calc I 4 1

Here we go.

- Cool; I was gettin' worried there.
- Me, too.

Okay. They're comin' in now.
I've got... 11.

I'll wait until noon, because that's
when I set this one to start.

And, um... then I'll begin.

This time, I didn't need
password to log in.

Good.

All right, I've got Manon; I've got...
Richard; I've got Causey.

Ross. Yeah.

I have Jordan...

Don't have Jordan.

[student] There you go. 15.

[student] Oh my gosh; I was
panicking for the last 30 minutes.

- Oh, well—never panic, because—
- [laughs]

I promise, I—even if there's
nobody here, I can record it...

You know, just by myself.
- Oh, sweet. Okay.

[instructor] I'm kind of waiting on Jordan,
because he was having trouble.

Is that Luke, Luke, Luke...?

Nope.

There's Jordan.

- [student] Finally.
- Heh. Do I have Audrey?

[student] So, what happened there?

[instructor] Your guess
is as good as mine.

Okay.

Um, so, it is 12:01, so I'm
gonna go ahead and start.

And hopefully you printed
out, or, you know,

copied the handouts that
have been on Blackboard.

Since we lost so much time today,

I really think we'll probably
just get through 4.1,

and then try to save the
2.8 until next time,

and maybe I can find a way to
combine 2.8 with 4.2. We'll see.

But 4.1, I'm glad you're here for me,

because this one is a section on
applications. It's called related rates.

And it... can be a little bit challenging;
but it's also really fun.

So let me get my screen-sharing
going for ya, here.

Wow.

Okay. That was a very long delay.

I don't know if it's just that there's
so many people on Zoom, and it's slow?

I don't know.

Hope this works.

So, here's where we are.

Today, I'm gonna definitely
get through 4.1, and like I said,

if I need to try to squeeze 2.8
into Wednesday, I'll do that.

All right. And here's our
handout for today.

So this one is related rates.

Try to work on my... focus.

And this is the famous balloon problem.

If I had a balloon, I would blow one
up for you; but I don't have a balloon.

And I didn't want to go
to the store to get one.

Hashtag: coronavirus.

So this is the famous balloon problem
which we shall just try to imagine.

It says, "Suppose I can blow up a balloon
at a rate of 3 cubic inches per second."

So, [puffing noises] blowing it up;
it's getting bigger.

This is a unit of volume.
So, (dv/dt) equals 3.

It's the derivative of volume
with respect to t,

because it's a rate of change in volume.

Both volume and radius are changing
as you blow that thing up;

the radius is increasing; well,
everything is increasing.

The radius; the diameter; the
circumference; the surface area;

the volume; everything.

The rate at which the radius is changing,
we're gonna call that (dr/dt).

Remember, a derivative
is a rate of change.

Now, at first, the radius grows quickly.

So imagine, you know, when you've got
the balloon, it's about this long.

And you put in that first couple
of [puff-puff-puffff].

And then it grows fast. Like, all of
a sudden, it's a round shape.

But as the balloon gets larger,
the radius grows more slowly.

So imagine you got a big balloon here,
and I put in a couple more puffs of air.

[puff-puff]

You hardly notice any change
in the shape of the balloon.

So, as the balloon gets larger,
the radius grows more slowly,

even though the rate at which the
volume is changing remains constant.

So, it's always 3 cubic inches per second.

It's just that that's more noticeable
when the balloon is this big;

less noticeable when
the balloon is this big.

Volume and radius are related
by the formula V = (4/3) π r³.

That's just the formula
for volume of a sphere.

And A says, "At what rate is the radius
increasing with respect to time,

when the radius is 2 inches?"

Okay. So here's my little handout here.

"At what rate is the radius increasing
with respect to time

when the radius is 2 inches?'

Okay.

So what we need to find, is this.

We need (dr/dt).

Because this is the rate
at which the radius is increasing.

We need (dr/dt).

Okay.

So. I'm gonna take the formula
that I have, which is V = (4/3) π r³;

and we're gonna differentiate
that with respect to t.

Not r, but t.

So, this is an implicit differentiation.

I want it with respect to t.

So I'm gonna find (dv/dt),

and that will be the derivative with
respect to t of the right-hand side,

which is (4/3) π r³.

So what I want you to notice here
is that r is the wrong letter.

t is the right letter;
r is the wrong letter.

So, with this implicit differentiation,

when I differentiate
that right-hand side,

I've got to follow it with an r prime,
'cause it's the wrong letter.

Now on the left-hand side, you know,
we just leave that, (dv/dt),

and here we go on the
right-hand side differentiating.

So we'll do 3 times (4/3) is 4.

And that's π r².

Now, the chain rule says, multiply
by the derivative of the inside.

And that inside is the r.

So since it was the wrong letter,
this is what we'd follow it with; r prime.

Now here's the deal.

So, instead of using r prime,
which is a perfectly fine symbol;

but instead of using that one—

- Professor?
- Uh-huh.

[student] Um, how did you get the r²?
The derivative function?

Minus one?
- [instructor] So, the 3 ti—mhmm.

3 times (4/3) is 4,
π, and then r ⁿ - 1. So r².

[student] Okay. Gotcha.

[instructor] Instead of
using this r prime,

we're gonna use this
notation for the derivative.

The only reason is because with
(dr/dt), it really makes it noticeable

that we're talking about a rate of change
of radius with respect to time here.

And with r prime, that's not
as obvious what the intent is.

So we'll follow this with (dr/dt).

So that's implicit differentiation
involving a chain rule.

Now, what we need is (dr/dt).

Well, we've got an equation here;
we're just going to isolate (dr/dt)

in this equation.

So if I isolate this equation...

Let's see. How can I do this?
I can... ummm... hmmm.

I can first substitute in
place of (dv/dt) the 3,

because I was told that
(dv/dt) equals 3.

So now, 3 equals 4π r² (dr/dt).

And I say we divide everything by the
4πr², and then we've got (dr/dt) all alone!

So we'll say (dr/dt) = (3/4 π r²).

Hmmmm.

What was the r in this problem?

What was the r.

See right here?

- [student] Two?
- When the radius is 2 inches.

So now we'll just substitute in
the 2; we'll have a number.

3 over 4 times π times r².

So, 2 squared is another 4.

This point, I'm gonna go to
my calculator; see what that is.

