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Interpreting change in exponential models | Mathematics II | High School Math | Khan Academy


5m read
·Nov 11, 2024

So I've taken some screenshots of the Khan Academy exercise interpreting rate of change for exponential models in terms of change. Maybe they're going to change the title; it seems a little bit too long. But anyway, let's actually just tackle these together.

So the first day of spring, an entire field of flowering trees blossoms. The population of locusts consuming these flowers rapidly increases as the trees blossom. The relationship between the elapsed time ( t ) in days since the beginning of spring and the total number of locusts ( l(t) ) — so the number of locusts is going to be a function of the number of days that have elapsed since the beginning of spring — is modeled by the following function:

[ l(t) = 750 \times 1.85^t ]

Complete the following sentence about the daily rate of change of the locust population. Every day, the locust population — well, every day, think about what's going to happen. I'll draw a little table just to make it hopefully a little bit clearer.

So let me draw a little bit of a table. We'll put ( t ) and ( l(t) ).

So when ( t = 0 ) (so when zero days have elapsed), well, there's going to be ( 1.85^0 ). This is going to be ( 1 ), so you're going to have ( 750 ) locusts right from the get-go. Then, when ( t = 1 ), what's going to happen? Well, then this is going to be:

[ 750 \times 1.85^1 ]

So it's going to be times ( 1.85 ). When ( t = 2 ), what's ( l(t) )? It's going to be:

[ 750 \times 1.85^2 ]

Well, that's the same thing as ( 1.85 \times 1.85 ). So notice, and this just comes out of this being an exponential function, every day you have ( 1.85 ) times as many as you had the day before.

We essentially take what we had the day before and we multiply by ( 1.85 ). Since ( 1.85 ) is larger than one, that's going to grow the number of locusts we have. So this is going to grow. I'm actually not using — I'm not on the website right now, so that's why normally there would be a drop-down here. So I'm going to grow by a factor of, well, I'm going to grow by a factor of ( 1.85 ) every day.

Let's do another one of these. All right, so this one tells us that Vira is an ecologist who studies the rate of change in the bear population of Siberia over time. The relationship between the elapsed time ( t ) in years since Vira began studying the population and the total number of bears ( n(t) ) is modeled by the following function. All right, fair enough. So we've got a little exponential thing going on.

Complete the following sentence about the yearly rate of change of the bear population. What's the thing about every year that passes? ( t ) is in years now.

Every year that passes is going to be ( \frac{2}{3} ) times the year before. I can do that same table that I just did just to make that clear. So let me do that. Whoops! Let me, let me make this clear.

So table — so this is ( t ) and this is ( n(t) ). When ( t = 0 ), ( n(t) ) — you're going to have ( 2187 ) bears. So that's the first year that she began studying that population. Zero years since Vira began studying the population.

The first year is going to be:

[ 2187 \times \left( \frac{2}{3} \right)^1 ]

So times ( \frac{2}{3} ). The second year is going to be:

[ 2187 \times \left( \frac{2}{3} \right)^2 ]

So that's just ( \frac{2}{3} \times \frac{2}{3} ). Each successive year, you're going to have ( \frac{2}{3} ) the bear population of the year before. You're multiplying the year before by ( \frac{2}{3} ).

So every year, the bear population shrinks by a factor of, by a factor of ( \frac{2}{3} ). All right, let's do one more of these.

So they tell us that Akiba started studying how the number of branches on his tree change over time. All right, the relationship between the elapsed time ( t ) in years since Akiba started studying history and the total number of its branches ( n(t) ) is modeled by the following function.

Complete the following sentence about the yearly percent change in the number of branches. Every year, blank percent of branches are added or subtracted from the total number of branches.

Well, I'll draw another table, although you might get used to just being able to look at this and say, well, look each year you're going to have ( 1.75 ) times the number of branches you had the year before. And so if you have ( 1.75 ) times the number of branches of the year before, you have grown by ( 75 ).

And I'll make that a little bit clearer. So ( 75 ) percent of branches — every year ( 75 ) percent of branches are added to the total number of branches. And I'll just draw that table again like I've done in the last two examples to make that hopefully clear.

Okay, okay, so this is ( t ) and this is ( n(t) ). So ( t = 0 ): you have ( 42 ) branches. ( t = 1 ): it's going to be:

[ 42 \times 1.75 ]

When ( t = 2 ): it's going to be:

[ 42 \times (1.75)^2 ]

( 42 \times 1.75 ) times ( 1.75 ). So every year you are multiplying times ( 1.75 ). So times ( 1.75 ). Something funky is happening with my pen right over there, but if you're multiplying by ( 1.75 ), if you're growing by a factor of ( 1.75 ), this is the same thing as adding ( 75 ).

Once again, you are adding ( 75 ). Think about it this way: if you just grew by a factor of one, then you're not adding anything; you're staying constant. If you grow by ten percent, then you're going to be ( 1.1 ) times as large. If you grow by two hundred percent, then you're going to be two times as large.

So this right over here — this right over here is — is going, or if, if you — let me be clear, careful what I just said. I think I just mistake that. If you grow by two hundred, you're going to be three times as large as you were before. One is constant, and then another two hundred percent would be another twofold, so that would make you three times as large.

Don't want to confuse you. My brain recognized that I said something weird right at that end. All right, hopefully you enjoyed that.

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