William Collins

2021-12-28

Elementary Applications of Differential Equations 1. A culture of bacteria consists of 1000 bacteria. After 2 hrs., the culture becomes 300 bacteria. a. How many bacteria remain after 4 hours? b. In how many hours will the culture become 50 bacteria?

zesponderyd

Step 1
(a) To find:
The number of bacteria after 4 hours.
Given:
The number of bacteria at the start is 1000 and after 2 hours the bacteria becomes 300.
Calculation:
The exponential decay function as, $P={P}_{0}{e}^{kt}$.
Here, P is final value, ${P}_{0}$ is initial value, t is time, k is constant:
Substitute in equation $P={P}_{0}{e}^{kt}$.
$300=1000{e}^{k\left(2\right)}$
$0.3={e}^{2k}$
Taking log both side:
$\mathrm{ln}0.3=2k\mathrm{ln}e$
$\frac{\mathrm{ln}0.3}{2}=k$
$-0.602=k$
Substitute in equation $P={P}_{0}{e}^{kt}$.
$P=300\cdot {e}^{-0.602×4}$
$=26.999$
$\approx 30$
Thus, the number of bacteria after 4 hours is 30.
Step 2
(b) To find:
The number of hours after that the number of bacteria is 50.
Calculation:
Substitute in equation $P={P}_{0}{e}^{kt}$
$50=300\cdot {e}^{-0.602×t}$
$\frac{1}{6}={e}^{-0.602×t}$
Taking log both side:
$\mathrm{ln}\frac{1}{6}=-0.602t\cdot \mathrm{ln}e$
$\frac{\mathrm{ln}\frac{1}{6}}{-0.602}=t$
$2.98=t$
Thus, the number of hours after that the number of bacteria is 50 is 2.98 hours.

Daniel Cormack

$\frac{3}{2}={e}^{r\cdot 2}$
$\mathrm{ln}\left(\frac{3}{2}\right)=r\cdot 2$
$r=\frac{\mathrm{ln}\left(3\right)-\mathrm{ln}\left(2\right)}{2}$
$r=0.20733$
so our population is modelled by,
$P=1000\cdot {e}^{0.20733\cdot t}$, t hours after start
so 2 days is 48 hours so subbing in,
$P=1000\cdot {e}^{0.20733\cdot 48}$
$P=16834112$

karton

(a) Given:
The number of bacteria at the start is 1000 and after 2 hours the bacteria becomes 300.
Solution:
$P={P}_{0}{e}^{kt}$
${P}_{0}$ is initial value, t is time, k is constant:
Substitute in equation $P={P}_{0}{e}^{kt}$.
$x300=1000{e}^{k\left(2\right)}$
$0.3={e}^{2k}$
$\mathrm{ln}0.3=2k\mathrm{ln}e$

$\frac{\mathrm{ln}0.3}{2}=k$
$-0.602=k$
Substitute in equation $P={P}_{0}{e}^{kt}$
$P=300\cdot {e}^{-0.602×4}$
$=26.999$
$\approx 30$
Answer: number of bacteria after 4 hours is 30.
(b) Calculation:
Substitute in equation $P={P}_{0}{e}^{kt}$
$50=300\cdot {e}^{-0.602×t}$
$\frac{1}{6}={e}^{-0.602×t}$
$\mathrm{ln}\frac{1}{6}=-0.602t\cdot \mathrm{ln}e$
$\frac{\mathrm{ln}\frac{1}{6}}{-0.602}=t$
$2.98=t$
Answer: number of hours after that the number of bacteria is 50 is 2.98 hours.

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