
English: 
- [Voiceover] So one common
type of problem that you see
in a number of multivariable
calculus classes
will say something to the
effect of the following:
find and classify all of
the critical point of,
and then you'll insert some
kind of multivariable function.
So first of all, this
idea of a critical point
basically means anywhere where
the gradient equals zero.
So you're looking for
places where the gradient
of your function, at some kind of input,
some specified input X and
Y that you're solving for,
is equal to zero.
And as I talked about in
the last couple of videos,
the reason you might want
to do this is because
you're hoping to maximize the function
or to maybe minimize the function.
And now the second requirement
of classifying those points,
that's what the second
derivative test is all about.
Once you find something, or
the gradient equals zero,
you want to be able to determine,
is it a local maximum,
is it a local minimum
or is it a settle point?
So let's go ahead and
work through this example.
The first thing we're gonna
need to do if we're solving
for when the gradient equals zero,
and remember, when we say equals zero,
we really mean the zero vector,
but it's just a convenient
way of putting it all

Korean: 
다변수 미적분학 시간에
자주 등장하는 문제 유형 중 하나는
다음과 같습니다
다음 함수의 임계점을 모두 구하시오
그리고 다변수 함수 하나가 주어집니다
임계점이란 것은 기본적으로
기울기가 0이 되는 모든 점을 이야기합니다
따라서 함수의 기울기가
어떤 특정 값에 대해
즉 특정 x와 y에 대해
0이 되는 지점을 찾으면 됩니다
지난 몇 개의 영상에서 이야기했듯
이러한 연산의 목적은
함수의 최댓값,
혹은 최솟값을 찾기 위함입니다
임계점을 분류하기 위해서는
이계 도함수를 이용할 수 있습니다
기울기가 0이 되는 점을 찾게 되면
이 점이 어떤 점인지,
즉 극대인지, 극소인지,
둘 다 아닌지 등을 구분할 수 있습니다
이 예제를 풀어봅시다
기울기가 0이 되는 점을
찾는 것이 목표입니다
여기서 0이라 함은
영벡터를 의미합니다.
보기 편하게 그냥 0이라 쓰는 겁니다

Korean: 
편도함수를 둘 다 구해봅시다
x에 대한 편도함수를 먼저 구해봅시다
첫 항은
3X^2 곱하기 Y를 X에 대해 미분해주면
2가 내려오면서 6 곱하기 X 곱하기 Y가 됩니다
Y세제곱은 Y를 상수 취급 하므로
Y세제곱 역시 상수입니다
-3X^2는
2가 내려오면서
-6X가 됩니다
써주도록 하고
그리고 여기 3Y^2 역시
Y가 상수인 관계로
상수가 되고 미분하면 0이 됩니다
X에 대해서는 말입니다
이제 Y에 대한 편도함수를 구합시다
첫 항의 계수는 상수인
3X^2입니다
이제 X가 상수니까요
거기 Y가 곱해져있습니다
따라서 미분하면 3X^2겠죠
두번째항은 -Y세제곱이니
미분하게 되면 -3Y제곱이
될 것입니다
-3Y^2

English: 
on one line, we take
both partial derivatives.
So the partial derivative
with respect to X is,
well, this first term,
when we take the derivative
of three X squared times
Y with respect to X,
the two hops down so we
have six times X times Y.
Y cubed, well, Y looks like a constant
so Y cubed looks like a
constant minus three X squared,
so that two comes down.
So we're subtracting off six times X.
Six times X.
And then again, this three Y squared term,
Y looks like a constant so everything here
looks like a constant
with zero derivative,
as far as the X direction is concerned.
And we do partial of f with respect to Y,
then this first term looks
like some sort of constant,
three X squared.
X looks like a constant so
some kind of constant times Y.
So the whole thing looks
like three X squared.
The second term, minus X,
minus Y cubed, excuse me,
looks like minus three
Y squared when we take
the derivative.
Minus three Y squared.

Korean: 
다음 항은 X로밖에 안 이루어졌으니
여기서는 상수로 취급합니다
마지막 항은 2가 지수에서 내려옵니다
Y제곱을 미분하는 거니까요
따라서 -6Y가 나옵니다
임계점을 찾을 때는
우선 이 두 식을 0으로 놓아야 합니다
첫 식을 0으로 놓으면
6X로 묶어 간단히 할 수 있죠
그러면 6X 곱하기 (Y-1)로 정리 가능합니다
이게 0이 되는 거죠
이게 의미하는 바는
6X가 0이거나,
즉 X가 0이거나,
Y-1이 0,
즉 Y가 1이라는 것입니다
둘 중 하나는 참이어야 합니다
이게 첫 조건입니다
밑으로 내려가서

English: 
And then this next term only
has an X so it looks like
a constant as far as Y is concerned.
And then this last term,
we take down the two
because we're differentiating Y squared,
and you'll get negative six
Y, negative six times Y.
So when we are finding
the critical points,
the first step is to set both
of these guys equal to zero.
So this first one, when we
do set it equal to zero,
we can typify a bit by
factoring out six X.
So this really looks like six
X multiplied by Y minus one.
And then that's what we're
setting equal to zero.
And what this equation
tells us is that either
it's the six X term that equals zero,
in which case that would
mean X is equal to zero;
or it's the case the Y
minus one equals zero,
in which case that would
mean that Y equals one.
So at least one of these
things has to be true.
That's kind of the first
requirement that we found.
Let me scroll down a little bit here.

