INSTRUCTOR 1: Hi, Virge.
VIRGE CORNELIUS: Hi, Mark.
Hi, everybody.
Thanks for joining us for session 5.
I'm Virge Cornelius.
I teach at Lafayette High School in Oxford, Mississippi, and--
MARK KIRALY: I am Mark Kiraly and I teach at Ryan High
School in Denton, Texas.
And let's go ahead and find one here
and as Virge said, if you want to pause this and download
and print these and try and work it
or you feel free to follow along and pause.
But we've got let f be the function
with a first derivative of f prime
of x equals 2 times cosine of x squared.
If f of 3 is 7 and 1/2, what is the value of f of 2?
And as we said, this is calculator active.
They've given us f prime, and they're
asking us to find values of f.
So in my mind, I immediately go to accumulation ideas.
And I like the accumulation model
of the fundamental theorem.
I don't know how many of you have seen this in your classes.
You definitely saw it in AP Daily videos,
and you're going to see it right now so--
This idea that if I take the integral
of the first derivative, I'll find out
the total amount of change.
I can add that to my value at a that'll find me the value at b.
And so what I'm trying to find, right?
I'm trying to find f of 2.
I know f of 3.
And Virge mentioned this actually last week
in one of the videos, but this might make you uncomfortable.
But 3 is where I'm starting, and I'm going to 2.
VIRGE CORNELIUS: Well, it's uncomfortable
because the lower bound is higher than the upper bound.
So if you haven't seen that a lot,
it's going to make you squeamish.
MARK KIRALY: So you can always say,
well, I'm going to take f of 3 and I'm
going to walk it back from 2 to 3 using those integral--
definite integral properties.
I trust the math.
I'm just going to do it this way,
but absolutely this is perfectly OK.
Also, a lot of people get here--
if you don't remember the accumulation model,
you can always come in with the definite integral.
And I'm going to spend a little more time on this problem
than some of the others just because there's
so many ways to look at it.
VIRGE CORNELIUS: And also not all students
have maybe seen this before, because it's just,
kind of, a gem of a question.
It's not like you need to practice on a lot
like you do substitution.
MARK KIRALY: Right.
So the fundamental theorem of calculus
is that if I integrate f prime from a to b, that's
equal to f of b minus f of a.
Well, I can put the 2 and 3 there.
Put my 3 and 2 there and then solve for f of 2, so to speak,
in the equation.
I would move that over and I end up
with something that looks similar to one of these,
and that's awesome because the math works.
I'm going to go ahead.
I'm going to take this, I'm going
to enter it in my calculator.
So the 7 and 1/2 is my known f of 3, right there
behind the calculator.
And I'm going to add to that the integral.
So we're going to go under our math functions
and I know where it's hiding.
It's under option 9 but you can scroll down and find it
on your device.
I'm integrating from 3 to 2 and I'm
doing 2 cosine of x squared.
Close my cosine.
Close my integrant, right?
Or do I not-- no, I don't have to do.
I have too many parentheses.
All right.
And with respect to x.
And I look-- I want to make sure I look at this
and I look at this because if they don't match,
we're not going to get very good response, right?
And I see 7.0 and 7.
I bubble C in my answer sheet.
I can write it down, down here, if it
makes you feel like you're completing something.
But for a time element, you don't
have to write it down, right?
We can just bubble our answer and move on.
Virge, you get C when you worked these?
VIRGE CORNELIUS: I did.
And I also want to say that if you
are in physics and your calculators in degree mode,
make sure you put it into radian mode
before you take your calculus exam.
So that's something that is important.
MARK KIRALY: --in front of the test booklet.
So if you're killing time, and they're coming out handing out
tests and they say, hey, read the directions
on the front of the test.
It does say that we use we use radians for our angles.
So this is number two.
The table shown gives values of a continuous function h,
and selected values of x.
Based on the information in the table, which of the following
must be true?
And so we've been told that at x, h of 3 is 8, at x equals 10,
h of 10 is 6 and then at x is 12, h of 12 is 0.
I'm going to go ahead and underline-- this is big.
Continuous is important.
And I have these points.
And that is-- what's the big word in this question that
should jump out at me, Virge?
VIRGE CORNELIUS: Must.
Must be true.
MARK KIRALY: --true, right?
Because they will ask for false.
So be sure you're answering the question that they're asking.
And we're just going to go through.
There's a lot of reading to do on this problem.
These do show up on the calculator active problem.
You may be why?
Well, as Virge mentioned, you get three minutes
per problem in this section.
You only get two in the non calculator portion.
