1. ## Rate of Change

The rate of change in the number of bacteria in a culture is proportional to the number present. In a certain laboratory experiment, a culture had 10,000 bacteria initially, 20,000 a time t1 minutes and 100,000 bacteria at (t1+10) minutes.
A)in terms of t only, find the number of bacteria in the culture a any time t minutes, t≥0.
B)How many bacteria were there after 20 minutes.
C)How many minutes had elasped when the 20,000 bacteria were observed

2. Hello, calc_help123!

The rate of change in the number of bacteria in a culture is proportional to the number present.
In a certain laboratory experiment, a culture had 10,000 bacteria initially,
20,000 a time $t_1$ minutes and 100,000 bacteria at $(t_1+10)$ minutes.

A) In terms of $t$ only, find the number of bacteria in the culture a any time $t \geq 0.$

Let $P$ = population of bacteria.

We have: . $\frac{dP}{dt} \:=\:kP \quad\Rightarrow\quad \frac{dP}{P} \:=\:k\,dt \quad\Rightarrow\quad \ln P \:=\: kt + c \quad\Rightarrow\quad P \:=\:e^{kt+c}$

. . . . $P \:=\:e^{kt}\cdot e^c \:=\:e^{kt}\cdot C \quad\Rightarrow\quad P \:=\:Ce^{kt}$

When $t = 0,\:P = 10,\!000\!:\quad 10,\!000 \:=\:Ce^0 \quad\Rightarrow\quad C \:=\:10,\!000$

The function (so far) is: . $P \;=\;10,\!000\,e^{kt}$

$\begin{array}{ccccc}\text{When }t = t_1,\;P = 20,\!000\!: & 10,\!000\,e^{kt_1} &=&20,\!000 & [1] \\

\text{When }t = t_1\!+\!10,\;P = 100,\!000\!: & 10,\!000\,e^{k(t_1+10)} &=& 100,\!000 & [2] \end{array}$

Divide [2] by [1]: . $\frac{10,\!000\,e^{kt_1}\cdot e^{10k}}{10,\!000\,e^{kt_1}} \;=\;\frac{100,\!000}{20,\!000}$

. . $e^{10k} \:=\:5 \quad\Rightarrow\quad 10k \:=\:\ln 5 \quad\Rightarrow\quad k \:=\:\frac{\ln5}{10}$

The function is: . $P \;=\;10,\!000\,e^{\frac{\ln5}{10}t} \;=\;10,\!000\,\left(e^{\ln 5}\right)^{\frac{t}{10}}$

. . Hence: . $P \;=\;10,\!000\cdot5^{\frac{1}{10}t}$

B) How many bacteria were there after 20 minutes?

When $t=20\!:\;\;P \;=\;10,\!000\cdot5^{(\frac{1}{10})(20)} \;=\;10,\!000\cdot5^2 \;=\;250,\!000$

C) How many minutes had elasped when the 20,000 bacteria were observed?

We have: . $10,000\cdot5^{\frac{1}{10}t_1} \:=\:20,000 \quad\Rightarrow\quad 5^{\frac{1}{10}t_1} \:=\:2$

Take logs: . $\ln\left(5^{\frac{1}{10}t_1}\right) \:=\:\ln2 \quad\Rightarrow\quad \tfrac{1}{10}\,t_1\cdot \ln(5) \:=\:\ln 2$

Therefore: . $t_1 \;=\;\frac{10\ln 2}{\ln 5} \;\approx\; 4.3\text{ minutes}$

3. thank you very much for your explanation but i just have one question. I followed you perfectly up until where you found $k$.

i do not get how $P \;=\;10,\!000\,\left(e^{\ln 5}\right)^{\frac{t}{10}}$ is true. and how did you get $P \;=\;10,\!000\cdot5^{\frac{1}{10}t}$?

thank you.

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### rate of change in number

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