00:01
For this problem, we are looking at an cylindrical oil tank.
00:04
We're going to be looking at volume and how the volume changes in some different scenarios.
00:09
Now, there's a lot of algebra in this.
00:11
We're going to take it nice and slow step by step.
00:14
First thing we want to do is find a formula for the volume of oil in my tank.
00:19
So i have a tank with a certain amount of oil, and we're going to call the height of the oil h.
00:25
And i have radius r for my tank.
00:29
And the length of this tank, it goes back.
00:31
This is only a cross section of the front.
00:33
It goes back l units.
00:37
So if i wanted to find the volume of oil that i have, that would be, if you imagine here, here's my oil, that blue bit, it's going to be the area of that shaded bit times the length of my tank.
00:57
So if i could figure out what my shaded area was, multiply it by the length, that would give me the volume of oil.
01:04
So let's see if we can figure out how to find the shaded area here.
01:09
I'm going to extend one more radius, and i'm going to call this angle that it makes right here theta.
01:16
So on my picture here, and i'll leave this in orange, this piece will be h.
01:21
So up here, that piece will be r minus h.
01:27
Now let's see if we can figure out the area of the shaded region.
01:31
I'm going to look at one piece.
01:33
I'm going to look at this bit right here, and then i'm going to multiply it by two since i have two of them.
01:40
So the shaded area, and i'm going to do this in blue since i've got it blue on my picture, area of my shaded bit is going to be that, and i've got two of them, so i'm going to put a two here so i don't forget.
01:52
It's going to be that sector.
01:54
Well, that sector is a certain fraction of the entire area of the circle, and in particular, it's going to be theta divided by two pi.
02:03
2 pi is the number of ratings in the whole circle.
02:06
I want theta, that fraction of the whole.
02:09
And that's the fraction of the area of the circle, which is pi r squared.
02:15
Now i'm going to subtract from that the white triangle on top.
02:18
I want the blue.
02:19
So i'm going to make that whole red piece minus that white triangle on top.
02:24
And the area of a triangle is 1⁄2 times the base times the height.
02:31
Well, i can find, and i'll do it in r.
02:35
Green since i've done the other ones in green here.
02:37
Using a pythagorean theorem, i can find the base of that triangle.
02:41
It's going to be the square root of the hypotenuse squared, which is r squared, minus the other side squared, which is r minus h squared.
02:54
Now before i go to plug that in, i can simplify this slightly.
02:58
I can say this is r squared minus r squared minus 2 r h plus h squared.
03:08
Or the r squared's cancel and i end up with 2 r h minus h squared.
03:15
That is that base of that white triangle.
03:19
So when i plug it into my area, the area of the triangle is one half times the base, times the height, which is r minus h.
03:32
Okay, let's see if we can simplify this.
03:36
Well first of all, if i distribute this two, it will cancel with a two in the the denominator in each of the terms.
03:44
Also, for my first piece, i have a pie on top and bottom.
03:48
So what i'm left with is pi is, i'm sorry, theta r squared minus.
03:55
I've got r minus h.
03:57
And under my radical, i have 2 rh minus h squared.
04:04
Now, i'm almost where i want to be.
04:06
Oh, sorry, it should be a capital r.
04:08
I don't know why i did a lowercase.
04:10
Okay, capital r is all the way through with my radius.
04:11
I don't want to have this in terms of theta.
04:15
I added theta so i could talk about this shape, but i want it in terms of things i can measure, my radius and my height.
04:22
So let's come back and look at this triangle one more time.
04:28
If i look at theta, cosine of theta, well, that's adjacent over a hypotenuse.
04:34
My adjacent is r minus h, and my hypotenuse is r.
04:39
So i can say the theta is the inverse cosine of that fraction.
04:45
And i can simplify that fraction slightly.
04:48
R over r is 1 minus h over r.
04:52
So that's my theta.
04:53
And i can plug that in over here.
04:57
I'll plug that in right for that value.
04:59
So i have my area, which means my volume, is going to be l times my shaded area.
05:08
And that shaded area, that's my blue, and i'm going to plug in my theta here, that's going to be r squared times inverse cosine of 1 minus h over r minus h times the square root of 2rh minus h squared.