And we'll clear.

Okay. So I want 3 divided by...

And I'm gonna put that
denominator in parentheses,

so the calculator understands
I'm dividing by all of this stuff.

So that would be a 16π down there.

And I get... (dr/dt) is
approximately 0.0597.

I just chose to round that
to four decimal places.

I need a unit, though.

So this was a rate of change of
the radius with respect to time.

So the unit for radius was...

-Inches per second.
[Prof] -Inches.

And for time, it was seconds.

So, I'll circle this by itself.

So, (dr/dt) equals
0.0597 inches per second.

The rate of change of the radius
with respect to time.

Okay, so that was the first example.

And that's the way all of
these are gonna work.

You're gonna have a formula.

Sometimes you're given the formula,

and sometimes you have to figure
the formula out; that's comin' up.

And then once you get that formula,
you're gonna be doing an, um—

implicit differentiation with
respect to time on all of these.

With respect to time. And then,
solving for an unknown.

Okay, so let's look at part B.
Still about the same problem.

It s-says—"It s-ssays."

It says, "Suppose I increase my
effort when r equals 2 inches,

and begin to blow air into
the balloon at a faster rate.

A rate of 4 cubic inches per second.

Well, how fast is the
radius changing now?"

Okay. So in this case, the (dv/dt)
is now equal to 4.

So, I had (dv/dt) from above,
so I can just write that down again.

Instead of reinventing the wheel.

And it was 4 times π
times r² times (dr/dt).

And now I'm going to substitute
the 4 in place of the (dv/dt).

So we've got that faster rate
of change of volume.

So 4 equals 4 times π.

That radius is still 2, as it says;
when r equals 2 inches.

So that's a 2² (dr/dt), and we just
need to solve for (dr/dt) again.

So, to do that, I'll divide both
sides by the 4π times 2²?

2² is 4.

4 times 4 is 16.

So, there's that 16π again.

So I'll bring out my calculator.

And... turn it on.

Ooh, I got a bad glare on that; sorry.

Okay, so this one'll be 4 divided by.

And then in parentheses, (16π).

That would also reduce to ¼π.

Dun't matter.

At any rate, this (dr/dt) is
approximately 0.0796.

And again, that would be
inches per second.

So you notice, when you
compare the two,

that your radius here is changing
at a faster rate than it did here.

Obviously; because you
were increasing your effort,

and the rate of change in
your volume was higher.

Okay. So, um.

Hold on one second, y'all.

I've got to mute you for a second
because my dog needs to go out.

[exaggerated whisper]
I'm so sorry.

[Student] Hey, I just came back.
My house just had a rolling blackout.

[Instructor] A rolling blackout?

[Student] Yeah, I've been gone
for, like, five minutes. [Chuckles]

[Instructor] Whoaaa.
So, where do you live?

[Student] Uh, right next to the
Crow Bar. On South Congress.

[Instructor] Oh, maaaan...

Okay. That's all we need,
are blackouts.

[Student] Yep.

[Instructor] How fun.

Sorry.

So, now, for part C it says, "At what
rate is the volume increasing

with respect to the radius, when
the radius is 1 inch or 3 inches?"

"At what rate is the volume increasing."

Okay. So, I'm gonna write a little note
here to be careful with this one.

You've got to read it carefully.

"At what rate is the volume."
So, now, underline that.

—"increasing with respect
to the radius."

So, what we want here, is (dv/dr).

This is the rate of change of the
volume with respect to the radius.

We need (dv/dr).

So, we're gonna start with the
formula we were given again.

And that was V = (4/3) π r³.

That was our formula for
volume of that sphere.

We need (dv/dr).

So, if you'll notice. When we get
(dv/dr), uh... r is the right letter.

So this one's not gonna require
an implicit differentiation.

This one's pretty straightforward.

(dv/dr). 3 times (4/3) is 4. Times π.
Times r², and there you have it.

That's the rate of change of volume.

Don't need to do—
following it by a (dr/dt),

because r was the correct
letter in the first place.

So, now we just need to evaluate this when
r is one inch, and when r is 3 inches.

So, let's see if I can get
a little more room here.

There we go.

So, at r = 1, (dv/dr) is equal to 4π.

Okay? If I'm looking for my units here,
this is a rate of change of volume

with respect to the radius.

The units of volume were inches cubed.

The units for the radius were inches.

Now. I don't want you to reduce
that to inches squared. [Chuckles]

Don't do that.

So, this is a rate of change of volume.

So what it says is that
when the radius is 1 inch,

that your volume is changing
at a rate of 4π cubic inches

for every 1-inch change in radius.

So, leave this be; it means something.

It's describing how the
volume is changing

with respect to how
the radius is changing.

Does that make sense to y'all?

[Student] Yeah.

All right. Let's try r = 3.

So, (dv/dr) in this case
would be 4π times 3².

3² is 9. 9 times 4.
This'll be 36π.

And I'm gonna leave it like that.

And my units, again,
are inches cubed, per inch.

Sounds better if I say "cubic
inches per inch," I think.

Okay.

So, there's example one.

And that was—

[Student] Is there a place that we can
get our... or find our graded tests?

[Student] -Like, you have—okay.
[Instructor] -Yes. Yeah.

[Instructor] So, um, when you
go to your gradebook,

and go down to, like, the row
that the test is on...

There should be a place where
you can see my feedback,

and that's where I uploaded
your graded test.

Can anybody else jump in
here; if you found it,

can you explain that
better than I just did?

[Student #2] Just next to the grade,
there's like, a little cloud thing in blue,

which has the comment.

[Student #1] -The little speech bubble.
-Yeah. And you can find there.

[Instructor] -Great. Thank you.
[Student #1] -Yes, uh, thanks.

[Instructor] Sure.

All right; so now, related rates procedure.

So we went through that first
example pretty slowly.

And so now I'm gonna show you;

this is just the general way
we're gonna handle all of these.

So the first thing is, we're gonna
draw a picture if we can.

Uh, I didn't really need to draw
a picture of the balloon problem.

I could have drawn a sphere,
I guess, if I wanted, but.

For some of these, you need
a diagram.

You're gonna need a picture,
and you'll need to label things.

And that's the second point,
is "label and assign variables."

Okay.

The third thing is, write down what
you know, and what you need to know.

