Stony Brook MAT 125 Spring 2015
Final Review Session
May 11, 2015

Start   As you saw, when we did the posted sample questions for MAT 125, those were from last year Hasn't changed that much.
I posted last years sample final, I just put up the solutions if you wanna check on Blackboard, make sure that didn't get screwed up.
The solutions should now be up there.
Um, I also posted my study guide.
If that helps any of you. So its got some things to concentrate on, review problems, etcetera.
0:38Ah, you can't see.
I can teach on this side.
That's not going anywhere.
That is definitely a flaw.
Ok, I posted that. I also posted the announcement.
So once again, if you were in my lecture, or if you were in recitation 1
1:02then you come here for the final on Wednesday.
If you are in any of the others they're all up there.
If you are in Put my glasses on.
Ok ladies, brace yourselves. I have glasses.
Grandma, ok If you're in recitation 6,7, or 8, that's Professor Sutherland's class, you go to 102.
There.
1:30If you're in 9, 10, 17, or 23 you go to Javits 110.
And all the rest go to ESS 001 and it's up on Blackboard so you don't need to write that down. Yes.
Do I? Do I really? Ugh, I hate when that happens.
I guess they all go to 110.
But we'll double check that. We'll fix that after class, ok?
2:02You know, you get old, you're lucky you're even wearing your pants. Alright.
Stuff you guys should be prepared for for the final.
Ok, for Part 1 Once again, by the way you all need to know how to do this. Just because you passed part 1 doesn't mean you can now forget this.
You should be able to take the derivative of limits, product rule, quotient rule
2:37chain rule, how to find the tangent line. Can everybody see if I write over here?
Ok.
We're gonna have a problem. Can you see if I write here? No.
Here? Sorta?
3:00So there. Alright.
You guys shouldn't have sat there.
You know, maybe that's the failure side over there. I don't know.
Alright, let's see. You should be able to find maximum and a minimum.
A point of inflection
3:31Ok, Part 2.
4:50Alright.
That looks like a nice assortment of things you should be able to do. We're gonna practice all of these.
Or you could take a midterm. It lasts longer.
5:08You good?
Alright.
Let's practice some antiderivative stuff.
So antiderivatives, you're doing the derivative backwards.
Ok, so you said you can't see this one? But we can see this one right?
Alright you guys have to lean a little bit.
5:31The antiderivative rules If you start with Remember, when you write an antiderivative you're always gonna put +c at the end.
6:01So if you see a problem like that on the exam, just remember when you're done, just write plus c.
Ok, why is that? Because when you take the derivative of something, there could always be a constant attached because the derivative of a constant is 0.
So if you know that the derivative of something is x^2 you say the original function is x^3, add one to the power, divided by 3.
But it could also be (x^3)/3 +1 Or (x^3)/3 +2. You don't know.
6:30Could be all sorts of things, that's why you have plus a constant.
Ok, I'm gonna erase this so I can start working.
Ok, if the derivative of sine is cosine, then the antiderivative of cosine is sine.
So let's just put a bunch of stuff. Let's say you had
7:05And I wanted to find the antiderivative.
Do a capital letter. Or you could say this is f prime and that would just be f, it doesn't really matter.
Very simple.
To do the antiderivative of x^4 you would do (x^5)/5.
So you raise the power by 1, divide by the new power.
7:30Ok, think about it, when you take the derivative of this the 5 comes down, and cancels with this 5 times x^4.
So the antiderivative, you go the other way.
You're gonna need this right away in 126. But I'll review it.
Ok what about the derivative, the antiderivative of cosine? Well, the derivative of what is cosine?
The derivative of sine is cosine.
So the antiderivative of cosine is sine.
When you do these always check.
8:00Say to yourself, I'm gonna differentiate this and I have to get back to here.
The derivative of ln(x) is 1/x so the antiderivative of 1/x is ln(x).
Plus c.
Convince yourself this is true, take the derivative.
Ok, look at this, take the derivative.
8:31The derivative of (x^5)/5 is x^4.
The derivative of -2sinx is -2cosx.
The derivative of ln(x) is 1/x.
The derivative of c is 0.
What if I had e^(6x) +5?
And I wanted to find the antiderivative.
9:02Well, what's the derivative of e^(6x)? It's e^(6x) times 6.
