• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Virus Modelling

Extracts from this document...

Introduction

IB HL Maths

Modelling the Course of a Viral Illness and its Treatment

Candidate Name: Sherul Mehta

Centre Number:002144

Candidate Number: CSY 114

Contents

Introduction - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg3

Modelling infection

• Part 1. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg4
• Part 2. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg7

Modelling Recovery

• Part 3. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg11
• Part 4. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg16
• Part 5. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg18
• Part 6. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg20

Analysing your models - - - - - - - - - - - - - - - - - - - - - - - -pg24

• Part7. - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - pg26
Introduction

Middle

13

2141151.922

14

1208116.419

15

158747.5758

16

-1021459.356

17

-2348817.615

Using this information I was able to plot a graph which would show the effect of medication when there were 8 million particles:

Model 3a: The effect of medicine after there were 8 million viral particles As it evident from the graph, the particles go below 0 after 15 hours. Therefore if a patient took medication when he/she had 8 million viral particles then he/she would be fully cured between 15-16hrs as all the particles are eliminate then and thus the medication would be effective.

Now I went through the same process again but this time testing whether the medication would work effectively if he/she had 9 million viral particles.

The results can be seen on the following page:

9 million particles

 A B 1 Hours No. Particles 2 0 9000000 3 1 =(B2*(1.6^0.25))-1200000 4 2 =(B3*(1.6^0.25))-1200000 5 3 =(B4*(1.6^0.25))-1200000 6 4 =(B5*(1.6^0.25))-1200000 7 5 =(B6*(1.6^0.25))-1200000 8 6 =(B7*(1.6^0.25))-1200000 9 7 =(B8*(1.6^0.25))-1200000 10 8 =(B9*(1.6^0.25))-1200000 11 9 =(B10*(1.6^0.25))-1200000 12 10 =(B11*(1.6^0.25))-1200000 13 11 =(B12*(1.6^0.25))-1200000 14 12 =(B13*(1.6^0.25))-1200000 15 13 =(B14*(1.6^0.25))-1200000 16 14 =(B15*(1.6^0.25))-1200000 17 15 =(B16*(1.6^0.25))-1200000 18 16 =(B17*(1.6^0.25))-1200000 19 17 =(B18*(1.6^0.25))-1200000 20 18 =(B19*(1.6^0.25))-1200000 21 19 =(B20*(1.6^0.25))-1200000 22 20 =(B21*(1.6^0.25))-1200000 23 21 =(B22*(1.6^0.25))-1200000 24 22 =(B23*(1.6^0.25))-1200000 25 23 =(B24*(1.6^0.25))-1200000 26 24 =(B25*(1.6^0.25))-1200000 27 25 =(B26*(1.6^0.25))-1200000
 Hours No. Particles 0 9000000 1 8922143.853 2 8834580.396 3 8736099.295 4 8625339.309 5 8500769.475 6 8360667.943 7 8203098.181 8 8025882.203 9 7826570.468 10 7602408.018 11 7350296.399 12 7066750.835 13 6747852.058 14 6389192.137 15 5985813.547 16 5532140.644 17 5021902.602 18 4448046.729 19 3802640.984 20 3076764.34 21 2260383.472 22 1342214.075 23 309564.8829 24 -851837.747 25 -2158047.135

Using the data above, I plotted a graph:

Model 3b: The effect of medicine after there were 9 million viral particles Again we can see from the graph that the medication would be effective when it is administered to a patient who has 9 million viral particles. Such a patient would be cured between 23-24hrs as the particles are below 0 then. Its effect would be slower than a patient who has 8 million particles but they would still be cured.

Now I did the same for a patient with 10 million viral particles.

10 million particles

 Hours No. Particles 0 10000000 1 10046826.5 2 10099491.46 3 10158722.82 4 10225339.31 5 10300261.72 6 10384525.65 7 10479295.83 8 10585882.2 9 10705758.05 10 10840580.34 11 10992212.63 12 11162750.83 13 11354552.19 14 11570267.86 15 11812879.52 16 12085740.64 17 12392622.82 18 12737767.88 19 13125946.54 20 13562524.34 21 14053535.82 22 14605767.91 23 15226853.77 24 15925378.25 25 16710996.62
 A B 1 Hours No. Particles 2 0 10000000 3 1 =(B2*(1.6^0.25))-1200000 4 2 =(B3*(1.6^0.25))-1200000 5 3 =(B4*(1.6^0.25))-1200000 6 4 =(B5*(1.6^0.25))-1200000 7 5 =(B6*(1.6^0.25))-1200000 8 6 =(B7*(1.6^0.25))-1200000 9 7 =(B8*(1.6^0.25))-1200000 10 8 =(B9*(1.6^0.25))-1200000 11 9 =(B10*(1.6^0.25))-1200000 12 10 =(B11*(1.6^0.25))-1200000 13 11 =(B12*(1.6^0.25))-1200000 14 12 =(B13*(1.6^0.25))-1200000 15 13 =(B14*(1.6^0.25))-1200000 16 14 =(B15*(1.6^0.25))-1200000 17 15

