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January 1st, 2018, 08:36 AM   #1
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Question 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 08:21 AM.
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January 1st, 2018, 02:16 PM   #2
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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.
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Last edited by skipjack; January 2nd, 2018 at 08:18 AM.
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January 1st, 2018, 05:05 PM   #3
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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
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January 1st, 2018, 07:22 PM   #4
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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.
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January 2nd, 2018, 12:04 AM   #5
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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 08:24 AM.
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January 2nd, 2018, 05:44 AM   #6
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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.
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Last edited by skipjack; January 2nd, 2018 at 08:15 AM.
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January 2nd, 2018, 08:22 AM   #7
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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.
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January 12th, 2018, 08:31 PM   #8
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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
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January 13th, 2018, 04:36 AM   #9
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Quote:
Originally Posted by Audiana View Post
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: $\displaystyle \frac{d ln(r)}{d ln(P_{co})}= 1- \frac{2K_{co}P_{co}}{1+ K_{co}P_{co}+ K_{o_2}P_{o_2}}$.
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January 13th, 2018, 07:09 AM   #10
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Quote:
Originally Posted by Country Boy View Post
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 ??
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