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## Calculus, all content (2017 edition)

### Unit 8: Lesson 1

AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
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- 2011 Calculus AB free response #1a
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- 2011 Calculus AB free response #5a
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# 2011 Calculus AB free response #5a

Using a tangent line to extrapolate a function from a known initial condition. Created by Sal Khan.

## Want to join the conversation?

- I have learned about the rules and techniques but I cannot answer these questions myself without Sal's help. What should I do?(11 votes)
- You should practice more, at the link I attached you can find a lot of problems from calculus. All answers are explained, so you can deduce which part is problematic for you !

http://tutorial.math.lamar.edu(14 votes)

- instead of finding the numerical value of the tangent and then framing the equation can't we just analyse that since the differential equation tells us the rate of change of the garbage in the landfill on a yearly basis, so for 3 months the rate of change of the garbage in the landfills would be one fourth of that in one year??

that is,

the slope of the function at W is 44. so at the end of three months the slope would be one fourth of 44 i.e. 11.

so the amount of garbage at the end of three months would be 1411 tons.(1 vote)- Yes, you can do almost that. You can realise that the slope is 44 at
`t=0`

, and since that means that the linear approximation would increase 44 in 1 year, then you can divide by 4 to get that it would increase by 11 in 1/4 of a year.

The only incorrect statement in your solution is that the slope will always be 44, it's the increase over 1/4 that's 11, the slope doesn't change.(2 votes)

## Video transcript

Problem 5. "At the beginning
of 2010, a landfill contained 1,400
tons of solid waste. The increasing
function W--" so I guess W constantly increases--
"the increasing function W models the total amount
of solid waste stored at the landfill. Planners estimate that W
will satisfy the differential equation." so the derivative
of W with respect to time is equal to 1 over 25 times
the quantity W minus 300, "for the next 20 years. W is measured in tons. t is measured in years
from the start of 2010." All right, let's get into this. So part a. "Use the tangent line
to the graph of W at t equals 0 to approximate
the amount of solid waste that the landfill contains at
the end of the first 3 months of 2010." So since time is in years, 3
months would be 1/4 of a year. So t equals 1/4. So it seems at first
this is really daunting. They gave us a
differential equation. We don't even know what
the actual function W is. How do we figure out
its tangent line? But we just have
to kind of think about what they're asking. So regardless of what
W looks like-- so let's think about it a little bit. So we don't know yet what
W actually does look like. But they tell us it's
an increasing function. So at time 0 it has
1,400 tons in it. They tell us that right up here. And that it increases. We don't know what the function
actually does look like. But let's say that
that is W. So what they're saying in
problem a is use a tangent line to the
graph at t equals 0. Let me draw a W a
little bit differently. So W might look like this. So what they're saying
is, find the tangent line, find the slope of that
tangent line at t equals 0. So there's some slope
right over there. And then we can use
this slope to create a linear approximation
of where we're going to be a quarter
of a year from then. So although we don't
know what W is just yet, we could take the slope
of this line out-- so this is our t axis--
to t is equal to 1/4. And wherever that line takes us
out after a fourth of a year, that will be at least
a decent approximation. We're extrapolating forward
from that first point and the current slope. So we really just have to figure
out the slope of this line, and just see where that line
is at t is equal to 1/4. And you say, wait. How do we know what
the derivative of W is at 0 without knowing W? And that's where
we can go straight to this differential equation. We could actually rewrite
this differential equation using slightly
different notation. This is the derivative
of W with respect to t. We could write that as W
prime of t is equal to 1 over 25 times the function
W, which is a function of t, minus 300. And when you look
at it this way, it becomes a little
bit clearer on how to figure out what the
derivative of W is at 0. The slope of that line
literally is just the derivative of W evaluated at 0. So let's literally take
the derivative of W and evaluate it at 0. So we have W prime
of 0 is equal to 1 over 25 times W of
0, which we know. We know this is 1,400
tons of solid waste. This is how much waste
there is at time 0. Minus 300. So this right over
here is 1,400. And so we have W prime
at time equals 0. So our slope at time equals
0, or our derivative at time equals 0, is equal to 1 over 25
times 1,400 minus 300 is 1,100. And 25 goes into
1,100 44 times, right? It goes four times
into each 100. We have 11 hundreds here. So it'll go 44 times. So it goes 44. So the slope of that line is 44. We could say m for slope. Or actually, let me just
write down the word. The slope of this line is 44. And I just ate some
peanuts or something. So my voice is a little dry. So bear with me. But the slope of
this line is 24. So how do we use that
to find an approximation for the amount of waste
that the landfill contains at the end of the
first three months? Well, let me zoom
in a little bit. And actually, my fingers
are all salty, too. So maybe I don't have a
proper grip on my pen. But I'll try my best. Let me just zoom in
a little bit more. So we're starting at 1,400 tons. So this is my W-axis. This is my time axis. We're starting at 1,400 tons. And we are increasing
from there. They're telling us that it
is an increasing function. So maybe it looks
something like that. And then our slope
right over here. And I don't know,
I haven't drawn W exactly that accurately. I'm just guessing what it
might look like at this point. The tangent line
has a slope of 44. So that is our tangent line. And what that's
saying is if we go out one unit of time,
which is one year, then we would have
gone up 44 in tonnage. So if we use this line as an
approximation, after one year, at this point, would
have 1,444 tons. But we're not trying to
approximate a year out. We're trying to approximate
1/4 of a year out. So we're trying to approximate. So this would be
half a year out. This is 1/4 of a year out. We're trying to approximate
that point right over there. So it's going to be 1,400,
this point right over here. And we could write down
the equation of this line, if we like. We could say this
line-- so we'll call this the W approximation,
because this isn't exactly our W function-- this is equal
to the slope of our line, 44, times time, plus
our W intercept, we could say, or plus our
initial condition, plus 1,400. So if you put time is
equal to 1/4 in there, you get it equal to 44. So let me write it this way. So our approximate W, at time
is equal to 1/4 of a year, is equal to 44 times
1/4 plus 1,400. I'm running out of space. And my salty hands are having
trouble writing this properly. And so 44 times 1/4, or 44
divided by 4, is 11 plus 1,400. And so you add them together. You get 1,411 or 1,411 tons. This is our approximation. We just took the slope
from our starting point and used that slope
as an approximation. It probably is not
the exact amount of tonnage based on
the actual function W. But it's an OK approximation. But that's our first answer
for Part A. 1,411 tons.