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abb0tBest ResponseYou've already chosen the best response.0
U ok? If it's a medical emergency, dial 911.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
I want to ask u about how to derive the 3rd Equation (On picture below), becomes the4th equation (on picture below) by using WLambert function ???
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
I used to know this when I took an enzyme kinetics course and this derivation is long. About half a page  1 page typed out.
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
Maybe you can read about it on the article you found it on? http://mvputz.iqstorm.ro/upload/LogisticEnzymeKinetics_Paper1.pdf
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
i have been reading this paper, and I did not find what i want.., would u kindly help me?? do you still have a reference about this derivation (complete with WLambert function)
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
I will look through my notes from my kinetics class. If I find it, I'll post it.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.1
hmm , i haven't heard of the W Lambert function before
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
do u have any idea @oldrin.bataku ??
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
W(x) is defined such that W(x) e^W(x) = x
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
yes.., i know.., i did read it on wikipedia. but I'm a little bit confused how to derive it (3rd equations) :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
I'm looking for some easy derivation of the above, step by step., :)
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
$$V_{max}t=([S_0][S](t))K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0][S](t)V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0][S](t)V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]V_{max}t}}{K_M}}{\frac{e^{[S](t)}}{K_M}}$$ $$[S](t)e^{[S](t)}=\frac{S_0}{K_M}e^{\frac{[S_0]V_{max}t}{K_m}}=\frac{S_0}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_m}\right)$$ $$[S](t)=W\left(\frac{S_0}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_m}\right)\right)$$
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
It's just algebra :p
 one year ago

abb0tBest ResponseYou've already chosen the best response.0
It is just algebra :) intense algebra.
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
i thank all of u guys :)
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
thank u so much @oldrin.bataku :)
 one year ago

oldrin.batakuBest ResponseYou've already chosen the best response.2
$$V_{max}t=([S_0][S](t))K_M\ln\left(\frac{[S](t)}{[S_0]}\right)\\\ln\left(\frac{[S](t)}{[S_0]}\right)=\frac{[S_0][S](t)V_{max}t}{K_M}\\\frac{[S](t)}{[S_0]}=\exp\left(\frac{[S_0][S](t)V_{max}t}{K_M}\right)=\frac{e^\frac{{[S_0]V_{max}t}}{K_M}}{e^\frac{[S](t)}{K_M}}$$$$\frac{[S](t)}{K_M}e^\frac{[S](t)}{K_M}=\frac{[S_0]}{K_M}e^{\frac{[S_0]V_{max}t}{K_M}}=\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)$$$$\frac{[S](t)}{K_M}=W\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)\right]$$$$[S](t)=K_MW\left[\frac{[S_0]}{K_M}\exp\left(\frac{[S_0]V_{max}t}{K_M}\right)\right]$$Sorry, I made a few really dumb errors the first time! http://en.wikipedia.org/wiki/Michaelis%E2%80%93Menten_kinetics#Determination_of_constants
 one year ago

gerryliyanaBest ResponseYou've already chosen the best response.1
ok.., no problem, thank u @oldrin.bataku ;)
 one year ago
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