# Math Help - Creating a Model

1. ## Creating a Model

Alright another question, I don't even know how to start this.

1955-650, 1958-2,180,
1960-5,300, 1961-8,200,
1962-12,400 1965-35,500
1968- 66,300 1975- 91,600 1980-92,
900 1986-92,800 1990-93,100

How do you create a model for this data using y=c(e)^kx

And if the model is created, will it be a good fit? Will it work for all portion of the data?

Like I have said, I don't know how to start this. Many thanks to each person that tries to help me!!!!

2. Your data is not formatted clearly. I cannot see which is the x data and which is the y data.

3. Originally Posted by BobCalc
[html]1955- 650 1958- 2,180
1960- 5,300 1961- 8,200
1962-12,400 1965-35,500
1968-66,300 1975-91,600
1980-92,900 1986-92,800
1990-93,100 [/html]
How do you create a model for this data using y=c(e)^kx
You'd probably need to use regression software, and it might be simpler to use smaller t-values. For instance, you could assign "t = 0" to indicate the year 1955, so 1958 would be t = 3, etc. If your class has been using the "STAT" features on graphing calculators, then this would probably be the way to go.

Originally Posted by BobCalc
And if the model is created, will it be a good fit? Will it work for all portion of the data?
That is a judgement call you'll have to make, once you've done the regression.

4. 1955- 650 1958- 2,180
1960- 5,300 1961- 8,200
1962-12,400 1965-35,500
1968-66,300 1975-91,600
1980-92,900 1986-92,800
1990-93,100

Is there any way to do this without graphing software? I had never
used one, and our class does not have graphing calculators.

At least a couple of hints how to start this properly, I tried doing it
in many different ways, but all successful.

5. Here is one way you can approach the problem. You don't need a graphing calculator but you will need a piece of graph paper and a pencil.

You will also conclude that it is not possible to provide a good model of the required form. However, it is possible to develop a good model for the first 6 points and a different model for the last 6 points.

Let's try to find a model of the form:

$y=ce^{kx}$ Where x is the (year - 1955).

Now take the natural log of both sides to get:

$log(y)=log(ce^{kx})=log(c)+kx$

now if we let y'=log(y), c'=log(c) and x=year-1955 then this becomes:

$y'=kx+c'$

This is just the equation of a straight line. So take out your pencil and graph log(y) against (year-1955). Then use a ruler to determine the line of best fit.

The y intercept equals c'=log(c) and the slope equals k. So calculate k and c and put them back into the model.

When you get around to sketching your graph you will find that you need to use one straight line for the first few points and a different straight line for the last few points. Therefore you actually need two different models if you want a reasonably good fit.

The first 5 points are modeled well by y'=0.4235x+6.4565. So you would use k=0.4235, $c=e^{6.4565}=636.8$ and your model for the first few points could be:

$y=636.8e^{0.4235*(year-1955)}$

The last few points can't be modelled so well by this equation.