So the antiderivative would be (e^(6x))/6.
Think about it, you have to do the chain rule.
You take the derivative of e^(6x), you get e^(6x) x 6, so dividing by 6 gets rid of that.
What's the antiderivative of 5?
Who's saying 0? You're all wrong if you say 0.
9:31It's 5x. Going the other way.
Because the derivative of 5x is 5.
Ok?
You have to be careful you don't confuse that with derivative.
We have the derivative of pi is 0. But the antiderivative of pi is pi(x).
Ok?
Plus c, of course.
I have other equally unpleasant ones.
10:34How do I do the antiderivative of that?
Well notice, cube root of x looks really scary but that's just x^(1/3).
So to do the antiderivative of x^(1/3) you're gonna add 1.
What's 1+(1/3)?
4/3.
So this would be 4/3.
Divided by 4/3.
You could leave it like that.
You don't need to show off how good you are at simplifying.
11:03What about 1/(sqrt(1-x^2))?
That is inverse sine.
Or arcsin.
I put that on my little review sheet which I posted on Blackboard.
Did anybody check Blackboard to make sure those things are active?
Are they up there?
One of you please check because I can't do it from here?
Alright, and what about the antiderivative of pi^3?
Pi^3(x).
Because the derivative of pi^3(x) would be pi^3.
11:32Can you see that? If you guys have any trouble seeing you might wanna squish in towards the center.
I don't know how professors teach here without a blackboard.
They don't use a blackboard, they do everything up there?
Oh yeah, that's not happening.
I mean, that would require me to prepare.
You guys all remember your derivatives, right?
12:00No?
Let's practice a couple more of these then we'll move on to something else.
12:46That's good enough. Ok, practice that for a second let's see if you can figure that one out.
Oh it is. Wonderful.
14:16Are you ok?
You sure?
So let's do this.
What's the antiderivative of cos((pi/2)x)?
Sin((pi/2)x) divided by pi/2.
14:36Why do I divide by pi/2?
Just think of it as the chain rule backwards.
Ok, because if I take the derivative, I'm gonna be multiplying by the derivative of the inside.
Multiplying by pi/2.
The antiderivative is divided by pi/2.
That's not that hard right?
This isn't chemistry.
Alright, the antiderivative of 3x^2 is (3x^3)/3.
15:02Or if you're clever, x^3.
Because the derivative of x^3 is 3x^2.
How about e^2x?
e^(2x)/2.
And what about 4/x? Well 1/x is ln(x), so this is gonna be 4ln(x).
15:31Plus c.
Don't forget the plus c.
How much fun was that?
Ok? Yes.
You don't understand why I divide by pi/2.
Take the derivative of sin((pi/2)x).
Take the derivative you get cos((pi/2)x)
16:03times pi/2.
But we only have cos((pi/2)x) we don't have pi/2.
So we get rid of that.
That's why we divide here.
Because now, if I had this over pi/2 those cancel.
Ok?
Because you don't need to use the quotient rule for that.
16:45Alright? Question.
Um, so, would you want us to like cross out 3x^3 /3?
You are not required to simplify anything.
You can leave it (3x^3)/3. Ok?
17:03The only need to simplify is to make your life simpler.
If it doesn't make your life simpler, don't bother.
Ok?
It's our job to figure out what you did, and if we're not sure we'll just mark you wrong.
Ok, that wasn't terrible.
Destress.
Alright, let's do some optimization problems.
We love those.
17:46Ok, the word problem stuff aren't fun.
Figure out how you're gonna set it up.
You're aware of this.
So let's do a couple of word problems.
19:12Ok.
Now for use of time, I'm gonna do this with you guys, because I know how much you wanna do it on your own, but we're gonna share, ok?
So we look and we say, we have a soda can and its gonna hold 240 cubic centimeters of soda. Is that a lot by the way?
19:33You guys have been taking chemistry. Is that a lot?
Yeah, about 8 ounces.
It's an 8 ounce soda can. That's not a lot.
8 ounces of Red Bull, but not 8 ounces of soda.
Ok, the top costs 6 cents per square centimeter because the top is more expensive.
The rest costs 4 cents per square centimeter.
So minimize the cost of the can.
So you look and you notice, pay attention to the units.
