User Name Remember Me? Password

 Pre-Calculus Pre-Calculus Math Forum

 January 1st, 2018, 07:36 AM #1 Newbie   Joined: Nov 2017 From: Malaysia Posts: 9 Thanks: 0 slope of a graph Hi there, I need help! Let's say you have an equation "y=(pq)/(rs)", then you calculate various values of "y" by changing the "p" Screen Shot 2018-01-02 at 0.32.54.png You will plot the ln p versus ln r graph. How can we calculate the slope of the graph by differentiation method? Thank you. Last edited by skipjack; January 2nd, 2018 at 07:21 AM. January 1st, 2018, 01:16 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I frankly don't know what you are talking about! You give an equation, $\displaystyle y= \frac{pq}{rs}$, and say "you can calculate various values of 'y' by changing the 'p'". How You still have q, r, and s. What about them? And then you give a chart of, not values of p and y, but values of p and r. "You will plot the ln p versus ln r graph". Okay, now what about y, q, and s? What do they have to do with this? Taking the values of p and r in the chart, (ln p, ln r) would be (0, 2.3026) (0.6931, 2.995) (1.099, 3.401) (1.386, 3.219) (1.609, 2.708) Since the first three pairs. (1, 10), (2, 20), and (3, 30), are r= 10p, the first three numbers on the ln graph are on the straight line ln r= ln p+ ln(10)= ln(p)+ 2.303. But the last three do not fall on that line. Thanks from Audiana Last edited by skipjack; January 2nd, 2018 at 07:18 AM. January 1st, 2018, 04:05 PM #3 Newbie   Joined: Nov 2017 From: Malaysia Posts: 9 Thanks: 0 Thanks Country Boy The equation that I posted is just an example, not represent anything. We calculate the value of "y" by changing the "p" and "q"; "r";"s" will have fixed value. only "p" will change. The table, my mistake, should have the value for "p" and "y". Then comes the "ln y" versus "ln p" graph. I need to find the slope of this graph (not a straight line) by differentiation method. But what should I differentiate? Thank You January 1st, 2018, 06:22 PM #4 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Okay, so y is a function of p while q, r, and s are constants. Then $y= \left(\frac{q}{rs}\right)p$ is a straight line with slope $\frac{q}{rs}$. However, in order to have y= 10 when p= 1 and y= 20 when p= 2, then $\frac{q}{rs}$ must be equal to 10. On that case, when p= 4, y= 40, not 25, and when p= 5, y= 50, not 50, not 15. What you are saying here is simply not possible. January 1st, 2018, 11:04 PM #5 Newbie   Joined: Nov 2017 From: Malaysia Posts: 9 Thanks: 0 Thank you Country Boy. What I plot is ln y versus ln p, not the y versus p. I calculated y by using y = pq/rs. Maybe the values are not possible (I got it wrong somewhere), but what I really need to know is that how or what I should differentiate to get the slope for ln y vs ln p. Thank you. Last edited by skipjack; January 2nd, 2018 at 07:24 AM. January 2nd, 2018, 04:44 AM #6 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Since the graph of y vs p is a simple straight line, I have no idea why you would want to graph "ln y vs ln p". If, as you say, $y= \left(\frac{qp}{rs}\right)$ then $\ln(y)= \ln\left(\frac{pq}{rs}\right)= \ln(p)+ \ln(q)- \ln(r)- \ln(s)$. The "y vs p" graph is a line, through the origin, with slope $\frac{q}{rs}$. The second is a line, not through the origin, with slope 1. Thanks from Audiana Last edited by skipjack; January 2nd, 2018 at 07:15 AM. January 2nd, 2018, 07:22 AM #7 Senior Member   Joined: Oct 2009 Posts: 783 Thanks: 280 As I understand it, the $y = \frac{pq}{rs}$ is NOT the equation, but just a random example? But do you actually know the right equation? If you don't know the correct equation, the differentiation you desire is not possible. Thanks from Audiana January 12th, 2018, 07:31 PM #8 Newbie   Joined: Nov 2017 From: Malaysia Posts: 9 Thanks: 0 Yes I know the actual equation r = (KcoPcoKo2Po2)/(1+KcoPco+Ko2Po2)^2 It is the Langmuir Isotherm with Kco and Ko2 are the adsorption constant for CO and O2 gas, while Pco and Po2 are the CO and O2 gas partial pressure. I need the derivatives of (d ln r)/(d ln Pco) I start with ln r = ln ((KcoPcoKo2Po2)/(1+KcoPco+Ko2Po2)^2) Thank You January 13th, 2018, 03:36 AM   #9
Math Team