Conclusion

If the patient was a young child than the models would really look different because from my research I found that a child’s immune system is not completely developed until he or she is 14 years old i.e. the immune system of a child is weaker than that of an adult. Therefore the models would have to be modified in several ways.

1. It says in the description that an adult’s immune system would reduce the replication rate from 200% to 160% every 4 hours. As a child’s immune system is weaker and therefore the rate will not decrease to 160%.
2. Also a child’s immune system will not be able to eliminate as much as 50000 particles every hour like the adult one does.
3. In the ‘Modelling Recovery’ section it states that together with the medication and the immune system, 1.2 million particles would be eliminated every hour. Again, as a child’s system is weaker, it will not be able to eliminate 1.2 million particles every hour with medication. Hence, for a child to make a full recovery, he/she must be given medicine before the particles reach 9624434 which is the figure for the last point of medication for a standard adult.
4. Also, since the medication would not be as effective in a child, he/she will need more than the minimum of 90 micrograms of medication.
5. If more than 90 micrograms of medication will be required than more dosage will be required to maintain that level.
6. If I child was infected, then he/she would have to take medication earlier than an adult would to make a full recovery.
7. Even when the medication was administered, it would take longer for a child to recover than an adult would.

As we can see from above, several figures would have to be altered and therefore all the models in my investigation would have to be modified.

http://www.whatreallyworks.co.uk/start/kidszone.asp?article_ID=559 (accessed on 28th Oct 2008)

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## MODELLING THE COURSE OF A VIRAL ILLNESS AND ITS TREATMENT

be: Since the replicate rate of the viral particles is cumulative, we could figure out the rate of 1.6 to the power of , where t is time. The decreasing number of a 4-hour period is 200000, however if we are to work out the decreasing number of viral particles

2. ## A logistic model

Growth factor r=2.3. Graphical plot of the fish population of the hydroelectric project on an interval of 20 years using logistic function model {7}. Table 4.3. Growth factor r=2.5. The population of fish in a lake over a time range of 20 years estimated using the logistic function model {9}.

1. ## Modelling the course of a viral illness and its treatment

The immune system is responding when the viral particles exceed 1 million and it will eliminate 50000 viral particles per hour. The replication will than decrease from 200% to 160%. (see figure 2-1). This means for every one hour, the replication will increase with a value of 160%/fourth hour =.

2. ## Creating a logistic model

to: y = Putting this in a graph, we have: Model for Growth Rate = 2.3 When we have a new initial growth rate, the ordered pairs i.e. (un, rn) have now changed to: (60000, 1) (10000, 2.3) To find the linear growth factor, we form the two equations: 2.3 = m(10000)

1. ## Modelling Probabilities in Tennis. In this investigation I shall examine the possibilities for ...

but in which, if both players have three points, deuce is called and the next point determines the winner. This scenario means that no game may go beyond 7 points. We can model games as arrangements of 4 A's and 4 B's in 7 spaces, thus: Number of possible games Some of these games finish earlier than 7 points, however.

2. ## The purpose of this investigation is to create and model a dice-based casino game ...

\$50), the casino can attract more players and make a larger profit. This investigation will now consider a game where each player rolls their dice twice. If the player?s largest roll is greater than the casino?s largest roll, then the player wins; in all other cases, the casino wins.

1. ## Develop a mathematical model for the placement of line guides on Fishing Rods.

Guide Number (from tip) 0* 1 2 3 4 5 6 7 8 n** Distance from Tip (cm) 0 10 23 38 55 74 96 120 149 230 *the guide at the tip of the rod is not counted **n is the finite value that represents the maximum number of guides that would fit on the rod.

2. ## This assignment aims to develop a mathematical model for the placement of lines on ...

equations into matrices: 1?1?1?|10 4?2?1?|23 9?3?1?|38 Take -4 x row 1 and add to row 2 to get a new row 2 Take -9 x row 1 and add to row 3 to get a new row 3 1?1?1?|10 0..-2..-3?|-17 0..-6..-8?|-52 Divide row 2 by -2 to get a new • Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to 