The number of bacteria in the early stages of a growing colony of bacteria can be approximated using the function: $N(t) = 450e^{0.065t}$ where $t$ is the time, measured in hours, and $N(t)$ is the number of bacteria in the colony at time $t$ - Leaving Cert Mathematics - Question 9 - 2020

To find the time when $N(t) = 790$, we use the equation: $790 = 450e^{0.065t}$ Dividing both sides by 450: $\frac{790}{450} = e^{0.065t}$ Taking the natural logarithm: $\ln\left(\frac{790}{450}\right) = 0.065t$ Solving for $t$: $t = \frac{\ln\left(\frac{790}{450}\right)}{0.065} \approx 8.7 \text{ hours}$ So, it takes about 8.7 hours.

The average number of bacteria from $t = 3$ to $t = 12$ is given by: $\text{Average} = \frac{1}{12-3} \int_3^{12} N(t) \, dt$ First, find $N(t)$ for both times: $N(3) = 450e^{0.195} \approx 450 \times 1.215 \approx 547.5$ $N(12) = 450e^{0.78} \approx 450 \times 2.182 \approx 981.9$

Now, we calculate the average: $\int_3^{12} N(t) \, dt = \left[\frac{450}{0.065} e^{0.065t}\right]_{3}^{12} \approx \frac{450}{0.065} \left(e^{0.78} - e^{0.195}\right)$ This integral evaluates to approximately 743.2, yielding an average of about 743 when rounded to the nearest whole number.

First, we find the derivative: $N'(t) = 450e^{0.065t} \cdot 0.065$ Evaluating this at $t = 12$: $N'(12) = 450e^{0.78} \cdot 0.065$ Calculating gives: $N'(12) \approx 29.25e^{0.78} \approx 63.8$ Thus, the colony is growing at approximately 63.8 bacteria per hour at $t = 12$.

To find the least value of $k$, we set up the inequality: $N'(t) > 90$ This translates to: $29.25e^{0.065t} > 90$ Dividing both sides: $e^{0.065t} > \frac{90}{29.25}$ Taking the natural log yields: $0.065t > \ln\left(\frac{90}{29.25}\right)$ Solving gives: $t > \frac{\ln\left(\frac{90}{29.25}\right)}{0.065} \approx 18$ Thus, the least value of $k$ where $k \in \mathbb{N}$ is 18.

To find when $N(t) = P(t)$, $450e^{0.065t} = 220e^{0.17t}$ Dividing both sides by $e^{0.065t}$ gives: $450 = 220e^{0.17t - 0.065t}$ This simplifies to: $450 = 220e^{0.105t}$ Dividing both sides by 220 and taking the natural log: $\ln\left(\frac{450}{220}\right) = 0.105t$ Solving for $t$: $t \approx 7 \text{ hours}$ Thus, both colonies will have the same number of bacteria after approximately 7 hours.