| Start | so we use the first and second derivative to give us information about what a curve looks like using that information we can then sketch a curve so first let's take a relatively simple curve I had one how about |
| 0:32 | of course i have to triple check that I got that right no, wait, nope hi okay so are we going to sketch that curve well were gonna need some information. were gonna want to know where the curve is going up, where the curve is going down, where the curve has a maximum and minimum. we gonna want to know where the concavity is concave down |
| 1:06 | or when its concave up so for those of you who were not here for some cavity that's the
second derivative okay
concavity, when I have a hole in my tooth. con cavity.
right so what do we you do? calculus class take the derivative okay and then take the second derivative |
| 1:48 | okay so now we have the first derivative and the second derivative and then we set them equal to zero because why not. 6x squared minus 6x minus 12 equals |
| 2:01 | zero and were gonna get some information out of that so lets divide it by six or factor out a 6 and factor so far so good? now we know where the first derivative is zero where the first derivative |
| 2:36 | iis 0 could be a max it could be a min or could be nothing all right we need a little more information but generally a maximum the curve is going up and then the derivative is zero and then the curve is going down and the minimum of the curve is going down, zero then going up. if you get a place where the first derivative is zero and it does |
| 3:02 | not change sign from the left side to the right side it's not a maximum or a minimum it's just a stopping place it just got flat for me and the curve kind of went like this. it's flat there but that not a maximum or minimum you need to go like this or like that okay then the derivative is flat right here and the slope is going up going down or from going down to going up okay so we're gonna figured it out by |
| 3:30 | sign testing.
negative 1, 2 thats our zeross and we signed test you take the number less than negative one and we plug into the derivative and see what's sign you get. so you take negative two and you plug it in and you can get a negative times a negative is positive. that |
| 4:00 | tells us the curve is going up in this zone. to the left of negative 1 now you take a number between negative 1 and 2 and you plug it in, like zero you plug it in the derivative. you get six times a negative times a positive that's a negative the curve must be going down and so that must be a maximum okay now we'll try a point to the right of 2, like three. plug that in you get six times one |
| 4:36 | times four those are all positive. the curve is going up here so that must be a minimum so we graph it, we know we have a maximum at negative one and we have a minimum at 2 x equals -1 and X equals 2 that's not the maximum value that what your x coordinate of the maximum is. okay now we get the second derivative and set it equal to 0 you get x is a half so we can test where its second derivative is positive or |
| 5:12 | negative thats a 2. all right take a number less than a half like zero plug it into the second derivative, you get a negative so the curve is going down there |
| 5:41 | now you pick a number greater than 1/2 like one and you plug it in and this is positive. so my bad. this is concavity so here it's concave down here its concave up which kind of makes sense if you look at what's going on here |
| 6:00 | and if 1/2 it's changing the concavity so that's a point of inflection a point of inflection is where the second derivative 0 and the curve switches from one concavity to the other concavity. now we can graph this except we need the actual coordinates of the maximums and minimums so to find the actual coordinates we have the x coordinates, we need the y cordinates |
| 6:32 | so let's see when X is negative 1. now how do we find the Y coordinates we plug into the original equation always into the original equation, because thats the equation y equals ok so if you want to find the Y coordinates you plug in negative 1 and you get 7 negative 2, minus 5 |
| 7:00 | minus 12 is 7, yes. you plug in two and you get 16 minus 12 is 4, minus 24 is minus 20 and we get a point of inflection at a quarter minus 3/4 thats a 1/2 minus 6 is minus 6 and 1/2 ok so those are the coordinates of your maximum, minimum and point of inflection. -1 to the 7th |
| 7:33 | a half, -13 halves and 2,-20 doesn't really matter. you got a maximum and minimum and notice how the curve switches from being concave up here to concave down there okay and that's your sketch the curve doesn't have to be perfect |
| 8:00 | this isn't graphing class, its sketching. okay alright now when you doing, when you did
this on monday we did relatively
straight graphs and I know you guys said can
you make them harder. Of course i can make them harder. how do I make them harder?
