BRYAN PASSWATER: Well, welcome back to AP Daily Practice
Sessions for 2023.
We're looking at session number 3 today.
And I am, of course, Bryan Passwater with my great friend
Tony Record.
And today, we're going to be looking
at another free-response problem.
And this time, it's going to be about parametric motion.
So when you get a free-response problem,
if it's parametric motion, number 1,
you should quietly just sit there and just celebrate
a little bit because these problems are
a great chance for you to just rack up a whole bunch of points
and then use that as a cushion for the rest
of free-response problems.
And Tony and I today are going to give you a lot of great tips
to do just that.
So, in this problem right here, we
have this information given to us about a particle moving
in the xy-plane.
And the position is going to be this parametric vector
x of t comma y of t.
And it's moving in a way such that x prime--
remember x prime is going to be the velocity
function in the x direction-- is t times e to the 2t.
And we also have some selected values but not
the actual equation of a differentiable and increasing
function y of t.
And we have some various times given to us,
but not the function itself.
And we're given the position of the particle at time 0
is at 4 comma 1.
And so we're going to use all that information
throughout these six parts of this question
that Tony and I have for you today.
So let's look at part a here right off the bat.
Part a is going to be to find the speed of the particle
at times 0.
And I love parametric problems because there's
a few formulas, a few equations that if you
memorize them and learn them, and understand them,
they're going to go a really long way on these problems.
They're going to get you a lot of points,
but you have to know your basic formulas.
And so speed is defined in parametric motion
as the square root of x prime of t squared,
the velocity in the x direction squared plus y prime of t
squared.
And so on this problem since we're
finding the speed at time 0, we know
we're going to be able to use that function.
But just plug 0 in for our time values.
And so how are we going to find x prime of 0 and y prime of 0?
Well, we have an equation for x prime.
And so using our equation for x prime,
we're going to plug in 0 to it, and we're
going to get this 0 times e to the 2 times 0 power.
And then our y prime-- we don't have an equation,
but we have a table of values.
And we can see from that table that our velocity in the y
direction is going to be negative 3.
And so, once we have this function,
we can simplify it a little bit, and we simplify,
and we end up getting 0 plus 9.
The square root of that is going to just end up being 3.
So our speed at time 0 is 3.
It is important to note on free-response problems--
on the AP exam on the free-response problems,
we're not required to simplify answers.
And so, as long as we have a numeric value that
could be typed into a basic scientific calculator,
we don't need to simplify on these problems.
So either of those earlier two answers
would have also been acceptable for full credit.
TONY RECORD: All right.
Now we're going to take a look at part b.
And the question reads, could the particle
be speeding up in that vertical direction at time t equals 0?
And we have to give a reason for our answer
using standard calculus approach.
So what else is new.
So the very first thing that we're
going to start thinking about here
is what is it that must happen when we see speeding up
in particle motion?
So we're going to visit our good friend the toolbox again.
And what we know is that the particle will speed up
traditionally when the velocity and acceleration have
the same sign, whether it kind of exhibiting
the same behavior.
So what we're going to do in this problem
is kind of isolate in on this vertical direction
this motion in its velocity form at times 0.
And so what is the velocity of this particle in the y
direction?
Well, it would be the derivative of y.
And so we think, well, what's the y prime of 0 value?
Well, that's going to be really easy to pull right
from the table.
And so we see that that's negative 3.
It's clearly going to be negative.
So we can make the declaration that velocity is indeed
negative.
Now the next part is just a little bit trickier
because we've given this information
about an increasing function y prime of t.
And that's going to help us figure out
this behavior of the acceleration, also known as y
double prime.
And so what do we know about increasing functions?
Well, we know that their derivative is very likely
positive, but you've got to understand
that you could run the risk of being just temporarily
equal to 0 for just a moment.
And so because of the language in this particular problem,
we don't want to just say y double prime of t
is strictly greater than 0.
We're going to say that it's greater than or equal to 0.
And that's still going to suffice in order to figure out
this speeding-up behavior.
And so what we're going to do is say
acceleration is non-negative instead of saying positive.
And so by virtue of the velocity being negative,
the acceleration being non-negative,
then it's perfectly acceptable for us
to say that this particle is not speeding up
in that vertical direction at that time
because we essentially have two different signs going on
with this particular problem.
So I'm going to move on to part c now.
