If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:9:50

Video transcript

land things up a little bit here alright so in the last video I set up the scaffolding for the quadratic approximation which I'm calling Q of a function an arbitrary two variable function which I'm calling F and the the form that we have right now looks like quite a lot actually we have six different terms now the first three were just basically stolen from the local linearization formula and written in their full abstractness it almost makes it seem a little bit more complicated than it is and then these next three terms are basically the the quadratic parts we have what is basically x squared we take it as X minus X naught squared so that we don't mess with anything previously once we plug in x equals x naught but basically we think of this as x squared and then this here is basically x times y but of course we're matching each one of them with the corresponding X naught Y naught and then this term is the Y squared and the question at hand is how do we fill in these constants the coefficients in front of each one of these quadratic terms to make it so that this guy queue hugs the graph of F as closely as possible and I showed that in the very first video kind of what that what that hugging means now in formulas the goal here I should probably state what it is that we want is for the second partial derivatives of Q so for example if we take the partial derivative with respect to X twice in a row we want it to be the case that if you take that guy and you evaluate it at the point of interest the point about which we are approximating it should be the same as when you take the second partial derivative of F or the corresponding second partial derivative I should say since there's multiple different second partial derivatives and you evaluate it at that same point and of course we want this to be true not just with the the second partial derivative with respect to X twice in a row but if we did it with the other ones like for example let's say we took the partial derivative first with respect to X and then with respect to Y this is called the mixed partial derivative we want it to be the case that when we evaluate that at the point of interest it's the same as taking the mixed partial derivative of F you know with respect to X and then with respect to Y and we evaluate it at that same point at that same point and remember for almost all functions that you deal with when you take this second partial derivative you know where we mix two of the variables it doesn't matter the order in which you take them right you could take it first with respect to X than Y or you could do it first with respect to Y and then with respect to X usually these guys are equal there are some functions for which this isn't true but we're going to basically assume that we're dealing with functions where this is so that's the only mixed partial derivative that we have to take into account and I'll just kind of get rid of that guy there and then of course the final one final one just to have it on record here is that we want the partial derivative when we take it with respect to Y two times in a row and we evaluate that at the same point there's kind of a lot of this is there's a lot of writing that goes on with these things and that's just kind of par for the course when it comes to multivariable calculus but you take the partial derivative with respect to Y at both of them and you want it to be the same value at this point so even though there's a lot going on here all I'm basically saying is all the second partial differential information should be the same for Q as it is for F so let's actually go up and take a look at our function and start thinking about what its partial derivatives are what its first and second partial derivatives are and to do that let me first just kind of clear up some of the board here just to make it so we can actually start computing what this what this second partial derivative is so let's let's go ahead and do it this first this partial derivative with respect to X twice what we'll do is I'll take one of those out and think partial derivative with respect to X and then on the inside I'm going to put what the partial derivative of this entire expression with respect to X is well we just take it one term at a time this first term here is a constant so that's that goes to zero the second term here actually has the variable X in it and when we take its partial derivative since this is a linear term it's just going to be that constant sitting in front of it so it'll be that constant which is the value of the partial derivative of F with respect to X evaluated at the point of interest and that's just a constant all right so that's there this next term has no X's in it so that's just going to go to zero this term is interesting because it's got an annex in it so when we take its derivative with respect to X that 2 comes down so this will be 2 times a whatever the constant a is multiplied by X minus X not that's what the derivative of this component is with respect to X then this over here this also has an X but it's just showing up basically as a linear term and when we treat Y as a constant since we're taking the partial derivative with respect to X what that ends up being is B multiplied by that what looks like a constant as far as X is concerned Y minus y naught and then the last term doesn't have any X's in it so that is that is the first partial derivative with respect to X and now we do it again now we take the partial derivative with respect to X and I'll maybe I should actually clear up even more of this guy and now when we take the partial derivative of this expression with respect to X F sub X of X naught Y naught that's just a constant so that goes to 0 2 times a times X that's going to we take the derivative with respect to X and we're just going to get 2 times a and this last term doesn't have any X's in it so that also goes to 0 so conveniently when we take the second partial derivative of Q with respect to X we just get a constant it's this constant 2a and since we want it to be the case we want that this entire thing is equal to well what do we want we want it to be the second partial derivative of F you know both times with respect to X so here I'm going to use the subscript notation over here I'm using the kind of Leibniz notation but here just second partial derivative with respect to X we want it to match whatever that looks like when we evaluate it at the point of interest so what we could do to make that happen to make sure that 2a is equal to this guy is we set a equal to 1/2 of that second partial derivative evaluated at the point of interest okay so this is something we kind of tuck away we remember this this is we have solved for one of the constants so now let's start thinking about another one of them like it's actually don't have to scroll off because let's say we just want to the mixed partial derivative here where if instead of taking it with respect to X twice we wanted to let's say I'll kind of erase this we wanted to first do it with respect to X and then do it with respect to Y then we can just kind of edit what we have over here and we say we already took it with respect to X so now as our second go as our second go we're going to be taking it with respect to Y so in that case instead of getting to a let's kind of figure out what it is that we get when we take the derivative of this whole guy with respect to Y well this looks like a constant this here also looks like a constant since we're doing it with respect to Y and no Y's show up and the partial derivative of this just ends up being B so again we just get a constant this time it's B not you know - previously it was two-way but now it's now it's just B and this time we want it to equal the mixed partial derivative so instead of saying F sub xx I'm going to say F XY which basically says you take the partial derivative first with respect to X and then with respect to Y we want this guy to equal the value of that mixed partial derivative evaluated at that point so that gives us another effect that means we can just basically set B equal to that and this is this is another fact another constant that we can record and now for C now for C when we're trying to figure out what that should be the reasoning is almost identical it's pretty much symmetric we did everything that we did for the case X and instead we do it for taking the partial derivative with respect to Y twice in a row and I encourage you to do that for yourself it'll definitely solidify everything that we're doing here because it can seem kind of like a lot in a lot of computations but you're going to get basically the same conclusion you did for the constant a it's going to be the case that you have the constant C is equal to one half of the second partial derivative of F with respect to Y so you're differentiating with respect to Y twice evaluated at the point of interest so this is going to be kind of the the third the third fact and the way that you get to that conclusion again it's going to be almost identical to the way that we found this one for X now when you plug in these values for a B and C and these are constants even though there's you know we've written them as formulas they are constants when you plug those in to this full formula you're going to get the quadrat approximation it'll have six separate terms one that corresponds to the constant to the correspond to the linear fact and then three which correspond to the various quadratic terms and if you want to dig into more details and kind of go through an example or two on this I do have an article on quadratic approximations and hopefully you can kind of step through and do some of the computations yourself as you go but in all of this even though there's a lot of formulas going on it can be pretty notationally heavy I want you to think back to that original graphical intuition here let me let me actually pull up the graphical intuition here so if you're approximating a function near a specific point the quadratic approximation looks like this curve where if you were to chop it in any direction it would be it would be a parabola but it's hugging the graph pretty closely so it gives us a pretty close approximation so even though there's a lot of formulas that go on to get us that the ultimate visual and I think the ultimate intuition is actually a pretty sensible one you just you're just hoping to find something that hugs the function nice and closely and with that I will see you next video