VIRGE CORNELIUS: Hey, everybody.
Thanks for joining us for session--
[GASPS]
MARK KIRALY: 7.
VIRGE CORNELIUS: 7?
MARK KIRALY: 7 or 8.
Yes.
VIRGE CORNELIUS: OK, I'm Virge Cornelius,
and I teach at Lafayette High School in Oxford, Mississippi.
MARK KIRALY: I am Mark Kiraly.
I teach at Ryan High School in Denton, Texas.
We're going to do an area volume problem question.
So, "Let's C be the region enclosed
by the graphs of m of x equals 2e to the negative x minus 1,
and v of x equals the square root
of x plus 2, the y-axis and the vertical line, x equals 3,
as shown in this figure."
And as Virge mentioned also, that's the STEM, all right?
So question-- very straight.
Very typical.
I think this is about as traditional area of volume
as you can get, isn't it, Virge?
VIRGE CORNELIUS: Yes, definitely.
MARK KIRALY: So again, we have this m of x and v of x.
They want to find the area of C. So traditional area between two
curves.
Hopefully, you have done this.
If not, quick way to find out how to do it.
We want to find the area under this top curve
and kind of subtract the area under, although "under" has
a different meaning here, right?
But we want to do top curve minus bottom curve.
And what haven't they told us, Virge?
VIRGE CORNELIUS: Which is which?
We don't know which one is m of x,
and we don't know which one is v of x.
By the way, we can't use our calculator to find out either.
MARK KIRALY: I hope this is going to be in part B,
and that question is 3, 4, 5, and 6.
And so that means we do have to have some knowledge
of our parent functions.
But I do notice that--
I think, you call this the cheat when we did.
I do know the y-intercepts are not the same.
So if I plug in x equals 0, it's e to the 0.
VIRGE CORNELIUS: Oh, that would be 1.
MARK KIRALY: 2 minus 1 is 1.
VIRGE CORNELIUS: 1.
MARK KIRALY: So this is m of x, which of course, I
can also realize.
This is an exponential decay model, shifted down 1.
So if you think of that swooping e to the negative x-graph
has been shifted down.
And likewise, I should note that square root of x-graph looks
like, right?
We should have that mental image of that graph.
It's been shifted up 2.
So v of x is indeed that graph.
And if I plug in 0, 0 plus 2 is a y-intercept of 2.
So I'm just going to do the integral v of x minus m of x,
and it is from 0 to 3.
They said x equals 3 is the borders.
0 to 3.
Unfortunately, because I don't have--
If calculator active, I would be good to go.
But we're going to have to do a lot of writing here.
Virge, your job is going to be to keep me honest
and on camera.
VIRGE CORNELIUS: OK, and also, make
sure you distribute your negatives.
MARK KIRALY: Yes, so it's all--
VIRGE CORNELIUS: Is that honest part?
MARK KIRALY: Yeah All the math is always there.
Your added responsibility-- this one
is because I'm going to be going up and back to the graph
that I remember to go back and forth.
VIRGE CORNELIUS: OK.
MARK KIRALY: Although, I don't want
to go back to the graph for part A here, I don't think.
I almost said number A because I'm a math guy.
So I'm going to go ahead and combine like terms here.
I could start finding antiderivatives,
but I'm going to rewrite that x as x to the 1/2.
I do have this minus negative 1 that Virge mentioned.
So that's a plus 3 minus 2e to the negative x.
Still integrating from 0 to 3.
And then how do I integrate x to the 1/2?
Well, it's the 3/2 when I multiply by the reciprocal
of that new power.
Antiderivative of 3 is 3x.
Oh, what do we got going on in this one?
VIRGE CORNELIUS: Hm.
So, I mean, you can think of this as u-Substitution,
but you could also think of it as the integral of e to the kx.
Yeah, Reverse Chain Rule.
That negative popped out because of Chain Rule.
So plus.
Yeah.
MARK KIRALY: Of course, if you're
doing it like this kind of by the seat of your pants
without using UD or whatever, it's just
take the derivative to check it.