So whatever the question's asking,
that's what you need to know.

And then what you know
is usually gonna be a formula

associated with the shape
that you're drawing.

Then you wanna find an equation or
a formula that relates the variables.

So, oftentimes, this is gonna be
a formula for volume, or for area.

It could be the Pythagorean theorem.

Just depends on the picture
that we end up drawing.

And then we're gonna use
implicit differentiation

to differentiate with respect to time.

And then the last thing is just
substitute in your known values,

and then solve for the unknown values.

So, we're gonna follow that pattern
on all of the rest of the problems.

'Kay, next up is the famous
sliding-ladder problem.

And I wish we were in a classroom,
because in a classroom,

I bring in my meter stick,
and pretend it's a ladder,

and then I prop it up against the wall,
and I pull it out slowly from the bottom,

and watch it slam down on the floor.

So, you know, the ladder
is sliding down the wall.

And when you see it in class, I just
think this makes a little more sense, but.

Darn it!

So this is the famous sliding-ladder problem.

Says, "A 10-foot ladder
rests against a wall."

So I'm just imagining a ladder
propped up against a wall.

If the bottom of the ladder
slides away from the wall

at a rate of 1 foot per second—

so that's steady, constant pulling the
bottom of that ladder away from that wall—

—"how fast is the top of the
ladder sliding down the wall,

when the bottom of the ladder
is 6 feet from the wall?"

So the first thing that we talk about is,

when that ladder is
propped up against the wall,

and you're pulling the bottom
of the ladder, pulling it out slowly,

the top of that ladder
is also falling down.

But would it fall at the exact same rate

at which you're pulling the
bottom of the ladder away?

I mean, it's all one ladder.

So, this is my ladder.
And this is my wall.

And I'm pulling the bottom away.

It seems like whatever rate
I'm pulling it away,

that the top should slide
down at that same rate.

But if you think about it...
I mean, really think about it.

If there really were a ladder there, and
you had a string tied around the bottom,

and you're pulling it out,

it's gonna slide down
the wall slowly at first,

but what happens when
it gets close to the floor?

[Student] Speeds up.

[Instructor] Yeah man, that thing
is gonna smack the floor so hard,

it's gonna damage the floor!
Unless it's on a carpet.

So, what really happens is,

even though we're pulling the
bottom out at a constant rate,

the rate at which the top is
sliding down is increasing.

Craziest thing about physics.

So let's try to draw this.

Says, "Draw a picture if you can."
That's the first bullet.

So when I draw my picture,
I'm just gonna draw a wall.

And then, here's my ladder
propped against the wall.

This is the floor.

Okay.

So, that ladder. One thing
I know about it is that it's 10 feet.

So I can label that "10."

And of course, you notice,
I just drew a triangle.

So, the hypotenuse of that
triangle is definitely 10.

Now, I can think of this as
being in a coordinate system.

And when I think of it that way,
then the base of this triangle is x,

and so what x really represents
is the distance from the wall.

And then y can be... Oh.

The height of the wall
where the ladder meets.

So, x is the distance from the base,

and then y is the height of the
ladder propped against the wall.

So, I've labeled it with what I know.

And I've assigned variables
to what I don't know.

Um. There is something else I knew.

It says, "If the bottom of the ladder
slides away from the wall

at a rate of 1 foot per second."

So, here's the bottom of the ladder.

It's being pulled away from the wall
at a rate of 1 foot per second.

That's one of our rates. And x is
the thing that's changing there.

So the distance from the base
is changing. That's actually (dx/dt).

So I'm gonna write down what
I know; that (dx/dt) equals 1.

So I really know two things about this:

I know the length of the ladder.
And I know (dx/dt) is 1.

So I wrote down what I know
and what I need to know—

Oh no, I didn't. What do I need
to know? What's it asking for?

How fast is the top of the
ladder sliding down the wall.

Well, as the top of that
ladder slides down—

[Student] (dr/dt).

[Instructor] True. It's y that's changing.

So what I want to know, what I
need to know, is (dy/dt).

Which is why it's called "related rates."

You're gonna have multiple
rates in the same problem.

(dx/dt) is given to us as 1.
We wanna find (dy/dt).

So now, the next thing says,
"Find an equation or a formula

that relates all the variables."

So, back to our draw-ring.

We have a 10, an x, and a y.

What's a formula you know
that relates these three numbers?

[Student] Pythagoras theorem.

[Instructor] Thank you, Pythagoras.

Pythagoras makes our lives easy.

"Py-tha-gor...as."

So thank you, Pythagoras.
And here's your theorem.

It says that x² plus y²
is equal to 10².

Okay.

So that's our equation.

And that's the thing
we need to differentiate.

So once we have that equation,

we use implicit differentiation to
differentiate with respect to time.

So, that means on the left-hand side,

we want the derivative with
respect to t, of x² plus y².

On the right-hand side, the derivative
with respect to t of 100.

So now on the left,
we're gonna split it up.

We want the derivative of that sum.

So I'm gonna write it as: the
derivative with respect to t of x²

plus the derivative with
respect to t of y², equals.

And then, what is the derivative
with respect to t of 100?

What is that?

[Student #1] -Zero.
[Student #2] -Zero.

It's the constant, so we're gonna
get a zero on the right-hand side.

Now, on the ladder problems, when
you know the length of the ladder,

you'll have the constant on that
side of the Pythagorean theorem,

and that derivative is
always going to be zero.

Now, on the left,
we need to differentiate.

So getting the derivative
with respect to t of x²?

x is the wrong letter.

So, we'll do our 2x, all right, but then
we've got to follow it by... (dx/dt).

And that's the chain rule.

Since x is the wrong letter, 2 times
x is the derivative of the outside.

This is the derivative of the inside.

Now for the y² term?
Same thing.

y is the wrong letter.

So we'll do 2y, followed by
(dy/dt) is equal to 0.

Great.

So now that we've differentiated, we're
gonna sub in the things that we know.

So, what do we know here?

Um, let's see. It says,
"10-foot ladder"...

A rate of 1 foot per second;
that was (dx/dt).

And we also are stopping this;
we're looking at this

when the bottom of the ladder
is 6 feet from the wall?

Okay. So it's 6 feet from
the wall right now. That's x.

So I'll say 2 times 6 times (dx/dt),
which is... what, now?

