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Okay and last section for today is on
implicit functions.

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00:00:09,720 --> 00:00:15,210
So, so far we've been expressing functions
of

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one variable in terms, so.
Functions.

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00:00:21,990 --> 00:00:22,750
What did I write here?

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00:00:22,750 --> 00:00:25,860
Functions have expressed one variable in
terms of another.

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Or, or, several, like we just started
today.

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00:00:29,730 --> 00:00:36,610
So, we had something like y is equal to f
of x, which is equal to.

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1 minus x squared.

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So this is sort of the, the nicest way of,
of presenting a function.

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It's like, you give

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me a value of x, I know exactly how to
tell you the corresponding value of y.

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So I have this, so I have something like y
equals sine x divided by x.

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So I have, oops, I have y on the left side
of the,

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of the equation and only x's on the right
side of the equation.

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So, an implicit function is defined by

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a more general relationship between the
variables.

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So I could say something, so this is the
equation of a circle.

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So I say, x squared plus y squared equals
1.

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And this is if I'm on a circle of radius
one.

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This is going to be true for every point
that's on that circle,

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and it's going to be false for every point
that's not on that circle.

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And this one, if you, if you think about

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what I've done up here, it at least has
kind

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of an easy way to go between the old

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or the, the, you know, specific,
definition of a function.

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So this is, you know, in high school, I

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had to do something called the vertical
line test.

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So, if I have a function, and I find the
intersection

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of that function with any vertical line,
it can have at

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most one point where they cross.

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If it, if it's like a circle, and it
crosses twice, then it's not a function.

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So, for a given x, there has to be a
unique value y.

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The circle's not going to have that
property

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except for two points one at either end.

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But it's not that bad either, it's very
easy to fix.

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I can just chop it into a top circle and a
bottom circle.

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this one we're going to see is not quite
so well behaved.

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So in some cases, you can just solve for
one function as another in any, any one

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of these functions, any function I can
write f of x equals, some expression of x.

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I could just move all of that to one side

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of the equals sign and say it's an
implicit function.

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and the way we're actually going to define
this is I'm going to say that, f of xyz

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is a function of three variables, and the
set of points that satisfy f of xyz.

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So I just choose x, y and z, it could be
just x and y.

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Or it could be many, many more variables
this is just kind of an example with

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three.
So, in this case, the implicit

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function is going to be the set of points
that satisfy f of

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xyz equals 0.
And in that case, I'm going to call that

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set of points where this condition holds.
The locus defined by f.

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00:03:27,100 --> 00:03:29,630
So it sort of like I mentioned before if I
have the

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equation of a circle, a x squared plus y
squared equals 1.

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If I on this circle, then this statement
is going to be true.

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So any point on this circle of radius one.

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Is going to satisfy this equation exactly.

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If I'm inside the circle x squared plus y
squared going to be less than 1.

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And if I'm outside the circle it's going
to be greater than 1.

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And I have this one set of points that
satisfies this.

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And so that's going to be my locus.

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It's got infinitely many points in it.

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because this is, it's a line, it's
continuous.

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For any two points on the circle, I can

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find in between them that's also on the
circle.

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But, in this case, I can fix it pretty
easily.

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So I can say y equals 1 minus x, square
root of 1 minus x squared.

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That gives me the blue curve which is the

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top semi circle and then I can say minus
square

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root of 1 minus x squared, and that gives
me

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the red curve, which is the bottom of the
semicircle.

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00:04:31,980 --> 00:04:35,969
This one on the other hand has a locus
that looks like this.

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And so this is more difficult.

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you know, maybe there's some hope that on
the left side of the y axis, so

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over here you can solve for, you know,

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at least here it passes the vertical line
test.

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So, maybe you could solve this and get an
expression y equals something.

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Well not maybe you can, but it's not very
much fun.

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00:05:00,580 --> 00:05:03,660
on the right side, until you get to this
point over here

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where this little loop ends, it doesn't
pass the vertical line test.

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So here for instance, a vertical line
would intersect the function three times.

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00:05:13,310 --> 00:05:17,800
But we might still want to be able to take
a derivative of either of these functions.

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00:05:21,450 --> 00:05:23,490
So suppose I want to find the slope of a

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tangent line to the unit circle, so, that
circle I

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just had a picture of that has radius one,
that's

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called the unit circle, because the radius
is one unit.

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So I have to split this up, if I want to
do this using Regular functions.

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I have to split this up into two cases.

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I have to consider the top semicircle and
the bottom semicircle independently.

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But I can use the chain rule to sort of
help me out here.

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So what I'm going to do is take the
derivative.

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I have y equals.
The square root of 1 minus x squared.

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And I'll take the derivative of both sides
of that equation with respect to x.

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And by the chain rule, when I'm taking the
derivative

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of y with respect to x, I have to
multiply.

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So I'm going to take the derivative of y.
But that's just going to be

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1, so there's kind of an invisible one

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over here that I didn't bother to write
down.

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Times the derivative of y with respect to
x

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and remember this is what I'm trying to
find.

