Proportionality with more than one variable?

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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?
 

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jbriggs444
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if x is direct, indirect or exponentially propotional to A and as well as B
If x is directly proportional to A then there is a constant k such that x = kA.
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.

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
Yes, we can write x = kAB.

[Editted to eliminate my first erroneous explanation]

It is tempting to multiply the two equations together to get x2 = k1k2AB

The problem is that the x=k1A is true only as long as one holds B constant. The value of k1 includes that constant value of B. Similarly, x=k2B is true only as long as one holds A constant. The value of k2 includes that constant value of A.

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)
It is the reaction rate that is proportional to the product of f(t) and F(t). So rather than f(t) being proportional to itself (a trivial tautology), it is f'(t) that is proportional to f(t).

So the short form would be "f'(t) is directly proportional to the product of f(t) and F(t)"
 
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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=constant1*A , x=constant2*B

when combined, x2=constant1+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
 
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jbriggs444
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if you answered x=constant*A*B could you explain why not x=constant*(A+B)
Suppose for a moment that the above formula held: x = k(A+B) for some constant k.
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.

and what about x=constant1*A , x=constant2*B

when combined, x2=constant1+2*AB
As I wrote, because that constant1 is not a constant. It is a function of B. Similarly, constant2 is not a constant. It is function of A.

What function of A can work for constant1? constant1 = k1B can work.

What function of B can work for constant2? constant2 = k2A can work.

What do you get when you multiply the two equations together now?
 
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