[Students] -One.
[Instructor] -One. Yeah.

And then plus 2 times yyyy.

[Student] You can find that
using Pythagoras' theorem.

[Instructor] Exactly right.

So, 2 times y, and then
times (dy/dt) equals 0.

We don't know what y is,
but we can find it.

So I'm gonna go back over here,
and rewrite my Pythagorean theorem,

which is x² + y² = 10².

I know what x is; x is 6.
So this is—

[Student] -Professor?
[Instructor] -Yes.

[Student] So the 2 times 6 times 1.
Is the 1 a derivative of the x?

[Ins.] Yes. That was the rate of change
of x with respect to time. It was a 1.

[Student] Okay.

So, down here, 6² + y² = 100.

That's y² = 100-36.

y² is equal to... uh...

Make that 64.

And y must be 8.

Negative-8 wouldn't make sense,

so we're going with the
positive square root of 8.

So now, I can plug in that unknown.

And this is somethin' that
commonly happens.

So, once you've got your equation,
you do your implicit differentiation;

you fill in the stuff you know.

A lot of times, there's another unknown
variable that you've gotta go find,

but you will be given the
information to find it,

and it's usually from your formula.

So you'll plug something in
to find something else;

then you can sub it all in, and finally,
just be left with that one unknown,

which is (dy/dt),
and that's what we want.

Well, 2 times 6 is 12.

12 plus 2 times 8 is 16;
times (dy/dt) equals 0.

Let's try to isolate (dy/dt),
so I have 16 (dy/dt) = -12;

and the last step is just
dividing both sides by 16.

(dy/dt) is equal to -12 over 16,
and that is negative...

-It's like, three-fourths?
[Student] - Yep.

So negative 0.75.
And then our units for this,

since this is a change in y
with respect to t,

is gonna be feet per second.

The units of y were feet;
the units of time were seconds,

so in (dy/dt), units are feet per second.
Man, I've almost run out of—

[Student] Would you have a
preference on fraction or decimal?

[Instructor] Oh no, I don't. Nah.

To me, on these kind of problems,
though, the decimals...

I guess I like 'em better because
I can imagine that better.

Like, I have an idea of -0.75 feet
per second, but -3/4—

Well, I guess it wouldn't matter.

I don't care.

Whatever makes you happy.

Okay. Now, that was the famous
ladder problem, and—

[chuckles] because, in every calculus
book since the history of calculus,

there has been a ladder problem.

And you will have more ladder
problems in your homework.

And you will most likely have
a ladder problem on your next test.

[Whispers] It's famous.

Okay, the last question says,
"How fast is the top moving down

when the ladder is
9 feet from the wall."

How about 9.9 feet.

How about 9.99999999999 feet?

So in other words:
the ladder's only 10 feet.

So, when you're pulling it out.
When it's 9 feet—

I mean, most of the ladder is down.
It only has another foot to fall;

so we're looking at the speed
at which it's falling at that point.

Okay.

So, let's go back to when x equals 9.

Because we need to figure out
what y is at that point.

'Cause, you know, if I'm
drawing a picture of it...

It now looks like that, right?

So it's almost all the way on the ground.

So when x is 9, let's
figure out what y is.

So using our Pythagorean
theorem, x² + y² = 10².

That is, 9² is 81.

Plus y² equals 100.

So then y² is 100 minus 81,
which would be 19,

and y will be the square root of 19.

So then, we'll go back to
our "(dy/dt) equals."

And our (dy/dt) was...

Oh, man. Do I have to reinvent that wheel?

Shoot. I do.

So, (dy/dt) would equal. I'm gonna go
back to this step so I can isolate (dy/dt)

before I've substituted in
numbers for x and for y.

(dy/dt) would be...

Will be -2x times (dx/dt),
and then that would be divided by 2y.

Think I got it.

(dy/dt) would be -2x(dx/dt) when
you subtract this from both sides,

and then to isolate the (dy/dt),
you're dividing both sides by 2y.

So. It looks ugly, but
this is what it looks like.

Now we'll substitute in
our new information.

So, our new x is a 9.
So this is -2 times 9.

(dx/dt) is still 1.

And then 2 times y would be
2 times the square root of 19.

Now, I plugged all that into my
calculator already,

and that was approximately
-2.06 feet per second.

So it sped up. Remember when it was
6 feet away,

the speed at which the top was falling
was -0.75 feet per second.

Now, it sped up to -2.06
feet per second.

[Student] -Professor?
[Instructor] -Yes, go ahead.

[Student] The 9.9, did you round it up?

[Student] The 10? The 10²?

[Student] Know when it says
x + y² = the 10². Is it from the—

[Instructor] Yes.

[Student] But it's a question,
or you just rounded it up?

[Instructor] So this is still going back
to my Pythagorean theorem.

[Student] Oh, okay.

[Instructor] I still have a hypotenuse
of 10 there; the base is 9.

And we were looking for y.

-Gotcha.
-Yeah.

Turned out to be...
the square root of 19.

That fits in there.

So then I would do it again for
9.9, and then for 9.9999999...

I don't have room, so I'm
gonna talk you through it.

So, when you get to 9.9. That
ladder's almost all the way down.

When you go through and calculate
the rate of change of y with respect to t,

when x is 9.9, your rate is then...

Uh, -7 feet per second.

When you go to 9.9999999,
it's approaching infinity.

It is negative, but so large,
it's incredible.

So, as it's slamming the floor, the rate
at which it's slamming the floor?

That rate is approaching infinity.

Can't make this stuff up.
It's really true.

That's why it damages the floor.

It's pretty darn fast.

All right, and that is another
famous sliding-ladder problem.

We'll take that one away.

And now I'm lookin' at
number 10 from the exercises.

This one's comin' up next.

Probably better also check what
time it is. 12:34? We're good.

So exercise 10 says: "A particle
moves along the curve; y = √(1+x³)."

"As it reaches the 0.23, the
y coordinate is increasing at a rate

of 4 centimeters per second."

That's (dy/dt).

"How fast is the x coordinate of
the point changing at that instant?"

Okay. So here, the graph that we draw
is the graph of the function.

So the curve is y = √(1+x³).

That's the graph we want to draw.

So I'm gonna draw my
coordinate system here.

Like so.

