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## Differential Calculus (2017 edition)

### Course: Differential Calculus (2017 edition) > Unit 11

Lesson 6: Applied rates of change# Applied rate of change: forgetfulness

AP.CALC:

CHA‑3 (EU)

, CHA‑3.C (LO)

, CHA‑3.C.1 (EK)

Let's face it, sometimes it feels like we forget much of what we learn. In this example we model that forgetful phenomenon. Created by Sal Khan.

## Want to join the conversation?

- Where did the modelling function come from? How is it known that this particular function models the forgetting of words?

This has popped up into my mind before... I always wondered where the parabolic path functions came from for physics, and I found that they are the results of diff eqs from manipulating the definitions of acceleration and velocity.

Is there some definition and diff eq that will yield this function as well?

In general, is there a class that teaches you how to model things? I can do it easily with geometric problems, however, it is not so easy for me with things like this, but once I figured out the diff eqs thing and the definitions, it was cake. I really would like to know how to make models of anything I see.(18 votes)- I'm almost certain that the modelling function in this question was made up purely for this question.

There are two main ways that science comes up with models/functions. The first empirical, which means you do an experiment. You get a large sample of people, teach them some words and then test their ability to recall then after certain time periods. Then you could plot the number of words over time on a graph and try to fit a function. For linear functions you can use linear regression to find the equation of the line that best fits the data. For non-linear functions it's more tricky, but there are techniques. Even then you often have to make a judgement call about whether you think the function should be polynomial (e.g. y = x^4) or exponential (e.g. y = 2^x) or something else. That's when you might want to think of a theory to explain the function.

The second way would be theoretically. Here you basically start with a function that you think is logical. Or you start with a simple model of the world and see what functions come out of that. That's tricky in this example, but more common in physics. Maybe you could imagine the brain is a bucket containing information and every day it randomly loses a proportion of the information in it. This would give an exponential decay type function (and probably isn't very accurate). Then you should go to the experiment approach and see if your model accurately predicts reality.

As for where the parabolic trajectories come from in physics: it's a result of gravity imposing an effectively constant acceleration on objects near the surface of Earth. A constant acceleration means that the downward velocity of an object increases linearly. And this means the position of the object above the surface decreases with the square of time. The velocity in the horizontal direction is assumed to be constant (i.e. there's no friction), so as the x-coordinate of an object changes linearly with time, the y-coordinate changes with the square of time.

I hope that makes some sense. don't think there's anything on Khan Academy that teaches this, but you could try: ://www.coursera.org/course/modelthinking(58 votes)

*The rate of change of the number of known words per day...*What does the word**per day**means?`w'(2) = -12.8`

,`w'(3) = -11.2`

and`w'(4) = -9.6`

. I know that the number of words (`w(t)`

) will keep decreasing but why**per day**?(8 votes)- We are talking about a unit rate that changes every day according to the curve we were given. Unit rates often include time as the independent variable, such as miles per hour, meters per second, words per day. In the case of this forgetful guy, the curve is a parabola that says he forgets fastest at first and slower later on. Here are some examples:

Day Loss rate

1 -16

2 -12.8

4 -9.4

5 -8

9 -1.6

10 0 (he has already forgotten all of them by then)

So, by the 9th day he is forgetting words at only 1.6 words per day. Too bad that wasn't his forgetting rate at the beginning--he would ace his test.(9 votes)

- I took the equation 80 (1-0.1t)^2 and expanded it and then took the derivative of that and got a totally different number. what's up with that?(3 votes)
- 80(1-0.1t)^2=

80(1-0.2t+0.01t^2)= (Square the equation in the parenthesis)

80-16t+.8t^2 (Multiply each term in the parenthesis by 80)

d(80-16t+.8t^2)/dx = -16+1.6t (Take the derivative and use the power rule)

-16+1.6t=-16(1-0.1t) (Factor out the -16)(2 votes)

- I am trying to solve the second derivative of

X^2+y^2= 6xy

Can you help(1 vote)- x² + y² = 6xy

2x + (2y)y' = 6*y + 6x*1*y'

