TONY RECORD: Hi, everybody, and welcome back
to our fourth installment of AP Daily Practice Sessions.
Bryan and I are really excited to be with you again.
And we've got a really interesting problem
planned for you tonight.
We don't really have a specific calculus concept
that we want to call this, so we're
going To Title this Communication and Notation.
And as you move through the parts of this problem,
you're going to see those various nuances that we're
trying to convey that will help you
score some of those points that deal with those two ideas.
So let's jump right to our rate in and rate out problem.
So the stem of the problem says, for the time interval 0 to 8,
the graph of f of x consists of four line segments,
as shown in the figure below.
The function of x is continuous and differentiable,
with g of x being greater than 0 for all values of x.
And selected values for x are shown in that table below.
We're going to let h be the continuous function defined
by h of x is the definite integral of f of t, z
of t with respect to t from 0 to x.
And we know that h 1 is negative 3.
It's a mouthful, and you're going
to find yourself going back to that
stem all throughout the particular parts
of this problem.
So what are we going to ask you in part A?
Well, how about to find h prime of 4.
And then we're going to ask if the graph of h
has a local minimum, local maximum, or neither at that
particular x equal 4.
And we'll give a reason for the answer
by just showing nice work throughout
the particular problem.
So first things first, finding h prime of 4.
Well, why don't we think about finding h prime of x first,
and then we can substitute our 4 n.
And so if you look back at the stem of the problem,
we see this function defined as an integral.
And so, if we want to take that derivative of that function,
we're going to use, well, one of the fundamental theorems,
whether you called it part 1 part 2, maybe you didn't call
it anything, that's the particular rule that
says that you're going to replace
all of the independent variables t with the boundary x.
And because the boundary is just an x,
we don't have to further do any chain rule.
And that's always going to work as long as the lower
boundary is a constant.
It doesn't have to be 0.
And so when we do that, we're just
going to get f of x times g of x, fairly simple.
And so then we're going to replace the 4 with x,
and we have f of 4 times g of 4.
Now, we have to start thinking, well, gosh, where are we
going to find these values?
Well, the f of 4 can be read easily from the graph,
because we had that graph of f of x.
Now, the g of 4 isn't very obvious to us.
We have a table of values for g, but we just
don't know what's going on at 4.
But turns out, we don't need to know because 0 times anything
is going to be 0, regardless.
And we're going to go ahead and draw attention to that answer.
We went ahead and circled that for you,
because that would be earning credit.
Because the question asked to find the H prime of 4.
So draw attention to that answer.
Now, the next part, does the graph have a local min, max,
or neither.
Well, notice that we're given some information, right?
This is where we're going to reread that stem.
So we have this notion that g of x is greater than 0
for all values of x.
g of x is positive, then, OK.
And so, basically, everything comes to fruition now.
Now, how does that lead to our answers so quickly?
Well, remember we know that f of 4 was 0.
Well, by the graph f of 4 is 0.
And what that does is it sets us up
to use the intermediate value theorem right here.
And we can see that the sine of f
is going to change from that positive to that negative
at that value 4.
And so by virtue of g always being positive, meaning he's
not going to have a lot of bearing
in the sign change of h prime.
So we can just focus on f sine's behavior.
And we see that positive to negative
invoking the idea of a relative maximum.
Taking a look at part B now.
BRYAN PASSWATER: Thanks, Tony.
So looking at part B, we're actually
going to utilize something that Tony
was sharing on the previous slide, which
is this value of h prime of x.
So remember that as we go through this part B.
So in part B, we're going to use Euler's method,
and starting at x equals 1 with two steps of equal size
to approximate h of 5.
So if you're in Calc BC, you can pretty much
expect at least once on the free-response of most exams
to have a question involving Euler's method.
And so it's pretty easy points as long
as you understand a couple of basic ideas.
and then understand how to show your work, right?
So looking at part B, we have this idea,
and remember what Tony had before about h prime of x,
is we're going to reuse this.
So with Euler's method, what's going to be important?
Like, what do we have to know?
Well, it's important for us to realize
how many steps we're doing.
And so since we're starting at x equals 1,
and we're going to approximate h of 5,
how are we going to get there in equal-sized steps.
But we're going to have to balance at 3 in between.
And so, our step size will end up being 2.
So anytime we do Euler's method, it's
important to understand based on how many steps you're using,
if they're equal size, how big each step will be.
That'll be a major part in our role,
because we're going to have to find the value when
x is 3 first on the way to finding
a way to approximate h of 5.
So what do we know?
We know that h of 1 is negative 3.
And so with Euler's method, it's not quite a tangent line
approximation, but it's going to use tangent lines that
allow us to maybe adjust halfway during our approximation.
And so I'm going to show you a little illustration
of the idea.
