| Start | There will be an optimization problem.
There will be a curved sketch. |
| 0:41 | I think thats it.
There won't be a definition of derivative, there won't be any basic differentiation, because thats what we just tested you on. However, of course, you are expected to be able to take derivatives. Ok? So I think that's whats pretty much going to be on it. I wrote the rough draft, Professor Sutherland is now going to go through and do his draft, |
| 1:03 | and he and I'll put our heads together and come up with the final draft. Ok?
Ok, after we write the final, which as I said will be in the next couple of days, We will then put up practice problems for everybody, because I don't want to put up practice problems that are a waste of your time. |
| 1:30 | This is intentional. I mean....
We're going to put up practice problems, they will probably be a couple of Professor Sutherland's old exams,
plus, I'll write some more practice problems.
When is the review session? Monday! 12:00-2:00 in Javits 100, everybody's favorite room. Ok? Nice big review. So I will be there, I will do sample practice problems for the exam. |
| 2:00 | I will be in my office at various times on Monday and Tuesday,
I don't know when yet, and when I figure that out I will post it.
Let's see... |
| 2:33 | Ok, here's a problem.
Alright let's do this one. Alright we love these involved product rule, because you guys all forget about the product rule. So let'd do this, the derivative of 3x^2 is 6x. The derivative of 4xy^3 you use the product rule. You have 2 functions, you have 4x and you and y^3. |
| 3:01 | So it's 4x times the derivative of y^3, which is 3y^2 (dy/dx).
Plus, the derivative of 4x, which is 4(y^3). You know we were originally going to have these on part one. But I took them off. We were going to have these. That would have been really entertaining... so. Given how well you did the first time through, we made the right choice. |
| 3:32 | This is 2y(dy/dx)-1.
If you can do this, you get half the points. Ok? Cause now its algebra. We're working on the calculus. So again, the middle term here is the tricky one, you have to do the product rule. You have 4x(3y^2(dy/dx)) + 4y^3. Alright let's simplify this. You have 6x +12xy^2 (dy/dx) + 4y^3 = 2y(dy/dx)-1 |
| 4:14 | And remember what we do, everything with (dy/dx) goes on one side, and everything without (dy/dx) goes on the other side.
So let's put all dy/dx on the left side. You get 12xy^2 (dy/dx) - 2 (dy/dx) = -1 -4y^3 -6x. |
| 4:41 | Or, you could have all of these be positive, and have 2(dy/dx)-12xy^2(dy/dx).
Ok? You can do either way. Now you factor out dy/dx, (dy/dx)(12xy^2 - 2) |
| 5:10 | Did I lose a y somewhere? Where is the y? Oh I lost a y!
Thank you. So you guys, I do that [?]. That's why god created partial credit. Where the sign would not forgive me. And last step, we divide, |
| 5:32 | You get (-1-4y^3-6x)/(12xy^2-2y).
Or, (1+4y^3+6x)/(2y-12xy^2). Ok? Either way. |
| 6:02 | How'd we do on that one?
Madison? Yes, ok, we got some thumbs up. Alright, let's do another problem. You guys remember continuity? Let's review continuity for a minute. And differentiability, because they kind of go together. A function in continuous if when you write it you never have to take your chalk off the board, or pencil off the paper. |
| 6:36 | So that means in order for something to be continuous, You have a function, if f(x) is continuous at x=a, then the limit when x approaches a from the minus side |
| 7:08 | will be equal to the limit when x approaches a from the plus side, = f(a) |
| 7:44 | Alright? Find the value of a that makes that function continuous at x=1.
This is not a hard problem. You've just forgotten what to do. Ok, thats why we're doing review. |
| 8:03 | How do you do this? The limit from the left has to equal the limit from the right, and they both have to equal the value of the function at 1.
Well f(1), you just plug in 1 and you get 6a-2. Now what is the limit when x approaches 1 from the minus side? |
| 8:32 | When x approaches 1 from the minus side, you plug in 1 and you get 6a-2.
And when x approaches 1 from the plus side, you get 1+4a+3, you get 4a+4. So in order for this to be continuous, this has to equal this. 6a-2 = 4a+4. |
| 9:05 | That's nice easy algebra, you get 2a=6, or a =3.
That was easy wasn't it? You just forgot how to do those. Now, what about differentiability? |
| 10:11 | Ok, now, find me the values of a and b that make this continuous and make this differentiable at x =1.
So again, now find the value of a and b that makes this function differentiable at x=1. |
| 10:36 | So in order for it to be differentiable, it has to be continuous, and the derivative from the left has to equal the derivative from the right. Ok?
