Register to reply 
Proportionality with more than one variable? 
Share this thread: 
#1
Jun1314, 06:42 AM

P: 12

if x is direct, indirect or exponentially propotional to A and as well as B
can we write x=kAB ? if we write the equation seperately, we have x=k1A, x=k2B when combined, x2=(k1k2)1/2 (AB)2 then x=k3(AB)1/2 to see the real complicate example EX.1 trypsinogen is converted to trypsin in the body where trypsin itself catalyzes its own reaction let f(t) = amount of trypsin at time t let F(t) = amount of trypsinogen at time t write differential equation satisfied by f(t) in short, f(t) is direct proportional to product itself f(t) and the substrate F(t) should i write f'(t)=k f(t)F(t) , =k ( f(t)+F(t) ) , = k (f(t)F(t))1/2 or something else? 


#2
Jun1314, 05:55 PM

P: 907

If x is inversely proportional to A then there is a constant k such that x = k/A. I have never encountered the terms "indirect proportionality" or a "exponential proportionality". Fortunately, those terms are irrelevant to the questions below. [Editted to eliminate my first erroneous explanation] It is tempting to multiply the two equations together to get x^{2} = k_{1}k_{2}AB The problem is that the x=k_{1}A is true only as long as one holds B constant. The value of k_{1} includes that constant value of B. Similarly, x=k_{2}B is true only as long as one holds A constant. The value of k_{2} includes that constant value of A. So the short form would be "f'(t) is directly proportional to the product of f(t) and F(t)" 


#3
Jun1514, 01:35 AM

P: 12

sorry for ambiguous writing it must be X=K(AB)^{1/2} and f'(t) is direct proportional to product itself f(t) and the substrate F(t)
However, your answer still hasn't answered my questions. if x is directly proportional to A and as well as B. How can we express x in terms of equations? if you answered x=constant*A*B could you explain why not x=constant*(A+B) and what about x=constant_{1}*A , x=constant_{2}*B when combined, x^{2}=constant_{1+2}*AB x=+K(AB)^{1/2} and x=K(AB)^{1/2} but we ignore the minus one so x=K(AB)^{1/2} 


#4
Jun1514, 04:28 AM

P: 907

Proportionality with more than one variable?
But we also know that x=k'A for some constant k'. Take A=1, B=2. Then x=3k by the one equation and x=k' by the second. So k'=3k. Take A=2, B=1. Then x=3k by the one equation and x=2k' by the second. So 2k'=3k. Clearly, the only way this can hold is if both k and k' are equal to zero. So x=constant*(A+B) cannot be right except in the degenerate case where x is always zero. What function of A can work for constant_{1}? constant_{1} = k_{1}B can work. What function of B can work for constant_{2}? constant_{2} = k_{2}A can work. What do you get when you multiply the two equations together now? 


Register to reply 
Related Discussions  
Proportionality and graphs  General Math  2  
Constant of proportionality  General Math  2  
Constant of proportionality  Introductory Physics Homework  1  
Proportionality  Introductory Physics Homework  6  
Proportionality of T to R, m1 and m2  Introductory Physics Homework  3 