20:00Cubic centimeters is volume, square centimeters is area.
So I'm gonna need to use volume and area to find a way to make this impossible soda can.
The cheapest soda can will be the soda can with the smallest area, factoring in the costs of the area.
That makes sense right? Because the area is the piece of metal that they need to bend to make it into a soda can.
So I have two pieces of information. I also have a cylinder.
What's the formula for the volume of a cylinder?
20:36I can see, somebody's raising their hand and asking "will we need to know that for the final?" Somebody's hand is up. We will give you geometry formulas if you need them.
We may give you more geometry formulas than you need, but we will certainly give you the formulas you need.
But let's be honest folks, you should know how to find the volume of a cylinder.
It's not that complex.
Surface area, well for surface area you need the area of the tube
21:06That's 2πRH Then you need the area of the top and the area of the bottom. That's both πr^2.
But we're not quite doing the surface area, we're doing the cost.
So it's a little different.
Alright, the volume of the can, πR^2H has to equal 240.
21:32And what's gonna be the cost of the can? Well let's see.
First, I need to put the top on the can.
The top is a circle, a solid circle.
So it's gonna cost 6 cents times πR^2.
Alright, that makes the circle on top.
So that's 6 cents times the area of the top.
The area of the bottom is only 4 cents times πR^2.
22:03Then the area of the tube of the cylinder, the cylindrical shell is gonna be taller than 126.
Its 2πRH, and that also costs 4 cents.
So the cost is the equation we're gonna minimize.
So that simplifies to 10πR^2.
22:32Plus 8πRH.
10πR^2 + 8πRH.
Now we take those pieces of information, put them together, take the derivative, set it equal to 0 and solve.
But hey, setting this up is worth a reasonable amount.
Ok, so learn to set up the problems.
If you do the correct set up you'll get something. I can't promise how much credit you'll get but you'll get something.
Alright, so we're gonna use this equation.
23:02It helps to get rid of one of the variables, because the problem we have right now is with 2 variables.
We only want one variable.
So let's rewrite this as H is 240/(πR^2) And that, you plug in as H in the cost equation.
The cost will now equal 10πR^2 + 8πR (240/(πR^2))
23:41Professor Kahn, what if we mess up the arithmetic?
We'll look at your derivative, that's why we have partial credit.
Ok, this is part 2 not part 1.
You wanna multiply 8 times 240? That's why you have paper and pencil.
Or a pen if you're cocky.
24:03So 8 times 240 is 1,920, one of the π's cancels and one of the R's cancels.
So you get 1920/R.
Ok?
So far so good, now you take the derivative.
I'll put the derivative over here.
The people on one side are gonna have trouble seeing it.
So C' is 20πR - (1920/R^2)
24:42And we set that equal to 0 of course.
So how do you set that equal to 0? Well if you multiply through by R^2 then you get 20πR^3 - 1920 = 0.
25:00How did I get that? I multiplied by R^2.
So that gives you 20πR^3 and the R^2 and the R^2 cancel.
I'm allowed to do that because I know R will not be 0.
Yes.
Sure.
25:33How's that?
Ok?
20πR^3 - 1920, you solve that You get R^3 = 1920/20π R, cube root of that.
26:02Now think about it, ok.
You're not gonna have a calculator on the exam. So we can't give you a number that would be very uncomfortable like this number.
Ok?
A nice, simple number.
Cause people would think 96/π is messy.
Right? So we wouldn't give you something that's quite that bad.
But once you've got R, then you plug it back in and you get H, and those are the dimensions that give you the minimum cost of the can.
26:32Ok?
So if we were to give you this we'd probably just ask for the radius.
Ok, of course we're not gonna give you this exact problem because that would be too easy.
Yes.
Why is that -1920/R^2?
Well, remember what the derivative of 1/x is.
What is it?
27:01So the derivative of 1920/R is -1920/R^2.
Got it?
You could also write it as 1920 times R^-1.
Ok, this is 1920R^-1 So the derivative is -1920R^-2 That is
27:37-1920/R^2.
Ok?
Good?
Let's do some more.
Yes.
What am I doing in which step?
Here and here? I told you guys, how did I get from here to here?
Ok, I multiplied everything by R^2.
28:01That makes this 20πR^3, and that R^2 cancels with this R^2.
