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Problem 25 Medium Difficulty

In Example 6 we considered a bacteria population that doubles every hour. Suppose that another population of bacteria triples every hour and starts with 400 bacteria. Find an expression for the number $ n $ of bacteria after $ t $ hours and use it to estimate the rate of growth of the bacteria population after 2.5 hours.


$n^{\prime}(2.5)=400 \cdot 3^{2.5} \cdot \ln 3 \approx 6850$ bacteria/hour.

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Video Transcript

here we have a bacterial growth problems so we can use the equation and equals and not eat the Katie. That's our basic exponential growth equation. We're told that the bacteria colony starts with 400 and it takes them one hour to triple, so we can use those pieces of information to find the value of K, the growth constant. So if you triple 400 you get 1200. So we have 1200 for end. We have 400 for and not and we have one for T one hour. So this is going to be to the K Times one or just k, just all that for K. Let's divide both sides of the equation by 400 we get three equals e to the K, and then we'll take the natural log of both sides. So K is the natural log of three so we can substitute that and get our model. We have n equals 400 times e to the natural log of three times T. Now we know that e to the natural log of three can be simplified and it's just three so we can write this model as an equals 400 times, three to the T, and what we want to do with that is use it to find the rate of growth and the rate of growth would be the derivative. So let's find the derivative of this in prime of tea would be 400 times three to the tee times natural log of three. And then we're going to find end prime of 2.5 to find the rate of growth after 2.5 hours and prime of 2.5, we would substitute the 2.5 in there and use a calculator, and you're going to get approximately 6850. That's the rate of growth. So we're saying that at that moment the bacteria is growing at that many bacteria per hour.

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