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?

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 \[ N(14)\approx8926 \text{ cells}. \]