Korean: 
2번째 식을 0으로 놓아주면
어떻게 풀지 곧바로 눈에 들어오지는
않는 것을 알 수 있습니다
하지만 위에서 얻은 결과를
대입해보면 뭔가 얻을 수 있을 겁니다
X=0을 대입해
어떤 결과가 나오는지 봅시다
3X^2는
0이 될 것이고 그럼
-3Y제곱 빼기 -6Y
=0이 남습니다
인수분해를 해줍시다
-3Y로 묶읍시다
-3Y를 앞으로 빼주면
첫 항에는 Y가 남고
둘째 항에서는 2가 남습니다
+2인 이유는
-3을 앞으로 빼냈기 때문입니다
이 식이 0이 됩니다
이게 의미하는 건
-3Y= 0, 즉
Y=0이거나,

English: 
And for the second equation,
when we set it equal to zero,
it's not immediately
straightforward how you would
solve for this in a nice
way in terms of X and Y,
but because we've already
solved one, we can kind of
plug them in and say, for
example, if it was the case
that X is equal to zero,
and we kind of wanna see what
that turns our equation into,
then we would have, well, three X squared
is nothing that would be zero
and we'd just be left with
negative three Y squared minus six Y
is equal to zero.
And that, we can factor out a
bit, so I'm gonna factor out
a negative three Y.
So I'll factor out negative three Y,
which means that first
term just has a Y remaining
and then that second term
has a two, a positive two,
since I factored out negative three.
So positive two and that equals zero.
So what this whole situation
would imply is that either
negative three Y equals
zero, which would be,
which would mean Y equals zero;

English: 
or it would be the case
the Y plus two equals zero,
which would mean that Y
is equal to negative two.
So that's the first
situation where we plug in
X equals zero.
Now alternatively, there's the possibility
that Y equals one, so we
could say Y equals one;
and what that gives us
in the entire equation,
we still have that three X squared
because we're kind of solving
for X now, three X squared,
and then the rest of
it becomes, let's see,
minus three times one squared,
so minus three, we're
plugging in one for Y
then we subtract off six,
plugging in that one for Y again,
and that whole thing is
equal to three X squared
then minus three minus six.
So I'm subtracting off nine.
So from here, I can
factor out a little bit,
and this will be three multiplied
by X squared minus three.
And what that implies then,
since this whole thing
has to equal zero, what that
implies is that X squared

Korean: 
Y+2=0, 즉
Y=-2라는 의미가 됩니다
이게 바로 X=0을 대입했을 때의
결과입니다
다음 고려해야 할 것은
Y=1인 경우입니다
식에 대입해주면
3X^2는 그대로 남습니다
이번엔 X에 대한 식이 나올 것 같습니다
나머지 부분은
-3 곱하기 1의 제곱은
-3이 되겠죠, Y에 1을 대입했으니까요
그리고 6을 빼줍니다
역시 Y에 1을 대입했습니다
그럼 결과적으로 3X^2에
빼기 3 빼기 6,
즉 빼기 9가 됩니다
공통부분을 묶으면
3에 X제곱 빼기 3을 곱한 것이 될 겁니다
이게 의미하는 건
이 식 전체가 0이 되어야 하니
X제곱 빼기 3이

English: 
minus three is equal to zero
so we have X is equal to
plus or minus the square root of three.
And maybe I should kind of specify,
these are distinct things that we found.
One of them was in the
circumstance where X equals zero,
and then the other was where
we found in the circumstance
where Y equals one.
So this give us a grand
total of three different
critical points because
in the first situation,
where X equals zero, the
critical point that we have,
well, both of them are going to have
an X coordinate of zero in
them, X coordinate of zero;
and the two corresponding
Y coordinates are zero
or negative two.
So you have zero or negative two.
There's kind of two possibilities.
And then there's another
two possibilities here,
where if Y is equal to one,
when Y is equal to one,
we'll have X as positive or
negative square root of three.
So we have positive square
root of three and Y equals one,
and then we have negative
square root of three
and Y equals one.
So these, these are the critical points,
critical points,

Korean: 
0이 되어야 함을 의미합니다. 따라서
X는 루트 3 또는 음의 루트 3이 됩니다
기억해야 할 건
이 둘은 서로 다른 경우들입니다
이건 X=0이라는 조건에서 얻은 결과고,
이건 Y=1이라는 조건에서 얻은
결과입니다
따라서 극점은 총 3개가 나옵니다
첫 상황에서,
즉 X=0에서 구한 두 극점은
모두 X 좌표는
0이 될 것입니다
그리고 대응되는 Y 좌표는 각각
0과 -2입니다
써놓도록 합시다
이렇게 두 경우가 나옵니다
그리고 여기 또 두 경우가 나옵니다
바로 Y=1인 경우들입니다
Y=1일 때
X는 + 혹은 - 루트 3이 나옵니다
X=+루트3과 Y=1
그리고 X=-루트3과
Y=1
이 점들이 바로

Korean: 
임계점들입니다
편도함수가 모두
0이 되는 점들입니다
다음 영상에서는 이 극점들을
이계도함수를 이용해
분류해보도록 하겠습니다

English: 
which basically means
all partial derivatives
are equal to zero.
In then in the next video, I
will classify each of these
critical points using the
second partial derivative test.