well let's see if you all can graph this one. e to the minus x squared over four so if |
| 8:42 | we'll give you a couple mins y=e to the minus x over 4 any time we want to screw you guys up we do e or the natural log trig or you could do e to the cosine, that would be really fun lets take the derivative of this |
| 9:01 | the derivative isnt to bad you do its e to the minus x squared over 4 times the derivative of negative x squared over 4 you dont need the quotient rule for that make that negative a quarter times x squared thats negative a quarter times 2x or -x over 2 i mean you can use the quotient rule but its kind of silly |
| 9:34 | because -x squared over 4 is just negative a quarter x squared okay so its negative 2/4 x alright were gonna set this equal to 0 heres a very important thing to remember the e term is always positive so this is never 0 the only thing you have to worry about is when x over 2 is 0, which is at 0 |
| 10:05 | okay we found where the first derivative is 0 that wasnt so bad of course its easier for me cause i stand in the front but not so bad. now lets look at the second derivative we have to use the product rule so we can take the first function times the derivative of the second plus the second function times the derivative of the first so first function |
| 10:31 | times the derivative of e minus x squared of a quarter. we just found that, thats is this right, e to the minus x squared over 4 times -x over 2 plus the derivative of the first function which is negative a half e to the minus x squared over 4 |
| 11:00 | this can be simplified to x squared over 4, e to the minus squared x over 4 minus 1/2 e minus x squared over 4 so were gonna set that equal to 0 x squared over 4 |
| 11:31 | e to the minus x squared over 4 minus 1/2 e to the minus x squared over 4 equals 0. again the e term is always positive so you can pull that out e term is always positive, so you pull that out and youre left with x squared over 4 minus -1/2 we can ignore that and you get x squared over 4 equals a 1/2 |
| 12:00 | or x squared equals 2
which means our 0's
at x quals plus or minus
the square root of 2
howd we do so far on this?
not so good, have trouble to get to the derivative? yea you should be able to take the derivative of something like this this is wouldnt be suprised i wouldnt ask this personally but if we put this on the exam in two weeks |
| 12:34 | we expect you to take the derivative of that or something like that you dont have to do the setting to 0 part, that wont e on the test but you should be able to take the derivative of it as i annouced on monday we will have practice problems before the exam i am presuming we will have practice problems before the exam lets do the sign test |
| 13:02 | first derivative is 0 at 0 now remeber e to the minus x squared over 4 is always positive so for a number less than 0 negative x/2 is going to become positive because youre putting a negative number into x but this is positive and this is negative again this is always positive so if you have a number less than 0, when x is negative this becomes positive so the whole thing becomes positive |
| 13:30 | and when you have a number greater than 0 this is negative so the whole thing is negative the curve is going up here and down there you have a maximum at 0 now for the second derivative plus and minus the squared root of 2 square root of 2 is somewhere between one and 2 correct cause the square root of 1 is 1 and the squared root of 4 is 2 1.414 but that doesnt matter |
| 14:01 | pick a number to the left f radical to like negative 2 and youre gonna be plugging it in here now the e term is always psoitive you just care where this is positive or negative so if you try negative 2 4/4 is 1 minus a half is positive its positive here so youre concave up the curve is going up and its increasing so its gonna have that kind of look to it |
| 14:32 | now you pick a number between negative square root of 2 and square root of 2 like 0 you plug in 0 here this is positive, remeber this is always positive 0 minus a half is negative so this is negative so the whole this is negative the curve is concave down here remeber changes from up to down here at 0, so theres gonna be a part of it where its going up and concave down |
| 15:02 | and a part where its going down and concave down, which makes sens because its a maximum there one more time. youre checking concavity at 0 this e term is always positive 0-1/2 is negative which is concave down here now you try a number bigger than the square root of 2 like 2 this is 4/4 which is 1, minus a half thats positive so its positive, so its concave up again |
| 15:34 | and now last we need some y coordinates so lets find a couple y coordinates i like to make a little t chart, personally i got 0 negative square root of 2 positive square root of 2, at 0 where do we plug it in, we only plug it into the originally equation e to the 0 is 1 |
| 16:02 | when i have negative radical 2, radical 2 squared is 2
2 over 2 is a half, so this is e to the 1/2
i dont know what e to the negative a half is
is a small number
its one over the square root of e
and when you plug in square root of 2 you get the same answer. you also get
e to te negative 1/2
thats some decimal, not important
so whats that going to look like?