Ok?
Yes.
Ok.
Where does the cost come from?
Across the top is πR^2, that's the area of the top, times 6 cents πR^2 is the area of the bottom times 4 cents.
28:302πRH is the area of the tube times 4 cents.
Ok?
Question.
What happened to the H?
You tell me.
What'd I do?
I took H and I made 240/πR^2 and I went to this equation right here and I replaced H with 240/πR^2.
29:06Ok?
And that reduces to this.
Ok?
I took H and I substituted 240/πR^2.
Ok.
Other questions.
You can watch it again on video.
29:30And again and again.
Maybe there's something that YouTube highlighted.
Only if I taught... then it would be the YouTube highlight.
Ok, when can I erase? I'm gonna erase in 5 seconds.
And I'm erasing.
32:22Ok.
A swimmer is located 300 meters from a straight shoreline.
32:30So, what does that mean?
When you do these word problems, break it down.
You've got a straight shoreline you've got the swimmer sitting somewhere there.
300 meters away.
1000 meters down here is the ice cream stand.
Yum, ice cream.
33:04I'm in the mood for ice cream. Alright.
She can swim at 4 meters a second and she can run at 6 meters a second.
So if you think about it, this is the fastest distance to the ice cream stand.
But she'd be swimming the whole way, she doesn't swim as fast as she runs.
She could get to shore as fast as possible and then run, because she runs faster than she swims.
33:30But now she'd be going the biggest distance.
Somewhere in the middle is the best place to land.
Ok, again.
She could go all the way from here to here, the hypotenuse.
That's the shortest distance. But, she swims slower than she runs.
She could do the maximum distance, which is swimming the minimum amount of, the minimum distance and then running the whole way cause she runs faster than she swims, but now she's doing the biggest distance.
34:03So somewhere in the middle is the best place to land.
Where is that place?
He's going for ice cream. Ok.
How are we gonna find the shortest time?
Well, remember that (rate)(time)=distance.
Or, time=distance/rate.
34:37So we could give you this problem with running, we could give you this problem with all sorts of variations.
But the basic idea is, where's the best place to be?
The hypotenuse.
So let's think about what the swimmer's gonna do.
She is going to first swim from here to x.
35:01So that distance is sqrt(300^2 + x^2).
So, she's gonna go the distance 300^2 + x^2.
And time is 300^2, is 300^2 + x^2 /4.
Now she's on land.
Now she's gonna run as fast as she can.
Which is 6 meters per second, which is not that fast.
35:32She's had too many ice creams in her life, ok.
So that distance is 1000-x And she can run at 6 meters/second.
When you put those together, and that's the time.
And that looks scary but it's really not that bad.
Can I erase the word problem part now?
Everybody got it? It's gonna be on video later.
36:02I can't hear you, what?
That's 1000-x All these people who criticize my handwriting, where were you when I needed a penmanship class?
Can you hear me when I talk to the board, by the way?
Ok.
Serina can, you can hear me. Ok.
Who said no?
36:30Who said they can't hear me? Now you can hear me.
Ok.
Don't worry you're not missing anything that great. Trust me.
Alright, so I'm gonna rewrite this. The time is This plus this. Now let's write this in a slightly better way.
Thats (1/4)(300^2 + x^2)^(1/2).
(1/4)(300^2 + x^2)^(1/2) is just this.
37:03Divided by 4 is the same as times 1/4.
You don't need to use the quotient rule.
This divided by 6, well that's 1000/6 minus x/6.
(1000/6)-(x/6) Am I projecting ok?
Can you hear me up there? Can you see me? Hi.
Ok. Let's take the derivative.
37:34T'=(1/8)(300^2 + x^2)^(-1/2) Times 2x.
That looks scary. Don't be scared by that.
What's the derivative of 1000/6? It's 0.
That's a constant.
38:00And what's the derivative of x/6? You do not need the quotient rule, it's 1/6.
Ok, you guys put a lot of energy to do the quotient rule.
On part 1 of the exam you don't need the quotient rule for this kind of thing.
Alright.
How will you solve that?
Well first of all the -1/2 means 1 over the square root.
So this is now 2x over 8 times the square root.
38:38Ok, the 2x goes in the numerator. This goes in the denominator cause its a -1/2.
