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Problem 2 Hard Difficulty

A common inhabitant of human intestines is the bacterium Escherichia coli, named after the German pediatrician Theodor Escherichia, who identified it in 1885. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 50 cell.
(a) Find the relative growth rate.
(b) Find an expression of the number of cells after $ t $ hours.
(c) Find the number of cells after 6 hours.
(d) Find the rate of growth after 6 hours.
(e) When will the population reach a million cells?

Answer

a) $r=3 \ln 2 \approx 2.079 /$ hour
b) $P(t)=50 \cdot 8^{t}$
c) 13107200
d) 27255656 cells/hour
e) 4.76 hours

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Video Transcript

in this problem. We're told that the cell divides every 20 minutes and the initial population is 50. When you look ahead to the other parts of the problem, you realize that they're all using units of ours. So let's convert 20 minutes into hours. So the cell divides everyone third hour. And the first thing we want to do is find the relative growth rate, which is what we call K. So we're going to find K in our equation. P of t equals p, not each of the K t. So let's substitute numbers into that equation. So we know that the initial value of the population is 50 and we know if 1/3 of an hour goes by, the population will be 100 because those cells are going to divide and will have double the amount. And then we have e raised to the K times 1/3 for 1/3 of an hour. We can solve that equation for K. So first will divide both sides by 50 and we get to equals e to the 1/3 K, and then we'll take the natural log of both sides. So natural log of two equals 1/3 K, and then we'll divide both sides of the equation by 1/3 or multiplied by three and we get K equals three natural log of to. That's the exact value of K, and it's important to use the exact value as we work through the other parts of the problem so that we don't lose accuracy. But you can also approximate this for now just to have something to write. You get 2.79 for the approximate value. Okay, okay, let's move on to part B and we're going to find the number of cells after t hours. So they should basically were writing this model with some numbers filled in so we would have p of t equals the initial value 50 times e to the K, which is three natural log of to times t. Okay, we could leave the model like that, but we could also simplify this part. So let me show you how to do that. So if we have e to the three natural log of two times T, that's the same as e to the natural log of to times three t and that's the same as he to the natural log of two to the power three t and each of the natural log of two is the same as two. So we have to to the three t to to the third. You can think of it as to to the third to the T and to to the third is eight. So all of that becomes a to the power teeth. So if we substitute that in, we have p A T equals 50 times eight to the T. And that's our model for the population at T hours. Okay, Now we'll move on to Part C where we're finding the number of cells after six hours. So let's take the model that we just wrote P of t equals 50 times, eight to the T and substitute a six in there. So p of six for six hours equals 50 times a to the sixth power. And that gives us approximately. We'll go ahead and put it down here approximately 13,107,200. Okay. And that's the number of bacteria in the intestines. At this moment in time for party, we are asked to find the rate of growth of the bacteria rate of growth would be rate of change, and that would be derivative. So we're finding P prime of tea. So let's take the derivative of P of tea. So we have 50 the constant multiplied by eight to the T multiplied by the natural log of eight typical way of finding the derivative of an exponential function. Now we're going to plug in six p. Prime of six would be 50 times a to the sixth power times natural log of eight. And that gives us approximately 27 1,000,000 255,656 is the rate of growth so that would be sells her our at that moment in time. That's how fast the cells are growing and finally, party. We're going to find the time when the population reaches a 1,000,000 cells. So let's take our model again. Pft equals 50 times eight to the T, and we want the p of T value to be one million. The population is one million we're going to solve for time. So let's go ahead and divide both sides by 50 and that gives us 20,000 equals a to the T and then we can take the natural log of both sides. Natural long 20,000 equals natural log of eight to the T. We can use the power property of logarithms and bring the tea down. So we have tee times natural log of eight and then divide both sides by natural log of eight. And that will isolate the tea and we get approximately 4.76 hours now. This makes sense if we think back to the previous parts of the problem where it took six hours to get to 13 million. So we know it's going to be less than six hours to get to one million. And 4.76 hours is definitely less than six hours.