|
| 16:33 | when at - and 1 is a maximum and it starts out, concave up its going up and its concave up and right here it switches to concave down and then here it switches back to concave up so your points of inflection are negative radical 2 |
| 17:00 | e to the negative 1/2
and radical 2
e to the negative a half
and your maximum is there at 0,1
anybody get this?
its okay, thats why your learning you should feel so accomplish right now today was the hard day remember, monday was the easy day |
| 17:32 | should we find one a little less difficult?
ill wait till everyboy got this copied down you understand what we did? one more time, you take the derivative and set it equal to 0 you can find where its 0 by ignoring the e term and find where the rest of it is 0. thats at 0 same with the second derivative the second derivative is not as messy as it looks remember the e to negative x squared over 4 factors out of everything you find where the rest of this is 0 |
| 18:01 | now that you go the coordinates you sign tyest. tells you where the graph is going up where the graph is going down where its concave up and concave down with that information you can you can find, when you know the coordinates, the actually y coordinates you can graph it so if we put this on the exam, this is a big question its a lot of parts to it lots of places to get partial credit and if you mess up somewhere, we will follow through with your thinking and try to figure out where you went wrong |
| 18:31 | and give you as much partial credit as possible maybe alright how about |
| 19:03 | okay how bout this one?
we had a webassign very similar to this lets see if we can graph it when i say that i mean you guys take the derivative because this is calculus class derivative of cosine is minus sine and i take the derivative of cosine squared, thats 2 |
| 19:31 | cosine theta to the 1
negative sine theta
howd we do on that?
did we miss a miss sine in there? entirely possible alright we have to set that equal to 0 lets take out a minus 2 sine theta first were left with 1 plus 2 cosine theta alright 1 plus cosine theta |
| 20:02 | that would also be easier to take the second derivative of this would be 0 where sine theta is 0 or when cosin theta is negative 1 sine theta is 0 at 0 and pi and cosine of theta is equal to -1 at pi you guys all get that yes, i see some puzzled faces |
| 20:33 | these are the same so you dont have to count twic
cause your critical points are 0 and pi
your critical numbers are 0 and pi
why did i not include 2 pi?
because i dont include 2 pi thats all i stopped there before 2 pi the second derivative |
| 21:00 | i could use either form but the factor form is a little easier so the derivative of minus 2 times theta or minus 2 sine theta sorry times the derivative of 1 plus cosine theta which is negative sine theta plus the derivative of negative 2 sine theta which is which is negative 2 cosine theta times |
| 21:30 | one plus cosine theta
and we have to find where that is 0
so to simplify that, thats
2 sine squared theta
minus 2 cosine theta
minus 2
cosine squared theta
thats our second derivative
anybody know how to solve that?
|
| 22:00 | you can take sine square and replace it with 1 minus cosine squared now you have 2 minus 2 cosine squared theta minus 2 cosine theta minus 2 cosine squared theta or |
| 22:32 | minus 2 cosine squared theta
minus 2cosine
theta plus two.
equals 0 thats why you learn trig identities we can divide that through by negative 2 and make this 2 cosine squared theta plus cosin theta minus one equals 0 |
| 23:01 | 2 cosine theta cosine theta plus one minue 1 so again you get the second derivative by taking the first derivative and using the product rule you can combined terms and you can take sine squared and replace it with one minus cosine squared you also couldve taken 2 sine squared and 2 cosine squared and use the double angle formula but i wouldnt advise that |
| 23:33 | i replaced those with 1-cosin squared you simplify it a bit and it factors very nicely so you get cosine theta is 1/2 or cos theta is negative 1. where is cos theta equal a half you love the unit circle pi over 3 and 5 pi over 3 and equals -1 at pi |
| 24:01 | so what do you do when the critical number shows up again in the second derivative you need that its gonna turn out not to be of any value in the first derivative. youre not gonna get a sign change there youll see alrighty should we do a sign test you guys wanna give up at this point? yea 0 and pi |
| 24:31 | lets try a number between 0 and pi. how bout pi over 2 plug into the first derivative sin of pi over two is one so this is negative 2 cosine of pi over two is 0 so this is one, so this is negative curves going down how bout bigger than pi, how bout 3pi/2 well lets see. the sin of three pi over two is negative one |
| 25:01 | so this would be positive
thats 0 again. so now its positive, so the curve is going up here
so thats a minimum
0 is a critical point. however 0 is also the end of the interval
so we really dont know whats going on
we dont really look at whats going on to the left of it, i mean we could
but dont need to.
