What is the purpose of the k constant in a logistics equation or any other exponential growth problem?

Question

What does it mean, and why is it there......


Also.......Are the equations for exponential growth, decay, etc.... that were used in elementary algebra in high school really just differnetial equations in disguise? or what do you think?

Answer 1

Do you mean like..
N=Ne^kt
'N' is the number in the population after a time
't' is the initial number,
'k' is the growth constant (if positive) or the decay constant (if negative),
'e' is the base of natural logarithms (approximately 2.71828)

K is like..the rate something grows or decays at. As for the differential equation stuff, i don't know. Is that calculus stuff? cuz' I haven't taken that yet, lol.

Answer 2

The k is there as a constant, as you said. It is necessary to the equation because without it the whole semblance would be only a relation. A constant is always required to make something an equation, otherwise the two variables will only be relating to each other, but with a constant k, which could even be 1, a factor is introduced which allows a direct relation to be converted to an equation as one variable can be modified to be equal to the other by making the constant a value to do so. You could say that exponentials and decays are diffeq's in disguise, but really you must consider how many dimensions the problem is interpreting, for example, usually exponentials only deal with two dimensions to start, and then way off in infinity become only 1D as the exponential reaches unknown bounds. Diffeq's tend to be a bit more powerful and therefore have a few more dimensions to them, whether it be space, time, or even a purely mathematical dimension, of which there are about 27 I believe.

Answer 3

The equation are the SOLUTIONS to differential equations. In fact, the growth equation is
dQ/dt = kQ
which says that, at any time, the rate of formation of Q is proportional to Q. k is the constant of proportionality. In the decay equation, the form is
dQ/dt = -kQ
where k is again the constant of proportionality.

To visualize this, make believe you are a ruthless prison camp commander. There are 1000 prisoners in your camp. You decide to execute 5 percent of your prisoners every day. On the first day, 50 are executed. On the next, 47 are executed (you can't execute half a prisoner). On the next, 45, since you have 903 prisoners on hand. That's how the decay equation works with a k=0.05/day

Answer 4

considering section affected = A and total section = 9 then section no longer affected = 9 ? A fee of exchange of A (dA/dt) is proportional ( = ok * ) to the made from floor section it quite is affected (A) and floor section that's no longer affected (9?A) dA/dt = ok A (9?A)

Answer 5

it's got somthing to do with the compond used to divide a fragment