(Population Growth) A bacterial population is growing continuously at a rate equal to 10% of its population each day. Its initial size is 10,000 organisms. How many bacteria are present after 10 days? After 30 days?

Question

(Population Growth) A bacterial population is growing continuously at a rate equal to [latex]10 \%[/latex] of its population each day. Its initial size is 10,000 organisms. How many bacteria are present after 10 days? After 30 days?

Accepted Answer

Since the percentage growth of the population is [latex]10 \%=0.1[/latex], we have

[latex]\frac{d P / d t}{P}=0.1, \quad ext { or } \quad \frac{d P}{d t}=0.1 P[/latex].

Here [latex]\alpha=0.1[/latex], and all solutions have the form

[latex]P(t)=c e^{0.1 t}[/latex]

where [latex]t[/latex] is measured in days. Since [latex]P(0)=10,000[/latex], we have

[latex]c e^{(0.1)(0)}=c=10,000, \quad ext { and } \quad P(t)=10,000 e^{0.1 t}[/latex]

After 10 days [latex]P(10)=10,000 e^{(0.1)(10)}=10,000 e \approx 27,183[/latex], and after 30 days [latex]P(30)= 10,000 e^{0.1(30)}=10,000 e^{3} \approx 200,855[/latex] bacteria.

Question 11.2.3: (Population Growth) A bacterial population is growing contin...

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