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A bacteria population is 3000 at time t = 0 and its rate of growth is 1000-6^t bacteria per hour...

Question:

A bacteria population is 3000 at time t = 0 and its rate of growth is {eq}1000-6^t {/eq} bacteria per hour after t hours. What is the population after one hour? (Round your answer to the nearest whole number.)

Exponential Growth Function

This problem presents us with the rate of exponential growth. We need to find a function that tells us the growth. Once we have the growth function we will use it to find the population after the given time period.

Answer and Explanation:


  • I wonder if the rate of growth function is properly written but I shall consider the one mentioned in the question.

Growth rate of the population is given by:

{eq}\displaystyle \frac{dp}{dt}=1000-6^t {/eq}

It can be written as,

{eq}\displaystyle dp=(1000-6^t)dt {/eq}

Integrating both the sides, we get,

{eq}\displaystyle \int_{p(0)}^{p(t)}dp=\int_{0}^{t}(1000-6^t)dt {/eq}

{eq}\displaystyle p(t)-p(0)=1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(t)=p(0)+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}

Plugging in the value of initial population, we get,

{eq}\displaystyle p(t)=3000+1000t-\frac{6^t}{\ln 6}+\frac{1}{\ln 6} {/eq}


Therefore, the population after {eq}1 {/eq} hour is:

{eq}\displaystyle p(1)=3000+1000(1)-\frac{6^{1}}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(1)=3000+1000-\frac{6}{\ln 6}+\frac{1}{\ln 6} {/eq}

{eq}\displaystyle p(1)=4000-\frac{6}{1.79}+\frac{1}{1.79} {/eq}

{eq}\displaystyle \boxed{\displaystyle p(1)=3997.208\approx 3997} {/eq}

So, the population of bacteria after 1 hour is 3997.


Learn more about this topic:

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Exponential Growth vs. Decay

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