So you get a little more time.
So we can take the time to read these.
H of x is 7 occurs exactly once in the interval of 3, 12.
Between 3, and 12.
So I do see right here, by intermediate value theorem,
there will be a 7 between that 8,
and 6 because we have a continuous function.
But integrating value 3 guarantees what, Virge?
VIRGE CORNELIUS: At least 1.
Not exactly 1.
MARK KIRALY: At least 1.
And this does look like it's descending,
so it might do that.
But just a quick throwaway graph here--
3, 8, 12, 0.
The graph might also do something like this.
I don't know, right?
It could-- 7 could be right there,
and I could cross it three times.
I just know I'm getting from one end
to the other in a continuous curve.
So while it's true that it hits at once, this is not a must.
So a is going to be eliminated.
This says h has a minimum at 12.
Well, the way I drew it, it does, but again--
VIRGE CORNELIUS: Oh, there goes the erasable pen.
MARK KIRALY: They haven't said the function's always
decreasing.
They haven't said anything about a number of critical numbers.
I can draw that and satisfy h maybe.
And so that is going to be--
then it would have a minimum down here.
So this one, I was--
I'm just going to say it's false.
It might be true but I don't want to declare it here.
It's-- maybe I should say possibly false.
But that definitely takes us out of the must be true.
The limit on C, should I deal with that, or should I skip it?
VIRGE CORNELIUS: Yeah, do it.
MARK KIRALY: OK.
So limits-- this isn't the table you typically
see when they talk about limits, but there's
something about limits and continuity that gets
tested on almost every AP exam.
An continuity at a point has three conditions, right?
VIRGE CORNELIUS: Mm-hmm.
MARK KIRALY: H of 10 would have to exist.
VIRGE CORNELIUS: It does.
MARK KIRALY: It does.
It's 6.
The limit, as x approaches 10 of h of x, has to exist.
Sometimes you'll see that second bullet is
the limit from the left has to equal the limit
from the right--
VIRGE CORNELIUS: Right.
Mm-hmm.
MARK KIRALY: That's also saying the limit exists.
And then the third bullet is those two have to be the same.
So the limit is h of--
as x approaches 10 of h of x, that
limit if this is continuous has to equal that function
value of 6, and hey, that must be true.
VIRGE CORNELIUS: Because we were told
this function is continuous.
MARK KIRALY: Because it's continuous, boom.
It's continuous at every point at its domain
and so that is true.
If we want to go ahead and check the third one,
it says h prime of c is 3.
Negative 3.
That really looks like a mean value theorem to me.
And if I check the slope between 10 to 12 and 6 to 0
is in fact-- the slope is actually negative 3.
But mean value theorem requires differentiability.
And if I erase this curve and draw a new version
one more time--
if I draw a zig-zag or something like that,
I might not hit that slope of negative 3 ever.
So while it is likely true, again, it's not must.
So I'm going to stick with C. So--
VIRGE CORNELIUS: I agree.
That's what I got.
MARK KIRALY: --tests.
Look how many things did that test.
It's like-- it tested two theorems, some understanding
of what conditions would have to be there in continuity.
That's a very rich question.
Let f be a continuous function such that the integral
from 1 to 10 is 7 and the integral
of our function from 5 to 10 is negative 8.
What is the value of the integral from 1 to 5 of 2 times
f of x plus 3?
Well, I need this 1 to 5 piece.
I have 1 to 10 and 5 to 10.
What do you think I should do, Virge?
Any suggestions?
VIRGE CORNELIUS: Yeah, use one of those properties
of definite integrals.
So integral from 1 to 5 of f of x, dx--
MARK KIRALY: I'm going to leave of x off--
VIRGE CORNELIUS: Yeah.
MARK KIRALY: --cleaner presentation, but--
VIRGE CORNELIUS: 5 to 10.
And then equals 1 to 10 of f of x, dx.
Yep.
MARK KIRALY: So I know that the interval from 1 to 5
plus the interval from 5 to 10 should equal 1 to 10
if it's a continuous function, which we are.
An integral from 1 to 5.
Well, that's what I'm trying to find.
This is my--
VIRGE CORNELIUS: Mm-hmm.
MARK KIRALY: 5 to 10 is negative 8 and 1 to 10 is 7.
So my integral from 1 to 5 effects with respect to x, add
the 8 to both sides is 15.
VIRGE CORNELIUS: Mm-hmm.
It's not an answer choice.
MARK KIRALY: Not an answer choice.
Good, it's not an answer choice because I want to find this.
