BRYAN PASSWATER: Welcome back to AP daily practice sessions
for AP calculus BC.
I'm Bryan Passwater and I'm joined, as always,
with my great friend Tony Record.
And in this video, we're going to look
through several different multiple choice problems that
are very common on the AP exam and we're
going to share some great strategies
that you can use to maximize your points and your AP exam
scores.
Of course, if you want to work through these problems with us,
just click on the link that's over the video now
and then you can download the PDF of all these questions
that will be featured in this session.
So our first multiple choice problem
is going to involve everyone's favorite, the logistic
differential equation.
And here we go.
Question number one, which of the following
is a logistic differential equation whose solution
has a carrying capacity of 50?
So remember, when we talk about logistic differential
equations, these are problems where the solution starts
to have some exponential growth ideas,
but it has this kind of hampering part that's
going to make it become logistic,
and it ends up with this idea of having a carrying capacity.
So in this case, how do we use this information
to find out which one is going to be our correct answer?
Well, we know that a very common form
of the logistic differential equation
is this idea of dy over dt equals some constant k times y,
times the quantity 1 minus y over L.
So it's really good to understand
the different forms for logistic differential equations.
And so knowing that form right there
is going to be really helpful because you
can identify which forms might have a similar shape to that,
right?
And so one thing to look out for is
this-- in the binomial expression, is that number 1.
That will be there in the first term.
And notice these last three terms or these last three
choices I'll have that 1 there, which is a good sign, right?
Choice A, by the m we can immediately, sort of,
get rid of because this has uninhibited growth.
This is just a exponential growth differential equation,
but not a logistic one.
So how else can we distinguish between these other three
choices, they all have a 1 in them?
Well, we're going to logistic for this L.
And L is the variable that we use to define our carrying
capacity, right?
L is like our limiting value.
So in this case, L is this idea of 50.
That's where you're carrying capacity would go.
And we look for the L in the denominator of the second term.
Well, if we look at these choices that we have,
left BC and D, we might notice that only that bottom one
has that 50 right there.
And so that ends up becoming our solution here.
B and C are common misconceptions, common mistakes
and part of it comes because there is an alternate form--
another form that's fairly common for the logistic
differential equation.
And that form is this idea of dy over dt
equals a constant times y times the quantity L minus y.
And the difference between the two forms
is in the original form that we looked at, the top form,
we had factored out that value of L
from the binomial expression to leave that 1 and the leading
term.
Whereas the second form, doesn't factor it out.
And so this is an alternate form that might work.
And you notice on choice C here on the right hand
side, that was almost the alternate form, right?
This is almost the correct form.
And in that box-- that y times 50 minus y--
that actually would have represented
differential equation that's logistic and has a carrying
capacity of 50.
So these can be a little bit tricky
and if you don't know these forms.
TONY RECORD: Bryan makes a great point.
The two differential equation forms
can both be used fairly equally on the AP exam.
So you're going to want to be very familiar with both
of them.
And so what we're going to do now is we're going to, kind of,
shift gears and look at a little bit of a different problem.
We're now going to take a look at an alternating series,
and we're going to try to determine
what the kind of convergence that these particular series
are going to have.
And so our problem looks a little something like this.
We've got, which of the following series
are going to be classified as conditionally convergence?
And the conditional convergence is
sort of a moniker that's used primarily
for these special alternating series.
And so we've got our choices there and so
now we're going to think through this process.
And so the first thing that we're going to want to do
is we're going to want to begin by considering
the summation of the absolute values of each
of these expressions that truly do alternate.
And so as we do that, we kind of notice choice A
and you'll notice that I'm not going
to take the absolute value of it and there's
a very special reason.
I'm going to let you think about maybe why I'm not
doing that for a moment as we kind of move on
through the rest of these parts.
Now, I'm fairly confident that I want
to take the absolute values of the expression in choice B
and I know that just removes the alternating component.
And therefore, I have this series.
This 2n plus 3 over 5n plus.
1 and then I know something about that series.
I know that series is going to diverge.
If I use the n-th term test, I'm going
to get that limit 2/5 that's not equal to 0
and I felt confident about the divergence.
What that means is I've only lost the chance of being
absolutely convergent.
So I still have to, kind of, keep my ear
to the ground on what's happening here with B
because there could be some conditional convergence.
I'm going to do the same thing with C.
We take the absolute value, it removes the alternator.
Now in this case, when you try to investigate
the behavior of the series, we'll
notice that this guy is going to diverge as well.
