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# A bacteria culture initially contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420.(a) Find an expression for the number of bacteria after $t$ hours.b) Find the number of bacteria after 3 hours.(c) Find the rate of growth after 3 hours.(d) When will the population reach 10,000?

## a) $P(t)=100(4.2)^{t}$b) 7409 bacteriac) 10,632 bacteria/hd) 3.2 hours

Derivatives

Differentiation

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##### Kristen K.

University of Michigan - Ann Arbor

##### Samuel H.

University of Nottingham

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### Video Transcript

the given information in this problem is that the initial population is 100 cells and at time, one which was one hour, the population was 420 cells and we're going to use our population growth model P of t equals p, not e to the Katie. And the first thing we want to do is find an expression for the number of bacteria after t hours. And basically that means fill in the model with the number four p not and a number for K. We don't know the number for K yet, so we're going to have to find it. So let's fill in the equation with all the other numbers we have. We know that the final population is 420 at time. One, the initial population is 100 weaken. Substitute those numbers in and we can use one for the time. So we have K Times one, and we can solve that for K. So let's divide both sides by 100 we have 4.2 equals e to the K, and then we can take the natural log of both sides. So K is the natural log of 4.2. Now we can substitute that into our model and we have p of t equals 100 times E to the natural log of 4.2 times T. Now we can simplify e to the natural log of 4.2. That's equivalent to 4.2. So our model is p of t equals 100 times 4.2 raised to the power T. Okay, let's move on to part B in this part. We went to find the number of bacteria after three hours. So we're using the model that we just came up with. P of t equals 100 times, 4.2 to the T and we're going to substitute three in there. So we're finding P of three 100 times 4.2 to the third power and put that in the calculator and we get approximately 7409 cells for part C. We want the rate of growth. Rate of growth is like rate of change. And so we're referring to the derivative. So we want the derivative of pft Pete prime of tea. So we have the derivative of an exponential function. So it's going to be 100 times 4.2 to the power tee times a natural log of 4.2, and we want the rate of growth at time. Three. So let's go ahead and substitute a three in there and approximated, and we get approximately 10,632 and that would be cells per hour. That's the rate of growth at that moment in time for Part D. We want to find the time when the population reaches 10,000. So let's use P of t equals 10,000 and let's sell for tea and the model again is pft equals 100 times 4.2 to the power teeth. So we can substitute are 10,000 in there for PFT, and we'll solve this for tea. Let's divide both sides by 100 we get 100 equals 4.2 to the T. Then we'll take the natural log of both sides. Natural log of 100 equals natural log of 4.2 times t. We'll use the power property of logarithms. Um, actually let me back up a second. The other side was natural log of 4.2 to the Power T. Now we use the power property of logarithms to bring that t down to the front. So the natural log of 100 equals t times the natural log of 4.2 and then to isolate T, we can divide both sides by natural log 4.2 to have natural log of 100 over natural of a 4.2. And then we put that in the calculator and we get approximately 3.21 hours 3.21 hours to reach population of 10,000 cells.