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Video: Evaluating Functions Involving Exponential Decay

Alex Cutbill

A population of bacteria decreases as a result of a chemical treatment. The population t hours after the treatment was applied can be modeled by the function 𝑃(𝑡), where 𝑃(𝑡) = 6000 × (0.4)^𝑡. What was the population when the chemical was first applied? What is the rate of population decrease?

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Video Transcript

A population of bacteria decreases as a result of a chemical treatment. The population 𝑑 hours after the treatment was applied can be modeled by the function 𝑃 of 𝑑, where 𝑃 of 𝑑 is equal to 6000 times 0.4 to the power of 𝑑.

So that’s describes the scenario in the question. This is a multipart question; there are two parts. The first part is β€œwhat was the population when the chemical was first applied?” The second part is β€œwhat is the rate of population decrease?” Naturally, we’re going to answer the first part of the question first: what was the population when the chemical was first applied?

Well, we have a function 𝑃 of 𝑑, which tells us the population 𝑑 hours after the treatment was applied. So asking for the population when the chemical was first applied, the initial population, we’re asking what was the population at 𝑑 equals zero. That is 𝑃 of zero, which is the population zero hours after the treatment was applied. And to find that, we just substitute into the expression we have for 𝑃 of 𝑑.

Substituting in zero, we get 𝑃 of zero equals 6000 times 0.4 to the power of zero. Anything to the power of zero is one, and so 𝑃 of zero is 6000. So the answer to the first part of the question β€œwhat was the population when the chemical was first supplied?” is 𝑃 of zero or 6000.

The next part of the question is β€œwhat is the rate of population decrease?” Again, we use the definition of the function 𝑃 of 𝑑. We’ve already seen in the first part of the question that the initial population 𝑃 of zero is 6000. 𝑃 of one is the population after one hour. Substituting this into the formula we have for 𝑃 of 𝑑, we get that 𝑃 of one is 6000 times 0.4 to the power of one, which is of course just 6000 times 0.4 or 2400.

The population of the two hours is 𝑃 of two or 6000 times 0.4 to the power of two. Again, we’re just substituting in to the expression we have, and this is 960. The actual values aren’t actually important here. What is important is that the population at a certain time is 0.4 times the population one hour before.

Another way of saying this is that the population is 40 percent of the population the hour before. And another way of saying this is that the population decreases by 60 percent every hour. So what is the rate of population decrease? It is 60 percent per hour.

So looking at our function 𝑃 of 𝑑, the multiplicative constant of 6000 tells us about the initial state β€” in our case the initial population. And the base 0.4 tells us about the change over time β€” in this case how much the- in this case how much the population decreases over time.