Let's do this one. So in order for this to be differentiable at x =1, you first must show it is continuous at x=1. |
| 11:01 | So how do we show that it's continuous? What we just did. Essentially, you plug 1 into the top and 1 into the bottom and try and make them match.
So the limit as x approaches 1 from the minus side of f(x) is 5a-2b+4. And the limit when x approaches 1 from the plus side of f(x) is 3b+6a-5. |
| 11:32 | So this, must be equal to this.
We'll put it right here. 5a-2b+4 = 3b+6a-5. Or, we'll put everything over here, a+5b=9. Ok, now we've figured out what we need in order for it to be continuous, now we need it to be differentiable. |
| 12:07 | So you take the derivative of this. You say, upper branch is 10a-2b and lower branch is 6bx+6a.
And they must also match. So when you plug 1 into the top, you get 10-2b, |
| 12:37 | And when you plug 1 into the bottom , you get 6b+6a, they have to be equal.
So 4a, did I make this come out well? Yes! 4a = 8b, or a = 2b. |
| 13:01 | Now we have 2 equations, so we get this equation by taking 1, plugging it into the top, plugging it in the bottom and setting them equal to each other.
We got this equation by taking the derivative of the top and bottom, plugging 1 into the top, 1 into the bottom and setting them equal to each other. Now we have a = 2b, and a+5b =9. So you can solve these rather easily, a+5b =9, a = 2b, so 2b +5b = 9. |
| 13:45 | 7b =9, or b =9/7.
And once you have b is 9/7, a = 18/7. Ok? |
| 14:44 | Alright, a related rate problem.
A conical tank has a height of 12m and a radius of 4m. It is filling with water at 36pi m^3/min. That's a lot of water by the way. |
| 15:01 | How fast is the height of the water increasing when the water is 8m deep?
Alright, you have a conical tank. That is a cone. That is not actually a cone, [?]. Alright. The great thing about a right circular cone, is that the radius and the height are always perpendicular to each other. |
| 15:35 | So when this is filling with water,
So when this fills with water, however much water is in the tank, the ratio of the radius of the water to the height of the water is the same as the ratio of the radius of the tank to the height of the tank.
|
| 16:01 | Ok? Those will be the same because you're given similar triangles.
Ok? So the radius of the tank to the height of the tank, is the radius of the water to the height of the water. So we know that the radius/height is always 4/12, also known as 1/3. So the radius will always be h/3. |
| 16:33 | You should be able to tell that (1/3)h and h/3 are the same thing.
This causes you lots of problems on part 1 of the final. A lot of you are doing quotient rule unnecessarily, its a lot of work and an excellent way to screw up an easy problem. Now we know how to find the ratio of the radius to the height. The volume of a cone is (1/3)pir^2h |
| 17:02 | And we know that the volume of the water at 36 pi.
We wanna know what is dh/dt when h=8? Thats the information that we have. |
| 17:32 | We could take the derivative of this with respect to t and start plugging things in. The problem is, we don't have dr/dt.
So we want to get rid of r. Well, we know that r is h/3. So we can take this equation, And rewrite it as (1/3)pi (h/3)^2(h) |
| 18:00 | Which simplifies to (pi*h^3)/27
3^2 is 9, and 9*3 is 27.
Now that is the same as (pi/27)h^3 for all you quotient fans out there. Let's take the derivative. dv/dt = (pi/27)(3h^2)dh/dt. Notice, no quotient rule. |
| 18:48 | Now you just plug in. You get 36pi = (pi/27)(3(8)^2)dh/dt. And now you just solve for dh/dt.
|
| 19:12 | Which is, (36*27)/(3*8^2) |
| 19:41 | Which is 81/32 m/min.
Now if this were an exam question and this were 10 points, oh I don't know, getting to here is worth 2 or 3 points, |
| 20:00 | Getting to here is worth 2 or 3 points, and getting to here is worth almost all the rest.
We really don't care if you can't quite get the right number here at the end. We'd like you to be able to do arithmetic, because you are in calculus, and arithmetic is something you finished in about 4th grade, But, we're not going to penalize you for not being able to do (27*36)/(3*8^2), might mess it up. |
| 20:32 | Do you have to show how you got pi/27 h^3? If you can get there right away that's fine.
The thing is remember, not every cone has a ratio of 1/3. It's only in this particular cone that the ratio is 1/3. So you need to start with this equation, or if I told you the ratio of the cone is 1/2, so you should understand the process. Which is, you find the ratio of the radius to the height, and then you can use it to get rid of either radius or height and simplify the equation. |
| 21:01 | Alright? As I said, the radius to the height is always 1/3, so I take the radius and replace it with h/3.