So here's what we know.
We have this 0.1 comma negative 3.
And I have it labeled here in this picture, right?
And we're wanting to approximate this function at 5.
We don't know a lot of information.
We don't actually know how to calculate h necessarily.
However, we could use a tangent line when x is 1.
So we're going to use a tangent line approximation when x is 1.
That's what you see is that line segment.
And then once we get to when x is 3,
you see a dot there, because at that moment,
we're going to readjust, we're going
to recalculate the slope of the original function when x is 3.
And then that way we can adjust our trajectory,
and then continue all the way until h is 5.
So we're going to use what seems like a couple of tangent lines.
Well, what do we need for tangent lines?
We need a point, and we need a slope.
And so we know our point is going to begin when x is 1
and h and 1 is negative 3.
And then our slope, will Tony did a nice job
on the part A of giving us a function for h prime of x.
Our function, h prime of x, is going
to be f of x times g of x.
So if I find the slope when x is 1,
it's going to be f of 1 times g of 1.
And looking at a combination of the table and the graph of f,
we see from the graph of f and the table for G,
that f of 1 times g of 1, is just
going to end up being 2 times 4, which gives us a slope of 8.
So we can create our tangent line equation.
Our tangent line is going to be negative 3,
our starting point, plus our slope x of x1 minus 1,
is our point slope kind of modified formula.
And remember what's our next coordinate?
This is why our step size is so important.
We're going to have to use when x
is 3 kind of intermediate step.
And so if we plug 3 in to our tangent line
to figure out what that second point would be,
we end up getting 13.
Negative 3 plus 16 is 13.
And so now we have this midway.
3 comma 13.
But I want to find the endpoint, so we're
going to end up having to, again, adjust our slope.
And so we're going to use h prime of 3. now.
Because our slopes can be a little bit different.
And so we're going to find f of 3 times g of 3.
And again, using our combination of our graph and our table,
we get 2 times 5.
We can see from the graph that f of 3 is 2.
And then g of 3, obviously, is 5.
And so, our slope now becomes 10.
So creating a linear function through that coordinate 3
comma 13 and then the slope of 10,
we get this second equation.
And then we're trying to use this to approximate h of 5,
and then we end up getting that our h of 5
is approximately what we would get
if we plugged 5 in here for x.
And simplifying this ends up giving us a nice value of 33.
And so if I kind of show us one more image here
of this original, the two kind of line segment I showed,
we have a line segment where the slope is 8,
and then we get to 3, we readjusted our slope to 10,
and then we can see that top point is going
to be at 33, which is hopefully a good approximation
of our answers.
TONY RECORD: Yeah, definitely looks like it is.
And I tell you, Bryan and I both agree that Euler's method
is a place where you can score some very easy points on the AP
exam.
It's just an iterative process that
really involves some basic ideas of calculus and arithmetic.
So great explanation on that.
We're going to go ahead and move gears here to question number--
or letter C, which looks to me like it's a limit problem.
So we're going to evaluate the limit
as x approaches 1 of h of x plus 3 of x all over the sine of pi
x, or we have to state that the limit might not exist.
And so, the problem is sort of keeping us on our toes,
and we've got to really keep careful track
about the inner workings of this particular function
and its limit.
So the first thing that we're going to do,
as we would probably want to do with all limits,
is let's see if we can take the easiest road out.
I sometimes call this the greenlight to my students.
Let's see if we could directly substitute that value
1 in for x.
Now, if you have any kind of desire to do this,
you have to make sure that the function that you're
going to try to replace the x with one is continuous.
Now we know that 3x is continuous.
h of x may not be so obviously continuous,
but if you read the stem, there in the third line,
we were told that it was.
And so that means we can just simply
replace the x's with ones.
We can try to decipher this from a bit of using the function
for h of x up above, right?
And we see that it is known that h of 1 is negative 3.
And then we're going to read 3 times 1 is 3
and obviously we have 0.
So that's our numerator.
I wonder what that's going to lead to?
Well, let's not make any assumptions.
Let's go ahead and take a look at that denominator.
And the sine of pi x is postulated to be continuous.
We don't have to state that.
And we see that if we directly substitute sine of pi
as the result, which is 0.
Now, we're on to something.
This is what we call an indeterminate form.
And because we've separated it, we've kind of not
violated any poor notation of communicating,
may be 0 divided by 0, which sometimes
can be a bit precarious.
And so we're just going to keep them separate.
And we're going to say we can now use L'Hospital's Rule.
And L'Hospital's Rule is going to allow
us to take the derivative of the numerator over the derivative
of the denominator, and then reapply that limit.
And both derivatives are fairly easy.
The denominator uses a little chain rule, as you can see.
And so let's try it.
Let's replace x with 1.
We find that we're looking at h prime of 1 now.
