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January 13th, 2018, 09:23 AM   #1
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determining equations parameter

Hi,
I'm new to this forum, maybe I should post my problem in some other topics.
I have problem in determining K and D in following Equation, it's experimental equation and I have 10 answer for "f" and "t", "s" and "v" are defined quantity
S=4.3589 , V=6.8557
F=[0.022081745 0.039747141 0.079494282 0.125865947 0.183278484 0.20094388 0.214192927 0.304728081 0.359932444 0.388638713 ];
t=[0.083333333 0.291666667 1 2 3 4 5 19 47 90 ];
Can anyone helping me to determine D and K? It's really important to me and I am in hurry,
I will attach equation.
I really appreciate.
Thanks.
Attached Images
 Capture1.PNG (3.1 KB, 2 views) Capture.PNG (4.3 KB, 6 views)

Last edited by skipjack; January 13th, 2018 at 11:52 AM.

 January 13th, 2018, 12:16 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 You have $f= \left(\frac{S}{V}\right)\sqrt{\frac{D}{K}}\left[\left(kt + \frac{1}{2}\right)\text{erf}(\sqrt{kt}) + \sqrt{\frac{kt}{\pi}}\exp(-kt)\right]$. Dividing both sides by $\left(\frac{S}{V}\right)$ and $\left(kt + \frac{1}{2}\right)\text{erf}(\sqrt{kt}) + \sqrt{\frac{kt}{\pi}}\exp(-kt)$, that becomes $\displaystyle \sqrt{\frac{D}{K}} = \frac{Vf}{S\left[\left(kt + \frac{1}{2}\right)\text{erf}(\sqrt{kt})+ \sqrt{\frac{kt}{\pi}}\exp(-kt)\right]}$. Squaring both sides, $\displaystyle \frac{D}{K} = \frac{V^2f^2}{S^2\left[\left(kt+ \frac{1}{2}\right)\text{erf}(\sqrt{kt})+ \sqrt{\frac{kt}{\pi}}\exp(-kt)\right]^2}$. Of course, you can't solve for both S and D from only one equation. Last edited by skipjack; January 18th, 2018 at 08:33 PM.
January 13th, 2018, 12:46 PM   #3
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Quote:
 Originally Posted by Country Boy You have $f= \left(\frac{S}{V}\right)\sqrt{\frac{D}{S}}[kt+ \frac{1}{2}erf(kt)+ \sqrt{\frac{kt}{\pi}}exp(-kt)]$. Dividing both sides by $\left(\frac{S}{V}\right)$ and $kt+ \frac{1}{2}erf(kt)+ \sqrt{\frac{kt}{\pi}}exp(-kt)$ that becomes $\sqrt{\frac{D}{S}}= \frac{Sf}{D[kt+ \frac{1}{2}erf(kt)+ \sqrt{\frac{kt}{\pi}}exp(-kt)]}$. Squaring both sides $\frac{D}{S}= \frac{S^2f^2}{D^2[kt+ \frac{1}{2}erf(kt)+ \sqrt{\frac{kt}{\pi}}exp(-kt)]}$. Now multiply on both sides by $D^2$ to get $\frac{D^3}{S}= \frac{f^2}{D^2[kt+ \frac{1}{2}erf(kt)+ \sqrt{\frac{kt}{\pi}}exp(-kt)}$. Of course, you can't solve for both S and D from only one equation.
fixed a tiny latex error.

 January 18th, 2018, 05:01 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 Thank you.
 January 18th, 2018, 08:25 PM #5 Global Moderator   Joined: Dec 2006 Posts: 19,951 Thanks: 1842 Hi Alirezap and welcome to the forum. In your equation, do you know the value (or values) of $k$, or was the use of $k$ a typo? Also, are $S$ and $V$ in your equation intended to be upper case?

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