Respuesta :
It will take 4 hours, this is because it said it adds 1000 after two hours, and it will add another 1000 two hours later which will give you your answer of four hours
The number of hours from the initial given time that will take for the bacteria to grow to 2500 number is 4.64 hours approx.
How to calculate the instantaneous rate of growth of a function?
Suppose that a function is defined as;
[tex]y = f(x)[/tex]
Then, suppose that we want to know the instantaneous rate of the growth of the function with respect to the change in x, then its instantaneous rate is given as:
[tex]\dfrac{dy}{dx} = \dfrac{d(f(x))}{dx}[/tex]
Assuming that the bacterial growth can be approximated by a continous and differentiable function y = f(x), where x represents the number of hours spent from the initial time, we're given that:
- [tex]\dfrac{dy}{dx} \propto y[/tex]
Supposing the proportionality constant be k, then we get:
[tex]\dfrac{dy}{dx} = ky[/tex]
Solving this differential equation, we get:
[tex]\dfrac{dy}{y} = kdx\\\\\text{Integrating both the sides without limits}\\\\\int \dfrac{dy}{y} = \int x dx\\\\\ln(y) + \ln(c) = kx\\\\\ln(yc) =kx\\ yc = e^{kx}\\y = \dfrac{e^{kx}}{c}[/tex]
where ln(c) represents the integration constant. (we took ln(c) because, firstly, ln's range is whole real number (which gives us the access to use it as integration constant), and secondly that it can merge with ln(y) to simplify the work)
Since we're given that:
At x = t (for some value of t in hours), we're given that y = 500,
and for x = t+2, y = 1000,
so we get two equations as:
[tex]\\500 = \dfrac{e^{kt}}{c}\\\\1000 = \dfrac{e^{k(t+2)}}{c}\\[/tex]
Thus, we get:
[tex]\dfrac{e^{kt}}{500} = \dfrac{e^{k(t+2)}}{1000} \\\\kt = \ln(0.5) + k(t+2)\\\\k = \dfrac{-\ln(0.5)}{2}} \approx 0.3465[/tex]
Thus, we get:
[tex]\\500 = \dfrac{e^{kt}}{c} \\\\c = \dfrac{e^{0.3465t}}{500}[/tex]
Thus, we get:
[tex]y = \dfrac{e^{0.3465x}}{\dfrac{e^{0.3465t}}{500}} = 500e^{0.3465(x-t)[/tex]
Let from the initial given time t, it takes h hours more for bacterias to be 2500, then we get:
[tex]2500 = 500 \times e^{0.3465 (t+h - t)}\\0.3465(h) = \ln(5)\\\\h = \dfrac{\ln(5)}{0.3465} = 4.644 \: \rm hours \: approx.[/tex]
Thus, the number of hours from the initial given time that will take for the bacteria to grow to 2500 number is 4.64 hours approx.
Learn more about differential equations here:
https://brainly.com/question/14658115
