# A bacteria culture grows with constant relative growth rate. The bacteria count was 2,560 after 4 hours and 2,621,440 after 12 hours.(a) What is the

###### Question:

% per hr

(b) What was the initial size of the culture?

bacteria

(c) Find an expression for the exact number of bacteria after t hours, (t).

y(t) =

(d) Find the number of bacteria after 5.5 hours. (Round your answer to the nearest whole number.)

bacteria

(e) Find the rate of growth in bacteria per hour) after 5.5 hours. (Round your answer to the nearest whole number.)

bacteria per hour

(f) How many hours did it take for the population to reach 104,000? (Round your answer to two decimal places.)

hr

## Answers

A) [tex]P = P_0 e^{rt}[/tex]

B) [tex]19564[/tex]

C) [tex]5.5[/tex] or [tex]6[/tex] hours

Explanation:

A) The formula that can be used for bacteria after "t" hours is

[tex]P = P_0 e^{rt}[/tex]

Where ,

[tex]P[/tex] represents the population after "t" hours

[tex]P_0[/tex] represents the initial population.

[tex]r[/tex] represents the rate of growth of population

[tex]t[/tex] represents the time taken by the population to grow

B) Here let us first find the rate of population growth.

[tex]9000 = 400 * e^{r*4}\\3.11 = r* 4\\r = 0.778[/tex]

Thus, the population of bacteria after 5 hours

[tex]P = 400 * e^{0.778 * 5} \\P = 400* 48.91\\P = 19564\\[/tex]

C)

[tex]30000 = 400 * e^{0.778 * t) \\\\4.317 = 0.778 * t\\t = 5.549\\or t= 5.5 or 6[/tex]

A) The expression for the number of bacteria is [tex]P(t) = 400e^{0.7783t}[/tex].

B) After 5 hours there will be 19593 bacteria.

C) After 5.55 hours the population of bacteria will reach 30000.

Step-by-step explanation:

A) Here we have a problem with differential equations. Recall that we can interpret the rate of change of a magnitude as its derivative. So, as the rate change proportionally to the size of the population, we have

[tex]P' = kP[/tex]

where [tex]P[/tex] stands for the population of bacteria.

Writing [tex]P'[/tex] as [tex]\frac{dP}{dt}[/tex], we get

[tex]\frac{dP}{dt} = kP[/tex].

Notice that this is a separable equation, so

[tex]\frac{dP}{P} = kdt[/tex].

Then, integrating in both sides of the equality:

[tex]\int\frac{dP}{P} = \int kdt[/tex].

We have,

[tex]\ln P = kt+C[/tex].

Now, taking exponential

[tex]P(t) = Ce^{kt}[/tex].

The next step is to find the value for the constant [tex]C[/tex]. We do this using the initial condition [tex]P(0)=400[/tex]. Recall that this is the initial population of bacteria. So,

[tex]400 = P(0) = Ce^{k0}=C[/tex].

Hence, the expression becomes

[tex]P(t) = 400e^{kt}[/tex].

Now, we find the value for [tex]k[/tex]. We are going to use that [tex]P(4)=9000[/tex]. Notice that

[tex]9000 = 400e^{k4}[/tex].

Then,

[tex]\frac{90}{4} = e^{4k}[/tex].

Taking logarithm

[tex]\ln\frac{90}{4} = 4k[/tex], so [tex]\frac{1}{4}\ln\frac{90}{4} = k[/tex].

So, [tex]k=0.7783788273[/tex], and approximating to the fourth decimal place we can take [tex]k=0.7783[/tex]. Hence,

[tex]P(t) = 400e^{0.7783t}[/tex].

B) To find the number of bacteria after 5 hours, we only need to evaluate the expression we have obtained in the previous exercise:

[tex]P(5) =400e^{0.7783*5} = 19593.723 \approx 19593[/tex].

C) In this case we want to do the reverse operation: we want to find the value of t such that

[tex]30000 = 400e^{0.7783t}[/tex].

This expression is equivalent to

[tex]75 = e^{0.7783t}[/tex].

Now, taking logarithm we have

[tex]\ln 75 = 0.7783t[/tex].

Finally,

[tex]t = \frac{\ln 75}{0.7783} \approx 5.55[/tex].

So, after 5.55 hours the population of bacteria will reach 30000.

A. 86.64%

B. 80

C. y(t)= 80[tex]e^{0.8664t}[/tex]

D. y(5.5) = 9388

E. y(t) = 8134 bacteria for every hour

F. 8.28 hours

Step-by-step explanation:

a. The bacteria count after 4 hours y(4) = 2560

The bacteria count after 12 hours y(12) = 2621440

equation 1: ye^4k = 2560

equation 2: ye^12k= 2621440

divide (2) by (1),

e^8k = 1024

Taking ln both sides:

ln(e8k ) = ln(1024)

8k = 6.931

k=0.8664 // after dividing both sides by 8

rate of growth k% = 0.8664*100 = 86.64%

b. substituting for k in equation 2:

ye^4*0.8664 = 2560

Initial bacteria count y(4) = 80

c. y(t)= 80e^0.8664t

d. the number of bacteria after 5.5 hours => y(5.5) = 80[tex]e^{0.8664(5.5)}[/tex] ≈ 9388

e. y'(5.5) = 80*0.8664e^0.8664(5.5) = 69.31e^0.8664(5.5) = 8134

f. for bacteria population = 104000

where ye^0.8664t = 104000

y = 80

80e^0.8664t = 104000 //divide both sides by 80

e^0.8664t = 1300

ln(e^0.8664t = ln(1300)

t = 7.17 / 0.8664 = 8.28 hours