Question

After t hours there are​ P(t) cells present in a​ culture, where ​P(t)equals2000e Superscript 0.3 t. ​(a) How many cells were present​ initially? ​(b) Give a differential equation satisfied by​ P(t). ​(c) When will the population​ double? ​(d) When will 8 comma 000 cells be​ present?

Answer 1

(a)2000 of cells were present initially.

(b)The differential equation of the given exponential model is

`(dP)/(dt)=0.3P`

(c)The population will be double after 2.31 hours.

(d)The number of cell will be 8,000 after 4.62 hours.

Explanation:

Given that,

`P(t)= 2000e^(0.3 t)`

where P(t) is number of cell and t is time in hours.

(a)

Initial time means t=0.

Putting t=0 i the given expression

`P(0)= 2000e^(0.3 * 0)`

`\Rightarrow P(0)= 2000`

2000 of cells were present initially.

(b)

The differential equation of population growth is

`(dP)/(dt)=kP`

The solution of the above equation is

`P(t)=Ce^(kt)`

Comparing the given exponential model to the above solution

So, k= 0.3 and C=2000

The differential equation of the given exponential model is

`(dP)/(dt)=0.3P`

(c)

If the population double then P(t) = 2× P(0) =2×2000

`P(t)= 2000e^(0.3 t)`

`\Rightarrow 2* 2000=2000e^(0.3t)`

`\Rightarrow e^(0.3t)=2`

Taking ln both sides

`\Rightarrow ln (e^(0.3t))=ln2`

`\Rightarrow {0.3t=ln2`

`\Rightarrow t=(ln2)/(0.3)`

`\Rightarrow t=2.31` h

The population will be double after 2.31 hours.

(d)

Now P(t)=8,000

`P(t)= 2000e^(0.3 t)`

`\Rightarrow 8,000= 2000e^(0.3 t)`

`\Rightarrow e^(0.3 t)=(8,000)/(2,000)`

`\Rightarrow e^(0.3 t)=4`

Taking ln function both sides

`\Rightarrow ln(e^(0.3 t))=ln 4`

`\Rightarrow t=(ln 4)/(0.3)`

`\Rightarrow t=4.62` h

The number of cell will be 8,000 after 4.62 hours.