And this is in the denominator cause it's a fraction.
Minus 1/6=0.
Got any ideas on how to solve that?
Put the 1/6 on the other side and cross multiply.
39:02Cross multiply and you get 12x =8 times sqrt(300^2 + x^2) The worst part of calculus is the algebra. It's not the calculus.
Ok, where do we go from here?
Well we could divide by 8.
Divide both sides by 4.
39:33I'm gonna erase this.
Divide both sides by 4.
Then you get 3x=2sqrt(300^2 + x^2) And now square both sides.
And you get 9x^2=4, don't forget to square the 2.
40:03Times 300^2 + x^2.
It is becoming easier than it looks.
9x^2 is 4(300^2) If you're not comfortable with what 4(300^2) is, leave it alone.
Plus 4x^2.
40:30Subtract 4x^2 and you get 5x^2 is 4(300^2) And now divide by 5.
(4(300^2))/5 So x is the square root of that. What's the square root of 4?
2. What's the square root of 300^2?
300.
Square root of 5 is square root of 5.
41:01Ok?
Boy it's a lot easier when you watch me do it, isn't it?
You're sitting there saying no way. Look, if you can get to this step and to this step, that's most of the work. Ok.
That's what we're looking for.
We're looking to see if you can set it up. We're looking to see if you can take the derivative properly.
I've got a chart with how our 10 point question should be, you've got 7 or 8 when you get to this step.
Then I wanna see if you can solve it. So it's worth something.
But it's not worth too much.
41:30Maybe I'll make it 7.
Yes.
Oh, why did I divide by 4 instead of 8? I didn't want a fraction.
Good.
You don't have to divide at all if you don't want to.
The key is, this looks really scary.
Ok, but this isn't really scary at all.
You have to get from the one to the other.
Ok, so that's a couple of optimization problems.
Now let's do a couple of related rate problems.
42:00We love related rates, right Diana? So much.
I don't know about you guys but all of this stuff is exhausting.
You have no idea what it's like to be tired and stressed.
Yes.
That would be, that's, no that's not second.
That's not a unit of time, that was the question.
42:322(300/sqrt(5)) is 40 something feet.
Where you land to get there at a certain time. Remember distance is the distance.
So you're landing at the place 600/sqrt(5) meters.
Ok?
You are landing at the spot 600/sqrt(5) meters.
43:00Then from there you run the rest of the way.
For those of you who do biathlons.
That gives you the shortest amount of time.
Ok? So you get to the ice cream stand the fastest, get your ice cream, get home, start eating.
Alright, should we move on to related rate problems? Because we love those.
So much.
Everyday is an opportunity for sarcasm in college. Everyday.
43:31Alright, can I start erasing?
Can I erase? Ok.
It'd be really nice if I could use the whole board.
Remind me never to teach class in this room.
Ok.
44:01Let's do a related rate problem. Let's see
44:58Ok.
A spherical balloon is increasing in volume at 60π cubic inches per minute.
45:07How do I know it's volume? You guys take chemistry, pay attention to units.
How fast is the surface area increasing when the radius is 6 inches.
Ok.
So volume of the balloon, volume of a sphere is (4/3)πR^3.
45:30The surface of a sphere is 4πR^2.
Next semester maybe you'll learn where those formulas come from.
Because I know you really were wondering.
Were you really wondering?
No. Yes?
Ok, I'll put it on the final next semester.
Ok, how fast the surface area is increasing.
So we're gonna need to relate
46:01the rate at which the volume is increasing to the rate at which the surface area is increasing.
That's why it's called related rate.
We also know that the volume dv/dt, the volume is increasing at 60π.
You wanna find ds/dt when R=6.
46:31That's what we're looking for.
So what do we do? Well it's calculus class, take the derivative.
How do I know it's dv/dt, all these dt's? Cause it's rates.
So I have volume is (4/3)πR^3
47:01The problem is volumes in terms of R.
And I ask you what the rate is at 6 inches.
But I wanna find how fast the surface area is increasing.
So what I could do is I could rewrite this as s/4π, square root, is R.
And plug that in there, I don't wanna do that.
Way too much work.
Some of you who write it very quickly are already erasing that.
47:30Serves you right for going so fast.
Or, I could do this part of the problem separate from this part of the problem.
