|Start||We did related rates last time lets do more related rates we do that for a while then we do some other stuff and this chalk is broken okay so related rates as you saw last time we give you some situation and you have to come up with an equation to figure out how the rate of something is changing is related to the rate of something else changing then you take the derivative in respect to t|
|0:33||db/dt, da/dt and you plug in its not as scary as they initially look the tricky part with related rates is the set up so lets practice some more the volume of a sphere is increasing|
|1:04||at 24pi in cubic inches per second how fast is the radius increasing. this is a basic one|
|1:31||when r=3 inches
so we did one exactly like this on monday
you should all be doing this, dont look at your note
see if you can do it. and i know by now you memorized
that the volume of a sphere
4/3 pi r cubed
so i dont even have to write that down
becuase you dont even have to look
but if you tell i was looking its over there
so what did we do?
|2:00||we got a sphere and the sphere is increasing
so you want to find how fast the radius is changing
when the volume changes
because the radius is related to the volume
and as the volume grows, the radius will grow at a slower rate becuase
that amount of volume
and it increase to there
of that amount of volume.
might only be that amount for the next increase and that amount for the next increase
|2:33||because each one is bigger each one is wider so the amuount of volume in this shell could be the same as in this shell and this shell. the radius can change at different rates and its related to the volume thats why we call it related rates so we have a way to relate the volume of the sphere and the radius of the sphere and that is why the radius is always 4/3 pi r cubed|
|3:03||and we know what the 4/3 pi r theory thats just a number think about 4.1 so really this is just the constant something r cubed so i want to know how fast this is changing and i know that the volume dv/dt is increasing at 24 pi i want to know how fast the radius is increasing|
|3:31||when r is 3 i shifted it from lower case and upper case so im going to take the volume of the equation and take the derivative dv/dt is 4/3 pi times 3r squared dr/dt okay so its implicated derivative all related rates are implicated derivatives|
|4:02||its also d the variable, dt
unless we can time
because you get dt squared, okay?
everything else is d variable dt so now we just plug in so notice we cancel those three to make the math easier and you get 24 pi is 4 pi times 3 squared, dr/dt
|4:32||the pi's cancel
and you get 2/3
equals dr/dt and
and 2/3 what? 2/3 of an inch per second
how do we know its an inch?
because the radius is a linear measurement its a line so its inches if we were doing the area itd be inches squared if we did volume itd be inches cubed you guys all know units caue you take high school
|5:30||okay everyone understands this, im going to do a slight tweak on this
to make it a little harder
can i start erasing?
so lets say the volume of the sphere is increasing at 48 pi cubic inches per second how fast is the surface
|6:01||area when r equals 4 and of course you dont need to be reminded that the surface area of a shpere is 4pi r squared so now as a clue what you could do isolate r in one of these equations and plug into the other and that will give you volume in terms of surface area|
|6:31||that would not be fun and you have soemthing to the 3/2 and something to the 2/3 you really dont want to do it that way what you want to do is set up two seperate relationships and the put them together that sounds scary, its really not that hard just start doing things and see what happenes this is going to be just like the other one we got the volume and the surface area the problem is we dont have an easy way to relate volume and surface area, as i said|
|7:03||you could go her and say
thats r and plug that in to volume but that would be very unpleasant
so instead what you do is work them separately
this has shown up on exams in the past, problems like this.
i dont know why i said that, thought id just through that out there
|7:31||you got volume and you got surface area so lets just start with volume we know that dv/dt is 48pi we are looking for ds/dt when r is 4 so lets plug in lets take the derivative of this, we get dv/dt|
|8:02||is 4/3 pi times 3r squared dr/dt sometimes what you should do is get in there and take the derivative plug in things and see if you stumble into the right answer that happens more often than you realize so you cancel the 3s if we plug in r we have dv/dt we can get dr/t why does that be useful well if you go over here|
|8:30||youd get ds/dt is 4 pi times 2r dr/dt we are trying to find ds/dt we know r and from here we can get dr/dt and then you plug in and get ds/dt so 48 pi is 4 pi times 4 squared dr dt|
|9:00||pi's cancel thats why i put 48pi makes your lives easier ill wait, people still catching up So You simplify this and you get 3/4 is dr/dt so far so good? this is simplifying when its comfortable you can write 48 60/4, itll work its just messy|
|9:31||okay now we go over to ds/dt this is what were looking for remeber, so ds/dt is 4 pi times 2 times 4 times 3/4 you do a little canceling and you get 24 pi zand what would the units be? thatd be square inches per second why sqaured inches? because were doing the area|
|10:03||i dont know if we'll expect you to appy the units on the exam, we might
howd we do on this one?
