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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?

User PgmFreek
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1 Answer

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Answer:

(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.

User Willis Blackburn
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