That's what I'm gonna do. I'm gonna do the volume separate from the surface area.
You guys can hear me? Good.
dv/dt is (4/3)π(3R^2 dR/dt)
48:03Notice, no product rule.
(4/3)π is just a constant. In fact, it's around 4.
Ok, so it's just that constant times the derivative of R^3 which is 3R^2 dR/dt.
And over here under surface area I get ds/dt is 8πR dR/dt.
48:35So I can go to this equation on the left solve for dR/dt, because I know dv/dt, I know R.
And you go to this equation, plug in R and dR/dt, then get ds/dt.
It's really not very hard.
After you do it So these 3's cancel to make my life easier.
And we said R is 6, so 60π is 4π(6^2) dR/dt.
49:14Pi's cancel.
And you get 60 divided by 4 is 15.
Divided by 36 is 5, divided by 12.
So dR/dt
49:30I guess I'll put it down here.
Is 5/12.
inches per minute So if you're just asked how fast the radius is changing you can go.
It's 5/12.
But we're not that nice.
So now we ask how fast the surface area is changing.
Well, all I have to do now is plug in.
8π(6)(5/12)
50:01That gives you 20π square centimeters per minute.
Ok? Not so bad?
50:33Alright, let's do another related rate problem. Yes.
Oh, yeah inches squared.
You're correct.
If you wrote cm^2 we wouldn't take off.
In fact, I don't think we're gonna ask you for the units. We could, but I don't think we will.
Ok?
I think it's important you get the units right, but this is math.
Principle is what matters.
51:00Save the units for what class? Chemistry.
Are there any opportunities for sarcasm in chemistry class?
Once in a while.
Can I do another related rate problem? Sure.
52:23Where are you two going?
52:48Off to get ice cream?
54:01I'll give you one second.
Ok.
A conical tank has a radius of 6 meters and a height of 18 meters.
It is filling with water at 12π cubic meters per minute.
How fast is the height of the water rising when the water is 6 meters high?
54:35So you have a conical tank.
You're dumping in water.
The radius of the tank is 6 meters.
And the height of the tank is 18 meters.
And you wanna find out how fast is that water level rising?
55:00Ok, what's the formula for the volume of a cone?
It's (πR^2H)/3.
What's the formula for the surface area of a cone?
Who cares? We didn't ask you for surface area.
Ok?
What's the formula for the volume of a cube? We didn't ask you that one either.
Alright.
So how fast is the height of the water rising?
55:31So we're gonna need to relate the rate that the height is increasing and the rate that the volume is increasing.
And you know that the volume is increasing at 12π.
We want how fast the height is increasing when the height is 6.
So if you look at this equation (πR^2H)/3, you have R, you have H.
56:01We wanna find dH/dt.
So if I take the derivative of this, the problem is that I have 2 variables. I have R and I have H.
Gonna have to use the product rule. And I don't know dR/dt.
So that's annoying.
So I'm gonna use the information about the cone to remove the radius from the problem.
56:30So the key to the cone problem is that the ratio of the radius to the height is always the same.
The radius divided by the height is always 6/18, is always 1/3.
Or R equals H/3.
So I can use that information now, and plug that into the formula for the cone.
57:02The volume of the cone is (πR^2H)/3, I'm gonna rewrite that as ((1/3)π), R^2, R is now H/3 squared times H.
Which can simplify to π, H^2 times H is H^3 3 times 3^2 is 27.
57:50Alright, dV/dt π/27 is just a constant
58:01times 3H^2 dH/dt.
Now I just plug in.
Cause I said H is 6, I know dV/dt, I just have to find dH/dt.
I get dV/dt, whoops, dV/dt I know now, sorry.
I get 12π=(π/27)(3(6^2))dH/dt.
58:39I hope you can all see that. I'll rewrite that over here.
That's too hard to see.
You get 12π is (π/27) times 3 times 6^2 dH/dt.
59:03Why do I put pi in the volume? We can cancel it.
We get 12 is 3 times 6^2 over 27, or 4.
So dH/dt is 3.
3 whats, it doesn't matter. It's 3.
59:37How are we doing so far?
Exhausted.
Ready for chemistry?
No? Ready for biology? Physics?
Nobody here is taking physics, right?
