MARK KIRALY: We are back with another AP Daily practice
session for AP calculus AB.
And we're going to do our third free response question.
And this is going to be a differential equation
problem, which are really cool.
VIRGE CORNELIUS: Those are my favorite.
I kind of like them all, but this is definitely my favorite.
I'm Virge Cornelius.
I teach at Lafayette High School in Oxford, Mississippi.
MARK KIRALY: I am Mark Kiraly, and I teach at Ryan High
School in Denton, Texas.
And again, this is AP calculus AB.
So if you're an AB student or a BC student,
you are welcome to stay.
So we're going to get right into our stem here.
As we mentioned, as Virge mentioned in video 3,
the stem is just the part of the question
that you use throughout.
And the stem for this is actually one sentence.
It says, consider the differential equation dy dx is
1/20 of x times y plus 1 cubed.
And then there's the slope field sitting underneath the stem
as well.
Part A-- a slope field for the given differential equation
is shown.
It says sketch the solution curve that
passes through the point 2, negative 1,
and sketch the solution curve that passes through the point
negative 2, 1.
So instead of going straight to the DotCam,
I'm actually going to bring in my little pointer here.
And this first one says we're going to look at the point 2,
negative 1.
So if I start at my origin, 2, negative 1, that is this point.
They've marked the dot for me.
It says sketch the solution curve.
So what I'd really like to do is I'm just going to follow this.
This is a little slope segment.
If the danger line exists there, the curve
is going to go this way and it takes me
to this tangent line, which takes me to this tangent line
and takes me to this stage line.
And it turns out in this case, I just stay along this line.
And it said the solution curve, but you
might remember from your algebra classes
that a solution curve is just a set of points
that solves the equation or makes the equation true.
So a line is a solution curve.
So I like that line right there for 2, negative 1.
And then for the point negative 2, 1 on this little segment,
I'm going to start off to the left here
and I'm going to use these guys like currents or wind markers,
like guidelines.
And I'm just going to try and stay in between them,
kind of like if I drifted off of this tangent line,
now maybe I'd come up between these two
and things are getting steeper.
So kind of a curve going off that way.
And then as I come down here, things are getting shallower.
And you notice along the y-axis, those are all horizontal.
So I'm horizontal and then I'm picking things up
and maybe I come back through these symmetrical points
and head up that way.
And that's what you're going to want to draw.
And this is generally one point for each curve.
The readers are not dramatically,
I would say, over the top on your accuracy.
They realize all you have at this point
is a pencil and maybe an eraser.
Your calculator is put away.
You're not allowed to have a straight edge.
So this sketch is decent, and that's what I want to see,
just a decent sketch and a line.
You do want to go from where you exit the slope
field to where you exit.
Or if there was an undefined point, I would draw a circle
and stop at an undefined point.
But in this case, the slope field was defined everywhere.
And for part b, let you read that, it
says let f of x be the particular solution
to the given differential equation
with the initial condition f of negative 2
is 1, which you might notice that is one of the points that
was on our slope field.
Write an equation for the line tangent to the graph of y
equals f of x and x equals negative 2.
And use your equation to approximate f of negative 1.5
All right, so same problem stem.
And I've got this f of negative 2.
I am going to rewrite them and say, hey,
that's the point negative 2, 1.
And they're saying x equals negative 2
and I want to approximate f of negative 1.5
with a line tangent.
So Virge is tired of this question
because we talk about these problems all the time.
But what do I need to write the equation of a line, Virge?
VIRGE CORNELIUS: I'm never tired of these kind of questions.
You're going to want your slope and you're
going to want your point.
So we have the point negative 2, 1 and the slope
is going to be found by using that differential equation
because that is the formula or the recipe for slope dy dx.
MARK KIRALY: The slope is going to be that dy dx evaluated
at negative 2, 1.
So 1/20 is being played by 1, 20. x
is being played by negative 2.
And y is being played by 1.
Now we've talked several times that you
don't have to evaluate.
And I can leave that as my slope going through the problem.
I may do that.
I can't, so I'm going to go ahead and work that out.
1 plus 1?
VIRGE CORNELIUS: 2.
MARK KIRALY: 2.
So this is negative 2 over 20 times
2 cubed, which is also known as--
VIRGE CORNELIUS: 8.
[CHUCKLES]
MARK KIRALY: And negative 16 over 20 or negative 4/5
if I reduce that.
Again, you can leave it.
If you want to just copy this down every time
because you're afraid you might make a mistake,
then you might make a mistake copying it.
So you don't have to weigh it, but definitely could
leave it as negative 16/20 as if you're
more comfortable with that.
So there's my slope.
There's my point.
What do you use for your tangent line, Virge?
You use this traditional point--
VIRGE CORNELIUS: You know what I use.
Yeah, y minus y1 equals m times x minus x1, yep.
MARK KIRALY: So y1 again is 1.