oka, yes no, maybe? Who go this one?
thats a very small number of hands the people who took calculus in high school
|10:30||not even people who took calculus in high school thats not a good sign, alright lets do a couple more|
|11:53||okay you got a cylinderical tank so a tank shaped like a cylinder and the cylinder tank|
|12:00||is filling with water at the rate of 4 cubic feet per min the radius of the tank is 20 feet how fast is the height of the water rising when each is 6 feet and you know the volume of a clinder is pi r squared h of course we know that cause its pi r squared at the circle times h, times how high it is okay lets give you a couple mins|
|12:38||the point is r is constant you have a couple ways to do this first let me erase while we talk the volume in pi r squared h|
|13:03||we know that dv/dt is 4 we know that r is 20 we want to find dh/dt its capital h dH/dt is what were looking for when h is 6 so you can either take the derivative and remeber you already keyed in 0, or you just plug in for r|
|13:34||cause the rdius doesnt change if you were gonna take the derivative youd use the product rule youd get pi r squared dh/dt plus h times 2 pi r dr/dt but the radius is constant so dr/dr is 0|
|14:02||so the volume. sorry thats dv/dt so the volume is pi times 20 squared dh/dt we actually dont care that the height is 6 now you just plug in 4 is 400 pi dh/dt then dvide|
|14:37||so 1/100pi by the way if you have a problem like this on webassign or something remember the pi is in the denominator thats the kind of thing you input incorrectly when doing be very careful its 1 over 100 pi, 100 pi is in the denominator that one what not so bad|
|15:00||we all kind of create the same kind of question lets do one more, now heres the type i really like to put on the exams, of course im not writing the exams but i can tell you the last time i did 125 it was on the exam so its similar to this, now its going to have a conical tank|
|16:41||a conical tank, a tank shaped like a cone is filling with water, so the water is coming in and the height of the tank is 24 meters and the radius of the tank is 8 meters|
|17:02||and its filing at 24 pi cubic meters per meter, thats fast by the way you dont want to be in there how fast is the height of the water rising when the water is 8 meters high and you guys know the bottom of a cone of course cause i wrote it down last class but just in case 1/3 pi r squared h now we can do this as a team or you can do it on your own|
|17:30||doesnt really matter ill let you take a minute to absorb whats going on we'll do it as a team so what do we know? we know that the height h is 24, the radius|
|18:02||is 8, why am i telling you that?
i am telling you that for a very specific reason okay in a cone okay thats a cone, thats the radius and the height at any moment the radium of the radius and the height is always the same so any radius divided by its height
|18:31||will equal any other radius divided by its height and so on, you get similar triangles so since the radius is 8 and the height is 24 that ratio is always going to stay 3 to 1 thats going to be important alright lets fill in dv/dt equals 24 pi we want to find dh/dt when h is 8|
|19:02||so why are we going to wan that information well lets go over here and look at the volume of the equation v is 1/3 pi r squared 8 both radius and height are changing if i take the derivative im going to get the product rule and im going to have a term with dh/dt in it and im gonna have a term with dr/dt in it the problem is i dont know what dr/dt is im only looking for dh/dt|
|19:30||i know dv/dt
i dont know anything about dr/dt
so how can i get around that?
i can use the facts that the radius and the height are related so rewrite the bottom equation and get rid of the radius cause i know that raduis over height is always 1/3 8/24 is a third so the radius will always equal h/3
|20:00||so now i can take that and i can plug it in here i get volume is 1/3 pi h/3 squared times h okay dont be scared off by that its not that hard so v little bit of mathematical alegbra|
|20:30||pi h cube over 27 3 squared is 9 9 times 3 is 27 h squared and h make h cubed take the derivative and you get dv/dt and you get pi/27 times 3 h squared|
|21:00||dh/dt now i just plug in so dv/dt is 24 pi equals pi over 27 times 3. h 8 is h, h squared dh/dt pi's cancel|
|21:36||and i get 27
meters per min
oh and by the way if you cant simplify that, thats okay
doesnt matter if you simplify that perfectly or not
thats not what were testing
we'll try and remind the ta's about that. Ta's are you watching?