And I graphed this on a graphing
calculator earlier to see what it looks like;

and you don't have to be exactly right,
but it looks something like that.

And then this point, I'm gonna
label this point right here at (2,3),

because the particle is moving along,

and at some point, it's
gonna reach that point.

Particle's moving along the curve.
As it reaches the point (2,3),

the y coordinate is increasing at
a rate of 4 centimeters per second.

So we know that (dy/dt) equals 4.

The question is, how fast is
the x coordinate of the point

changing at that instant?

So, what we want is (dx/dt).

We know (dy/dt);
we want (dx/dt).

So if I look at, you know, my little
bullets, and see where I'm at.

I drew a picture.

It says, "Label and assign variables."

Well, I guess I kind of did.
I've got the point labeled,

and I wrote down
what (dy/dt) is, and...

I wrote down what I don't
know, which is (dx/dt).

So then I find an equation or formula
that relates all of these variables.

Well, that equation or formula
is the y = √(1+x³).

That's relating x and y.

We wanna use implicit differentiation
now to differentiate with respect to time.

And then, we'll substitute in what we
know; solve for what we don't know.

So now I need to find the
derivative with respect to t.

So I want derivative with respect
to t of the left-hand side.

I want derivative with respect
to t of the right-hand side,

which I'm going to rewrite
as (1+x³) to the ½ power.

Just makes it easier
for me to differentiate.

So now on the left, it is just (dy/dt).

And then on the right, derivative
of that (1+x³) to the ½.

So bring my ½ down in front.

(1+x³) to the -½ power.

Now multiply by the
derivative of the inside.

Okay, now. Your inside is this (1+x³).

Derivative of (1+x³) is...

3x².

But now, chain rule says,
"Do it again"; it's a double chain.

Now we need to multiply by the
derivative of x with respect to t.

Because x was the wrong letter.

t's the right letter;
x is the wrong letter,

so I've gotta follow it
with that (dx/dt).

Now I've differentiated implicitly;
now it's time to sub in what I know.

So, I do know that (dy/dt) is 4.

That's 4 equals ½ times 1 plus...
What's x at this point?

[Student] -2.
[Instructor] -2.

So that's a 2³, to the -½.
And that's times 3 times a 2².

And then that's times (dx/dt).

(dx/dt) is the unknown;
that's what I need to solve for.

All right, so this is 4 equals.

Ummm, 2³ is 8.
8 plus 1 is 9.

9 to the -½, so it's like 1 over
the square root of 9, is... 3, I think.

So this would be 1 over, 2 times...

8+1 is 9; square root of that is 3.

So that's 1 over 6.

And then 3 times 2²;
that's 4 times 3; that's 12.

(dx/dt).

So, this is 4 equals. 2 goes into 12
six times; 6 over 3—

that's just a 2 times (dx/dt).

I think I'm ready to isolate my (dx/dt)
by dividing both sides by 2.

And (dx/dt) is 4 over 2, which is 2.

And the rate is in centimeters per second.

Okay. So sometimes, I guess, solving
the equation after you substitute in

your known values can get
a little tricky, but you know;

just take it one step at a time,
and you'll get there.

So let me know how that one went.

[Student] Can you just go over
what happened to, uh, 12?

[Instructor] Yeah, sure.

So, the ½, times the 12? Is 6.

So I just canceled the 2 with
the 12, leaving me a 6 on top;

but 6 over 3 is 2.

Good?

[Student] Yeah.

All right. So, you guys are so quiet.
I don't—I don't like that about Zoom;

it's different than being in a classroom;
in a classroom, you know...

We can see each other's eyeballs,
and you can just ask a question;

or sometimes I'll look at you,
and I know you have a question,

and I'll say "What's up."

Um, jump in there; really. Stop me
any time you wanna stop me.

Don't be shy or embarrassed about it.
Stop me, and ask your question.

Because the most important thing
is that you guys continue to learn.

Exercise 4 says, "The length of a
rectangle is increasing at a rate

of 8 centimeters per second."
Got a rectangle.

"And its width is increasing at a rate
of 3 centimeters per second."

"When the length is 20,
and the width is 10,

how fast is the area of
the rectangle increasing?"

Okay. So the first thing we're
gonna do? Draw a picture.

So, I've got a rectangle here.

Here we go. And I'm gonna
label this thing.

So it says, "The length of the rectangle
is increasing at a rate

of 8 centimeters per second;

width is increasing at a rate of
3 centimeters per second."

"When the length is 20,
and the width is 10,

how fast is the area of
the rectangle increasing."

So, like, right now, the area is
20 times 10, or 200, but.

We're gonna be increasing the length
and the width,

and looking at how fast
that area is changing.

So I'm gonna write down the things
that I know. I've given a lot in this problem.

It says the length is increasing at a rate
of 8 centimeters per second.

So, that would be the derivative
of l, with respect to time.

That is 8.

It says, the width is increasing at a
rate of 3 centimeters, so.

dw, the change in width,
with respect to time. That one is 3.

We know that we're kind of stopping this
when l is 20, and when w is 10.

So, there are four things that I know.

What do I not know? What do I need.

I need, or want to know...
how fast the area—

[Student] -(da/dt)?
[Instructor] Yeah. How fast the area.

Derivative of area with respect to time.

I need the rate of change of
the area with respect to time.

So, if I'm looking for the
rate of change of area,

then I want to use
the area formula here.

Area of a rectangle?

Length times width.

So there's my formula
relating all of my variables;

it's time to differentiate implicitly.

So now we'll get the derivative
with respect to t of the left-hand side.

And the derivative with respect
to t of the right-hand side.

Now on the left, there's your (da/dt).

This is the very thing we're lookin' for.

So then on the right, we need to get
the derivative of length times width.

So I said it: length times width.
This is a...?

[Student] -Product rule.
[Instructor] -Product rule.

So we want the first function, l. Times
the derivative of the second function.

Okay now, remember: t's the right letter.
Everything else is the wrong letter.

So when I do first times
the derivative of the second,

I don't know what the
derivative of the second is,

so I have to write (dw/dt).

So, l times (dw/dt).

And then plus the second, which is w,
times the derivative of the first,

which has to be (dl /dt).

So now I'm gonna go back up here,

where I was given these
four pieces of information.

I'm gonna substitute them in.

(da/dt) = l, at this moment
in time, is 20.