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This is the, the derivative and on the
right hand side I'm now going to take the

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derivative with respect to x and this
should be

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pretty easy because it's just a function
of x.

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So I'll rewrite the square root as
quantity to the one

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half power, so it's easier to use the
power rule.

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And then, cheat a little bit with the
chain rule again.

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So here, the derivative of the inside of
this quantity.

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The derivative of one is zero because it's
a constant.

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The derivative of minus x squared is going
to be minus 2 x.

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So the chain rule pops that minus 2x out
here

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and if I simplify so the two is going to
cancel

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out the one half that popped out here
because of the power rule.

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So I end up with minus x divided by this
square root of 1 minus x squared.

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And then case two is going to be very
similar

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except there's going to be a minus sign
here.

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So I'll just skip down to the bottom.

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And we get exactly the same thing except
now I have the minus sign down here.

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And when I don't bother cancelling out a
minus

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sign, it's because I have something clever
for you coming in the future.

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So just keep this bit in mind.

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That this is minus x divided by the square
root of 1 minus x squared.

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And this is minus x, divided by minus the
square root of 1 minus x squared.

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[COUGH]

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And

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so, now I'm going to try doing this using
the idea of an implicit function.

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So.

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If I take the, derivative of a function of
y squared, by the chain rule, I can take

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the derivative of this with respect to y
and then multiply by dy, dy.

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So, this is the same thing I did on

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the previous slide, except now, I have a y
squared

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here instead of a y.
So, it doesn't just go away.

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So instead of having my invisible one, I
end up

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with the derivative of this, 2 y times dy
dx.

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let's see what I'm going to get down here.
Oh, okay.

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So this is my little result.

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And now I'm going to try and take the
derivative of.

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The equation of a circle.

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The implicit equation of a circle, with
respect to x.

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So, I still have the linearity property.
So here I want to do

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the derivative of a sum, so that's just
going to be

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the sum of the derivatives.

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And on the right hand side I'm just
going to take the derivative

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of one, one is a constant so hopefully we
get zero for there.

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00:09:01,638 --> 00:09:02,998
And we do.

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And when I split this sum up into two
terms I

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end up with the derivative with respect to
x of x squared

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so that's going to be easy to do because I
have, I've

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got my function already in the right
units, in the right variable.

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And then when I

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want to take the derivative of y squared
with res-, with respect to x, that's when

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I'm going to have to, oops, use this
result that I, I went through up here.

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So I'm going just end up with 2y, dy dx.

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So I end up with this expression here, 2x
plus 2y, times dy dx.

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00:09:35,480 --> 00:09:37,080
But remember I'm trying to find the

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derivative for this tangent line touching
the circle.

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And d y d x is going to be the slope of
that tangent line.

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So d y d x is what I'm trying to get.

155
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So I'm just going to solve for d y d x
now.

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So I subtract 2 x from both sides, and
then divide by 2 y.

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So I end up with minus x over y.

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00:10:00,020 --> 00:10:05,480
And now remember when I did my two cases,
I had

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to have the top half of the circle, that
was y

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equals the square root of 1 minus x, and
so when

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y is greater then zero, this is what I
would get.

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And when y is less than zero I get the,
the bottom half.

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But in both

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cases, oops I guess I can't highlight just
the denominator here, in both cases the

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denominator is just the value of y
appropriate for where I am on the circle.

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So if I don't insist on getting rid of Y
entirely I can actually

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express both of these as this one formula,
dy, dx equals minus x divided by y.

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00:10:44,560 --> 00:10:45,990
And so that the only caveat,

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the, so the thing you have to be careful
of, is that this is

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only going to hold four points in the
locus So, you know, I can still

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00:10:54,840 --> 00:10:59,010
evaluate this at the point 1,1 and get
negative one but that has no

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interpretation for what I've done because
the point 1,1 is not on the circle.

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00:11:08,120 --> 00:11:11,150
Okay, and then just for fun I did this, so
this.

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Funny shaped thing with the, the loop in
it is called a Folium which I

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00:11:14,590 --> 00:11:17,470
guess is Latin for leaf and somebody
thought

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00:11:17,470 --> 00:11:18,990
that that little loop looked like a leaf.

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00:11:20,430 --> 00:11:23,640
So here it's going to be a little bit
trickier, but we're still going to

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use all of the properties exactly the way
that we had been doing it.

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00:11:28,030 --> 00:11:29,820
So I'm just going to take the derivative
of

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00:11:29,820 --> 00:11:33,470
both sides Of this equation with respect
to x.

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00:11:34,940 --> 00:11:38,350
Since here I have a project.
So it's x times y.

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00:11:38,350 --> 00:11:39,720
I'm going to have to, woops.

183
00:11:42,070 --> 00:11:44,120
Yeah.
So I need to treat x and y.

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00:11:44,120 --> 00:11:47,100
So I'm kind of imagining y is a function
of x.

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00:11:47,100 --> 00:11:49,270
Even though I can't actually write it down
that way.

