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, since the colony started to grow, 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 the number of bacteria is 790, set ( N(t) = 790 ):

$790 = 450e^{0.065t}$

Dividing by 450 gives:

$\frac{790}{450} = e^{0.065t}$

Taking the natural logarithm:

$\ln\left(\frac{790}{450}\right) = 0.065t$

Solving for ( t ):

$t = \frac{\ln\left(\frac{790}{450}\right)}{0.065} \approx 8.7$

Therefore, it takes approximately 8.7 hours.

To find the average number of bacteria from ( t = 3 ) to ( t = 12 ), we use the formula:

$\text{Average} = \frac{1}{b-a} \int_{a}^{b} N(t) \, dt$

where ( a = 3 ) and ( b = 12 ). Calculating the average:

$\text{Average} = \frac{1}{9} \int_{3}^{12} 450e^{0.065t} \, dt$

Evaluating the integral gives:

( = \frac{450}{9} \left[ e^{0.065t} \right]_{3}^{12} \approx 743 )

Thus, the average number of bacteria during this period is approximately 743.

To find the rate of change, we first differentiate ( N(t) ):

$N'(t) = 450 \cdot 0.065 e^{0.065t}$

Now substitute ( t = 12 ):

$N'(12) = 29.25 e^{0.78} \approx 63.8$

Thus, the rate at which ( N(t) ) is changing at ( t = 12 ) is approximately 63.8 bacteria per hour.

We need to find ( k ) such that:

$N'(t) = 29.25e^{0.065t} > 90$

Solving:

$e^{0.065t} > \frac{90}{29.25}$

Taking the natural logarithm:

$0.065t > \ln\left(\frac{90}{29.25}\right) \implies t > \frac{\ln\left(\frac{90}{29.25}\right)}{0.065} \approx 17$

Thus, the least value of ( k ) is 18.

To find when ( N(t) = P(t) ):

Set the equations equal:

$450e^{0.065t} = 220e^{0.17t}$

This leads to:

$\frac{450}{220} = e^{0.17t - 0.065t}$

Taking the natural logarithm:

$\ln\left(\frac{450}{220}\right) = (0.17 - 0.065)t$

Solving for ( t ) gives:

$t \approx 7 \text{ hours}$

Therefore, the colonies will have equal bacteria counts after approximately 7 hours.