05:28
That is a fairly nasty looking formula here, but that is indeed what we were trying to prove per our question.
05:39
So if i know h and i know r and i know the length of my tub, i can plug them all in and find out the volume of my oil.
05:48
Okay.
05:49
Now, we're going to take this equation.
05:51
And i'm going to come down a little bit.
05:52
We need lots of room.
05:54
Next thing what we want to do is find out what the derivative of v is.
05:59
Derivative of v with respect to h.
06:02
So r is a constant.
06:05
The radius of my tank doesn't change.
06:07
L is a constant.
06:09
The length of my tank doesn't change.
06:11
The only variable i have here is h.
06:14
So let's find dvdh, first derivative of this formula.
06:21
So l is a constant.
06:23
So whatever it is, i'm just going to multiply by that constant.
06:28
R squared.
06:29
R is a constant.
06:30
So this will be r squared.
06:33
Now the derivative of inverse cosine is negative 1.
06:38
So it's going to be a negative 1 over.
06:40
Over the square root of one minus whatever's in those parentheses squared.
06:47
So that's going to be 1 minus h over r squared, times the derivative of what's in the parentheses, which is going to be negative 1 over r.
07:02
Remember, r is a constant, h is my variable, so that's my derivative.
07:07
Okay, minus.
07:09
Now here i have a product.
07:11
So i need to use the product rule.
07:14
So that is the first function times the derivative of the second.
07:18
Since it is a radical, the denominator will be two times what's under the radical.
07:25
The numerator will be the derivative of what's under the radical sign.
07:29
So that's going to be 2h.
07:31
I'm sorry, 2r minus 2h.
07:35
Okay.
07:36
So first times the derivative of the second plus the second times the derivative of the second, the first.
07:46
And since r is a constant, that piece is zero, so i just have negative one.
07:53
Okay.
07:54
Now it's just a matter of algebra.
07:56
So we're going to clean this up as best we can.
07:59
Again, that l throughout this whole thing, that l is just going to be along for the ride.
08:03
We're going to be focusing on all of the rest of it.
08:05
So that l is just going to stay here.
08:08
Okay, so i have r squared times.
08:13
Okay, on top, i have negative one times a negative that's a positive, and that means i'm going to have 1 over r.
08:25
So actually, what i can do is i can take that r and cancel it with one of the r's in my r squared.
08:32
So that r is going to cancel one of those r's, and i'll do it in red so it stands out here.
08:37
So what i really am going to have is just an r out here.
08:41
The negatives cancel.
08:42
That becomes a positive.
08:44
Now, in the denominator, what do we have? let's go back to black.
08:49
Do we have? we have one, minus.
08:51
We have to expand this binomial.
08:54
One minus two h over r plus h squared over r squared.
09:01
That's all under that radical.
09:04
We have a one on top.
09:06
Okay.
09:07
We'll come back.
09:07
We're going to just keep moving through and we'll keep simplifying on the next step.
09:13
Okay.
09:14
Next, we have a negative so i can change this plus to be a minus, just to get rid of that piece, i don't need that anymore, that negatives there.
09:26
I'd like to be able to add these.
09:28
So i'm going to multiply top and bottom over here with a 2rh minus h squared over square root of 2rh minus h squared.
09:45
That will give me a common denominator that i can start simplifying with.
09:49
Also, on this side, 2r minus 2h is the same as 2 times.
09:54
Times r minus h, and these twos can cancel.
09:59
So what do we have? do you have to keep those brackets since i'm subtracting? i have r minus h squared.
10:06
That is r squared minus 2rh plus h squared, minus radical times a radical.
10:16
I just get what's underneath 2rh minus h squared, all over my common denominator, which is 2 rh minus h squared that square root.
10:31
Okay, come down a little bit.
10:34
Next step.
10:38
Again, that l is just out here hanging on.
10:40
So, r in the numerator.
10:43
Denominator, those ones are going to cancel, and i will have 2h over r minus, because i have to distribute that negative sign, minus h squared over r squared.
10:57
All of that's under that square root.
10:59
Now i would like to be able to put these together.
11:03
So i am going to multiply both top and bottom of that first piece by an r.
11:09
And so what that's going to give me, and i'll just come down and do this on my next level here.
11:15
It's going to be square root...