My slope is the negative 4/5 I just found.
And then I'm going to write minus negative 2 is x plus 2.
So that's my tangent line.
VIRGE CORNELIUS: And you don't have to box it, but yeah--
MARK KIRALY: You do not have to box it.
I'm boxing it to separate it from-- this
is the way I write my tangent lines.
I put the y1 over on the other side.
It's tomato, tomato, right?
It's just another way to organize things.
The reason I like that though, that is a function notation
because now they've asked me to approximate f of negative 1.5.
So you don't have to put squiggles here.
You could put equals.
We know you're finding an approximation,
but formally, it should be an approximation.
1 minus 4/5.
And my x now is negative 1.5 plus 2.
And I can stop there, right?
VIRGE CORNELIUS: Not a bad idea.
MARK KIRALY: No, not a bad idea at all.
I believe it does all work down to 6/10.
VIRGE CORNELIUS: It does, 0.6 is your final answer.
MARK KIRALY: This is half, 1 minus 4/10,
which is 6/10 or 3/5.
Now, if you're like me and you really
want to know what the number is, what's really cool is,
what happens if you cross stuff out, Virge?
VIRGE CORNELIUS: What do you mean if you cross stuff out?
MARK KIRALY: If you're working through a response and you
make a mistake and you just cross it out.
VIRGE CORNELIUS: Oh, then you just--
that doesn't get graded.
MARK KIRALY: So I can just do that.
VIRGE CORNELIUS: Yeah.
MARK KIRALY: Remember, this is my final answer.
VIRGE CORNELIUS: We, we, they, we--
[CHUCKLES] because we both read AP exams,
we don't read crossed-out work.
So you don't have to go to the trouble of erasing it.
If you don't want it to be read, just cross it out.
MARK KIRALY: You're watching a video.
You get to pause read but-- so 1 minus 4/5,
negative 1.5 plus 2 is a perfectly good answer
as is 3/5, 6/10, 0.6, any equivalent value,
which brings us to part C, the final part of this problem.
So C says, find-- which there's a reason, right?
There's a lot to do here.
Find the particular solution y equals
f of x to the given differential equation
with the initial condition negative 2, 1.
So there's that point negative 2, 1 again.
But this time, they actually want me to solve this.
So they want me to find out what y is.
So got any ideas for me?
VIRGE CORNELIUS: Separate.
MARK KIRALY: Seperate.
This is called separation of variables.
And we have an independent and dependent variable
named x and y.
I'm definitely going to bring that y plus 1 cubed.
I'm going to divide that over to this side.
You could bring it over his y plus 1 to the negative 3.
That might actually be easier for doing an antiderivative,
but again, this is a little more traditional of the division.
I see this x here.
Do you like the 1/20 on the x?
Do you care?
VIRGE CORNELIUS: I've worked this problem several times,
and usually, if there's a coefficient like that,
I usually leave it on the right hand side
with the independent variable.
So I'm going to have to anti-differentiate it
and it's going to become a 1/40.
I can tell that already.
So I don't know.
MARK KIRALY: I don't think it matters.
I think when I did it the first time, I kind of--
I'm more comfortable working with just x.
So I put the 20 on the other side.
But it does it 1 way, 6 the other, right?
VIRGE CORNELIUS: Mm-hmm.
MARK KIRALY: I had mentioned I could
have done that as y plus 1 to the negative 1/3 over there.
And this is negative 2.
And so negative 1 and 1/2 out front.
And there's a plus c there, but 1 plus c
will take care of the entire equation.
So I'm going to save that for the right side.
This is what I'm going to do a little more in my head.
This is 1/20 time's 1/2.
That's at 1/40 that Virge alluded to.
And then on that plus c, there's a constant on both sides.
But if I gather constants, I have a new constant.
I call that one c.
So at this point, I can start cleaning things up,
start solving for y.
Personally, I want to know what this c is.
So I know we're not allowed to use multi-color pens
on the test, but for some clarity
here, I'm going to bring in my negative 2, 1.
I'm going to plug 1 in for y and negative 2 in for x.
I said 1/40 squared.
So 1 plus 1, that's 2.
2 squared is 4.
The negative means we're dividing.
Negative 2 squared is 4.
4 over 40 is 1/10.
I think we should have kept it as 40th in hindsight.
Negative 1/2 times 1/4 is negative 1/8.
I get that 1/10 plus c.
And you can tell I really had to concentrate because I
stopped talking for a second.
VIRGE CORNELIUS: [CHUCKLES]
MARK KIRALY: If I get the common denominator, this is 5/40.
I'm going to subtract that over as 4/40.
So C is negative 9 over 40.
And I'm going to scoot over now.
I know you can't see the original problem anymore,
but I think it's more important for me
to have some real estate.
So I come back and I'm going to start rewriting some stuff.