Derivative e to the negative x is e to the x.
Change our negative e.
We got it right.
VIRGE CORNELIUS: Yehey.
MARK KIRALY: And 3x and so on.
OK, I'm going to plug in my 3 for my upper bound.
And I'm going to plug in my lower bound, which
is more important than usual.
VIRGE CORNELIUS: Mm-hm.
Just because you plug in 0 doesn't
mean it's going to be a 0.
[LAUGHS]
MARK KIRALY: Our very--
VIRGE CORNELIUS: If you have--
MARK KIRALY: I think with polynomial integration,
we kind of get a little lazy sometimes like,
oh, it's going to be a bunch of 0.
And it's mostly 0.
But e to the 0, as we mentioned before, is 1.
3 times 0.
0 to the 3/2 is 0, but we are going to get a minus 2
out of that part.
This is 2 over e cubed.
Here's a 9, and that's a power in a root.
Honestly, on the exam--
[LAUGHS]
--stop there, but that is a square root of 27
because 3 cubed, which is 9 times 3.
And I think that whole thing ends up being a 6, doesn't it?
I think.
VIRGE CORNELIUS: Yeah, yeah.
MARK KIRALY: Oh, no.
It becomes--
VIRGE CORNELIUS: Well--
MARK KIRALY: The 9 comes out.
The 3's cancels.
It's 2 root--
VIRGE CORNELIUS: I actually left it as 3 raised the 3/2,
but the 3's do cancel.
Yeah.
MARK KIRALY: I believe that's the correct answer.
VIRGE CORNELIUS: It is,
MARK KIRALY: But, yeah, that's [LAUGHS] a free response
question on the exam.
I definitely would stop it at the point
where it's a numeric answer.
I can't stop here because we still have x's in it,
but as soon as we are numeric, I think
I would bail for time too.
Not just because I'm necessarily uncomfortable, but just time.
All right, so resetting for part B,
region C is now the base of a solid.
So we're looking down at a solid.
There's a solid sitting on top of my page.
"For that solid, at each x, the cross section
perpendicular to the x-axis."
So our cross sections are going this way.
I'm just going to draw a couple line here.
And they are rectangles with height x.
So this is a rectangle that I'm looking at top down.
And they have heights coming out of the paper at me, right?
But do not evaluate an integral expression that
gives the volume of the solid.
I love drawing these things 3D if I can give it a shot,
but I'm not going to do that in a free response
from the AP exam.
I do not have time for that.
Those shenanigans are for the exploration times.
But I do want to make a note that to find the volume,
I'm going to integrate area, and I'm
going to use a dummy variable of dw from A to B.
And so I always think of these as, how would
I slice out these cross sections?
As I'm coming down, how would I slice them out?
And I would slice them vertically,
but my blade is moving along the x-axis.
And where do I first hit the interval?
I first started hitting it 0, and I slice out rectangle,
slice out rectangle, slice out rectangle until I get to 3.
So they integrate from 0 to 3.
My area of rectangle is based times height.
I know my height is x, and now, I
need to figure out what my bases.
We set he v's on top and m is down here on this lower
function.
And this distance right here is v minus m.
So my base is v of x minus f of x.
So there's my area.
I'm going to bring that into my integrand.
0 and 3 are values of x's, so I have
dx on my variable of integration, right?
But do not evaluate.
I've done what I've been asked to do.
VIRGE CORNELIUS: You're finished.
MARK KIRALY: Yehey.
VIRGE CORNELIUS: Yehey.
MARK KIRALY: And we mentioned this
in one of the other videos.
These are named functions.
I could easily put square root of x plus 2 here minus 2e
to the negative x minus 1.
But as Virge mentioned, then I have
to keep track of my negative signs and all
that because they're named.
I can use their names.
So--
VIRGE CORNELIUS: You could even rename stuff if you wanted to.
If you had like a y equals something and a y
equals something else, you could call 1 f of x and 1 g of x.
Just rename it, and show it, and then
bring it into your work like that if you want.
MARK KIRALY: Absolutely.
