# Thread: Population problem..desperately need help!

1. ## Population problem..desperately need help!

The population of bacteria in a culture grows at a rate proportional to the number of bacteria present at any time. After 3 hours it is observed that there are 400 bacteria present After 10 hours there are 2000 bacteria present. What is the initial number of bacteria?

just how to answer this question? i'm totally blurry..

2. Hi bleu90,

Let's define $B~=~ Number~of~Bacteria ~,~t~=~ Time~in~hours$

We are told that the bacteria is growing at a "rate proportional to the number of bacteria present at any time", thus;

$\frac{dB}{dt}~\propto~B$

$\implies \frac{dB}{dt}~=~kB$ for some $k \geq 1$ (Since we are told that the rate of bacteria is growing)

Now solve the differential using the change of variable method.

$\int{\frac{dB}{B}}~=~k\cdot \int dt$

$\implies \ln|B|~=~kt+C$ where $C$ is the constant retained from both integrals.

Now rearrange for $B$.

$B~=~Ae^{kt}$ where $A~=~e^C~~~(*)$

Firstly we are told that "after 3 hours...there are 400 Bacteria present", hence subbing into $(*)$ we get;

$400~=~Ae^{3k}~~~(**)$

Secondly we are told that "After 10 hours there are 2000 bacteria present", thus subbing into $(*)$ we get;

$2000~=~Ae^{10k}~~~(***)$

Now we have to solve equations $(**)$ and $(***)$ simultaneously. Dividing equation $(***)$ by equation $(**)$ we obtain;

$5~=~e^{7k}~\implies~k~=~\frac{1}{7} \cdot \ln 5$

Now we can work $A$ by subbing $k$ into either of the 2 equations. Note that $A$ is the number of bacteria present initially since if $t~=~0$ then $e^{kt}~=~1$

3. how about this question? do i need to use the same method as the above? i mean, is using derivative the only method of solution?

question:

Initially there were 100 milligrams of radioactive substance present. After 6 hours the mass decreased by 35. If the rate of decay is proportional to the amount of the substance present at any time, find the amount remaining after 24 hours.

4. Yes, except that we are told "the rate of decay is proportional too" which gives;

$\frac{dR}{dt}=-kR$ for some $k \geq 1$ where $R$ is the amount (in mg) of radioactive substance present.

5. Originally Posted by Sean12345
Yes, except that we are told "the rate of decay is proportional too" which gives;

$\frac{dR}{dt}=-kR$ for some $k \geq 1$ where $R$ is the amount (in mg) of radioactive substance present.
however, my friend said that derivative is not the only method. he told me that i can only use exponential decay and will get something like this;

y = A e^(-kx)

how can i proceed with that, if i want to use this method?

6. Oh, my apologies.

So we are told "the rate of decay is proportional too". Now the general equation for exponential decay is quite rightly as you put it;

$R~=~Ae^{-kt}$ where $k \geq 1$ and where $R$ is the amount (in mg) of radioactive substance present. Call this equation $(*)$

Firstly we are told "initially (when $t=0$ ) there were 100 mg of radioactive substance present", hence subbing into equation $(*)$ gives;

$A~=~100$ since if $t~=~0$ then $e^{-kt}~=~1$

Equation $(*)$ now becomes $R~=~100e^{-kt}$

Seconldy we are told "After 6 hours the mass decreased by 35mg", thus subbing $65$ ( $=100-35$ ) into equation $(*)$ gives;

$65~=~100e^{-6k}~\implies~k~=~-\frac{1}{6} \cdot \ln {\frac{13}{20}}$

Now subbing this value of $k$ into $(*)$ and rearranging a bit gives a new equation $(*)$ ;

$R~=~100\left\{ \frac{13}{20}\right\}^{\dfrac{t}{6}}$

We are then told to find the "amount remaining after 24 hours" so sub $t=24$ into $(*)$ to find the value of $R$

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# bacterial growth) The population of bacteria proportional to the number of bacteria present at a time. After 3 hours, 400 bacteria are present. After 10 hours 2000 bacteria are present. What is the initial number of bacteria?

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