Problem

The half-life of bismuth-210, $ ^{210} Bi $, is 5…

03:02

Problem 30 Hard Difficulty

A bacteria culture starts with 500 bacteria and doubles in size every half hour.

(a) How many bacteria are there after 3 hours?
(b) How many bacteria are there after $ t $ hours?
(c) How many bacteria are there after 40 minutes?
(d) Graph the population function and estimate the time for the population to reach 100,000.

Answer

a) $A=32,000$
b) $A=500 e^{2 \ln (2) t}$
c) $A=1259.921$
d) $t \approx 3.82193$

Missouri State University

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Video Transcript

all right in this problem, we have a bacteria population that's doubling, doubling every half an hour so we could make a little chart that represents the time that has elapsed on the number of bacteria. So what time? Zero. We're starting with 500 bacteria, half a Knauer goes by and now it's doubled to 1000. Another half on hour goes by and now there's 2000 and so on. It keeps doubling. We want to know the number of bacteria when three hours have elapsed. So we've doubled until we got to three hours and there are 32,000 bacteria. So that takes care of part A. Now what we want to do is generalize so that we can say how many bacteria there would be after t hours. So what we want to do is look at the numbers in the table and find the pattern. So the first number was just 500. The next number was 500 times two or we could say times two to the first. The next number was 500 times to the second power. Then we have 500 times, two to the third, etcetera, continuing on that pattern. So where do we get this number, This exponents here? Well, that would be the number of compounding Zor the number of double ings we've had. And if three hours have gone by and it doubles twice every hour than there have been six double ing's so we can generalize this and we come up with a formula 500 times to raise to the to T, where T is the number of hours so that will tell us how many bacteria we have at time. T. So that helps us with part B because Part B asks us how many bacteria are there after t hours, and the answer would be 500 times two to the power to teach or to use an hours. Now, let's use that formula to answer part. See how many bacteria are there after 40 minutes, while we're going to have to convert 40 minutes into hours. So 40 minutes is 2/3 of an hour, and now we can substitute that into our expression above 40. So we get 500 times to raise to the 2/3 times two and 2/3 times two is 4/3. We put that in a calculator and round, and we get approximately 1260 bacteria. And lastly, we want to graft this and look for the time when the population reaches 100,000. So using a calculator, we can go to y equals and we can type are function into the y equals menu, and then we can also type y equals 100,000 into the Y equals menu, and that will give us a horizontal line at a height of 100,000 that stands for 100,000 bacteria. So a window I've chosen for this is going from negative 1 to 6 on the X axis and negative 211,000 on my Y axis. You can choose different numbers here. Just fiddle with your window values until you find something you like. Make sure you go at least to 100,000. So here's the graph. The blue graph is that exponential curve, the one that models the number of bacteria and the red line is that line at a height of 100,000 and now we're going to look for where they intersect so we can go into the calculate menu. We can choose Number five Intersect. We can put the cursor on the first curve press, enter cursor on the second curve, press enter, and then we can move the cursor closer to the intersection point and press enter and looking at the bottom of the screen, we see that it's 3.82 so that would be about 3.82 hours until the population reaches 100,000.

Heather Z.

Oregon State University

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Problem

The half-life of bismuth-210, $ ^{210} Bi $, is 5…

Problem 30 Hard Difficulty

Answer

a) $A=32,000$
b) $A=500 e^{2 \ln (2) t}$
c) $A=1259.921$
d) $t \approx 3.82193$

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a) $A=32,000$b) $A=500 e^{2 \ln (2) t}$c) $A=1259.921$d) $t \approx 3.82193$

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a) $A=32,000$
b) $A=500 e^{2 \ln (2) t}$
c) $A=1259.921$
d) $t \approx 3.82193$