It diverges primarily because we could compare that
to our nice harmonic series, 1 over n.
And when I go to D--
remove the alternator through the absolute value process--
I find that I have a series that can compare, very favorably,
to 1 over n squared, which we know that to be a convergent p
series.
So what does all this mean?
Well, I know that I don't have any chance
to conditionally converge at D. And so we're going to go ahead
and we're going to remove that choice.
Now, remember what I said earlier.
Choice A doesn't alternate and so
I don't want to analyze it by taking the absolute values.
I want have just look at choice A
as it's just sort of a negative comparable series
to the harmonic series.
And so we can pretty much eliminate him from contention
for being able to be conditionally convergent.
So you only have two left that you have to investigate 0
Which one are we going to go with?
Well, when we look a little closer at choice B,
we notice that we've got an alternating series test that's
going to require us to analyze that limit one more time,
but, sort of, by way of the n-th term test for divergence,
we're going to see once again that, that limit is not
going to equal the 0 that we want it to be.
And so therefore, we know that this is going to diverge.
Now, obviously, that leaves us with choice C.
If we look at that a little bit more closely,
we'll see that demonstrates that behavior
to have the convergence with the absolute values--
the divergence, I'm sorry, with the absolute values.
But in and of its own, using the alternating series test,
we're going to get those two criteria that
are going to be met.
Besides, it's the only choice left.
BRYAN PASSWATER: Well done, Tony.
I think we're going to go ahead and keep
on this theme of some series questions here.
And so we're going to look at another multiple choice
question.
But in this time, instead of looking at just a series test,
we're going to do everyone's favorite thing, power series
and the idea of interval of convergence.
But instead of making this question strictly
about a computation, it's more of a theoretical conceptual
type idea.
So in this problem, we're going to have
this power series given to us.
So we have this power series of some a sub n times the quantity
x minus 5 to n power-- that's our power series.
And we're given the fact that it converges when x equals 9.
For which of the following pairs of x values must--
the key word there is going to be must in this case--
the power series also converge?
So we a couple of things, right?
Number one, you want to get really
comfortable with the fact that when
you're given a power series in this x minus C type form,
that we know the series is centered at x equals 5, right?
We should be able to look at that
and immediately see that from the form it's written in.
And since it converges that 9, then we know that--
if it centered at 5 and it converges at 9,
then essentially our radius becomes at least 4, right?
So the distance from 5 to 9 is going to be 4.
And so we know that's going to converge everywhere
between there.
So one of the cool things about a power series
is if you know the center of it, we
know that whatever your radius is
will be the same amount in either direction.
And so since it converges at 9--
number one, we know that it will converge
at least until right before at least it gets to the number 1
going forward in the other direction.
And we know that it has to converge
for every number between 5 and 9.
It will always converge for all values inside of its radius.
Now, going in the other direction,
we know it can go forward to the left from the center
and so we know it must converge from the open interval 1
to the closed interval 5.
And notice we have that open interval on the left hand side
because just because the power series might converge
on one end of it's interval, does not
guarantee that it has to converge on the other end,
right?
So we know that converge is everywhere in between,
but we can't really make any conclusions
about what happens at 1.
It might converge there, it might not.
We just don't know.
So what else do we know?
What about numbers beyond the 1 and the 9?
Just also we know that this might
or it might not converge past when x is 9, right?
We just don't know what's happening there.
So let's take a look at these choices.
X is 1 and 7 is option A. 2 and 10 for option B. 2
and 8 and 0 and 6.
And think about where these values
fall on this interval, right?
So I'm going to place all of these values that
are kind of options and we can see
anything inside of that interval within that radius
must converge.
It has to.
So anytime we have a 2 or a 6, or 7, or 8
given to us on options, we know they converge.
Anything that goes beyond those values like the 0 and 10,
it might, it might not but we can't guarantee it either way.
And we also don't know what's happening at 1.
So using those ideas, we know the only option
that we can guarantee for sure is
going to be option C. X equals 2 and x equals 8.
So let's take a look at another multiple choice problem
but this time let's jump over to the world of polar areas.
So in this problem, we have this function r of theta
equals theta over 2 times sine of theta
and we know its graphed between 0 and 2pi.
And we're trying to find this shaded region
R that you see on your screen.
That's in between the inner and outer loops.
And so when we have a problem like this,
we really have to think about how polar functions work
and think about how it's going to be sweeping out
this area between those two functions.
