# Sars Math Portfolio 1.

Extracts from this document...

Introduction

Sars Math Portfolio

1.

Day (0 = March 28) | Cumulative number of deaths |

0 | 53 |

7 | 89 |

14 | 119 |

21 | 182 |

28 | 293 |

35 | 435 |

42 | 526 |

49 | 623 |

56 | 696 |

63 | 764 |

70 | 784 |

77 | 801 |

84 | 809 |

91 | 812 |

The result of plotting the data above using a scatter graph is shown below. The main observations that can be made from this graph is that the start (day 0 to day 30 – approximately) has a constant (approximate again as this is only description of graph) gradient that is lesser than the constant gradient from day 30 to day 60. This part we will call the ‘middle’. The ‘end’ of the graph also has a constant gradient that is similar to the ‘start’. Therefore from the graph we can see that the death toll rises constantly for 30 days, then the rate increases significantly for duration of 30 days then decreases back to its lowest level for the last 30 days. Also notice that the graph begins from the y-intercept (which is not 0).

2.

y = mx + c

We will use the equation above to figure out a whole equation for the first two weeks of data.

53 = c (first data set. Since x = 0)

since we are looking to find the equation for the first two weeks we will use the data for the second week to find out the equation

hence, 119 = m(14) + 53

thus, 66 = 14m

therefore 33 = 7m, which means m = 33/7 = 4.714285 (recurring)

The linear function that has emerged from my calculations is y = 4.71x + 53.

Note – only 3 significant figures have been written as ‘m’ but in calculations the exact number will be used.

The

3.

Middle

- quadratic function (y = ax2 + bx + c)

- cubic function ( y = ax3 + bx2 + cx + d)

- exponential function ( y = abx)

- logarithmic function ( y = n logxy)

The first type of function that will be used is the exponential function.

y = abx

We will have to solve for a and b using simultaneous equations.

We start by inserting figures for day 28 and day 0.

53 = ab0

293 = ab28

therefore if we divide the second equation by the first equation we get

293/53 = b28

We get b28 because ab0 is simply a(1) and b28/1 = b28

Hence from our calculations b = 1.062970309

We do not have to solve for a because the first equation (53 = ab0) already gives us the value of a which is 53.

Therefore the exponential equation that can be used to model the first four weeks of data is y = 53(1.062970309)x

This model was first developed from the exponential function y = abx. At first I inserted the values for day 7 and day 28 because I assumed ab0 = 1 therefore we could not figure out b. After discovering a function to model those figures I realized that we were suppose to actually develop a model for the first four weeks so I went and inserted the values for day 0 and day 28 and developed the model shown above. There are no modifications to this function because it mirrors the results almost exactly (as shown in the graph below)

Conclusion

This decision that I have made is backed up by data. Day 28, number of deaths – 111. Day 35, number of deaths – 142, Day 42, number of deaths – 91. Although there are points in the second set that the gradient increases, compensation must be made so that the function for that set (the second set) can be created. The decision as to the choice of the function has also been made and that is that the main function that will be used for both data sets will be the exponential function because I value a low uncertainty more than the function actually passing through data points.

Exponential function from day 0 to day 35.

y = abx

53 = ab0

435 = ab35

Divide the second equation by the first equation to get

435/53 = b35

b = 1.06198989

And we know a = 53 because of the first equation therefore the function for the first set of date is

y = 53(1.06198989)x

Exponential function from day 35 to day 91.

y = abx

Because we are developing a new function for a constructed set of data, we can ‘reset’ day 35 and make it day 0 so that it is easier to develop the function.

Therefore, day 35 = 0. Day 91 = 56.

435 = ab0

812 = ab56

Divide equation two by equation one.

812/435 = b56

Solve for b

b = 1.011207956

a = 435. This is calculated from using the first equation.

Therefore the function for the second data set is

y = 435(1.011207956)x

As we can see, the graph does not model the data very well. This means that we have to explore other possible functions.

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

## Found what you're looking for?

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