Physics is really hard.
No?
Well I can keep going for a little while longer.
Yes.
Can we go over part 1?
60:00I'm not gonna go over part 1 here.
I did that at the review on Monday. It's on video.
One more related rate?
You guys want one more related rate? I'll give you a graphing problem.
Oh velocity, acceleration stuff.
No, no just basic antiderivatives.
60:32I mean I'm not gonna get to cover everything, sorry.
I could sketch a curve.
Want me to sketch a curve? Or should I do something else?
What are you voting for though?
I'm holding.
I'm not necessarily gonna do what she wants, I can tell you that.
5, 4,
61:023, 2, 1 I can't hear you. So let's make sure you guys can do a curve.
Maybe a curve involved in a logarithm.
L'Hopital? I'll do one of those in a minute.
Bye ladies.
Say hi to the sisters for me.
61:49Alright how about a nice, not too crazy graph.
62:41This is draw this graph.
Alright, we're gonna sketch this curve. Boy, I turn around and the room gets really empty. Alright
63:01I feel bad, don't you feel bad?
You don't listen anyway right?
We never like those people.
This is when I give away all the secrets to the final.
I just need a few more of you to leave.
And then I'm gonna start saying "this will be problem 5 on the final." Maybe, I might do that.
I want you to be rewarded for coming to class.
63:30Yeah, and it's not because of you.
I'm saying next semester as far as right now.
Video, whatever that is.
Alright.
125, this is 125.
126, as far as I know they're not gonna be recording my class.
There might be a place where I can turn on Echo.
I might, but I might not. You should show up once in a while.
Some of you come to every single class.
You should be rewarded.
64:00Some of you don't come to every single class.
You shouldn't be punished, but you should not be rewarded.
That's that.
Alright, let's take the derivative. Why do you take the derivative?
Cause this is calculus class.
We get 6x^2-42x-48.
64:32We set that equal to 0.
Divide through, factor out a 6.
And that factors very nicely.
So that means we have critical numbers at 8 and -1.
65:08Now what do we do? We do a sign test.
Pick a number to the left of -1, like -2.
Plug in the derivative.
And you'll get a positive value.
Take a number between -1 and 8, like 0.
Plug it in, you get a negative value.
Take a number greater than 8, like 9.
65:30Plug it in, you get a positive value.
The curve is going up then the curve is going down curve is going up. How are we doing so far?
Which part?
The sign testing? Yeah.
You pick a number to the left of -1.
Such as -2.
Plug it into here, the derivative. Or any form of the derivative.
You get a positive answer, so that tells you that the curve is increasing here.
You pick a number between -1 and 8, like 0.
66:01You plug it into the derivative. You get a negative value.
This comes out negative. The curve is going down.
You pick a number bigger than 8, like 9. Plug it in, you get a positive value. Curve's going up.
Ok? It's going up, down, it's going up.
Ok, now we need to do the second derivative.
The second derivative is 12x-42.
Set it equal to 0.
66:40That's You get 7/2, also known as 3.5.
But since most of you are even worse with decimals than you are with fractions you'll probably leave it as 7/2.
Ok, sign test again.
67:03Pick a number less than 7/2, like 0. We love 0.
Plug it in. You get a negative value.
The curve is going down, it's concave down here.
Pick a number greater than 7/2, like a billion.
Plug it in, you get a positive value.
It's going up there. It's concave up there.
Ok?
67:31Alright, now we know the shape of the graph.
We know where the maximum and minimum point of inflection are. We have to figure out the actual values.
So we have a maximum here.
At x=-1 because curves going up then it's coming down.
And a minimum at x=8.
And you have point of inflection at x=7/2.
Now look, on the exam, it's entirely possible that you'll screw up when you plug these values back in to get the y-coordinates.
68:07Show your work.
Ok? If you have the shape of the graph right and you have the x-coordinates right, we're not gonna punish you if you can't get the y-axis exactly right.
But, you should be able to plug -1 in successfully.
You get -2-21 is -23 +48 is 25 +10 is 35.
68:31Here?
Point of inflection. Oh.
Person of interest in the FBI world.
Point of inflection for the rest of us.
8 is a huge number, we're not gonna ask you to find it. Ok?
Cause you could all do 8^3. That's 512, 1024.
