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Parameter vs. Temperature Formula

Date: 10/28/2002 at 21:29:53
From: Carl
Subject: Parameter vs. temperature formula

I am trying to figure out a formula to work out a problem. The 
intervals that I have to work with are:

Frequency vs Temperature
  1.5           0.2
  2.25          0.37
  3.00          1.40

The spacing between the frequency had to meet the Delta T of the 
temperature. The problem is to determine how accurate the spacing has 
to be to meet the above requirements.

I need to know how to derive the delta T (f) function equation. I was 
told that the function equation is exponential in nature and the 
general form is Delta T = A + Be^(Cf), Where Am, B, C are constants. 

Thaks for your assistance,

Date: 10/29/2002 at 01:37:18
From: Doctor Jeremiah
Subject: Re: Parameter vs. temperature formula

Hi Carl,

You have three values for f and T so you can make three equations from 
T = A + Be^(Cf)  If you do that, you can solve for the constants A, B, 
and C.

The three equations would be:

    0.20 = A + Be^(C 1.5)
    0.37 = A + Be^(C 2.25)
    1.40 = A + Be^(C 3)

We can rearrange then to make them easier to solve like this:

    ln[(0.20-A)/B] =  1.5 C 
    ln[(0.37-A)/B] = 2.25 C 
    ln[(1.40-A)/B] =    3 C 

and with a bit more manipulation of the first one: 

    ln[(0.20-A)/B] =  1.5 C
    2 ln[(0.20-A)/B] = 3 C
    ln[(0.20-A)^2/B^2] = 3 C

which gives us these three:

    ln[(0.20-A)^2/B^2] = 3 C
    ln[(0.37-A)/B]     = 2.25 C 
    ln[(1.40-A)/B]     =    3 C 

Now we can equate the first and third:

    ln[(0.20-A)^2/B^2] = ln[(1.40-A)/B]
    (0.20-A)^2/B^2 = (1.40-A)/B
    (0.20-A)^2 = (1.40-A) B^2/B
    (0.20-A)^2 = (1.40-A) B
    (0.20-A)^2/(1.40-A) = B            call this equation A

If we manipulate the first one again:

    ln[(0.20-A)/B] =  1.5 C
    1.5 ln[(0.20-A)/B] = 2.25 C
    ln[(0.20-A)^(3/2)/B^(3/2)] = 2.25 C

Now we can equate the first and second:

    ln[(0.20-A)^(3/2)/B^(3/2)] = ln[(0.37-A)/B]
    (0.20-A)^(3/2)/B^(3/2) = (0.37-A)/B
    (0.20-A)^(3/2) = (0.37-A) B^(3/2)/B
    (0.20-A)^(3/2) = (0.37-A) B^(1/2)
    (0.20-A)^(3/2)/(0.37-A) = B^(1/2)
    [(0.20-A)^(3/2)/(0.37-A)]^2 = B    call this equation B

Now we can manipulate equation A and B:

    (0.20-A)^2/(1.40-A) = [(0.20-A)^(3/2)/(0.37-A)]^2
    (0.20-A)^2/(1.40-A) = (0.20-A)^3/(0.37-A)^2
    (0.37-A)^2 (0.20-A)^2 = (0.20-A)^3 (1.40-A)
    (0.37-A)^2 = (0.20-A) (1.40-A)

And solving for A:

    (0.37-A)^2 = (0.20-A) (1.40-A)
    0.37^2 - 0.74A + A^2 = 0.28 - 1.60A + A^2
    0.37^2 - 0.74A = 0.28 - 1.60A
    1.60A - 0.74A = 0.28 - 0.37^2
    0.86A = 0.1431
    A = 0.1431/0.86
    A = 1431/8600

Now we can take equation A and solve for B:

    (0.20-A)^2/(1.40-A) = B
    (0.20-1431/8600)^2/(1.40-1431/8600) = B
    83521/91237400 = B

And now we can use the third equation to solve for C:

    ln[(1.40-A)/B] = 3 C 
    ln[(10609*91237400)/(8600*83521)] = 3 C 
    ln[9679375766/7182806] = 3 C
    ln[9679375766/7182806]/3 = C

So these are the three constants:

    A = 1431/8600
    B = 83521/91237400
    C = ln[9679375766/7182806]/3

Represented as numbers:

    A = 0.16639534883720930232558139534884
    B = 9.15425034032096486747759142632e-4
    C = 2.4027629321651825832225861370353

Let me know if you need more details about how I got these.

- Doctor Jeremiah, The Math Forum

Date: 10/30/2002 at 07:13:31
From: Carl
Subject: Thank you (Parameter vs. temperature formula)

Thank you very much; you were very helpful in helping me understand 
the problem. Thanks again.
Associated Topics:
High School Functions

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