we'll try to remind you of that when the time comes to score
|22:06||alright lets do one more
how'd we do on this one?
we did it as a team so i think you did okay
|23:40||you are standing three miles from a launch pad
you want to stand pretty far, on your own
a rocket is launched, takes off and goes up at 4 miles per second
thats pretty fast right?
yea well once they get going they move pretty fast after two seconds how fast is the angle of elevation increasing
|24:02||so you have the rocket over there
blasts off, straight up and youre watching it
how fast is the angle of acceleration increasing
were going to to give you a couple mins.
the rocket is launched it takes of. by the way that cloud that comes out under neath the rocket, you dont want to be anywhere near that its very hot and very big cloud it tends to end things very quickly
|24:32||and you are standing three miles away, you are here and the rocket is going to go up like that you really want that to be pythagorean theorem dont you? you really do but unfortunately its not. you are three miles away and this, we'll call this distance x ds/dt is 4 miles per second|
|25:00||and we'll call this theta because we love greek letters so d theta/dt is what we are looking for and we want to find that when two seconds have passed so x is 8 once the rockets get going theyre going very fast takes them a few seconds to buil up some speed, the acceleration and all that|
|25:31||have you do physics you'll realize how fast they have to go pretty fast so do we have a relationship between x and theta yes we do not sine tangent soh cah toa, so opposite over agacent itd be theta will always equal x over three or if you dont like fraction which a lot of you dont|
|26:04||3 tan theta equals x. you dont have to do that is you dont want to but you can take the derivative 3 secant squared theta d theta/dt equals dx/dt|
|26:34||okay so we can plug in we know dx/dt we have to figure out the second of theta is and we know we have a triangle thats 3 and thats 8 you do a little pi thag and you get square root of 73 so the cosine would be 3 over the square root of 73 so secant is the square root of 73 over 3|
|27:00||its hypothenuse over adjacent this is 3 square root of 73 over 3 squared d theta/dt equals 4 you do a little arethmatic and you get 73 over 3|
dy/ is 4
is 12/73. now what are the units
radians per.. wait that says minutes? oh this should say minutes no thats the old problem radian per second
|28:04||remember never degrees
we like that one
that was tricky?
alright ill give you one more
|29:47||this presumes a little cultural knowledge a runner on first base is heading towards second base so what does that mean? for those who dont know what im talking about|
|30:03||in baseball you have a diamond these bases are 90 feet apart as opposed to pitchers mond which is 60 6 inches, dont forget the inches you got a runner whose standing on first that is a perfect example of a runner very life like and the runner is heading this way|
|30:33||okay and at 2 seconds how fast is the distance between home plate and the runner changing and by the way these are all right angles this is not hard think about it for a minute you can do it alright lets look at this for a minute|
|31:11||lets call the distance between the runner distance between first and second base, lets call that x and that distance x is growing because the runner is heading from first to second so dx/dt is how fast that runner is going and thats about 24 feet per second|
|31:36||ill tel you know youre running at 24 ft/s youre going to be out you have to get closer to 30 f/s if you want to have a shot we want to know how fast this is changing at t equals 2 so when x is going to be 48 b because|
|32:01||youve gone 24 feet per second for 2 seconds and this is the distance between home plate and the runner do we have a relationship between these two well i dont know if they have a relationship but we can relate the distances we know that x squared plus 90 squared equas y squared|
|32:31||right thats pythagorean theorem so lets take the derivative 2x dx/dt plus 0 equals 2y dy/dt thats pretty easy relationship okay cancel 2's, you dont have to but its only makes the arithmetic easier and we know x is 48|
|33:00||we have to figure out what y is
so we can do a little pythag
and 48 squared plus 90 squared
is y suqared
oh 48 squared.
thats not fun to do 2304, 2304? okay
|33:33||theres a trick to that so thats 10404 y is the square root of 10404 we could ask you to do 48 squared on the exam you just take a piece of paper and pencil and write 48 and 48 and you multiply alright so we got 48|
|34:00||times dx/dt which is 24, equals the square root of 10404 dy/dt, this problem is a lot easier in meters so 48 times 24 over the square root of 10404 is your answer, i dont really care what that simplifies to this is actually very straight forward|
|34:33||so far so good?
alright thats related rates, i want to get a little start on the next topic to give you guys an advantage heading into the homework for the next week okay, do you understand this one?