what do you think is going on, on the other side of it probably going up okay now lets do the second derivative |
| 25:36 | pi over 3 pi, 5 pi over 3 so you take a number less that pi over 3, like 0 and you plug it in and you get negative 4 minus 2 plus 2, its negative here |
| 26:00 | concave down
plug in a number between pie over 3 and pi like pi over 2.
cosin pi over two is 0 its positive, son its concave up try a numer between pi and 5pi/2 like 3 pi over 2 cosine of 3pi/2 is also 0 so its concave up again does not change the concavity thoguh and 5 pi over three. so lets try |
| 26:30 | well you know 2 pi isnt in the interal, you might as well test 2pi and see what happens and you get back to where you started so like that and now we just have to find the y coordinates thats actually theta so lets see we go t0 pi pi/3 5pi/3 |
| 27:03 | 2 cos 0 is 2 cos square root of 0 is one so thats 3 two cos pi is negative 2 plus 1 that negative 1 and pi over 3 you get two times a half is one plus a half squared which is a quarter, so you get 5/4 and 5pi over 3 you get 2 times cos pi over 3 is 1 |
| 27:30 | and its same value this should be entertaining to graph pi/3 and 5 pi over three lets see we have a minimum here at pi pi,-1 we got |
| 28:02 | what else is happening? lets see its going something like that concavity, concave down then it switches to concave up, stays concave up and then switches to concave down and if you kept going these are probably maximums and minimums but were not sure we dont care, cause we only want the interval from 0 to 2pi technically theres a hwole there okay |
| 28:30 | trigonometry we'll give you more practice i know im tired too, but we have to keep going we couldnt stop i know youre begging me, i can read your minds just make the pain end but we need to do a little more now lets talk about absolute maximums and absolute minimums. so far |
| 29:00 | weve just been talking about maximum and minimums
okay. any questions
no, i know ou guys have questions
but before i start erasing, anything i should go over?
no what i miss? no questions? alright remember you can watch the video tape many times |
| 29:40 | suppose you have one of those nice simple polynomials
like that
actually im gonna exaggerate that a bit, hang on.
do it again like that |
| 30:04 | so right now, if this was infinite if i asked you wheres the maximum youd say well thats a maximum you say thats a minimum but thats reall a local maximum its a maximum in its zone and thats a minimum in its zone, its area but thats actually not that maximum value, this goes up to infinity and thats not actually the minimum value it goes down to negative infinity |
| 30:30 | so we talk about absolute maximum things are a little bit different. now suppose i only wanted in from the interval here to here. so this part of the curve doesnt exist anymore well now thats a absolute maximum and thats and absolute minimum if instead lets say i took the interval from here to here even though thats a relative maximum |
| 31:01 | thats an absolute maximum even though, sorry, thats a local minimum thats an absolute minimum\ so sometime you need to do more than just the critical numbers to determine the absolute maximum and the absolute minimum for example |
| 31:49 | notcie they give you an interval this time one of them clues one of the clues when we ask for absolute maximum and minimum will be there will be an interval |
| 32:01 | and an interval, what that means is you have to test whats going on in the function at the end points of the interval so lets take the derivative y prime is 6x squared minus 6x minus 12 set it equal to 0 youve seen this one before, pull out the 6 |
| 32:31 | x squared minus x minus equals 0 that was an exact problem on the final last year this is the exact problem i did on the final thats not a clue, im just telling you so i like this problem so you can factor this and get x minus 2 x plus1 = 0 and you get x equals 2 x equals -1 |