(dw/dt) is 3.

Plus w at this moment is 10.

And (dl/dt) is 8.

Okay, this one's gonna be easy.

Don't have to isolate anything;
just multiply and add.

So, that's gonna be 60 plus 80,

and 80 plus 60 would be 140.

Now, units. What are the
units for the area?

[Student] Centimeters
squared per second?

[Instructor] Mhm.

So, units for area are
centimeters squared.

The units for time are second.
So it says that,

at this point in time, when our
rectangle is this big, and it's increasing?

That the rate of change in the area is
140 square centimeters for every second.

Okay. Almost done.
That was an easy one.

Maybe we should've done that one first.

Okay. The last one on this handout,
I believe. Exercise 32. Exercise 32.

Oh, but this is a good one.

So, exercise 32 says, "Two sides of
a triangle have lengths 12 meters

and 15 meters."
Two sides of a triangle.

It didn't say a right triangle.
Just said "a triangle."

"The angle between them is increasing
at a rate of 2 degrees per minute."

"How fast is the length of
the third side increasing

when the angle between
the sides of fixed length is 60?"

[Exaggerated shriek]

If it were a right triangle, this
would be so much easier to draw!

But it didn't say that; and it's not;
and it's changing; so, man!

Let me just go for it.

So I'm gonna draw a triangle.

Maybe something like—I'm gonna
make this pretty big. [Chuckle]

Something like that.

And your triangle doesn't have to
look exactly like mine, but.

I'll be danged if that doesn't
look like a right triangle.

That looks like a right angle right
there. I just couldn't help myself.

It's not. Not a right triangle.

So I'm gonna label my sides.

I'm gonna call that one 12, and
this one 15, because it looks longer.

And then there's an angle between them,
and that angle between them we'll call θ.

So, if that's θ, and here
are the two sides.

What's happening is,
that is opening up.

So as that opens up, we're looking to
see how that third side is changing.

It's obviously growing; it's getting
longer. We're looking for that.

"How fast is the length of
the third side increasing

when the angle between the sides
of fixed length is 60 degrees."

So guess let's start writing down
the things that we know here.

So, we know that two sides of the
triangle are 12 and 15... OK, got that.

The angle between them is increasing
at a rate. Ah. This is a rate that we know.

And it's the rate of change of that
angle with respect to time.

So, we know (dθ/dt). (dθ/dt).

And that is a rate of 2 degrees
per minute.

So (dθ/dt) is 2.

"How fast is the length of
the third side increasing

when the angle between the
sides of fixed length is 60?

So, it's telling us that we're kind of
stopping this,

looking at when that angle
is 60 degrees right then,

how fast is the third side changing.

Well, we need to give a name
to that third side.

Hmm, I don't—what do you wanna
call that third side? Anybody?

Any variable?

[Student] -x.
[Instructor] -Why not?

So we'll call that third side x.
Works for me.

Now we need a formula.

We need a formula that relates
what's going on here.

So, look at your picture.
Your knowns; your unknowns.

Does a formula come to mind—

and it cannot be Pythagoras,
because this is not a right triangle.

[Student] This is the
double-angle thing? I mean...

It's sine over hypotenuse
equals sine over hypotenuse?

[Instructor] Not that one.

[Student #2] Is this a sine-angle-sine
problem? Or a side-angle-side problem?

[Student #3] -Is it the law of sines?
[Instructor] -Yes. Yes, it's SAS.

So, think trigonometry.

[Student #3] It's not the
law of sines or anything, is it?

[Instructor] Keep thinkin'. You're close.

[Student] Law of cosine?

[Instructor] Yeah, that might
help if I write that in there.

So when you know two sides and the
included angle, that's a law of cosines.

And we know two sides. And we know
the included angle at this moment is 60°.

So definitely a law of cosines.

Yayyy, I love the law of cosines!

That part of trig was so fun; solving
for the triangles using the law of sines

and the law of cosines.
I loved doing those problems.

Remember the vector problems?
They were great.

Okay, now, what does the
law of cosines say?

Well, the law of cosines says this:

That your side opposite,
which we're calling x.

We're gonna square it.
x² is equal to.

And it's the sum of the squares
of the other two sides;

so it starts out kind of looking
like the Pythagorean theorem.

But then it's minus 2 times a
times b times the cosine of θ.

This is the law of cosines.

So that's our formula relating everything.

Ummmm, what do we next?

Implicit differentiation.

So, we want derivative
with respect to t.

Of the left-hand side, which is x².

And the derivative with respect
to t of the right-hand side,

which is (a² + b² − 2abcosθ).

Well, I say before we get this derivative,
maybe we substitute in what we know,

with the sides 12 and 15?

Ummm, I can do that.

Or not; I don't have to.
I can live with it like this.

Y'all, give me a preference.

Do you want me to substitute in those
numbers now, or get the derivative first?

If I get the derivative first, you know,
these will just be zeros,

because there are only constants.

Weigh in with your preference here.

[Student] Put the numbers?

[Instructor] -Put the numbers in?
[Student] -Yes.

So, derivative with respect to t of x²
equals the derivative with respect to t;

and we'll put those numbers in.

So, the a? I guess I'll just
call a the base; 15.

That would be 15². Plus the other
side squared; so that's 12².

Minus 2 times 15 times 12.
Times the cosine of θ.

And then we can clean that up a bit.

This is the derivative
with respect to t of x².

Notice I'm not taking the derivative
yet; I'm just cleaning this up a bit.

Equals derivative with respect to t of.

If I do 15² + 12².

Go into my calculator here.

15² plus 12².

Okay. That is 369.

So that would be 369 minus.

Now, the 2 times 15 times 12?

That is 360.

Sitting in front of the cosθ.

Okay. Now, let's differentiate.
Let's do it now.

So then on the left-hand side,
remember that x is the wrong letter.

So when I get the derivative of x²,
it's 2x, but follow it by...?

(dx/dt).

On the right-hand side.

The derivative of 369 is just 0,
so we won't worry about that.

So now let's look at the
derivative of -360cosθ.

Well, that constant in front
just hangs out.

What's the derivative of cosine?

[Student] Negative-sine.

[Instructor] So since it's
negative-sine, then we can do...

That.

So 360sinθ.

Now, θ is the wrong variable,
so what do we follow this by?