I've got negative 1 over 2 times y
plus 1 squared equals that was 1/40.
x squared minus 9/40.
That look good?
VIRGE CORNELIUS: Yeah, it looks great.
MARK KIRALY: At this point, your own work might just
split a lot from mine.
My work may look different than Virge's.
There's a lot of ways to start isolating that y.
So I'm going to go the ways that makes sense to me,
but realize that the readers actually
have a list of equivalent correct answers at the end.
So if you manipulate your fractions a little different,
as long as everything's done with good algebra, we're OK.
VIRGE CORNELIUS: And I might also
say that your work in between might have even
looked a little different.
I don't know.
I mean--
MARK KIRALY: Very true.
VIRGE CORNELIUS: Yeah.
So there's a lot of equivalent answers
that look very different here.
So you can always ask your teacher
if you had done it correctly if you did it before we
started going over it.
MARK KIRALY: Yeah, there's just there's so many things, right?
If you solve for C later after you've done some moving around,
your c might look a little different.
But I'm actually going to flip the whole thing.
I'm going to reciprocate both sides,
call this negative 2y plus 1 squared.
And called this side 40 over x squared minus 9.
I'm going to divide my negative 2.
This is negative 20, x squared minus 9.
I think the first time I did this,
I actually divided that whole right side by negative 1
to get 20 over 9 minus x squared.
I don't think I'm going to do that this time.
I'm going to write it like this.
It just seems more natural at this point.
When I get rid of that squared, what
do I have to consider when I do this, Virge?
VIRGE CORNELIUS: Well, when you take
the square root of both sides, you're
going to get that plus or minus.
So that algebraic--
MARK KIRALY: Plus or minus.
So I'm going to leave it for now.
VIRGE CORNELIUS: That happens with absolute value as well.
MARK KIRALY: Yes, if you have a-- especially
if you have a natural log and that
normally has an absolute value inside it that
has to be considered.
So they asked for--
VIRGE CORNELIUS: I think you need a square root.
MARK KIRALY: I do.
They asked for a function and we don't have a function
right now.
VIRGE CORNELIUS: Nope.
MARK KIRALY: This plus or minus means I have two y's for each
of these x's.
So I'm going to come back with my point negative 2, 1.
And this one, I'm just going to do a little more mentally.
What's negative 2 squared, Virge?
VIRGE CORNELIUS: 4.
MARK KIRALY: At 4 minus 9.
VIRGE CORNELIUS: Hold on a minute.
MARK KIRALY: That's OK.
VIRGE CORNELIUS: So wait a minute,
I was thinking negative 4, but I'm having a moment.
So it should be negative 5 in the denominator, yeah.
Yeah.
MARK KIRALY: So negative 20 divided by negative 4--
VIRGE CORNELIUS: Is going--
MARK KIRALY: Negative 5 is 4.
Square root of 4 is 2.
I need this to be a positive 1.
I need that to be plus 2 there.
So then it's really what I'm going for, is do I need this?
I need that positive coming out of this.
So this has to be negative 1 plus the square root
of negative 20 over x squared minus 9.
VIRGE CORNELIUS: Check your input, output when
you have that plus or minus--
MARK KIRALY: I think I'm going to box at this time
because we've said I don't have to block the answers.
Again, you might have 20 over 9 minus x squared.
You might even have a complex fraction in there, depending on
how--
VIRGE CORNELIUS: Very complex.
MARK KIRALY: Yeah, that's fine.
But that is a perfectly good function.
If I plug negative 2 in, I do get one back.
And it has that derivative, so I'm pretty happy with that.
So remember, when you're sketching slope fields
and solutions, make sure to use points given on the graph.
So Virge mentioned the other style
where they give you some points and you draw those slope field
segments.
Make sure you use the points they give you.
VIRGE CORNELIUS: Yeah, sometimes you
look at something like that slope field.
And if you haven't covered it in class,
it seems like, ugh, if I can't do part A, I can't do anything.
If it feels awkward, just read all the parts.
The way the committee designs these is that they--
you can enter on any part--
A, B, C, or D. Sometimes actually C or D
is an easier part or you've just had more practice with it,
so it seems easier.
So read the whole thing.
MARK KIRALY: Yes, definitely, keep reading.
Don't give up on a problem.
Make sure you take your time, show any support for your work.
So again, my tangent line, I want
to show where my slope came from.
Use the information you were given.
Sometimes the stems are longer, shorter.
They're giving you information in those stems.
And if you don't know what to do, keep your pencil moving.
Like Virge said, take a derivative, do an integral,
do the stuff you know, right?
VIRGE CORNELIUS: Yep.
MARK KIRALY: All right, well, thank you
for a good first week.
VIRGE CORNELIUS: Yeah.
This has been a great four sessions.
And we've got four more sessions planned for you for next week,
so come on back.
MARK KIRALY: Bye, Virge.
VIRGE CORNELIUS: Bye.