So I'm going to break this up into--
I can use those-- it's all these properties, right?
VIRGE CORNELIUS: Yep.
MARK KIRALY: --integral from 1 to 5.
Integral from 1 to 5.
And then this constant-- well, this is a sum--
if that integral is just a sum of parts,
I can factor that multiplier out front.
There's that 1 to 5 f of x, dx I found.
2 times 15 plus--
I could totally thought bubble this as 3x from 1 to 5
and then 3 times 5 minus 3 times 1.
I prefer to think of this as a rectangle.
That is a horizontal line at 3.
From 1 to 5, that is 3 times 4.
And 30 plus 12 is 42.
There you go, Virge.
VIRGE CORNELIUS: Yep, I like it.
I also think of it as a rectangle,
but there's that geometry coming in again
to evaluate a definite integral, which we talked about
in a previous session.
MARK KIRALY: That's true.
All right.
Number four.
Our last one.
And you'll notice that I could've
used the calculator for some arithmetic up there,
but we did really only use the calculator for calculus
on one problem on this.
The temperature in a gym at 6:00 AM, at t equals 6,
is 65 degrees Fahrenheit.
Over the next 16 hours, the temperature changes
at a rate measured by this differentiable function
h, where h is measured in degrees Fahrenheit per hour
and t is measured in hours since midnight.
Which of the following is the best interpretation of 65
plus the integral from 6 to 18 of h of t, dt?
You'll notice that this says 18 and that's a 16
but that is the next 16 hours from 6.
That doesn't mean that function works up until 22 or 10:00 PM,
I guess.
And so this is the hours we're concerned
about inside that interval.
It's a subinterval.
We're fine.
If I haven't thrown away my papers,
that looks a lot like this.
VIRGE CORNELIUS: Oh, it sure does.
MARK KIRALY: Not a coincidence.
Not a coincidence.
H of t is a rate.
In fact, it's the rate at which the temperature is changing.
So this is the total-- so this part right
here, if we integrate temperature change,
is the total change in temperature from 6 to 18.
T equals 6.
But then it's being added to the 65 degrees, which
was our starting temperature.
So-- wow, that does not look like the word start.
That's a little better.
OK, so if we start at 65 degrees and add the temperature change
over the next 12 hours or is that-- it's 12 hours--
14.
18 minus 6-- yeah, OK.
So this is the temperature of the gym in the degrees
Fahrenheit at 6:00 PM.
This feels good, because 12 hours after 6:00 AM
would be 6:00 PM.
Average temperature.
What would average temperature look like, Virge?
VIRGE CORNELIUS: That would look like 1 over b minus a--
well, I mean, with context.
To me, that sounds like average value.
MARK KIRALY: It's actually a different function
even because it's the temperature.
VIRGE CORNELIUS: Mm-hmm.
MARK KIRALY: --would be t.
We have rate of change of temperature here.
It's the wrong format, and we don't even
have a function for temperature.
We could build one but we don't have it.
C-- C is a pretty good wrong answer.
C is just that integral.
See what have just been the integral from 6 to 18.
And D would have been the rate at which the temperature
in the gym is changing.
Well, how do we measure how the rate of the gym is changing?
That's actually h of t.
So that would have been h of 18.
So I'm going to go back and choose A. All right.
So let's go back to our takeaways, Virge.
VIRGE CORNELIUS: OK.
What are they?
MARK KIRALY: --did a good job of illustrating this.
That just because the calculator's allowed,
doesn't mean we're going to use it.
We had a 1 to 4 ratio here.
It's more like-- we said 7 or 8 out of 15 on their actual exam,
but you're not going to use it for every problem.
Take your time to read the problem.
Figure out what is being asked.
We went through these really quick.
If we went too, quick go back, look at some screenshots,
pause it.
Read everything.
It's there for a reason.
And keep your pencil moving.
Don't get stuck staring at it.
Write stuff down.
Mark stuff up.
Underline, circle.
All those good things.
And like Virge said, if you don't know what to do--
VIRGE CORNELIUS: Do what you know.
And Mark does a really good job of crossing stuff
out and moving his pencil and trying-- writing an integral
or whatever.
So that's exactly what we want you to do
when you take that AP exam.
MARK KIRALY: Thank you all very much.
We'll see you tomorrow as we join our free response
questions again.
VIRGE CORNELIUS: Yeah, we're going
to do one on particle motion with calculator actually.
So we'll see you then.
Thanks for joining us.
MARK KIRALY: Bye, Virge.
VIRGE CORNELIUS: Bye, Mark.