(dθ/dt).

And that's the chain rule.

So if you have a cosθ,
derivative is -sinθ (dθ/dt).

That's the derivative of the inside.

Okay, great.

So now we're ready to substitute in
things that we know;

and we're solving for...

What are we solving for? I didn't
write down what we needed to know.

We need...

And it says, "How fast is the length
of the third side increasing?"

We need (dx/dt), the rate of
change of x with respect to t.

Okay, got it. So I'm solving for (dx/dt).

Well, then on the left-hand side,
I'll have 2 times x.

Ummm. How are we gonna find x here?

How we gonna find x.

I'm bringing my picture right down
in front of your face, there.

If I'm looking for this side...

[Student] Is it a [inaudible] equation?

[Instructor] We're gonna plug it
into the law of cosines

to find out what this
third side would be

when the two sides are 12 and 15,
and, at this moment, that angle is 60°.

So we're going to go back to the law of
cosines just to determine this unknown.

Remember, we had to do this
before on one of the ladder problems.

Okay. So then, using a law
of cosines, it would say...

I'll try to do this over here on the side.

It would say that x²
is equal to a² + b².

So that's 15² + 12² again.

Minus 15 times 12 times the 2;
times the cosine of 60°.

So our x² equals.
That 15² +12² ?

That was the 369.

And then 15 times 12 times 2,
that was the -360cos60°.

So x² is 369 minus 360 times.

And the cosine of 60°
is one that we know.

[Student] -One-half?
[Instructor] -Is what?

[Student] -One-half, I think?
[Instructor] -One over two.

One-half is right.

So this is ½.

x² is 369 minus...
I guess that'd be 180?

And then 369 minus 180 is—
[goofy voice] I unno.

[Instructor] -I got—
[Student] -189.

[Instructor] 189. So x would be
the square root of that.

Which is not real pretty;
it's 13.7-ish.

So I'm just gonna leave it at 13.7.

More decimal places would be better;

but I kinda messed myself up by not
giving myself very much room to write

any number in here at all, sooo,
I'm gonna have to just round it off.

So 13.7.

And that equals the 360
times the sine of θ.

Oh, but the sine of θ is the sine of...
60°, times (dθ/dt), which was 2.

Hold on. I can fix this.

My Calc 2 student told me on Wednesday,
"So why don't you just use a pencil?"

"Then if you mess up, it's no big deal!"

Well.

I've got a... Wite-Out tape here.

So let's fix all that.

Like, really? You're not gonna work?

[Cries] Why is my life so hard?!

All right. So I'll just rewrite it.

2 times 13.7, times (dx/dt).

Equals 360 times the sine of 60°;

times (dθ/dt), which was 2.

There we go.

Now, 2 times 13.7—aw, heck.
You know what I'm gonna do?

Say (dx/dt) is equal to.
2 times 360 would be...

720. Sine of 60°.
I know that one, too.

That would be...

[Student] -Square root of 3 over 2.
[Instructor] -Square root of 3 over 2.

And then, let's divide that
by 2 times 13.7.

All right, I'm going to my
calculator to figure this one out.

720. Times the square root of 3.

Divided by 2.

And then that divided by
2 times 13.7.

Hey y'all; if I didn't fat-finger this,
I got approximately 22.7.

And now I need a unit for that.

This was a rate of change
of x with respect to time.

And x was measured in meters,
and time was measured in minutes.

So, 22.7 meters per minute.

Okay.

[Student] Umm.

[Instructor] Yes?

[Student] -Oh, nothing; I just said "Wow."
[Instructor] -Oh, okay. Wow!

So, yeah. This was a pretty
challenging section; but also doable.

So if you look at those bullets,
draw the picture; label;

write down what you know and what you
don't know; what you need to know.

Find a formula that relates everything.

If you try to go through that
step-by-step, I think you'll be just fine.

And I tried to pick problems—
most of them—

are like the ones that we did...
in class today? On Zoom today?

And so, hopefully you'll have an example
for almost everything in the homework.

But definitely stop by
during office hours; um...

I've got that all figured out now,
and I'm in Blackboard from 10 to 11;

so, you know, before your
class for an hour, stop in.

Or this afternoon from 3:45 to 4:45. I'm
on Blackboard again, and Collaborate.

So, there's a link in your Blackboard.
Just go to it; and I'll be there,

and I can help you with
homework problems, so.

Especially some of you who were used to
coming by when we were still at Hays.

Come on by! I wanna still be able
to help you, even though it's not...

quite as effective this way?
It's still better than nothing.

Umm. And then the tutoring labs,
the learning labs,

have gone online using Brainfuse.

They were supposed to send out
an email about that.

I never got one. I'm hoping
that the students did.

Somebody let me know if
you've got anything about that?

[Student #1] -I didn't.
[Instructor] -Ahh.

[Student #2] I think I remember seeing
something about [inaudible]...

[Student #3] So, I've been
talking to a tutor. Um.

And we've been meeting on Zoom.
But the way that she sees it

is that it's almost like a ticketing system,
kinda like how Highland works now;

where you send in a ticket
for a singular question,

and then they can reach
out to you and help?

[Instructor] Do you do it
from the website?

[Student #3] Yeah, I believe so.

There was a post on it on the front
page; I don't know if it's still there.

[Instructor] Oh, okay. So, maybe
just go to austincc.edu.

And if there's not anything
on the front page,

maybe do a search for "learning lab,"

and then hopefully their page
will come up with information.

I did not recieve anything about it.

I just—I heard from someone who works
there that they were gonna do Brainfuse.

So yeah, the ticketing system,
that would be okay, I guess;

and just kinda wait until it's your turn.

[Student #3] What problem set
are we working on?

[Instructor] Sorry?

[Student #3] What problem set are we
going to be working on, for next class?

[Instructor] Oh, um. So this is
Section... uhh, what is this? 4.1.

So you'll have homework from 4.1.

And then on next class,
we're gonna try to do...

2.8, which will go fast.

And then, 4.2.

We'll try to do two sections. We'll see.

[Student] So the homework
is 4.1, right? And 4.2?

[Instructor] -Yeah.
[Student] -Okay.

[Student #3] Wait, so um...

We're doing the homework
for 4.1 and 4.2 for next class?

-We're not doing the—
-[Instructor] No-no-no-no-no, no-no.

