My Math Forum  

Go Back   My Math Forum > College Math Forum > Calculus

Calculus Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
January 13th, 2018, 08:23 AM   #1
Newbie
 
Joined: Jan 2018
From: iran

Posts: 1
Thanks: 0

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
File Type: png Capture1.PNG (3.1 KB, 2 views)
File Type: png Capture.PNG (4.3 KB, 6 views)

Last edited by skipjack; January 13th, 2018 at 10:52 AM.
Alirezap is offline  
 
January 13th, 2018, 11:16 AM   #2
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,191
Thanks: 871

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 07:33 PM.
Country Boy is offline  
January 13th, 2018, 11:46 AM   #3
Senior Member
 
romsek's Avatar
 
Joined: Sep 2015
From: USA

Posts: 1,975
Thanks: 1026

Quote:
Originally Posted by Country Boy View Post
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.
romsek is offline  
January 18th, 2018, 04:01 PM   #4
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 3,191
Thanks: 871

Thank you.
Country Boy is offline  
January 18th, 2018, 07:25 PM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 19,054
Thanks: 1618

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?
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Calculus

Tags
determining, equations, parameter



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Parameter help! eglaud Calculus 12 February 11th, 2016 06:28 PM
parameter representation Zman15 Geometry 2 March 11th, 2015 12:42 PM
what is parameter? helloprajna Number Theory 6 July 9th, 2013 09:46 PM
Determining equations of growth functions xpoisnp Calculus 0 February 8th, 2012 04:37 AM
parameter integral DeliaDS Calculus 1 June 20th, 2010 07:03 AM





Copyright © 2018 My Math Forum. All rights reserved.