Chapter 8: Q22P (page 399)

Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant rate by the removal of bacteria for experimental purposes

Short Answer

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Answer

The rate of growth of bacteria N is $N (t) = N_{0} e^{k t} - \frac{R}{K} (e^{k t}) - 1$ .

Step by step solution

Given information

It is given that the rate of increase is proportional to the number present, but the population is being reduced at a constant rate by removing bacteria for experimental purposes.

Definition of differential equation

A differential equation contains at least onederivative of an unknown function, either an ordinary derivative or a partial derivative.

Find N as a function of t

It is given that the rate of increase is proportional to the number present, but the population is being reduced at a constant rate by removing bacteria for experimental purposes.

Let removal rate = R

Therefore,

$\frac{d N}{d t} = K N - R \int \frac{d N}{K N - R} = \int d t b n (\frac{K N - R}{K}) = t + N (t) = \frac{α_{2} e^{K t} + R}{K}$ .

Here, $α_{2} = e^{α k}$ .

Now, at $t = 0, N = N_{0}$

$N (0) = N_{0} = \frac{α_{2} + R}{K} \Rightarrow α_{2} = N_{0} K - R$

Thus,

$N (t) = N_{0} e^{k t} - \frac{R}{K} (e^{k t} - 1)$ .

Therefore, the number N of bacteria as a function of time t is $N (t) = N_{0} e^{k t} - \frac{R}{K} (e^{k t} - 1)$ .

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Solve the equation for the rate of growth of bacteria if the rate of increase is proportional to the number present but the population is being reduced at a constant rate by the removal of bacteria for experimental purposes

Short Answer

Step by step solution

Given information

Definition of differential equation

Find N as a function of t

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