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A type of algae grows continuously so that its population doubles in 3 days. Given a beginning population of 100 algae cells per milliliter of water, to the nearest whole number, how many algae cells would you expect at the end of 10 days?

I start off by dividing 10 by 3 giving me 3 doubles in population: $$100\rightarrow200\rightarrow400\rightarrow800$$

Now I have to deal with the extra day. I get $800+\frac{1}{3}\cdot800\approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?

Thanks! Your help is appreciated!

Max0815

Max0815
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  • If the algae doubles in three days then the equation is $y(t)=y(0)\cdot 2^{\frac{t}3}$, where $y(0)$ is the initial population. For instance, if $t=3$ the equation becomes $y(3)=y(0)\cdot 2^{\frac{3}3}=y(0)\cdot 2$, – callculus42 Mar 23 '19 at 18:37

2 Answers2

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You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = \sqrt[3]{2}$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.

If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is $$ A = 100 \times 2^{t/3}. $$ Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.

You can tell that will give the same as the first method since $$ 2^{10/3} = 2^3 2^{1/3} = 8 \sqrt[3]{2}. $$

Ethan Bolker
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As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=k\cdot P(t)$, where $P(t)$ is the population of the algae cells.

First of all we solve this differential equation by the method of Separation of variables.

$P'(t)=k\cdot P(t)$

$\frac{dP}{dt}=k\cdot P(t)$

Dividing the equation by $P(t)$ and multiplying the equation by $dt$.

$\frac{1}{P(t)} \ dP=k \ dt$

Integrating both sides

$\int \frac{1}{P(t)} \ dP=\int k \ dt$

$\ln(P)=k\cdot t+c$

To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.

$e^{\ln(P)}=e^{k\cdot t+c}$

$P=e^{k\cdot t}\cdot e^c$

Replacing $e^c$ by $C$

$P=C\cdot e^{k\cdot t}\quad (*)$

To determine the constants C and k we use the following information:

  1. Given a beginning population of $100$ algae cells per milliliter of water ...

The equation is $P(0)=100\Rightarrow P(0)=C\cdot e^{k\cdot 0}=C\cdot 1=100\Rightarrow C=100$

  1. ... so that its population doubles in 3 days

That means after 3 days the population is $200$: $P(3)=200$

$P(3)=100\cdot e^{k\cdot 3}=200$

$e^{k\cdot 3}=\frac12$

Taking $\ln()$ on both sides.

$3\cdot k=\ln\left( \frac12\right)$

$k=\frac{\ln\left( \frac12\right)}{3}$

Now we can use (*) to obtain $P(t)$

$P(t)=100\cdot e^{\ln(2)\cdot \frac{t}3}=100\cdot \left(e^{\ln(2)}\right)^{ \frac{t}3}=100\cdot 2^{\frac{t}{3}}$

callculus42
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  • I see. Thank you for this method! – Max0815 Mar 23 '19 at 20:05
  • @Max0815 It is nice to hear that the answer was not for the wastebasket. You´re welcome. My intention was to show the whole calculation which is needed to obtain the equation. – callculus42 Mar 23 '19 at 20:10