Minus 21(8^2) which is
69:011344 So now you're down to -210.
Minus 48 times 8 which is -384 And I forgot what the number was already. Anyway, it's a big number.
It goes way down. So What does this graph look like?
69:30Something like that.
Ok?
That's x=-1, that's x=7/2.
That's x=8. We're not that concerned with what the y axis does. Ok?
Wow. Handwriting is really hard.
Ok? If you make it look like that that's fine.
And I'll tell the TAs on the final.
Or, we'll make the numbers come out easy.
70:02Best graphs possible.
Alright, who wanted to go to the hospital. Was that Lola?
You wanted the hospital rule?
Also known as L'Hopital.
We like to say it in French.
70:51What if I have to find that limit?
71:08Well, I plug in 0. And what do I get?
I get 0 on top, I get 0 at the bottom.
That makes me really sad.
So I write the limit does not exist and I get no credit.
Or What do I do actually?
71:31What do I do?
L'Hopital. What's L'Hopital's rule say?
Correct. Derivative of the top, derivative of the bottom. Perfect.
Not the quotient rule.
The derivative of the top, the derivative of the bottom.
Now I plug in 0.
72:01Gotta love the unit circle. What's the cosine of 0?
1.
So this is 10 times 1 over 7 times 1.
Also known as 10/7.
72:38What if I wanna find this?
The limit as x goes to infinity, lnx /3x^2.
Well, when I plug infinity into the top I get infinity.
73:02When I plug infinity into the bottom I get infinity.
So I use L'Hopital's rule.
Alright, what's L'Hopital's rule?
Take the derivative of the top and the derivative of the bottom.
There you go.
Same limit.
Derivative of log of x is 1/x.
Derivative of 3x^2 is 6x.
73:35That's the same as, think of that as 6x/1 1/(6x^2) Ok.
Whats the limit as x goes to infinity of 1/(6x^2)?
74:010.
You guys could do this stuff.
You could pass this final.
You're gonna do well, could've went out for ice cream when my back was hurting.
[inaudible] Right?
[inaudible] Let's do some more.
74:30I don't think too much more, you've got 10 minutes.
You guys are running out of steam. And frankly, so am I.
[inaudible] Yes. L'Hopital's rule will be on the final.
75:04Oh, yes, let's do an [inaudible] problem.
75:56Oh, you know what we need to review? We need to review the implicit derivative.
76:24We should do an implicit derivative. Very important.
77:12Alright. Why don't you guys do it on your own?
79:29Alright. Pretty sure we're ready.
79:33How do we do this?
What's the derivative of x^3?
3x^2.
Ok.
Now for the fun part.
This is the part you will mess up.
We have to do the middle.
And that's the product rule. And notice the minus sign.
I love to use the minus sign.
Product rule. Derivative of 3x^2y^2, well Take the derivative. Take the 3x^2, leave it alone.
80:01Multiply by the derivative of y^2 which is 2y dy/dx.
Plus y^2 times the derivative of 3x^2, which is 6x.
close parenthesis That minus sign is gonna get distributed.
This is gonna be minus and this is gonna be minus.
Take the derivative of 4y^3, is 12y^2
80:38dy/dx, and that equals 0.
Getting to there is most of the way.
Now, you have to isolate dy/dx.
So, distribute the minus sign.
3x^2-(6x^2)y dy/dx.
-6xy^2 + 12y^2 dy/dx
81:07equals 0.
All I did was distribute the minus sign to clean it up a little bit.
Now what we do? We group the dy/dx terms on one side and the non dy/dx terms on another side.
To get 3x^2 - 6xy^2 = (6x^2)y dy/dx
81:35-12y^2 dy/dx.
Now I could factor out dy/dx.
82:07And I get 3x^2 - 6xy^2 = (6x^2)y - 12y^2 all times dy/dx, then I divide.
82:30Ok?
How was that?
Alright, I sense the natives are getting restless. Yes.
You can keep the negative when you're doing product rule, sure.
As long as you get the right answer, there's more than one way to do the algebra. I didn't get the right answer.
Ah well, then watch your minus sign.
Alright, I've had enough. So, study hard.
I put up my office hours.
Otherwise, study, there's lots of practice material online.
83:00We'll fix the recitation room thing.
Alright.