I'm going to ask about the x-coordinate.
We haven't been given him a little love here so much.
So we're going to do that right now.
What is the x-coordinate of the particle's position at time t
equal 3?
Our toolbox here is just going to revisit our good friend
and adaptation of the fundamental theorem
of calculus.
Sometimes you call this the net change formula,
but it's basically saying our ending position is just
simply going to be our starting position plus any kind
of displacement that we might undergo.
And so, for this particular problem,
we're going to say that x of 3 would be x of 0.
We know a lot about what's going on at times 0.
So that's why we want to use that lower boundary
plus this definite integral.
So what do we see?
Well, we're going to extract the x of 0 from not the table.
The table is talking about y.
But if you look at the bottom line there of the stem,
we see that the position of the particle
is at the x-coordinate of 4 times 0.
Now we can just integrate the te to the 2t from 0 to 3.
Probably a little easier said than done.
We want to recognize this as another toolbox tool.
It's going to be the integration by parts formula
because you're taking the integration, the antiderivative
of a product.
And so we set up integration by parts where we find a u value.
In this case, we're going to use the algebraic portion t.
The dv would be the e to the 2t with respect to t.
We take our derivatives.
We take our integrals like we're trained,
and then we just assemble it.
So we have our 4 plus the u times v
minus the integral of vdu.
I'm trying to be real careful using good notation the fact
that I have boundaries that still must apply here.
And then we take that last antiderivative of 1/2
e to the 2t.
Notice it is a bit of a u substitution
where I'm going to have an extra factor of 1/2 out in front.
And then we just start plugging in our 3.
We plug in our 0.
We just carefully keep track of the values
that are coming about.
It's very important that you have this minus
sign right here because that's going
to have an impact on your final answer there.
And hopefully, this all comes out
in the wash to be equivalent to that value there.
And, of course, you could pause the video
if you didn't get that answer and maybe look
to see where your mistake may have been.
BRYAN PASSWATER: Tony, I think that's a really great example
of another reason why under free-response, we
should take advantage of not simplifying answers.
The multiple choice is going to give you
plenty of opportunities to show your skills when that--
by the end of this exam when you get the FRQs,
use the moment to get a numerical answer
and let the readers, the graders sort it out
to make sure it's accurate.
You don't have to worry about that.
So we're going to take a look now at part d of this problem,
and we're going to find another position.
But in this case, we want to find the y-coordinate
of the position at time 4.
However, since we don't have an actual function in equation
for y prime, we're going to end up having to approximate it.
So AP example-- if you see a FRQ that has a table of values,
there are certain things that tie well to that.
A lot of things that are very common that you might see.
One of those things is going to be a Riemann sum
approximation of some sort.
So we're going to use a left Riemann sum.
Always pay very close attention to whether you're
doing a left Riemann sum, a right Riemann sum, a midpoint,
or a trapezoid approximation that could certainly
affect your answers and how many intervals you have.
So we have three intervals, which make sense.
And we're going to use that to approximate the position.
So using the same idea of this toolbox,
we still have the same idea that the position
is going to be found using an initial position
plus the integral from the starting time
to end time of y prime.
And so we know that y of 4 is going to equal whatever y of 0
is, our starting position, plus the integral from 0 to 4 of y
prime.
Now we also know that from the given information that y of 0
is just 1 in the stem of the problem.
And then the rest of this, we don't have a way
to calculate the exact value of the integral of y prime,
which is why we're going to have to use a Reimann sum.
And so we do our Riemann sums.
We're going to have these the bases in the first interval
from 0 to 1.
We see the width is 1.
And then we have two options to choose from are height.
We have the negative 3 value we could use
or we have the positive 2 value.
Well, we're going to choose the one on the left.
And so, if we follow this pattern
by looking at each interval finding the width,
and then choosing the left side height of our function,
we're going to end up getting a Riemann sum, something
like this.
The first interval has a base of 1, height of negative 3.
Then our middle interval is going to have a base of 2
from 1 to 3.
And then the height is going to be 2
because we're choosing that number 2 instead of the 5.
And then finally, our last interval
is going to have a base of 1 and a height of 5.
And again, we could have left it just like this.
But if we simplify this a little bit,
we end up getting 1 minus 3 plus 4
plus 5, which ends up being 7.
Now, of course, the answer 7, but we use approximation sign
because we're using a Riemann sum.
