A culture of bacteria in a particular dish has an initial population of 300 cells grows at a rate of $$ \mathrm{N}^{\prime}(\mathrm{t})=40 e^{0.29863 \mathrm{t}} $$ cells/day. a. Find the population $$ \mathrm{N}(\mathrm{t}) $$ at any time $$ \mathrm{t} \geq 0 $$. b. What is the population after 14 days?

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We are given that the rate of change of the bacteria population is
$$
N'(t)=40\, e^{0.29863\,t} \quad \text{(cells/day)}
$$
with an initial population of
$$
N(0)=300.
$$

### a. Finding $$ N(t) $$

To find the population at time $$ t $$, we integrate the derivative:
$$
N(t)=\int N'(t)\,dt = \int 40\, e^{0.29863\,t}\,dt.
$$

Using the formula for the integral of an exponential function,
$$
\int e^{at}\,dt=\frac{1}{a}e^{at}+C,
$$
with $$ a=0.29863 $$, we have
$$
\int 40\, e^{0.29863\,t}\,dt = 40\cdot\frac{1}{0.29863}e^{0.29863\,t}+C = \frac{40}{0.29863}\,e^{0.29863\,t}+C.
$$

Now, apply the initial condition $$ N(0)=300 $$:
$$
N(0)=\frac{40}{0.29863}\,e^{0.29863\cdot0}+C = \frac{40}{0.29863}+C = 300.
$$
Thus, we solve for $$ C $$:
$$
C=300-\frac{40}{0.29863}.
$$

The expression for the population at time $$ t $$ is therefore
$$
N(t)=\frac{40}{0.29863}\,e^{0.29863\,t}+\left(300-\frac{40}{0.29863}\right).
$$

### b. Population after 14 Days

To find the population after 14 days, substitute $$ t=14 $$ into $$ N(t) $$:
$$
N(14)=\frac{40}{0.29863}\,e^{0.29863\cdot14}+\left(300-\frac{40}{0.29863}\right).
$$

For an approximate numerical value, first compute:
$$
0.29863\times14\approx 4.18082,
$$
and then
$$
e^{4.18082}\approx 65.43.
$$

Next, calculate
$$
\frac{40}{0.29863}\approx 133.93.
$$

Now, substitute these approximations:
$$
N(14)\approx 133.93\times65.43+\left(300-133.93\right).
$$

Evaluate each part:
$$
133.93\times65.43\approx 8760,
$$
and
$$
300-133.93\approx 166.07.
$$

Thus,
$$
N(14)\approx8760+166.07\approx8926.
$$

### Final Answers

a. The population at time $$ t \ge 0 $$ is
$$
N(t)=\frac{40}{0.29863}\,e^{0.29863\,t}+\left(300-\frac{40}{0.29863}\right).
$$

b. After 14 days, the population is approximately
$$
N(14)\approx8926 \text{ cells}.
$$
a. The population at any time $$ t $$ is $$ N(t) = \frac{40}{0.29863} e^{0.29863t} + \left(300 - \frac{40}{0.29863}\right) $$. b. After 14 days, the population is approximately 8,926 cells.

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