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

Mathematics IA - Particles

Extracts from this document...

Introduction

In this investigation I will be studying the case of an infection of particles. I will be looking into, and analyzing, how the particles work when they first enter the body, what effect the response of the immune system has, how medication is delivered and maintained, as well as death and recovery. Furthermore, I will be altering my investigation models to cater to a young child as opposed to an adult.

First I will look at the initial phase of the infection – the part where the particles enter the body and replicate yet none are expelled because the immune system hasn’t responded. To determine how long it will take before the immune system responds I need to create a basic formula:

a(rn)

In this formula, a represents the initial amount of particles and r represents the ratio at which they multiply every 4 hours – just like they are used in sequences. n represents how many times they multiply, which is once every 4 hours.

Considering the case of a young adult male, I presume that he is initially infected with 10,000 particles and that they double every hour. I also presume that the immune system responds when the particle count reaches 1,000,000. Therefore in order

...read more.

Middle

÷ 60 = 0.0416)% of the medicine is eliminated every minute. This enables to create the following equation:

D(0.9996240) + D(0.9996239) + … + D(0.99961) + D

This formula helps determine the dosage because D represents the dosage per minute, and multiplying with 0.999583 is the equivalent of removing 0.0416%. This number has a power of 240 because 0.0416% is removed 240 times by the end. For the next dose injected a minute later, this takes place 239 times because it is in the body for 1 minute less. Similarly, the dose injected 1 minute before the end of the 4 hours only has a power of 1 because 0.0416% is only removed one time before the end.

This equation can then be rearranged to put the smallest powers at the beginning and the largest powers at the end but either way I have a geometric sequence. I then use the formula to find the sum of a geometric sequence:

a(rn – 1) ÷ (r – 1) where a is the initial term (D, if the equation is rearranged) and r is the ratio of multiplication (0.9996, if the equation is rearranged) and n is the number of times the ratio is multiplied (240).

This equation must multiply out to 90 so that 90 micrograms are ensured in the system at the end of the four hours. Plugging the values in I get the following:

D(0.9996240 – 1) ÷ (0.9996 – 1) = 90

D(-0.095) ÷ (-0.000416) = 90

D(-0.095) = 90 × (-0.000416) = -0.038

D = (=0.038) ÷ (-0.095) = 0.39 micrograms

...read more.

Conclusion

9 instead of 1012.

I will now explain how I derived the general formula. First I presume that I are just finding a general formula for the time period after the immune system kicks in so I get the following:

{[1,000,000(1.60.25) – 50,000](1.60.25) – 50,000}(1.60.25) – 50,000 …

Opening this up I get the following:

1,000,000(1.60.25)(1.60.25)(1.60.25)–50,000(1.60.25)(1.60.25)–50,000(1.60.25)–50,000 …

For the first part I can immediately make out that the number of (1.60.25) attached to the 1,000,000 are equal to the number of hours passed making it, where n is the number of hours passed:

1,000,000(1.60.25n)

For the second half we can factorize 50,000 out leaving us with:

X – 50,000{(1.60.25)(1.60.25) + (1.60.25) + 1}

X – 50,000{(1.60.25(2)) + (1.60.25(1)) + (1.60.25(0))}

I then see a geometric equation forming with a being 1 because (1.60.25(0)) = 1 and r being (1.60.25). Using the sum of a geometric formula:

a(rn – 1) ÷ (r – 1)

1[(1.60.25n) – 1] ÷ [(1.60.25) – 1]

From this we can derive the general formula after the immune system kicks in to be:

1,000,000(1.60.25n) – 50,000[(1.60.25n) – 1] ÷ [(1.60.25) – 1])

We  can then use this format to derive a general formula that would work for any numbers:

P(r0.25n) – e[(r0.25n – 1) ÷ (r – 1)]

In this formula P is the starting number of particles (so 10,000 if we start from when he was first affected, or 1,000,000 if we start from when the immune system kicks in). r is the rate of multiplication every 4 hours, and e is the number of particles eliminated from the body every hour. From this formula we can plug in any values for any of the factors and determine the number of particles.

...read more.

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math Studies I.A

    2209000000 6093.3636 Uzbekistan 2,600 64.98 168948 6760000 4222.4004 Venezuela 13,500 73.7 994950 182250000 5431.69 Virgin Islands 14,500 76.86 1114470 210250000 5907.4596 West Bank 2,900 73.46 213034 8410000 5396.3716 Yemen 2,400 62.5 150000 5760000 3906.25 Zimbabwe 200 39.5 7900 40000 1560.25 Total= 1707100 7914.39 130013258 51470030000 559261.8111 r = ,,??-????.-,,??-?? .??-??

  2. Math Studies - IA

    To do this, a potential model for correlation will be calculated, including relevant measures, such as the strength of that correlation. The question then arises whether, if a correlation becomes apparent, is indeed true. The Chi-squared test can be used to test if the two events are independent and if a potential dependency is due to chance.

  1. Math IA - Logan's Logo

    - represented diagrammatically by the circled red dot. We can now compare my graph to see what the horizontal shift should be. The starting point for the original sine curve is (0, 0) while the starting point for my curve is (-3.0, 0.35). Thus, I determined the value of c=3.0.

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    The points in contention therefore be 3, 4, 5 and 6 since 1 point, the winning point is fixed in each game. Therefore the probability of winning would be: For 4-0 result. For 4-1 result For 4-2 result For 4-3 result Therefore, the probability of Player C winning the game

  1. IB Mathematics Portfolio - Modeling the amount of a drug in the bloodstream

    I assume that because the starting amount of the drug at 6 hours is different than the amount of drug starting at zero hours. For a person's body, I think the decay would not be the same rate if the initial amount of drug already in the body is increased.

  2. A logistic model

    A harvest of 8x103 fish is made per annum. Year (after Population Year Population attaining stability) 1 3.00?104 16 3.94?104 2 3.20?104 17 3.95?104 3 3.30?104 18 3.96?104 4 3.39?104 19 3.97?104 5 3.47?104 20 3.97?104 6 3.55?104 21 3.98?104 7 3.62?104 22 3.98?104 8 3.68?104 23 3.99?104 9 3.73?104

  1. Gold Medal heights IB IA- score 15

    Algebraically approaching the function Amplitude (a) Period (k) The year with lowest height is 1932, and highest height is 1980. Therefore, multiplying the difference by two will give the distance of one cycle (between two successive max and/or min). Horizontal stretch/ compression (b) Sub in value of k This will not be converted to degree in order to maintain exact value.

  2. MATH Lacsap's Fractions IA

    Also, 0 rows does not make practical mathematical sense so it can also be regarded as a limitation to this task. During the investigation, the ?1?s? at the beginning and end of each row did not count when making calculations, as they were not proper fractions.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work