Mathematics Question about Power Trend lines?

snowbeard

New Member
Joined
Mar 13, 2012
Messages
7
hello all, I am trying to fit a cooling graph with a trendline and I have found the power line to fit the best, as well as corresponding with Newton's law of cooling. However, I am not savvy enough to discuss how similar or dissimilar the two equations are! :confused:

the resulting equations from two different curves are as follows:

y=4305 x^-0.87733

y=4025 x^-0.91004


both fits have an R^2 of 0.996.

I am analyzing two scenarios of cooling the same object with three variables, one intended (the extraction of energy using water flow in one case, and no extraction in the other) and the others not under my control: different starting temperatures and different ambient temperatures. I need to understand if these are generally the same curve, or how very different are they? is there any statistical way to represent their relationship?



Thanks in advance for any help deciphering this!!!
 

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shg

MrExcel MVP
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May 7, 2008
Messages
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Office Version
  1. 2010
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Well, they are both power functions, so they are the same in that sense. You could calculate and plot the ratio between them:

y1 = a * x^b

y2 = b * x^c

y1/y2 = a/b * x^(b-c)

A​
B​
C​
D​
1​
T​
y1​
y2​
y1/y2​
2​
100​
75.7​
60.9​
1.24​
3​
200​
41.2​
32.4​
1.27​
4​
300​
28.9​
22.4​
1.29​
5​
400​
22.4​
17.2​
1.30​
6​
500​
18.5​
14.1​
1.31​
7​
600​
15.7​
11.9​
1.32​
 

snowbeard

New Member
Joined
Mar 13, 2012
Messages
7
hmmm, that makes about as much sense to me as what I had before.. but I appreciate the help!! ;)

What I really want to know is if the two curves are significantly different from each other, or are they basically (statistically) the same? I want them to both be the same type of fit, but my real question is is the "slope" or "rate of cooling" different between the two?

Could you explain how I can use the ratio plot to assess my question?

Thanks tons for the help!!
 

shg

MrExcel MVP
Joined
May 7, 2008
Messages
21,782
Office Version
  1. 2010
Platform
  1. Windows
In my example, I used the variable b in both equations; it should have been just in the first (a and b for the first, c and d for the second).

I'm still not sure what you're asking. You can differentiate both equations to compare cooling rates.
 

snowbeard

New Member
Joined
Mar 13, 2012
Messages
7
well, I'm a dunce at math, I made it thru my algebra courses years ago, but calculus kicked me in the head... I did very well in trig and geometry, but the laniguistics of advanced mathematics have escaped me. How would I differentiate the equations? :confused:
 

shg

MrExcel MVP
Joined
May 7, 2008
Messages
21,782
Office Version
  1. 2010
Platform
  1. Windows
If y = a*t^b, then dy/dt = a*b*t^(b-1). So,

A​
B​
C​
D​
E​
F​
G​
1​
T​
y1​
y2​
dy1/dt​
dy2/dt​
2​
100​
75.7​
60.9​
-0.66​
-0.55​
B2: =4305 * A2 ^ -0.87733
3​
200​
41.2​
32.4​
-0.18​
-0.15​
C2: =4025 * A2 ^ -0.91004
4​
300​
28.9​
22.4​
-0.08​
-0.07​
D2: =4305 * (-0.87733) * A2 ^ (-0.87733 - 1)
5​
400​
22.4​
17.2​
-0.05​
-0.04​
E2: =4025 * (-0.91004) * A2 ^ (-0.91004 - 1)
6​
500​
18.5​
14.1​
-0.03​
-0.03​
7​
600​
15.7​
11.9​
-0.02​
-0.02​
 

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