(2y)y' = 6y + (6x)y' -2x

(2y)y' - (6x)y' = 6y - 2x

y'(2y - 6x) = 6y - 2x

y' = (6y - 2x) / (2y - 6x)

y' = [2(3y - x)] / [2(y - 3x)]

y' = (3y - x) / (y - 3x)

y" = [(y - 3x)*(3*y' - 1) - (3y - x)*(1*y' - 3)] / (y - 3x)²

That's our second derivative, but we have y' in it. We do know what y' is so we can substitute y' = (3y - x) / (y - 3x) in for y'. Then just simplify it. Note I do all this on the computer and with all the nesting, it's very confusing, so there is a chance that I made some arithmetic error. But the point is substitute in the first derivative for y' and simplify.

y" = [(y - 3x)*(3*y' - 1) - (3y - x)*(1*y' - 3)] / (y - 3x)²

y" = { (y - 3x) * {3 [(3y -x) / (y - 3x)] - 1} - (3y - x) * { [(3y - x) / (y - 3x)] - 3} } / (y - 3x)²

y" = { [(9y - 3x) - (y - 3x)] - [((9y² - 6xy + x²) / (y - 3x)) - (9y - 3x)] } / (y - 3x)²

y" = { (9y - 3x - y + 3x) - [((9y² - 6xy + x²) - (9y² - 30xy + 9x²)) / (y - 3x)] } / (y - 3x)²

y" = {8y - [(9y² - 6xy + x² - 9y² + 30xy - 9x²) / (y - 3x)] } / (y - 3x)²

y" = { [8y (y - 3x) / (y -3x)] - [(24xy -8x²) / (y -3x)] } / (y - 3x)²

y" = { [(8y² - 24xy) - (24xy - 8x²)] / (y -3x) }/ (y - 3x)²

y" = [(8y² - 24xy - 24xy + 8x²) / (y -3x)] / (y - 3x)²

y" = (8y² - 48xy + 8x²) / (y - 3x)³

y" = [8 (y² - 6xy + x²)] / (y - 3x)³

y" = 8(x² + y² - 6xy) / (y - 3x)³

Note that our original equation is x² + y² = 6xy, we can substitute in 6xy for x² + y².

y" = 8(6xy - 6xy) / (y -3x)³

y" = 8(0) / (y -3x)³

y" = 0(5 votes)

- what is the meaning of the rate of change number itself? it can be probably used to draw a secant on the graph and to decide if the value on the graph is decreasing or increasing but besides that?(3 votes)
- I do not understand what is meant by w' at 2 days is -12.8 word par day, would he mean in average, he forget 2*-12.8 words in two days?(2 votes)
- So the derivative is the instantaneous rate of change so you cannot say "average, he forget 2*-12.8 words in two days"(1 vote)

- what is the formula of rate of change probllems(1 vote)
- Take the derivative once of the function given in the problem(2 votes)

- why is the constant not subjected to the derivative, shouldn't the derivative of a constant always be zero? At1:17(1 vote)
- Product Rule for derivatives: for a constant c*f(x), d/dx(c*f(x)) = c*f'(x)

Here's proof:

W(t) = 80(1 - 0.1t)²

= 80(1 - 0.1t)(1 - 0.1t)

= 80(1 - 0.2t + 0.01t²)

= 80 - 16t + 0.8t²

dW/dt = 0 - 16 + 1.6t

= -16 + 1.6t

dW/dt = 80 * d/dt (1 - 0.1t)² Product Rule

= 80 * 2(1 - 0.1t)(-0.1) Chain Rule.

= 16 * (-1 + 0.1t)

= -16 + 1.6t

http://jwilson.coe.uga.edu/EMAT6680/Horst/derivativeconstantmult/derivativeconstantmult.html(1 vote)

- y=(x-1)(x^2-6x=2)

given that z=y^2 and that z is increasing at the constant rate of 10 units per second. Find the rate of change of y when x=2.

plzzzzzz help.(1 vote)- I'm not sure we heard that equation correctly because of the = within the ( ) Will you restate the question please?(1 vote)

- can we use the gradient formula to get the rate instead of using the derivative.(1 vote)
- That only works for straight lines, where the slope is constant. We need a broader tool, like the derivative, to handle any other case.(1 vote)

## 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.