And so a really good place to start on a problem like this
is to figure out when this graph is going
to cross the pole, the origin.
And so we know that happens when r is 0.
So in this case, theta over 2 times sine of theta equals 0.
And as we solve this equation, we
know that means that either theta has to be 0
or sine of theta has to be 0 and the only solution between 0
and 2pi is going to be 0 pi and 2pi.
What that means is this curve is going
to create that loop between 0 and pi
and then make a second loop between 0 and 2pi
and we just have to figure out which one
is going to be graphed first.
When we you think about this function r of theta,
it has this theta over 2.
And so as theta gets bigger and bigger and bigger, and bigger,
we're going to end up getting larger and larger and larger
values of r as sine cycles through.
And so we can tell by looking at that,
then it's going to have to cycle the inner loop first
between 0 and pi.
And then on that second loop between pi and 2pi,
we know is going to graph this outer loop, right?
So we know now it's going to the loop first
and the outer loop second.
So how do we figure out what that area
is between those two things?
Well, looking at these choices, we
can see that if we just go from 0 to 2pi using our area
formula--
one half times the integral of r theta squared--
now if we go all the way around between 0 and 2pi,
that's going to end up doing the inner loop and the outer loop
both.
So it's not going to give us the difference.
It's going to give us way too much area so we know
that choice B isn't going to work.
And we just look at the last slide
that if we just were to sweep out between 0 and pi,
that's going to end up being just the inner loop, right?
So option A ends up being the inner loop only.
And so we're down to B or C and we can see this subtraction,
kind of, makes sense.
The area between two curves that makes
sense that you might be subtracting some things.
So how do I decide which one is going to be correct?
Well, look at option D. Option D goes from 0 to 2pi.
So it's going to graph the whole big loop,
and it's going to graph the inner loop both
and then it's going to subtract just the big loop.
Well, what are you going to be left with.
If you grab both, and then take away the outer loop,
you're going to be left with just
that part on the inner sine, which was the same as option A.
But option C, in this case, is going
to work just fine because we're going
to have the outer loop minus the inner loop
and we're left with the correct expression.
So these are certainly some tricky problems
but I think that we'll get better at them with practice.
TONY RECORD: Oh, Bryan, you said it.
Polar's typically going to be one
of the pieces of the AP exam that's going to give students
the most trouble.
But stick around, we've got some more videos
coming up later on in this series that's going to help you
quite a bit with that.
Now, to round things out, we've got a really short and sweet
problem here about parametric and taking
the second derivative.
And so the problem reads, x of t is
3t squared minus 5 and y of t is t to the third minus 40 plus 2.
We know that time, t, if that is actually what the reference is.
It doesn't have to be but we know
that t is going to be positive.
Our directions are to find the second derivative.
So what we're going to do in this particular case is
understand that the derivative of y, with respect to x,
is kind of at play here, but we have
to write that derivative so that we are referencing it
in terms of t.
And so we use the standard dy over dt all divided
by dx over t.
Well, those two derivatives are going
to be pretty easy to compute if you think about it.
We just have a couple of polynomials.
3t squared minus 4 in the numerator and what?
6t to the first in the denominator.
Now, this is where the fun starts.
Before we begin our second derivative,
we're going to want to simplify this much like we simplify
things with monomial denominators.
We're going to split it apart.
Just do a little reducing so then the next derivative is
a little easier.
Now that we've done that, the second derivative has
a little bit of a unique feel to it.
We're taking the derivative of our dy, dx with respect to t.
And when I say dy, dx, I know I'm
referring to the derivative of y with respect to x.
But I have to, sort of, invoke this other derivative
of x with respect to t terms in order
that our notation works out for this second derivative feel.
And so the process is to take that derivative
of that dy over dx you originally
had and then divide by the original derivative of x
with respect to t.
Now this is going to clean up pretty easily, if you
multiply everything through by 6t squared over 6t squared.
And then luckily, we have an answer choice
but we want to point out that the most common mistake
for a student to make is to forget
that little denominator of 6t.
It's vitally important and if you forget it,
you're going to land on choice A. Try that on your own.
Simplify and see if that's indeed what you discover.
So I think that wraps it up here for our first session.
If you want more review, be sure to check out our other seven
videos in this series, and keep in mind that you can also still
check out the in-depth reviews from 2021 and 2022
that Bryan I had so much fun doing in those past few years.
And they're still available on AP Classroom
and on Advanced Placements YouTube channel.
Until next time.
Take care, everybody.