A great suggestion.
All right, so for C--
we have this area C. "Write, but do not evaluate,
an integral expression that gives the volume of the solid
generated when C is rotated about the horizontal line
negative 2."
And I am going to draw this one just
to get a little bit of going on.
So I actually have a cheat sheet over here
because I want you to be able to see what I'm trying to draw.
These drawings do not have to be great.
This drawing is not going to be great,
but it's going to be enough for me to just pull off
some information.
VIRGE CORNELIUS: It helps you to think about it,
I think and get into the problem if you can try and sketch.
MARK KIRALY: The square root curve coming off of 2.
This exponential coming off of 1.
That may not be a perfect mirror of C,
but it's a nice approximation.
This is coming in a little bit above 1 or negative 1 actually.
So here's negative 2, negative 3.
Let's see.
There's 2, 1, 2, 3, 4.
Got to go on points here.
So 1, 2, 3 there.
1, 2, 3 is going to mirror that.
And again, this is not a perfect drawing.
It's kind of intentionally not a perfect drawing.
I don't want to spend a lot of time on it.
We kind of get like a bundt cake pan feeling out of this.
VIRGE CORNELIUS: Totally.
Oh, my gosh.
Yeah.
Can you draw your horizontal axis of revolution?
MARK KIRALY: There's our little shaded area, right?
And there's my--
VIRGE CORNELIUS: Yeah, I always put that in there.
Yeah.
MARK KIRALY: Yeah, I usually do too.
I don't know why that didn't happen.
VIRGE CORNELIUS: Hey, Mark.
It should be y equals negative 2.
MARK KIRALY: Oh, well, you know.
VIRGE CORNELIUS: It pretty much-- yeah.
MARK KIRALY: That's why I didn't draw it, I guess.
I don't know.
So what I want to look at-- the way
I want to look at this is I want a single washer,
and this is a washer if you remember that phrase.
And I know I'm usually multi-color here,
and you might not have that option on the AP exam,
but you can use a pen and a pencil
or going thicker with your pencil.
But I get this big R, and I've got this little R.
And for washers, I'm looking at this idea, right?
Looking at [INAUDIBLE].
I want that shaded region.
So my area is pi R squared minus pi r squared,
if you can hear the difference in the pronunciations.
[LAUGHS]
Pi times big R squared minus-- and I factor the pi off.
And again, when I find volume, I want to integrate my area.
And how are we slicing this one?
Again, my blade is a vertical blade
to make these cross-sections come out.
So I'm moving along the x-axis.
So these are going to be dx.
Or you could do more than one thing at a time,
but I'm just kind of piecing my way through this.
I start hitting at 0 again.
I stopped hitting at 3, and now, what's R, right?
So big R is the difference between--
this was v of x--
v of x and negative 2.
So v of x is the distance to the x-axis and then two more.
So v of x minus negative 2 or v of x plus 2.
So that's v of x is the distance here plus the distance
2 back to that line.
And little r is also going to be m of x minus negative 2
or m of x plus 2, I think.
And that's m of x minus negative 2
here, then when we're up above.
It's mx plus 2.
I'm going to insert those minor.
I'm still on screen, Virge?
VIRGE CORNELIUS: Yes, looking good.
MARK KIRALY: And this is going to be the v of x plus 2.
Oh, I need another parenthesis.
So many parenthesis.
I apologize.
Minus m of x plus 2 squared dx.
And did they say not to evaluate?
Thank goodness.
They said do not evaluate.
[LAUGHS]
VIRGE CORNELIUS: "They" is us, by the way.
MARK KIRALY: Yeah, yeah right.
Don't worry about the order on these.
Some people will shift the graph up and do some things.
And you might end up with plus v of x and 2 plus f of x.
That's fine because we're going to square it,
but the outer radius has to be subtracting the inner radius.
That's very important.
You don't want to reverse those.
Pi shows up because these are circular shapes, right?
There was no pi in our rectangle,
but there's a pi for our washer, pi
for our circular cross-sections.
Just remember that you're integrating area,
and a lot of stuff will take care of itself.
