 And then you can move on to showing that this is the highest or when the pressure is the highest.

02:15 So let’s try this out on this question.

02:18 We need to differentiate this, so we need to do dP/dt and we need to equal it to 0 to get the stationary points.

02:30 Also, a reminder of how to differentiate trig.

02:34 So we know that sine and cos and cos minus sine.

02:38 This direction to differentiate, this direction to integrate.

02:44 So let’s start to differentiate now, so I’ll say that dP/dt equals to any constant differentiates to 0.

02:53 So that goes away.

02:54 We have the 15, which can stay, because it’s just a number that’s multiplying.

02:59 Sine, look at sine, differentiates to cos.

03:03 So this goes to cos of 2.5 pi t.

03:08 But there is an important thing that you shouldn’t forget.

03:11 And that is to observe that this is a function of t inside of a function.

03:17 So you have to remember how to differentiate using the chain rule.

03:21 So we’ve differentiated the entire function sine of something goes to cos of that thing.

03:27 And then you have to multiply it with the differential of the inside.

03:31 So the differential of 2.5 pi t.

03:34 Remember pi is just a number, so it’s just the same as 10t.

03:38 So if you were differentiating 10t, you just get an answer of 10.

03:42 If you’re differentiating 2.5 pi t, you just get an answer of 2.5 pi.

03:50 So we’re now saying that our gradient is 15, let’s put that here.

03:55 2.5 pi cos of 2.5 pi t.

04:02 And remember that at stationary points, that needs to equal to 0.

04:09 You can rewrite your 2.5M as a fraction, so you can write this as 15, 5 over 2 pi, cos of 2.5 pi t equals to 0.

04:21 So we can work this out.

04:23 So if you times that through, we’ve got 15 multiplied by 5, which gives you 75 over 2 pi.

04:30 And then we’ve got cos of 2.5 pi t equals to 0.

04:36 We can now start to deal with this fraction here.

04:39 We’ve got 2 at the bottom, so because it’s dividing, you can multiply it on the other side.

04:44 And when you multiply 0 times 2, that will give you 0.

04:48 So we can say that just disappeared, I’ll do that here.

04:50 So we’ve got 75 pi cos of 2.5 pi t equals to 0.

04:58 You can then take this 75 pi and since it’s multiplying here, it’s going to divide on the other side, which will also give you 0, leaving you with cos 2.5 pi t equals to 0.

05:10 So remember that basically that entire term here just goes to the other side, and because it’s multiplying and dividing with 0, it just goes to zero.

05:21 Continuing on from here, so we now need to figure out what t is, so we can take cos to the other side, so we can rewrite this as 2.5 pi t equals to cos inverse of 0.

05:36 Cos inverse of 0, if you’re using your calculator, will give you an answer of 1.

05:41 Or you can remind yourselves what that equals to just by drawing a cos curve.

05:45 So a cos graph looks like this.

05:49 So you know that when cos equals to 0, so when we’re here, the answer is 1.

05:54 So you will get 2.5 pi t equals to 1.

05:59 Just make some more space for myself here.

06:03 So when I rearrange this equation, I’ll get t equals to 1 over 2.5 pi, 2.5 is just 5 over 2, so you can write this as 5 over 2 pi.

06:13 And if you bring that 2 up, you get 2 over 5 pi is your time.

06:19 And you can work that out, let’s just put that as t is 2 over 5 pi in seconds and you can work that out as a decimal and round up or you could keep it in its whole number form as it is now.

06:31 So we’ve just concluded that because that’s a stationary point and that gives the time of when the pressure is highest is when t equals to 2 over 5 pi seconds.

06:43 And so we've now done some medical examples using lots and lots of calculus.

06:48 Now, remember no matter what field you’re planning of going into, whether it’s a medical field or any other field, maths will make you a better student.

06:57 It will make you a better problem solver because you’ll always be able to look at a problem and look at it systematically and go through it method by method.

07:06 Practicing mathematics is like exercising for your brain.

07:09 The more you do it, the better you’ll get at it.

07:12 So my advice if you’re sitting in any type of math exam is to practice as much as you can.

07:18 Don’t leave it until the last day or the last week, just consistently practice.

07:22 So that you’re mind will become sharper and you’re becoming better with all the methods.

07:27 Enjoy the challenges, that's what you’re doing maths for.

07:31 You’re doing it to look at problems and to try and solve them.

07:34 Enjoy the kind of questions that are a lot harder to solve, that take a lot more thinking, that take a lot more methods and algebra to solve.

07:42 Because even if you don’t use it in your tests or your exams, it’s making you a better student, it’s making you a better person at solving different problems.

07:52 So once you’re geared at certain methods, push yourselves with slightly harder questions and challenge yourselves further to look at questions that need a more advanced level of algebra.

08:04 Employers and degrees are increasingly wanting maths and are impressed with students that do maths only because it’s considered to be a difficult subject because it makes you think in a logical way that lots of other subjects perhaps don’t manage to develop that skill as well as mathematics does.

08:22 It makes you think in a methodical way so that you look at a problem, you think of everything that you know, think of all the things that you need and you put them together to find your solutions.

08:34 Make sure you sleep well for mathematics because it’s important that your brain is well rested.

08:39 It’s not like reading a book.

08:41 It is your brain continuously working and remembering.

08:45 And remember when things get easy and you feel that you’re just doing the same thing over and over again and you’re really good at it, that’s the time to move forward to a further challenge and to a question that’s of more difficulty.

08:57 Your analytical skills are developing because you’re looking at results and you’re looking at data and you’re looking at equations and you’re understanding what they’re saying.

09:07 When you look at the rate of change of population, so dP by dt, you know what that means now.

09:13 You understand that as population changing with time, you’re actually analyzing more information that you were able to before.

09:21 You’re able to visualize what these functions look like, so you can look at a quadratic and you know what it looks like in XY space or in geometric space, which again helps you analyze data and helps you understand whether your numbers or your function is growing or whether it’s becoming smaller.

09:40 You're understanding gradients whether they’re positive or negative.

09:44 What kind of rate of change can you see in the growth or decay of an object? Is it getting bigger? Is it getting smaller? And it’s all analytical skills that you’re developing.

09:53 It’s all problem solving skills that you’re developing.

09:56 So my advice at the end is try and enjoy mathematics.

10:00 When you look at a challenge, don’t panic, think about it.

10:03 Let your brain do the work.

10:05 Think of all the processes.

10:06 Think of everything that you know.

10:08 Think of things that you might need to learn further and put it all together.

10:12 Stay calm and do it one step at a time because remember doing maths is learning how to solve problems.

10:18 And if it isn’t a problem, it isn’t really too much of math or too much of challenge.

The lecture Mathematics in Medicine: Exercise 3 by Batool Akmal is from the course Calculation Methods: Exercises.

### Included Quiz Questions

1. dp(t) / dt = 28π cos(2πt)
2. dp(t)/dt = 85 + 28π cos (2πt)
3. dp(t) / dt = 14π cos (2πt)
4. dp(t) / dt = -28π cos(2πt)
5. dp(t) / dt = -14π cos (2πt)
1. t = 1/4 s
2. t = 1 s
3. t = 0 s
4. t = -1 s
5. t = 1/2 s
1. 99
2. 100
3. 85
4. 98
5. 71
1. t = 3/4 s
2. t = 1/4 s
3. t = 4/3 s
4. t = 2/3 s
5. t = 1 s
1. 71
2. 78
3. 85
4. 76
5. 69

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