2) A mathematical representation of the data
Graphically
Algebraically
The equation for this data is 3.429801*1.126417^x. To get the equation I used the Graphics calculator and find the equation in the data matrix editor. Although when I was trying to validate it the computer gave me a different equation (this is using excel). The equation given was 3.4298*0.119^x. but both of the equation fits the data nearly perfectly. So therefore I don’t know which one is the right equation, but the equation I will be using will be the first one which is 3.429801*1.126417^x.
3) Validating
I used Excel to Validate. I added a trendline to my equation
Part B
Yi+1 – Yi versus X graph
This is the graph of Yi+1 – Yi versus X. as you can see from the graph you can tell that this is going to be an exponential function. Interpreting the graph you can see it is almost the same as the original graph. There is a gradual increase in the last section of the graph the reason for this is because of the huge increase in time between 25 cm and 22.5 cm. The equation for this graph is 1.961621*1.088371^x, I found this equation using the Graphics Calculator, but putting the equation to the graph the equation will not be able to fit the last number since there is huge increase at the end of the graph
Yi+1-Yi versus Y Graph
This is the graph of Yi+1-Yi Versus Yi. From this graph you can say this will be a linear equation. To find the equation for this there is two ways of doing this, firstly by pen paper method the second is using the Graphics Calculator. The equation will be
0.19975x+1.8367. I had to take out the last point in this data, because it was a unusual point therefore there is a point missing in this graph
Part C(Simple Applications)
We can find the instantaneous rate of change using the equation
F(x+h)-F(x)/h, h=0
I have made up a table showing this, at the end of each row the number at the end of this is heading toward a certain number, that is the instantaneous rate of change. I have only picked out some points from my data 2 from the front 2 in the middle and 2 from the end.
This table is for X values
H
0.01 0.001 0.0001 0.00001 Instantaneous Rate of Change
X
2.5 0.55 0.545 0.545 0.545 0.545
5 0.7408 0.7404 0.7404 0.7404 0.7404
12.5 1.809 1.808 1.808 1.808 1.808
15 2.436 2.435 2.435 2.435 2.435
22.5 5.949 5.946 5.946 5.946 5.946
25 8.011 8.007 8.007 8.007 8.007
Now we have to graph this against the initial X values
The graph will look something like this
To find a equation to fit this graph you can use the Graphics Calculator. Using the Math function on the Data Matrix Editor. It will give you equation of 0.406336*1.126693^x.
Part D(Complex Applications)
Looking at the equation from part b and c there is not really thing that is similar thing. Both of the equation have not really much in common except that for ‘b’ is greater than 1.
The reason for a big gap in the equation between part c and b is because that the instantaneous rate of change is only added by a small number (e.g. 0.0001) and therefore does not make much of a difference, but in part b which you have to find the Yi+1-Y1 which is greater which means the initial value (which is a)will be greater than b.
In conclusion you can draw it is possible to model a exponential function on this assignment.