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00:11:51,420 --> 00:11:56,820
So what I did here I on the left hand side
I just use linearity

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00:11:56,820 --> 00:12:01,700
to write the sum as a derivative of a sum
as the sum of the derivative

188
00:12:04,290 --> 00:12:09,900
and on the right hand side I had to use
the product rule so I used 6x times y.

189
00:12:12,080 --> 00:12:15,600
And so I get the derivative of the first
function with respect

190
00:12:15,600 --> 00:12:19,770
to x is 6 times y, so that's where this
came from.

191
00:12:19,770 --> 00:12:24,210
And then I'm also going to take the
derivative of the second function,

192
00:12:24,210 --> 00:12:29,620
which is y, so I'm going to take
derivative with respect to x.

193
00:12:29,620 --> 00:12:32,580
Of y times the first function which is 6
x.

194
00:12:32,580 --> 00:12:34,710
So that's where this sec, second term
comes from.

195
00:12:34,710 --> 00:12:35,740
It's just the product rule.

196
00:12:41,870 --> 00:12:46,820
But the derivative of y with respect to x,
so that's, that's what I'm looking for.

197
00:12:46,820 --> 00:12:48,020
Again, that's d y d x.

198
00:12:48,020 --> 00:12:53,960
So technically it's The chain rule, so I
take the derivative of y with respect

199
00:12:53,960 --> 00:12:59,700
to y times dy dx, but really, you can just
think of this as going straight to dy dx.

200
00:13:01,530 --> 00:13:07,570
But now I've got this dy dx on both sides.
But I'm just going to use algebra to

201
00:13:07,570 --> 00:13:09,720
move both of those terms to one side of

202
00:13:09,720 --> 00:13:11,820
the equation, and everything else to the
other side.

203
00:13:14,300 --> 00:13:19,040
So I'm going to end up with, so here I had
a positive 3y squared.

204
00:13:19,040 --> 00:13:20,060
That's where this came from.

205
00:13:20,060 --> 00:13:22,440
And then I had a 6x on this side, but when

206
00:13:22,440 --> 00:13:25,620
I move it to the other side it'll become
minus 6x.

207
00:13:25,620 --> 00:13:30,080
So I have 3y squared minus 6x times dy dx,
and then everything

208
00:13:30,080 --> 00:13:34,610
else, so 6y minus 3x squared on the right
hand side.

209
00:13:34,610 --> 00:13:40,690
And, then, I'm just going to divide both
sides by 3y square minus 6x and

210
00:13:40,690 --> 00:13:43,930
that gives me an expression for my
derivative.

211
00:13:43,930 --> 00:13:50,440
And, again, so I think, okay, let's see,
2y minus x so, you have to be a little bit

212
00:13:50,440 --> 00:13:56,170
careful with this again because, again,
it's only going to be defined when I'm On

213
00:13:56,170 --> 00:14:00,160
the curve, so if I take any other point,
I'll still be able to evaluate this, and

214
00:14:00,160 --> 00:14:01,810
it will give me a number, but that

215
00:14:01,810 --> 00:14:05,570
number has no interpretation for what I've
just done.

216
00:14:05,570 --> 00:14:10,540
This only gives me the derivative for
points that are on this curve.

217
00:14:10,540 --> 00:14:16,700
And then there's still some, one small
problem so if you look at

218
00:14:16,700 --> 00:14:22,470
Where the curves cross I actually should
have two derivatives somehow here.

219
00:14:22,470 --> 00:14:26,540
I have one flat derivative and one
infinitely steep

220
00:14:26,540 --> 00:14:30,720
derivative and so the way you're going to,
capture that

221
00:14:30,720 --> 00:14:35,780
is that's at the origin so that's at x
equals zero, y equals zero.

222
00:14:35,780 --> 00:14:39,850
And if x and y are equal to 0, then this
is zero and this is zero.

223
00:14:39,850 --> 00:14:42,410
So I get one of these numbers that do not
exist.

224
00:14:45,320 --> 00:14:45,910
And so,

225
00:14:48,170 --> 00:14:51,110
in that case you just have to examine the
function

226
00:14:51,110 --> 00:14:53,220
and try and figure out what its behavior
is here.

227
00:14:53,220 --> 00:14:55,240
But since it's crossing the derivative
it's

228
00:14:55,240 --> 00:14:57,490
not going to be differential at that
point.

229
00:14:57,490 --> 00:15:01,050
Is that the whole idea doesn't really make
any sense.

230
00:15:01,050 --> 00:15:02,440
So it's, you know, I couldn't say

231
00:15:05,240 --> 00:15:07,890
you know, if I wanted to add h to this.

232
00:15:07,890 --> 00:15:11,720
You know, would it be h going up this way
or would it be h going out that way?

233
00:15:11,720 --> 00:15:14,660
I, I can't do that with just one h so
you'd have to

234
00:15:14,660 --> 00:15:19,138
do something more complicated if you
wanted to extend the idea of a derivative.

235
00:15:19,138 --> 00:15:21,250
To this one point where the curves cross.