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Calculus, all content (2017 edition)
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Lesson 1: AP Calculus AB questions- Sal interviews the AP Calculus Lead at College Board
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2011 Calculus AB free response #5b
Using the second derivative to judge whether an approximation with the tangent line is an overestimate or underestimate. Created by Sal Khan.
Want to join the conversation?
- Isn't it posssible that the second derivative switches back and we have an overestimate?(12 votes)
- That would a possibility, except that we know for sure that W(t) is concave up for the whole interval from 0 to 1/4. This is because W''(t) is always positive on this interval.(12 votes)
- When we used the 2nd derivative equation to check if it was an under/over-estimate, why did we plug in t as 0 instead of 1/4, because in the problem it says when t=1/4(4 votes)
- Strictly speaking, to rely on this test we need to be able to determine that the 2nd derivative is either positive for the entire interval or negative for the entire interval. If it changes sign during the interval we can't use it to determine whether the estimate is high or low. In this case, we can see that if W'' is positive at t = 0 it will always be positive because W is increasing and the only way for W'' to become negative would be for W to fall below 300.(4 votes)
- why do we plug in 0 for t and not 1/4?(3 votes)
- Because we are estimating what t=1/4 might be based on the tangent we found at t=0 because we only know (and are given) the initial conditions. It is enough that the second derivative at 0 tells us if the function's tangent is increasing or decreasing or 0, so as to tell us if our tangent estimate is over, under or about right.(4 votes)
- at the beginning of the video sal makes a comment about wiping salt off his fingers? what's this about?(2 votes)
- he was probably eating something like chips or salted nuts before filming this video lol(3 votes)
- Can I explain the question by saying how W(t) concaves upward?(2 votes)
- What is this wierd notation
d^2W/ dt^2
? If it's a second derivative then why notd^2W/ d^2t
?(2 votes)
Video transcript
Now ready for part b. And I've wiped all the
salt off of my fingers, and I've had a glass of water. So now I'm ready
for some business. Find the second derivative
of W with respect to t in terms of W. Use
this second derivative to determine whether your answer
in part a is an underestimate or overestimate for the
amount of solid waste that the landfill contains
at time t equals 1/4. So let's do the
first things first. Let's find the second
derivative in terms of W. And we already have the
first derivative over here. And in part a, I rewrote it
just with slightly different notation. But I'll just use this just
because I need to write down here and I can still
refer to this thing. So let's just take
the derivative of both sides of this
differential equation with respect to t. So you take the derivative
of this left-hand side with respect to t, you
get the second derivative of W with respect to t as a
function of t is equal to-- now you take the derivative
of this with respect to t. This is the same thing
as 1/25Wt minus 12. And so when you
take the derivative, that constant part drops
off, and you're just left with 1/25 times
the derivative of Wt. Let me make that clear. This is the same thing as 1/25W
as a function of t minus 12. 1/25 of 300 is 12. And you take the
derivative of this, you get 1/25 times the
first derivative of W and then the derivative of this
with respect to t is just 0. A constant, obviously, does
not change with respect to t. And so we get this
right over here. Now this is the
second derivative in terms of the
first derivative. But the question asks us,
write the second derivative in terms of W. But
lucky for us, we know how to express
this as a function of W. They gave that to
us in the problem. This was given. This is just me rewriting
it in a different notation. So this is going to
be the same thing as 1/25 times the derivative
of W. And the derivative of W is this. The differential
equation literally tell us the derivative
of W is this over here. So 1 over 25 times Wt, the
function W as a function of t, minus 300. And so we can say the
second derivative-- let me write it a little
smaller so I don't waste space. The second derivative
of W as a function of t is equal to 1/625 times
W as a function of t, which is a function
of t, minus 300. So we've done the first part. We've found out the
second derivative of W in terms of just W. Now let's try to address the
second part of their question. So we did the first part. Now use what we just