We understand that the actual position of y at time 4
is most likely not going to be exactly 7,
but it's going to be something hopefully near that.
So we do Riemann sum or any kind of approximations.
We do sometimes ask these questions
about how do you know whether that answer could
be too much or too low?
Is it an over-approximation or under-approximation?
And so we're going to look at part e now,
which is going to be, does our answer from part d that we just
got-- that 7--
would that overestimate or underestimate the coordinate?
Well, most of the time, when you estimate things,
you don't know if it's over or under because you
think you would have to know the right answer to determine that.
How could I know if it's too big or too small
as I know what the answer was.
Well, if we knew what the answer was,
why are we approximating it?
So normally, you won't always know.
However, if you're given certain information,
there is a way to tell.
So remember, we used a left Riemann sum to do this.
And from the stem of the problem,
if you go back up there, it tells us
that y prime is actually an increasing function.
And that's actually a really, really important piece
of information.
I don't know what it looks like exactly,
but it must be increasing.
And so when I do these over-under approximation-type
problems, I like to just sketch a little picture.
So if y prime is increasing, I'll
just draw a little line going up.
It's increasing.
And then I'll draw a couple of rectangles inside of it
based on the problem.
We used a left Riemann sum.
And so when I see this, I can see that the ones I drew,
of course, are always going to be
a little bit under because our heights are fixed.
As soon as we fix our height of our little rectangles,
the graph keeps going up.
So the graph is always going to go above those.
And so that's not going to be an actual explanation,
but it helps me remember what goes with what.
And so we know from calculus that a left Riemann
sum will always be an underestimate
if the function that we're using is increasing.
And then conversely, if--
it would be an overestimate if our function was decreasing,
but I always draw the picture.
I don't like to memorize that phrase.
I want to kind of sketch it out, look at it,
and kind of think about it intuitively.
So how would we write that on the exam?
Well, when we talk about things being over underestimate
with Riemann sums, we want to talk
about the behavior of the function
that we're using to find the area.
And so here we can just say it's an underestimate
because y prime is increasing.
And that is it.
That simple explanation is a perfect and concise and clear
explanation to get those points.
TONY RECORD: So we move now to the final part.
You might think, well, what else could Bryan and I ask you
about this particular question?
Well, we're going to ask, must there
be a time c on the open interval 0, 4 such
that the vertical acceleration of the particle is 3?
And, of course, we have to justify the answer.
And so you think about what kind of tool
do we have in our toolbox that's going to help us with this?
And it all boils down to that very important mean value
theorem that you saw about halfway
through maybe the first semester of the course.
I believe we talk about it in unit 5 in the course and exam
description.
Well, if you're ever asked to find or discuss
the existence of a derivative, your go-to
is going to be the mean value theorem.
Not like, say, if you're asked to find
the existence of a function value, that's going to be
the intermediate value theorem.
And so the mean value theorem is basically
just saying that your slope is equal to the derivative value.
And so when we think about this, we're
going to find that slope first.
We're talking about acceleration--
the acceleration is indeed the derivative of our given y
prime in the table.
And so, if we find the slope of this y prime at the value
0 and 4-- now, why do we have to use 0 and 4?
Well, we don't.
Maybe you try other combinations of pairs of points
and you don't quite get the results that you want.
But we're going to try about every permutation
that we can here, and we're going
to find that if we use the 0 and 4,
obviously, we can pluck the values from the table for the y
prime of 4 and the y prime is 0.
We get our 9 minus negative 3 on top, simplifies to 12/4
and that, indeed, is that 3 that we were asked for.
And so that would take on the value of y double prime.
Thus, our acceleration.
Now we need to be a little bit more thorough when
we put the finishing touches on this,
and I want you to look at this very carefully.
I know it's a little wordy, but we know that the mean value
theorem applies because the function y prime is
given to us as differentiable.
If you read the stem, you'll see that in the second line.
It's going to be very important that you
state that because the differentiability is
there that implies that the function y prime is also
continuous.
And that's going to open the door.
Continuity on that particular interval
it's going to open the door for us
to be able to apply this mean value theorem
and, therefore, get the answer of 3.
So I think that wraps up another one.
Hopefully, you stick around.
We've got five more great videos to share with you
that are going to help you solve some more problems
in the world of AP Calc BC.
So be sure to stick around for those.
Until then, take care of your body.