Anything you want to add on this one?
VIRGE CORNELIUS: I think you're good.
I like the bundt cake analogy.
MARK KIRALY: [INAUDIBLE]
VIRGE CORNELIUS: Or if that's an analogy.
I don't know.
Yeah, I want to make one now.
MARK KIRALY: Oh, yeah, and one section left.
Part D. "Write a function d of x to express
the vertical distance between m of x and v of x."
We've actually-- I think I've done this a couple of times.
That's the vertical distance between f of x and m of x.
[INAUDIBLE] Something just occurred to me
and didn't before.
VIRGE CORNELIUS: What's that?
MARK KIRALY: I'll say it at the end.
VIRGE CORNELIUS: OK.
MARK KIRALY: Another way I could have written this.
So d of x is v of x minus m of x.
It says, "Find the rate at which d of x
is increasing at x equals 1."
And I can definitely say, as we move along the x-axis,
this is increasing.
They are getting farther and farther apart.
So not a news flash that they're increasing,
but it was nice of them to let me know.
So rate at which this is changing,
what does that imply, Virge?
VIRGE CORNELIUS: That sounds like a derivative to me.
MARK KIRALY: It sounds a lot like a derivative.
And I think you did this the other day.
I'll go ahead and do that left side real quick.
That's d--
[LAUGHS]
VIRGE CORNELIUS: Good job.
Oh, now, we have to take the derivative of a difference.
MARK KIRALY: I'm actually need the expressions.
OK, so square root of x plus 2 minus 2 e
to the negative x minus 1.
All right.
And so you know what?
I'm going to postpone this because--
VIRGE CORNELIUS: You're going to combine like terms?
MARK KIRALY: x to the 1/2.
Yeah, minus 2e to the negative x plus 3.
We've done this before as well when
we cleaned up the other one.
Minus the negative 1 plus the 2 is the 3.
Now the derivatives are a little more palpable.
1/2 x to the negative 1/2 just to get old power rule.
We tested this earlier when we did the antiderivative
because they're so close.
Derivative e of the x is e to the negative x.
2 is along for the ride.
Times that chain plus and plus 0.
And so d find that 1 is 1/2, 1 to the negative 1/2
plus to e to the negative 1.
Perfectly good answer.
If you really want to clean it up,
I guess, you could make it 1/2 plus 2 over e,
but that's how fast we're increasing at 1.
No, it just occurred to me.
The only thing I thought was when you check the derivative,
you wouldn't be able to change anything.
But it just said, "Write a function
to express the vertical distance."
I could have just written m minus v inside absolute value,
but then when I go to take the derivative,
I'm going to have to reconcile that anyway, and figure out
which one's on top.
But for the first part of the problem,
I could have used an absent value.
VIRGE CORNELIUS: Yep.
MARK KIRALY: All right, so our takeaways?
VIRGE CORNELIUS: Yes.
MARK KIRALY: Area or volume?
Integrate, I think.
VIRGE CORNELIUS: Yeah, big application of the integral
is area-volume.
Like huge--
MARK KIRALY: Part D, they asked for a rate.
Then we differentiate it.
I think you saw me do this.
I drew all over this thing.
I drew it.
I redrew it.
Like I said, I would have drawn the rectangle one
if it was in my class and I had some extra time,
but we don't in these situations.
And keep your pencil moving, you know.
I know washers.
A lot of students have trouble trying
to visualize that big R, little r,
but it's just subtract the axis that you're rotating around,
right?
And start writing those things down and get them in there.
It's R squared minus r squared and all that good stuff.
7 out of 8, Virge.
VIRGE CORNELIUS: Yeah, and this area-volume
was a typical free response question.
However, if you take the exam or when you take the exam,
if you don't see area volume on free response,
it was probably had high profile in the multiple choice.
So it will be tested one way or the other for sure.
So it's a good thing you watch this video.
And join us next time because we're
going to have our last session, session 8.
MARK KIRALY: Bye, Virge.
VIRGE CORNELIUS: Bye, Mark.