A Bacteria Culture Grows with Constant Relative Growth Rate. The bacteria count was 400 after 2 hours and 25,600 after 6 hours.
a) What is the relative growth rate? Express your answer as a percentage.
Using this formula $P(t)=P_0e^{kt}$
$t=time$
$k=growth\ rate$
$p_0=initial\ amount$
How do I calculate a formula with the information I have?
b) What was the initial size of the culture?
c) Find an expression for the number of bacteria after $t$ hours
d) Find the number of cells after 4.5 hours
e) Find the rate of growth after 4.5 hours
f) When the population reach 50,000?
This is a self study question. I have Calculus II next semester and really would like to have a clear method as to how to solve a problem such as this.
Thanks in advance.
$\endgroup$2 Answers
$\begingroup$So let $t$ be the time in hours.
You know two things:
$$P(2) = P_0 e^{2k} = 400$$ $$P(6) = P_0 e^{6k} = 25600$$
This means you have two equations and two unknowns ($P_0$ and $k$). To solve for $k$, you can divide the second equation by the first:
$$\frac{P_0e^{6k}}{P_0e^{2k}}=\frac{25600}{400}$$
$$e^{4k} = 64$$
$$(e^k)^4 = 64 = 2^6$$
$$e^k = 2 \sqrt{2} = 2^{1.5}$$
$$\ln (e^k) = \ln 2^{1.5}$$
$$k = \frac{3 \ln 2}{2} \approx 1.0397$$
Now that we know the value of $k$, we can substitute it into the first of our original two equations to find the value of $P_0$:
$$P_0e^{2k}=400$$
$$P_0e^{3\ln 2} = 400$$
$$P_0 (e^{\ln 2})^3 = 400$$
$$8P_0 = 400$$
$$P_0 = 50$$
Note that $P_0$ is the same thing as $P(0)$, the bacteria count at time $t=0$ hours.
I think you should be able to answer all of the parts of your problem now. Good luck :)
$\endgroup$ $\begingroup$You have two equations with two unknowns. So you can calculate as a first step $k$. That is $k=\ln(P_1/P_2)/(t_1-t_2)$. Then you calcultate $P_0$. Then, c) you have with $k$ and $P_0$. The rest is to calculate directly.
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