Joined: Jan 2015
From: Alabama

Posts: 3,264
Thanks: 902

Quote:
 Originally Posted by Audiana Yes I know the actual equation r = (KcoPcoKo2Po2)/(1+KcoPco+Ko2Po2)^2 It is the Langmuir Isotherm with Kco and Ko2 are the adsorption constant for CO and O2 gas, while Pco and Po2 are the CO and O2 gas partial pressure. I need the derivatives of (d ln r)/(d ln Pco) I start with ln r = ln ((KcoPcoKo2Po2)/(1+KcoPco+Ko2Po2)^2) Thank You
Subscripts would make that easier to read:
$\displaystyle r= \frac{K_{co}P_{co}K_{o_2}P_{o_2}}{(1+ K_{co}P_{co}+ K_{o_2}P_{o2})^2}$

By the properties of logarithms:
$\displaystyle ln(r)= ln(K_{co})+ ln(P_{co})+ ln(K_{o_2})_+ ln(P_{o_2})- 2ln(1+ K_{co}P_{co}+ K_{o_2}P_{o_2})$.

Now, take the derivative with respect to $\displaystyle ln(P_{co})$. The derivative of function f(x) with respect to function g(x) is, by the chain rule, given by $\displaystyle \frac{df}{dg}\frac{dg}{dx}= \frac{df}{dx}$ so that $\displaystyle \frac{df}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}$. Here $\displaystyle g(x)= ln(x)$ so that $\displaystyle \frac{dg}{dx}= \frac{1}{x}$ and $\displaystyle \frac{dr}{dg}= \frac{\frac{df}{dx}}{\frac{1}{x}}= x\frac{df}{dx}$.

In this problem, "x" is $\displaystyle P_{co}$ so that $\displaystyle \frac{df}{dx}= \frac{d ln(r)}{dP_{co}}= \frac{1}{P_{co}}- \frac{2K_{co}}{1+ K_{co}P_{co}+ K_{o_2}P_{o_2}}$.

$\displaystyle \frac{d ln(r)}{d ln(P_{co})}$ is $\displaystyle x= P_{co}$ times that: $\displaystyle \frac{d ln(r)}{d ln(P_{co})}= 1- \frac{2K_{co}P_{co}}{1+ K_{co}P_{co}+ K_{o_2}P_{o_2}}$. January 13th, 2018, 06:09 AM   #10
Newbie

Joined: Nov 2017
From: Malaysia

Posts: 9
Thanks: 0

Quote:
 Originally Posted by Country Boy Subscripts would make that easier to read: $\displaystyle r= \frac{K_{co}P_{co}K_{o_2}P_{o_2}}{(1+ K_{co}P_{co}+ K_{o_2}P_{o2})^2}$ By the properties of logarithms: $\displaystyle ln(r)= ln(K_{co})+ ln(P_{co})+ ln(K_{o_2})_+ ln(P_{o_2})- 2ln(1+ K_{co}P_{co}+ K_{o_2}P_{o_2})$. Now, take the derivative with respect to $\displaystyle ln(P_{co})$. The derivative of function f(x) with respect to function g(x) is, by the chain rule, given by $\displaystyle \frac{df}{dg}\frac{dg}{dx}= \frac{df}{dx}$ so that $\displaystyle \frac{df}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}$. Here $\displaystyle g(x)= ln(x)$ so that $\displaystyle \frac{dg}{dx}= \frac{1}{x}$ and $\displaystyle \frac{dr}{dg}= \frac{\frac{df}{dx}}{\frac{1}{x}}= x\frac{df}{dx}$. In this problem, "x" is $\displaystyle P_{co}$ so that $\displaystyle \frac{df}{dx}= \frac{d ln(r)}{dP_{co}}= \frac{1}{P_{co}}- \frac{2K_{co}}{1+ K_{co}P_{co}+ K_{o_2}P_{o_2}}$. $\displaystyle \frac{d ln(r)}{d ln(P_{co})}$ is $\displaystyle x= P_{co}$ times that: $\displaystyle \frac{d ln(r)}{d ln(P_{co})}= 1- \frac{2K_{co}P_{co}}{1+ K_{co}P_{co}+ K_{o_2}P_{o_2}}$.
Thank You very much!!

Then if we plot this (d ln r)/(d ln Pco) as "y" exist versus ln Pco ; will it give similar graph with ln r versus ln Pco ?? Tags graph, slope Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post GIjoefan1976 Algebra 12 March 17th, 2016 08:36 PM avnerg Applied Math 0 September 18th, 2013 06:03 AM taylor_1989_2012 Trigonometry 1 February 7th, 2013 09:07 AM johnyjj2 Applied Math 0 December 28th, 2010 02:49 PM Singularity Applied Math 1 September 18th, 2010 08:35 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      