Question

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 71 cells. (a) Find the relative growth rate. (Assume t is measured in hours.) k = Incorrect: Your answer is incorrect. (b) Find an expression for the number of cells after t hours. P(t) = Incorrect: Your answer is incorrect. (c) Find the number of cells after 8 hours. cells (d) Find the rate of growth after 8 hours. (Round your answer to three decimal places.) billion cells per hour (e) When will the population reach 20,000 cells? (Round your answer to two decimal places.) hr

Answer 1

Answer:

$r=1.57$, $P(t)=71e^{3t\ln2}$, $P(8)=1191182336$,$P'(8)=2476994033$  $t=2.713\; hr$

Step-by-step explanation:

$(a)$ Find the relative growth rate.

$P=Ae^{rt}$ where given $A=71, P(t)=120\;\;and \;\; t=\frac{1}{3}$

Find the value of $r$

$\Rightarrow 120=71 e^{\frac{r}{3} }$

$\Rightarrow \frac{120}{71} =e^{\frac{r}{3} }$

$\Rightarrow 1.69=e^{\frac{r}{3} }$

$\Rightarrow \ln1.69=\frac{r}{3}$

$\Rightarrow 3\ln1.69=r \;\;\; or \;\;\; r=1.57$

$(b)$   Find an expression for the number of cells after t hours.

$P(t)=71e^{3t\ln2}$

$(c)$  Find the number of cells after 8 hours.

$P(8)=71e^{3\times 8\times \ln2}$

$\Rightarrow P(8)=1191182336$

$(d)$  Find the rate of growth after 8 hours.

  $P(t)=71e^{3\times t\times \ln2}$

Now differentiate w.r.t. $t$

$\Rightarrow P'(t)=71\times 3\ln2\times e^{3t\ln2}$

$\Rightarrow P'(8)=3573547008\ln2$

             $=2476994033$

$(e)$  When will the population reach 20,000 cells?

     $20,000=71e^{3t\ln2}$

$\Rightarrow \ln\frac{20000}{71} =3t\ln2$

$\Rightarrow \frac{ \ln\frac{20000}{71}}{3\ln2} =t$

S0, $t=2.713 hr$