Question

I am unsure about how to do this problem, pictured below.

Answer 1

Given that a culture of bacteria grows at a rate proportional to its size. Where the culture starts with 50 cells, then grows 150 after time, "t" equals 2 hours.

We are asked to:

> (a) Find an expression, P(t), to model the number of cells present after "t" hours.

> (b) Determine the time at which the population is at 500 cells.

For part (a):

Since the culture of bacteria grows at a rate proportional to its size, we can model it as the following differential equation.

`\Rightarrow (dP)/(dt)=kP; \ Where \ P =P_0 \ at \ t=0 \ and \ P=3P_0 \ at \ t=2`

Solve the first-order separable differential equation with the given initial condition.

`\Longrightarrow (dP)/(dt)=kP \Longrightarrow (1)/(P)dP=kdt \Longrightarrow \int\limits {(1)/(P) } \, dP=\int\ {k} \, dt \Longrightarrow ln(P)=kt+c`

`\Longrightarrow e^(ln(P))=e^(kt)+e^(c) \Longrightarrow P=ce^(kt)`

Plug in the initial condition.

`\Longrightarrow P_0=ce^(k(0)) \Longrightarrow P_0=c(1) \Longrightarrow \boxed{c=P_0}`

`\Longrightarrow P=ce^(kt) \Longrightarrow \boxed{P=P_0e^(kt)}`

Use the second initial condition to find "k."

`\Longrightarrow 3P_0=P_0e^(k(2)) \Longrightarrow 3=e^(k(2)) \Longrightarrow 3=e^(2k) \Longrightarrow ln(3)=ln(e^(2k))`

`\Longrightarrow k=(ln(3))/(2) \Longrightarrow \boxed{k \approx 0.5493}`

Thus, the equation to model the situation is,

`\boxed{\boxed{P(t)=50e^(0.5493t)}} \therefore Sol.`

For part (b):

`P=10P_0`

`\Rightarrow 10P_0=P_0e^(0.5493t) \Longrightarrow 10=e^(0.5493t) \Longrightarrow ln(10)=ln(e^(0.5493t))`

`\Longrightarrow ln(10)=0.5493t \Longrightarrow t=(ln(10))/(0.5493) \Longrightarrow \boxed{t=4.192 \ hrs}`

Thus, the time it takes the population to reach 500 is approx. 4.192 hours.

Answer 2

Explanation:
150/2 = 75
500/75 = 6.6666666667 hours

it will take 6.2/3 hours to reach 500 bacteria cells

NOT SURE ABOUT EXPERSION BUT YOU COULD TRY THIS

p = 75t