| 33:01 | now i ask you whats the absolute maximum and absolute minimum or we can sign test this we sign test this and you can find the local maximum and minimum the absolute max and absolute min is actually easier so lets do both first lets just do a sign test pick a number less than negative 1 like negative 2 plug it in the derivative |
| 33:30 | and you get positive value you have a curve going up here pick a number between -1 and 2 like 0 plug in you get 6 times -2 times 1 thats negative, the curves going down and pick a number greater than 2 like 4 plus in. you get 6 times 2 times 5, thats positive |
| 34:01 | its going up again so thats a minimum we dont know if its and absolute maximum or relative maximum. or if its maximum min or relative min so now would we test, well remember i just need the y values lets take oour critical number and you also take the end points and now you find the value of y. you get the value, you get the values, its the absolute values. the absolute maximum and the small value is the absolute minimum |
| 34:39 | take negative 1, you plug in and get -2 minus 3 minus 5 plus 12 is 7 plus 1 is 8 you take 2 and you get 16 minus 12 is 4 minus 24 is minus 20 plus 1 is minus 19 |
| 35:03 | you take negative 2, you get -16 minue 12, minus 28 plus 24 is minus 4 the -3 and you plug in 3 and get 54 minus 54 is 0 minus 36 plus 1 is minus 35 |
| 35:30 | thats not right sorry 27 minus 9, minus 8 this is your absolute max and this is your absolute minimum |
| 36:02 | suppose however i took the exact same thing and i did it on a different interval so now, i already know where the critical values are same equation as before there negative 1 and 2 and negative 4 and 5 |
| 36:34 | now i just have to plug in and test what happens at -4, what happens when i plug in 5 when i plug in -4, i get -128 minus 48 is -196 plus 48 is 128 plus 1 is 129 doesnt sound right did i do a minus sign there |
| 37:00 | minus 128 thats minus 120 minus 127, my bad you can check the video tape later now i plug in 5 and get 250 minus 75, is 175 i get 116 this time this is the maximum and this is the minimum |
| 37:30 | you get the idea all that means after you get your critical values you also have to test the end points that will help you figure out what the absolute maximum and absolute monimum lets have you guys do one |
| 38:39 | you dont have to graph it but what are the coordinates of the absolute maximum and absolute minimum now, all we ask for is the absolute maximum and absolute minimum, you dont have to graph it you dont have to tell me the local maximum and minimum you just have to tell me whats the absolute maximum and absolute minimu is |
| 39:00 | so take the derivative
you get 3x squared minus 12x
plus 9, set it equal to 0
divide everything by 3, you get x squared minus 4x plus 3
equals 0, so this is 2 0s
critical numbers, x equals 1, x equals 3.
|
| 39:32 | how we doing so far?
now, i dont have to actually sign test this i just need to know the absolute maximum and where the absolute minimu is trying 4 values to find the y values and the biggest one is the absolute maximum and the smallest one is the absolute minimum i plug in negative, -1 minus 6 is negative 7 |
| 40:01 | minus 9 is negative 14
64 minus 96, is minus 32
plus 36 is 4 plus 2 is 6
how we doing? okay?
plug in 1 thats 6 thats okay you can get two 6's and plug in 3 |
| 40:30 | and i get
2
a little slower on that one
the absolute maximum is 6
and the absolute minimum is -14
now if we would actually graph this, this isnt a regular graph graph?
we sign test and at -2 for example this is positive so this is going up |
| 41:01 | and it comes down and then it goes up
so if you were gonna graph it
did i do that right?
sorry positive one that says 6 right and it has a minimum here at 3,2 |
| 41:30 | and all its telling you is at 4 it got back to the level it was at 1 and if we found the seond derivative we find the point of inflection right there cause it goes from concave up, down, to concave up tis turns out to be the absolute maximum, absolute maximum in two different spots if you went beyond 4, 5 and beyond 4.01 or whatever, anything beyond 4. thats the absolute maximum |
| 42:07 | well spend a little more time on this on monday |