So, all you need to be working
on right now is 4.1.

[Student] Okay. So we'll just
do the normal homework.

So we're not doing problem
sets... between Mondays

-and Wednesdays anymore?
-Oh, oh; I see what you're saying.

So I'm not giving you a problem set
this week, because you just had a test.

So I don't really have anything
to problem-set you over.

-I appreciate that.
-[laughs] You're welcome.

But we will next Monday. Next
Monday, you'll get a problem set.

Um, I was gonna say that in my
Calc 2 class, some of those students

are also meeting on Zoom,
to work on homework together.

So, they had study groups going, and
they're just keeping those going on Zoom.

So, I'm gonna throw that out there.
If any of you guys had study groups.

You know, continue to do that.

[Student] Um, before we go.

Is this meeting going to be
posted in Recorded Meetings?

[Instructor] Yes, it is.
It just takes a while.

So, once I'm finished, it has to convert
it, or something? I don't know.

And that can take hours.

So, hopefully... hopefully by
tonight I'll have it posted?

But in the morning, as a last resort.

[Student] Okay. Heard.

Okay. Well, so, we are early.

And I'm gonna let you go; so if you wanna
go, just go ahead and exit the meeting.

I'm gonna just stay here for a minute,

in case anybody wants to
talk or ask me a question.

And if you're leaving, bye—

[Student] -I actually have a question.
[Instructor] -Sure. You can hang out.

So, I actually was struggling with the
particle moves along a curve equation.

[Instructor] Okay.

I would just like you to break down
a little bit more what you did,

um, a couple steps through it.

[Instructor] Sure. Let me turn
my screen-sharing back on.

It's really taking forever.

[Student] Uh, you said this is
already recorded already, right?

[Instructor] Mhmm.

[Student] How do we go to view it?
Do we just go on Blackboard, then...

[Instructor] Yeah; so. Once it's ready
to post, then you'll see a link.

I think the link's already
there, right? That says...

Zoom Recordings, or Recorded
Meetings. I forgot what I called it,

but. "Recorded" is in it.

So you'll just click there,
and then you'll see them.

[Student] -All right. Thank you.
[Instructor] -Sure.

Okay. So here's my particle problem, again.

So the deal was, um.

Little dude is moving.
Little particle is moving.

And the curve, the y = √(1+x³)?

That is the formula that relates
your variables, x and y.

And then, let's see.

The y coordinate was increasing
at a rate of 4; so as it's moving,

the rate of change of
the y coordinate is 4.

And what we wanted here was the
rate of change of dx coordinates.

So we want (dx/dt).

Okay, so my formula was y = √(1+x³)?

And one thing that's nice
about these problems is

you don't have to find the formula,
or think about it,

or figure out what it is, because
it's just handed to you.

It's whatever the equation is.
It's kinda nice, really.

So we're always differentiating
with respect to t;

so I wrote (d/dt) of both sides.

But what I did over here was,
I just rewrote it in the form of

a rational exponent, because it makes
it easier for me to differentiate.

So my left is (dy/dt).

And... here's differentiating
on the right-hand side.

½ down in front. Rewrite 1 +x³.

Decrease by 1, so -½.

This is the derivative of the inside.

The inside is the (1 +x³).

But the derivative of (1 +x³)
is 3x²(dx/dt).

Any time it's the wrong letter,
gotta follow it by that (dx/dt).

And then I just substituted
all of the stuff in.

(dy/dt) was 4.

½ times 1 plus.

x was 2, because that's the
point we're kind of lookin' at here.

Here's 3 times 2²;
and then (dx/dt).

And then I did a bunch
of stuff in my head.

Maybe that's where...

[Student] That's the part
I had a problem with. Yeah.

[Instructor] Sorry about that.

I could see that I didn't have much
room left; and so, that's why I did that.

So the 2³ plus 1 is 9.
But it's 9 to the -½.

That's 1 over the square root of 9.

So that's ⅓.

And that's why I wrote this 3—
oh, it paused on me—

That's why I wrote this 3 down here.

So my ½ is one over two.

This, when I bring it downstairs, is 3.

And then, 2² is 4. Times 3;
there's the 12.

And then (dx/dt).

Did that help?

[Student] -Uh, just give me a moment.
[Instructor] -Sure.

So you're multiplying ½ by -⅑ ?

So, it's more like this. I'm gonna show
y'all the steps out here to the side.

The 1 + 2³, that's a 9.

But it's a 9 to the -½.

Well, that's the same as
⅑ to the positive ½.

And then times 3 times 4.

So that's ½ times ⅓; 'cause 9 to the
½ is square root of 9, and that's 3.

And that's times 12.

So, this is 12 over 6, which is 2.

And that's where that comes from.

-Thank you.
-Sure.

-I really needed to see that.
-Good.

Glad to help.

Anybody else still in the room,
you have a question; go ahead.

[Student] So, is this considered as
a multivariable calculus, or...

This is just single-variable?

[Instructor] Uh, yeah; actually,
that's a good question.

Uh, no. It's not considered
multi-variable calculus. It's not.

[Student, softly] Okay.

There you'll get into, you know...

xyz, and all this cool stuff
called partial differentiation,

and you'll have double and
triple integrals. It's good.

So, no; this is not that.

-Can't wait to learn that.
-Yeah, thank you for today; I'm leaving.

Oh, you're welcome. Bye-bye.

-Manon, you said you can't wait?
-Yeah, I can't wait to learn that.

Yeah, it's beautiful stuff.
You'll love it.

I was looking at one of the
hardest challenging problems

in the mathematics right now,
which is Riemann's data function,

-that they can't prove it.
-They did? Oh wow.

Yeah; they can't, um.
There's no proof of it.

We know the answer, like...

The numbers, but. We can't
prove it, basically, right now.

Yeah. So, if you can't prove it,
then you don't know it.

We can graph it, and look at the values,
and we know where it's approaching.

The answer, but.

Yeah, there's a $1,000,000 prize
on it, if someone solves it.

-Oh ho-ho, nice.
-Yeah.

Anybody else?

[Student] All right, I'm gonna leave.
Thank you for the lecture.

Okay. Bye, Manon!

-Yeah. Stay safe.
-You, too.

All right, everybody;
I'll end the meeting, and um...

I'll see ya Wednesday. Bye!