# Another Mathmatical Problem..

• Aug 7th 2006, 11:11 AM
c00ky
Another Mathmatical Problem..
Problem:

The following experimental data values of x and y are believed to be related by the law y = mx^2 + c, where m and c are constants. By plotting a suitable graph, verify that this law does indeed relate the values and find approximate values of m and c.

x 2.3 4.1 6.0 8.4 9.9 11.3

y 13.9 31.2 60.2 111.8 153.0 197.5
• Aug 7th 2006, 11:40 AM
galactus
I used Excel to plot this graph using your data and a quadric regression. I hope I didn't get it too small. Note the coorelation coefficient R^2. It's 1. That means it's as good as it gets as far as regressions go.

Incase you can't make it out, the equation is $\displaystyle 1.4938x^{2}+0.0809x+5.8118$
• Aug 7th 2006, 11:52 AM
c00ky
Could you breifly explain how you solved this problem? I'm a little confused.

Once i understand this problem, i shall try some similar ones.

Thanks for your help so far.
• Aug 7th 2006, 12:24 PM
ThePerfectHacker
Do not beg for answers! You can wait for a response! :mad:
-=USER WARNED=-
• Aug 7th 2006, 12:25 PM
galactus
All I done was to use Excel. It did all the work. If you want to perform a quadric regression by hand....good luck. That's what technology is for. Do you have a TI-83?. I believe it'll do regression: linear, quadric, cubic, quartic, sine, logistic, exponential, etc....
• Aug 7th 2006, 12:53 PM
CaptainBlack
Quote:

Originally Posted by c00ky
Problem:

The following experimental data values of x and y are believed to be related by the law y = mx^2 + c, where m and c are constants. By plotting a suitable graph, verify that this law does indeed relate the values and find approximate values of m and c.

x 2.3 4.1 6.0 8.4 9.9 11.3

y 13.9 31.2 60.2 111.8 153.0 197.5

Plot a graph of y against x^2 (x^2 the independent variable and y the
dependent), it should be a straight line, its slope will be m, and the intercept
on the x^2=0 axis will be c.

RonL
• Aug 7th 2006, 02:01 PM
galactus
Yes, I am sorry, but I was wrong. A linear regression is more suited, as Cap,N said. I saw the x^2 and got carried away.

If you want to perform a linear regression 'by hand' you can use the following formulae:

y=mx+b, n=number of data points.

$\displaystyle m=slope=\frac{n\sum{xy}-(\sum{x})(\sum{y})}{n\sum{x^{2}}-(\sum{x})^{2}}$

b=y-intercept=$\displaystyle \frac{\sum{y}}{n}-m\frac{\sum{x}}{n}$

Personally, I would use my calculator to do it, unless, you're forbidden.

Just thought you'd be interested.
• Aug 7th 2006, 02:58 PM
topsquark
Quote:

Originally Posted by c00ky
Problem:

The following experimental data values of x and y are believed to be related by the law y = mx^2 + c, where m and c are constants. By plotting a suitable graph, verify that this law does indeed relate the values and find approximate values of m and c.

x 2.3 4.1 6.0 8.4 9.9 11.3

y 13.9 31.2 60.2 111.8 153.0 197.5

Actually, it's fairly easy to figure out the equations for the coefficients of a least squares (polynomial) regression curve. (The problem for me is the formulae for the error estimates. I don't know how to do those.) If you like I can show you the general method. I should warn you that it involves a little Calculus (not much though). I won't post it unless you ask, since coding it will take a bit of time, but I would be happy to do so.

-Dan
• Aug 7th 2006, 06:23 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
I should warn you that it involves a little Calculus (not much though).

It actually involves quite a lot of calculus,...mutli-variable funtions,....partial diffrenciation,....hessian matrix.
• Aug 8th 2006, 03:59 AM
c00ky
Thanks Guys, I followed captain blacks steps and i think i've cracked it!