so all of these is the same problem you just have to take the trouble word problem into an equation, once in the equation its very simple
|35:11||so this is gonna be able maxima and minima] which is plural maximum and minimum think about a graph remember the derivative tell us the slope in graphs|
|35:32||you look at the slope of the graph and you say what do we know? we dont know much we know for example the graph is going up here the graph is going up in this first peak the derivative here would be positive and we did a little bit of this before now and we know the derivative here would be negative and the derivative here would be positive again and right here the derivative is 0|
|36:01||those are horizontal tangent lines so if we look at a graph we know where it could have maximums or minimums, maxima or minima try looking at the derivative and see if it is 0 there now these are what we call relative maxima and minima because this is a minimum for this dome but its not actually lower than this number so thats called a relative|
|36:30||maxima or relative minima you all have what we call an absolute maxima and an absolute minima okay thats were the real maximum is because nothing is less or bigger than the real minimum we could use calculus to find these. but why would we care|
|37:01||well thats a good question you care about it for all types of reasons in the business world you have your profit equation and want to know your maxima profit or minima cost in biology world you might want to know miximum growth okay and in chemistry you got all these crazy rates going on physics you launch a projectile, want to know when it hits the top knowing my maxima and minima are is actually very important thats kinda what we use calculus for, lots of problems|
|37:33||use calculus for them for example lets say we have your favorite quadratic equation very straight forward like that and i want to know where the maxima and minima is of that equation by the way thats a probable and its opening up so its going to have a minimum, not gonna have a maxima|
|38:02||so what do i do? i take the derivative and find where its 0 so remember what i told you before this is calc class so when and doubt take the derivative now its give you a cuaral area to that once you take the derivative the thing else to do is find where the derivative is 0 maybe that would be worth something you take 2x plus 8|
|38:31||you it equal to 0 and you get x x=-4 thats called a critical number or a critical value to the book so your technique is try to find critical numbers once we know thats negative 4 we just need to figure out if this is a maximum or minimum and we happen to know in this case but you kind of want to show it thats when things say justify your answer|
|39:02||what you do take a number line and put the critical value on the number line critical number sorry now take a number less than -4 and plug it in to the derivative and take a number greater than -4 and plug it into the derivative see what happens inside to take a numbe rless than -4 like -5 and you plug it in and get -2 so thats a negative, that means the curve is going done here|
|39:33||now take a number gretaer than -4 like 0 when and doubt you want to plug in 0 very easy to plug in 0 you get 8, thats positive the curve is going up here, that must be a minimum so your minimum curves at x=4 -4 now how do we find the y coordinate|
|40:00||you go back to your original equation and you plug it in and you get 16 32 minus 16 plus 12 is -4 its just a coincident your minimum is at the point -4,-4 so we could ask you wnat value of x does this have a minimum we could say what is the minimum value or we could say what were the coordinates of the minimum|
|40:31||unfortunately i picked one way were they come out the same now ill pick one where they dont come out the same|
|41:00||alright try something like that i want to find minimums and maximums rule numbe rone take the derivatove f of x is x cubed minus6x suqared plus 9x minus 5 take the derivative and you get 3x squared minus 12x plus 9|
|41:32||take that derivative and set it equal to 0 that factors quit nicely you get x equals 1 and x equals 3 you dont have calculators so im not giving you anything to dramatic|
|42:15||okay lets sin test this take a value less than 1 like 0 plug it into the derivative you get 0-1 is -1|
|42:32||0-3 is -3 two negatives multiply together make a positive the graph is going up to the left of one remember e asked you in the previous example where its incnreasing and decreasing its the same thing now pick a number between one and three like two you get 1 times negative one, thats a negative numbe so the graph is going down here so this is where the maximum is|
|43:02||now you pick a numer greater than 3 like 4 and you plug it in and you get thre times 1 is a positive number the graph is going up again so thats the minimum we have a maxima at x=1 and the minmuma at x=3 now you just need the y coordinate so when x equals 1 f of 1 is 1 cubed minus six times one squared|
|43:32||plus nine times one minus five is negative 1 you plug in three and you get three cubed minus 6 times three squared plus nine times 3 minus 5 is negative 5 therefore you have a maximum|
|44:03||at 1 comma -1 and a minimum at 3,-5 of course you can graph this. we love graphing were gonna ask you to do that just so you know of course it'll make your life terrible by asking you to graph something and trigonometry are equal awful or maybe no|
|44:31||okay thats enough for one day|