Main content

## Advanced interpretation of exponential models

Current time:0:00Total duration:4:21

# Interpreting change in exponential models: with manipulation

CCSS Math: HSA.SSE.A.1, HSA.SSE.B.3, HSA.SSE.B.3c, HSF.IF.C.8, HSF.IF.C.8b, HSF.LE.B.5, HSF.LE.B

## Video transcript

- [Voiceover] Ocean
sunfishes are well-known for rapidly gaining a lot of weight on a diet based on jellyfish. The relationship between the
elapsed time, t, in days, since an ocean sunfish
is born, and its mass, M of t, in milligrams, is modeled
by the following function. All right. Complete the following sentence about the daily percent change in the mass of the sunfish. Every day, there is a blank percent addition or removal from
the mass of the sunfish. So one thing that we know
almost from the get-go, we know that the sunfish gains weight, and we also see that
as t grows, as t grows, the exponent here is going to grow. And if you grow an exponent on something that is larger than one,
M of t is going to grow. So I already know it's going
to be about addition to the mass of the sunfish. But let's think about how
much is added every day. Well, let's think about it. Well let's see if we can rewrite this. I'm going to just focus
on the right-hand side of this expression. So 1.35 to the t/6 plus five. That's the same thing as 1.35 to the fifth power, times 1.35 to the t/6 power. And that's going to be equal to 1.35 to the fifth power, times 1.35, and I can separate this
t/6 as 1/6 times t. So 1.35 to the 1/6 power, and then that being
raised to the t-th power. So let's think about it. Every day, as t increases by one, now we can say that we're gonna take the previous day's mass, and multiply it by this common ratio. The common ratio here isn't
the way I've written it. Isn't 1.35. It's 1.35 to the 1/6 power. Let me draw a little table here to make that really, really clear. And all of that algebraic
manipulation I just did is just so I could simplify this, so I have some common
ratio to the t-th power. So t and M of t. So based on how I've just written it, when t is zero, well if t is zero, this is one, so then we just have our initial mass, it's going to be 1.35 to the fifth power. And then when t is equal to one, it's going to be our initial mass, 1.35 to the fifth power times our common ratio, times 1.35 to the 1/6 power. When t equals two, we're
just gonna multiply what we had at t equals one, and we're just gonna multiply that times 1.35 to the 1/6 again. And so, every day, we are growing by our common ratio,
1.35 to the 1/6 power. Actually, let me get a calculator out. We're allowed to use
calculators in this exercise. So 1.35 to the, open parentheses, one divided by six, close parentheses, power, is equal to, I'll say 1.051, approximately. So this is approximately 1.051. So we can say this is approximately 1.35 times 1.051 to the t-th power. So every day, we are growing
by a factor of 1.051. Well growing by a factor of 1.051 means that you are adding
a little bit more than 5%. You're adding 0.51 every day of your mass, so you're adding 5.1%. And if you're rounding
to the nearest percent, we would say there's a 5% addition to the mass of the sunfish every day.