Answer:
Initial size of the culture: [tex]\sf \bold{478.19} [/tex] bacteria
Doubling period: [tex]\sf 18.19 [/tex] minutes
Population after 115 minutes: [tex]\sf \bold{38265.94} [/tex] bacteria
The population will reach 10000 [tex]\sf \bold{79.78} [/tex] minutes after the initial time
Step-by-step explanation:
To solve these questions, we can use the exponential growth formula for bacterial count, which is modeled as:
[tex]\large\boxed{\boxed{\sf N(t) = N_0 e^{kt} }[/tex]
Where:
- [tex]\sf N(t) [/tex] is the population size at time [tex]\sf t [/tex],
- [tex]\sf N_0 [/tex] is the initial population size,
- [tex]\sf k [/tex] is the growth rate constant,
- [tex]\sf e [/tex] is the base of the natural logarithm.
Finding the initial size of the culture [tex]\sf (N_0) [/tex]:
Given:
- [tex]\sf N(10) = 700 [/tex]
- [tex]\sf N(30) = 1500 [/tex]
Using [tex]\sf N(t) = N_0 e^{kt} [/tex], we have two equations:
- [tex]\sf 700 = N_0 e^{10k} [/tex]
- [tex]\sf 1500 = N_0 e^{30k} [/tex]
Divide equation (2) by equation (1):
[tex]\sf \dfrac{1500}{700} = \dfrac{N_0 e^{30k}}{N_0 e^{10k}} [/tex]
[tex]\sf \dfrac{1500}{700} = e^{30k - 10k} [/tex]
[tex]\sf \dfrac{1500}{700} = e^{20k} [/tex]
Taking natural logarithm on both sides:
[tex]\sf \ln\left(\dfrac{1500}{700}\right) = 20k [/tex]
[tex]\sf \ln\left(\dfrac{1500}{700}\right) = 20k [/tex]
[tex]\sf k = \dfrac{\ln\left(\dfrac{1500}{700}\right)}{20} [/tex]
[tex]\sf k \approx \dfrac{\ln(2.1429)}{20} [/tex]
[tex]\sf k \approx \dfrac{0.762140052}{20} [/tex]
[tex]\sf k \approx 0.0381070026 [/tex]
Now substitute [tex]\sf k [/tex] back to find [tex]\sf N_0 [/tex]:
[tex]\sf 700 = N_0 e^{10 \times 0.0381070026} [/tex]
[tex]\sf 700 = N_0 e^{38.1070026} [/tex]
[tex]\sf N_0 = \dfrac{700}{e^{38.1070026}} [/tex]
[tex]\sf N_0 \approx \dfrac{700}{1.463850109} [/tex]
[tex]\sf N_0 \approx 478.1910357 [/tex]
[tex]\sf N_0 \approx 478.19\textsf{(in 2 decimal places)} [/tex]
Therefore, the initial size of the culture ( [tex]\sf N_0 [/tex] ) is approximately [tex]\sf \bold{478.19} [/tex] bacteria.
Finding the doubling period:
The doubling period [tex]\sf T_d [/tex] is related to [tex]\sf k [/tex] in exponential growth. For exponential growth, the doubling period is given by:
[tex]\sf T_d = \dfrac{\ln(2)}{k} [/tex]
[tex]\sf T_d = \dfrac{\ln(2)}{ 0.0381070026} [/tex]
[tex]\sf T_d \approx \dfrac{0.6931471806}{0.03819} [/tex]
[tex]\sf T_d \approx 18.18949624 \textsf{ minutes} [/tex]
[tex]\sf T_d \approx 18.19 \textsf{ minutes (in 2 decimal places)} [/tex]
Therefore, the doubling period is approximately [tex]\sf 18.19 [/tex] minutes.
Finding the population after 115 minutes
Using [tex]\sf N(t) = N_0 e^{kt} [/tex] with [tex]\sf t = 115 [/tex] minutes:
[tex]\sf N(115) = N_0 e^{0.0381070026 \times 115} [/tex]
[tex]\sf N(115) \approx 478.1910357 \times e^{4.382305299} [/tex]
[tex]\sf N(115) \approx 477.28 \times 80.02229625 [/tex]
[tex]\sf N(115) \approx 38265.94472 [/tex]
[tex]\sf N(115) \approx 38265.94\textsf{(in 2 decimal places)} [/tex]
Therefore, the population after 115 minutes is approximately [tex]\sf \bold{38265.94} [/tex] bacteria.
Finding when the population will reach 10000:
We want to find [tex]\sf t [/tex] when [tex]\sf N(t) = 10000 [/tex]:
[tex]\sf 10000 = N_0 e^{0.0381070026t} [/tex]
[tex]\sf \dfrac{10000}{N_0} = e^{0.0381070026t} [/tex]
[tex]\sf \ln\left(\dfrac{10000}{N_0}\right) = 0.0381070026t [/tex]
[tex]\sf t = \dfrac{\ln\left(\dfrac{10000}{478.1910357}\right)}{0.0381070026} [/tex]
[tex]\sf t \approx \dfrac{\ln(20.91214442)}{0.0381070026} [/tex]
[tex]\sf t \approx \dfrac{3.040330063}{0.0381070026}[/tex]
[tex]\sf t \approx 79.78402539 \textsf{ minutes} [/tex]
[tex]\sf t \approx 79.78 \textsf{ minutes (in 2 decimal places)} [/tex]
Therefore, the population will reach 10000 bacteria approximately [tex]\sf \bold{79.78} [/tex] minutes after the initial time [tex]\sf t = 0 [/tex].
