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## Non-motion applications of derivatives

Current time:0:00Total duration:5:00

# Applied rate of change: forgetfulness

AP.CALC:

CHA‑3 (EU)

, CHA‑3.C (LO)

, CHA‑3.C.1 (EK)

## Video transcript

I studied for an
English test today and learned 80 vocabulary words. In 10 days, I will have
forgotten every word. The number of words that I
remember t days after studying is modeled by-- so W of t, so
this is the number of words I have in my head as a function of
time is going to be equal to 80 times 1 minus 0.1t squared
for t is between 0 and 10, including the two boundaries. That's why we have
brackets right over here. What is the rate of
change of the number of known words per day two days
after studying for the test? And I encourage you
to pause this video and try it on your own. So the key here is we come up
with this equation for modeling how many words have retained
in my brain every day after I first
memorized them, after I got the 80 of them into my head? And that's this expression here. And they want to know
the rate of change two days after studying. Well, the rate of change,
I can take the derivative of this with respect to time. So let's do that. So let's take the derivative. The derivative of
the number of words I know with respect to time is
going to be equal to-- well, we have this 80 out front. That's just a constant. And now I can apply the
chain rule right over here. So the derivative
of 1 minus 0.1t, the whole thing squared with
respect to 1 minus 0.1t is going to be-- so I'm essentially
taking the derivative of this whole pink thing, this whole
expression squared with respect to the expression. So that's going to be
2 times 1 minus 0.1t. And now I can find
the derivative of this inner expression
with respect to t. So the derivative of this inner
expression with respect to t is just going to be 0 minus 0.1. So it's just going
to be negative 0.1. And of course, we can
simplify this a little bit. This is going to be equal
to-- if we take 80 times 2 is 160 times negative 0.1,
that's going to be negative 16. 160 times 0.1 is 16, so
negative 16 times 1 minus 0.1t. And if we want, we
could distribute the 16, or we could just
leave it like this. But we're ready now to
answer our question. We could write this
as the rate of change of the number of words we
know with respect to time. Or we could use the
alternate notation. We could say this
is W prime of t. Either way, it's going to
be equal to this thing. Let me do that same color. It's equal to negative
16 times 1 minus 0.1t. So what's this going to be? What is the rate of change
of the number of words known per day two days
after studying for the test? Well, we just have to evaluate
this when t is equal to 2. So W prime of 2 is going to be
equal to negative 16 times 1 minus 0.1 times 2
close parentheses. And that's going to
be equal to-- well, let's see, what is this? This is 1 minus essentially 0.2. This is going to be 0.8. So this is going to be
equal to negative 16 times-- is that right? 1 minus 0.2 is-- yep, it's
going to be times 0.8. And what is that going to be? If I were to multiply 16
times 8, it would be 128. It's 2 times 8 times
8 so 2 times 64, 128. So this is going to
be negative 12.8. And just to really hit the
point home of what we're doing, the rate of change is
negative 12.8 words per day. So if you believe this
model for how many words we know on a given day,
this is saying on day two, right at the day
two point, right after exactly two days after
studying for the test, right at that moment, I am essentially
losing 12.8 words per day. The number of words
I know is decreasing by 12.8 words per day.