figured out to determine whether your answer in
part a is an underestimate or an overestimate of
the amount of solid waste that the landfill
contains at time 1/4. So in part a, we
found the slope. We found the slope of the
tangent line at time equals 0, and we used that slope
to extrapolate out to 3 months, or time
equals 1/4, 1/4 of a year. Now if the function
W's slope over that 1/4 was exactly the same as the
slope of the tangent line, or if that slope did not
change, then our extrapolation would be exactly right. If W's slope is
increasing over that time, then our estimate would
be an underestimate of the correct thing. And if the slope is
decreasing over that time, then our estimate would
be an overestimate of the actual amount
in the landfill. And to figure out whether
the slope is increasing or decreasing, we
just have to look at the value of the
second derivative. If the second
derivative is positive, that means our
slope is increasing, which means that our
extrapolation would be an understatement. Let me just be clear here. Let me draw this. Let me make it very clear. So let me just draw
a random function. So let's say that's W. So this
is a case where W's slope is increasing faster,
or W's slope is increasing from
that starting point. So our starting point--
this was our slope. And then W's slope keeps
increasing from there. And in this case,
our estimate is going to be underestimate
of where W actually is after 1/4 of the year. If W's slope is exactly the same
as our function over the course of the first three months, it
would look something like that. Maybe it diverges later on. And in this case,
our approximation would be really good. It would probably
be exact for W. And if W's slope,
for whatever reason, goes negative after that point,
and they already tells us it's an increasing function,
so that is not likely. Or it could-- it doesn't
have to even go negative. If W's slope decreases, it
could still stay positive. And the way I drew it, it
doesn't-- let me draw it like this, just to make it clear. So let's say we find out that
the slope looks something like this. This is the slope of the
tangent line at time equals 0. If W's slope increases
from that point, then w might look
something like that. And then our answer to part
a would be an understatement, would be an underestimate
of where W actually is after a fourth of a year. Let me draw it and make it
clear to you what I'm doing. So this is my w-axis. So this is right at our
initial condition of 1,400. And then this right over
here is our time axis, and this is that 1/4. So in the last
video, we said, hey, this is sitting right at 1,411. If W's slope increases
from that point, then this is an underestimate. If W's slope stays
the same, then this is actually a very good
estimate because then we're going to hit that
point directly with W. And if W's slope decreases--
so you can imagine maybe W looks something like this. It has that slope of the
tangent line when it starts, but then the slope decreases. And it's still an
increasing function, but the slope is decreasing. And in that case, we would
have an overestimate. And this is a situation--
so this first situation, slope is increasing. That means W prime
prime is positive. The slope is increasing. The second derivative
is positive. This would be a
byproduct of or this would cause the second
derivative to be negative. Our slope is decreasing. And this would be our
second derivative. Our second derivative is 0. If your slope isn't changing--
if your slope is constant, if your first
derivative is constant, your second derivative
is going to be 0. So let's just see what
our second derivative is at our initial
condition, and then we'll have a pretty good
sense of whether we have an overestimate or an
underestimate for part a. So let's just figure out
what our second derivative at time 0. It's going to be equal to 1/625
times W of 0, which we know, minus 300. Well W of 0, the amount of
waste we have at time 0, they told us in the
problem, is 1,400 tons. 1,400 minus 1,300 is 1,100. And then 1,100 minus--
or 1,100 divided by 625 is a small number. It's 1. something. But it is a positive number. And that's the
important thing here. So this thing, all of this
business, it is positive. So the second
derivative is positive, which means the
slope is increasing. At least right at our starting
point, our slope is increasing. Since our slope is increasing
from that point-- and frankly, the fact that it's
increasing at all and that we know the W is
an increasing function-- tells us that our estimate in
part a is an underestimate. So this is the case
that we're dealing with. So our estimate in part
a is an underestimate of the actual amount of solid
waste in the landfill at time t equals 1/4.