And boy, we've given h prime a lot of attention tonight.
In three consecutive problems, we're
going to use f of x times g of x or f prime--
I'm sorry, f of 1 times g of 1.
Got tongue-tied there.
Add the 3, and then in the denominator
of the cosine of pi, of course, is going to be negative 1.
And so we just whittle away at this and we can read the f of 1
and the g of 1 value directly from our table and our graph.
And, of course, that's going to result in this negative 11
over pi for our final answer to our limit.
BRYAN PASSWATER: Yeah, great job, Tony.
I think we're learning a lot in this video,
especially about how you're going
to show your work on the exam.
So the AP exam is a great opportunity for us
to think and act like mathematicians.
And so, we have to make sure that we're not just
getting the correct answers, but all of our work
has to tell a story.
Someone grading our paper has to see what we're writing,
and understand what we're thinking,
how we're approaching it, why we're doing what we're doing.
And so, it's really important that we show clear and concise
work, and that our work is going to be mathematical and correct.
What we do in L'Hospital's rule is another example
of an opportunity for us to show how
we can be mathematicians ensuring correct work, not just
simple work, but with good notation as well.
So we're going to jump into the last part of this problem, Part
D. And now in part D, we're giving some more information
that this capital G of x is going
to be the antiderivative of our function G.
And we're given the fact that capital G of 0
is going to equal 5.
And then even more information.
If the integral from 0 to 1 of capital G times f prime of x dx
is equal to 24, use integration by parts to find g of 1.
So we have a little bit of a hint
here, right, that we're going to have
to use integration by parts.
So immediately when I see something
like that on the exam, I just write down the formula
immediately, because I want to be able to see it
and it helps me process it, understand
what it is I'm trying to solve for here.
And so I'm just trying to do it all in my head
and try to make a plan up front.
And so whenever we do integration of our parts,
we saw one in a previous video that Tony did for us.
And he did a really great job of laying out the information.
We had to decide what's going to be our u, what's
going to be our v prime.
And then we lay that out here, and it's here
we're going to make you going to be
capital G of x, because I know that I'm going to need u prime.
And so I wouldn't want to take capital G's derivative to turn
it into little g, that would be an advantageous selection,
right?
And then I can make v prime be our f prime, because we're
going to have to take the antiderivative of our v prime.
And so, since it's f prime, I know
it happens to be a little f.
And so we kind of lay that information out
and then we figure out what our u prime would be.
So capital G's derivative becomes little g.
And then if v prime is f prime, then
v is going to be the antiderivative or f of x.
And so, like Tony says, we're just going to take this values,
we're going to plug them right into our formula.
And so I'm just going to write this out and say
the integral of capital G times f prime is going to equal,
using our formula, u times v. So we get capital G times f,
just plugging things right in, minus the integral
of u prime v, which is going to be g times f of x.
So we get this expression here.
We're not quite sure how we're going to use it yet.
But let's jump to a clean slide here,
so we can see this with a little bit more space.
And we want to evaluate this, right?
So we're going to take our first expression,
and we have to evaluate it from 0 to 1.
And so, I'm going to plug in 1 to those functions,
and we get capital G of 1 little f of 1 minus capital G of 0, f
of 0, using our fundamental theorem of calculus,
and I still have the second piece here.
But at least we can take the first part,
and we can find some of these values.
Of course, we may not know all the values upfront.
I don't know what g of 1 is.
That's literally the entire question we're solving.
But I can look at the graph and find f of 1.
F of 1 is going to end up being 2.
And then I know that capital G of 0 is 5.
That was given to us in the stim Then I can look at the graph
and find f of 0, which gives me 1.
And so the last piece was that integral from 0 to 1 of g times
f.
And it turns out, if you look at that, that is just
the definition of h of 1.
Look at the original problem h of x is the integral from 0
to x of f times g.
Well, h of 1 would just be the integral
of from 0 to 1 of f times g, which
is what we're asking here.
And so, that literally is h of 1, which we happen to know,
is given to us to be as negative 3.
So from here, it's just a matter of simplifying,
because we're told in this problem
that the whole function has to equal 24.
And if the whole function equals 24,
just doing a little bit of arithmetic,
we can solve and then quickly see that g of 1
must end up being 13.
That's a very whirlwind of a Part D,
and a whirlwind of an entire FRQ,
but Tony and I are trying to come up
with some questions that would give you
a whole broad array of good questions that you can do.
So if you can believe it, we somehow
managed to come to the end of our fourth session
halfway through the sessions.
We're excited to bring you sessions 5 through 8.
So don't give up yet.
Keep plugging along, looking at our videos,
and getting prepped for these AP exams.
So Tony and I are looking forward to it.
We thank you, as always, for joining us.
And we'll see you in the next video.