And then I can simplify from there. Before you take the derivative, you want to try to get the equation into as simple a form as possible. Let's do another related rate, since you guys love related rates so much. What happened to the pi? Well I have a pi on the left side, and a pi on the right side so that's why I can cut pi. |
| 21:34 | [?]
Alright, let's do another related rate problem. Because we like related rates, don't we?
|
| 22:02 | Should I give you a really nasty one? Nah.
|
| 23:16 | Ok, a 25 ft long ladder is sliding down a wall at 4 ft/sec.
How fast is the bottom of the ladder sliding out when the bottom is 20 ft from the wall? |
| 23:31 | This is a nice, straight forward, related rate problem.
Alright, a 25 ft ladder is sliding down a wall, so we have a wall, and a ladder. The ladder is 25 ft long, lets call the distance on the ground, from the bottom of the ladder to the wall x, And the distance from the top of the ladder to the ground, y. |
| 24:03 | And we know it is sliding down at 4 ft/sec, so dy/dt is 4.
How fast is the bottom sliding out, means we want to find dx/dt. By the way, I'm giving you points for setting up and putting down knowing what you're looking for. I think that's worth at least a point. How fast is the bottom sliding out, so how fast is x changing when x is 20? |
| 24:40 | Ok, do we know a relationship between x and y? We do! x^2 + y^2 = 25^2.
Ok? We now take the derivitave, we get 2x(dx/dt)+2y(dy/dt)=0. |
| 25:12 | We can then divide through by 2, make our algebra a little bit easier, we're trying to find dx/dt, we know dy/dt, and know x, we only need y.
And then we can solve this. So how can we find y? Well, we also know that x^2 + y^2 =25^2. |
| 25:30 | So 20^2 + y^2 =25^2. Doing a little Pythagorean, you realize it's a 3,4,5 triangle, so y will come out to be 15.
So now you just plug in. You get 20(dx/dt)+15(4)=0. |
| 26:01 | You get dx/dt is -3 ft/sec.
That wasn't very hard was it? Not a hard problem, but if we put that on the exam, a lot of you would mess it up. Why do I plug which values in? These values in? Oh here, well y = 15 ft, and x=20ft. Right, y=15, when x=20. And you know y is sliding down at 4 ft/sec. |
| 26:39 | If you made y negative, the x would come out positive. Doesn't really matter.
Ok, let's do another type of word problem. |
| 28:19 | Ok, a corral, as we say out west, consists of 3 rectangular, identical pens,
As in this picture, so you have 3 holding pens, that are the same size, and they share these walls.
|
| 28:40 | You have 800 ft of fence, maximize the area of the corral.
Ok, see if you can do it. This is not a related rate problem. One of the things thats going to happen on the final, is there will be word problems. |
| 29:01 | There will be optimization problems, there will be related rate problems. Try not to confuse them.
It will be very entertaining when you try to do this as a related rate problem because the walls are not moving. Ok? So let's see, why don't we call these vertical walls y, you can call them x if you want to, and these horizontal walls x. |
| 29:32 | So the amount of fence that we have is 2x+4y=800.
We can divide through by 2, so we know x+2y=400. If we want to maximize the area, the area is simply x(y). So now you can take this top equation, isolate one of the variables and plug it into the bottom equation. |
| 30:05 | So for example, I could say x = 400-2y.
And come over here and say A = (400-2y)(y) Or, A=400y-2y^2. Now I take the derivative and set it equal to zero. So A' = 400-4y =0, we set it equal to 0 because thats what we do in calculus class, |
| 30:39 | We take derivatives and set the equal to 0. We get y=100 ft.
x=400-2(100) or x=200 ft. And therefore the corral, we want to find the maximum area of the corral, so A = 100(200), or 20,000ft^2. |
| 31:18 | So again, we have 4 vertical pieces, thats 4y, 2 horizontal pieces, thats 2x.
We know we have to have a total of 800 ft, so we can say 2x+4y=800, which allows for x = 400-2y. |
| 31:37 | Then we take the area equation which is x*y, get rid of a variable, take the derivative and set it equal to 0.
That'll give us the maximum value of y, and then you go back and find x because x=400-2y. So once we know y = 100, we know x=200. And then the area is just 100*200. Ok? |
| 32:04 | Alright folks, so we've gotten to the end of another semester.
Thank you very much, I will see you at the review session, those of you that show up. And a disclaimer for the rest of you, your nightmare is almost over! |