# dP/dt = kP(m - P)

• Aug 20th 2010, 03:36 AM
ulow
dP/dt = kP(m - P)
dP/dt=kP(M-P) find the function for P
• Aug 20th 2010, 03:55 AM
Prove It
$\displaystyle \frac{dP}{dt} = kP(M-P)$

$\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$.

Now applying Partial Fractions we have

$\displaystyle \frac{A}{P} + \frac{B}{M-P} = \frac{1}{P(M-P)}$

$\displaystyle \frac{A(M-P) + BP}{P(M-P)} = \frac{1}{P(M-P)}$

$\displaystyle A(M-P)+BP=1$

$\displaystyle AM - AP + BP = 1$

$\displaystyle (B-A)P + AM = 0P + 1$

So $\displaystyle B - A = 0, AM = 1$.

This means $\displaystyle A = \frac{1}{M}$ and $\displaystyle B = \frac{1}{M}$.

Therefore:

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{MP} + \frac{1}{M(M-P)}\right]$

$\displaystyle \frac{dt}{dP} = \frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)$

$\displaystyle t = \int{\frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

$\displaystyle t = \frac{1}{kM}\int{\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

$\displaystyle t = \frac{1}{kM}\left(\ln{|P|} - \ln{|M-P|}\right) + C$

$\displaystyle t = \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} + C$

$\displaystyle \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} = t-C$

$\displaystyle \ln{\left|\frac{P}{M-P}\right|} = kMt - kMC$

$\displaystyle \left|\frac{P}{M-P}\right| = e^{kMt - kMC}$

$\displaystyle \left|\frac{P}{M-P}\right| = e^{-kMC}e^{kMt}$

$\displaystyle \frac{P}{M-P} = \pm e^{-kMC}e^{kMt}$

$\displaystyle \frac{P}{M-P} = Ae^{kMt}$ where $\displaystyle A= \pm e^{-kMC}$

$\displaystyle \frac{P-M+M}{M-P} = Ae^{kMt}$

$\displaystyle \frac{P-M}{M-P} + \frac{M}{M-P} = Ae^{kMt}$

$\displaystyle -1 + \frac{M}{M-P} = Ae^{kMt}$

$\displaystyle \frac{M}{M-P} = Ae^{kMt} + 1$

$\displaystyle \frac{M - P}{M} = \frac{1}{Ae^{kMt} + 1}$

$\displaystyle M - P = \frac{M}{Ae^{kMt} + 1}$

$\displaystyle P = M - \frac{M}{Ae^{kMt} + 1}$

$\displaystyle P = \frac{AMe^{kMt} + M - M}{Ae^{kMt} + 1}$

$\displaystyle P= \frac{AMe^{kMt}}{Ae^{kMt} + 1}$.
• Aug 20th 2010, 04:50 AM
bugatti79
Quote:

Originally Posted by Prove It
$\displaystyle \frac{dP}{dt} = kP(M-P)$

$\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$

Why do you invert dp/dt? Do you not integrate P wrt t?

Just wondering
Thanks
• Aug 20th 2010, 05:03 AM
Prove It
Because your function is given in terms of $\displaystyle P$. That means you will need to integrate with respect to $\displaystyle P$.
• Aug 21st 2010, 12:03 PM
bugatti79
Quote:

Originally Posted by Prove It
Because your function is given in terms of $\displaystyle P$. That means you will need to integrate with respect to $\displaystyle P$.

ok, fair enough, but can you not just integrate p wrt t and then group the p terms at the end and make it the subject of the expression?
• Aug 21st 2010, 05:43 PM
Prove It
Because you don't know what function $\displaystyle P$ is in terms of $\displaystyle t$.
• Aug 22nd 2010, 11:20 PM
ulow
Quote:

Originally Posted by Prove It
$\displaystyle \frac{dP}{dt} = kP(M-P)$

$\displaystyle \frac{dt}{dP} = \frac{1}{kP(M-P)}$

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$.

Now applying Partial Fractions we have

$\displaystyle \frac{A}{P} + \frac{B}{M-P} = \frac{1}{P(M-P)}$

$\displaystyle \frac{A(M-P) + BP}{P(M-P)} = \frac{1}{P(M-P)}$

$\displaystyle A(M-P)+BP=1$

$\displaystyle AM - AP + BP = 1$

$\displaystyle (B-A)P + AM = 0P + 1$

So $\displaystyle B - A = 0, AM = 1$.

This means $\displaystyle A = \frac{1}{M}$ and $\displaystyle B = \frac{1}{M}$.

Therefore:

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{P(M-P)}\right]$

$\displaystyle \frac{dt}{dP} = \frac{1}{k}\left[\frac{1}{MP} + \frac{1}{M(M-P)}\right]$

$\displaystyle \frac{dt}{dP} = \frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)$

$\displaystyle t = \int{\frac{1}{kM}\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

$\displaystyle t = \frac{1}{kM}\int{\left(\frac{1}{P} + \frac{1}{M-P}\right)\,dP}$

$\displaystyle t = \frac{1}{kM}\left(\ln{|P|} - \ln{|M-P|}\right) + C$

$\displaystyle t = \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} + C$

$\displaystyle \frac{1}{kM}\ln{\left|\frac{P}{M-P}\right|} = t-C$

$\displaystyle \ln{\left|\frac{P}{M-P}\right|} = kMt - kMC$

$\displaystyle \left|\frac{P}{M-P}\right| = e^{kMt - kMC}$

$\displaystyle \left|\frac{P}{M-P}\right| = e^{-kMC}e^{kMt}$

$\displaystyle \frac{P}{M-P} = \pm e^{-kMC}e^{kMt}$

$\displaystyle \frac{P}{M-P} = Ae^{kMt}$ where $\displaystyle A= \pm e^{-kMC}$

$\displaystyle \frac{P-M+M}{M-P} = Ae^{kMt}$

$\displaystyle \frac{P-M}{M-P} + \frac{M}{M-P} = Ae^{kMt}$

$\displaystyle -1 + \frac{M}{M-P} = Ae^{kMt}$

$\displaystyle \frac{M}{M-P} = Ae^{kMt} + 1$

$\displaystyle \frac{M - P}{M} = \frac{1}{Ae^{kMt} + 1}$

$\displaystyle M - P = \frac{M}{Ae^{kMt} + 1}$

$\displaystyle P = M - \frac{M}{Ae^{kMt} + 1}$

$\displaystyle P = \frac{AMe^{kMt} + M - M}{Ae^{kMt} + 1}$

$\displaystyle P= \frac{AMe^{kMt}}{Ae^{kMt} + 1}$.

Thanks if M= 197.3 how would i go about to find K?