Okay and last section for today is on
implicit functions.
So, so far we've been expressing functions
of
one variable in terms, so.
Functions.
What did I write here?
Functions have expressed one variable in
terms of another.
Or, or, several, like we just started
today.
So, we had something like y is equal to f
of x, which is equal to.
1 minus x squared.
So this is sort of the, the nicest way of,
of presenting a function.
It's like, you give
me a value of x, I know exactly how to
tell you the corresponding value of y.
So I have this, so I have something like y
equals sine x divided by x.
So I have, oops, I have y on the left side
of the,
of the equation and only x's on the right
side of the equation.
So, an implicit function is defined by
a more general relationship between the
variables.
So I could say something, so this is the
equation of a circle.
So I say, x squared plus y squared equals
1.
And this is if I'm on a circle of radius
one.
This is going to be true for every point
that's on that circle,
and it's going to be false for every point
that's not on that circle.
And this one, if you, if you think about
what I've done up here, it at least has
kind
of an easy way to go between the old
or the, the, you know, specific,
definition of a function.
So this is, you know, in high school, I
had to do something called the vertical
line test.
So, if I have a function, and I find the
intersection
of that function with any vertical line,
it can have at
most one point where they cross.
If it, if it's like a circle, and it
crosses twice, then it's not a function.
So, for a given x, there has to be a
unique value y.
The circle's not going to have that
property
except for two points one at either end.
But it's not that bad either, it's very
easy to fix.
I can just chop it into a top circle and a
bottom circle.
this one we're going to see is not quite
so well behaved.
So in some cases, you can just solve for
one function as another in any, any one
of these functions, any function I can
write f of x equals, some expression of x.
I could just move all of that to one side
of the equals sign and say it's an
implicit function.
and the way we're actually going to define
this is I'm going to say that, f of xyz
is a function of three variables, and the
set of points that satisfy f of xyz.
So I just choose x, y and z, it could be
just x and y.
Or it could be many, many more variables
this is just kind of an example with
three.
So, in this case, the implicit
function is going to be the set of points
that satisfy f of
xyz equals 0.
And in that case, I'm going to call that
set of points where this condition holds.
The locus defined by f.
So it sort of like I mentioned before if I
have the
equation of a circle, a x squared plus y
squared equals 1.
If I on this circle, then this statement
is going to be true.
So any point on this circle of radius one.
Is going to satisfy this equation exactly.
If I'm inside the circle x squared plus y
squared going to be less than 1.
And if I'm outside the circle it's going
to be greater than 1.
And I have this one set of points that
satisfies this.
And so that's going to be my locus.
It's got infinitely many points in it.
because this is, it's a line, it's
continuous.
For any two points on the circle, I can
find in between them that's also on the
circle.
But, in this case, I can fix it pretty
easily.
So I can say y equals 1 minus x, square
root of 1 minus x squared.
That gives me the blue curve which is the
top semi circle and then I can say minus
square
root of 1 minus x squared, and that gives
me
the red curve, which is the bottom of the
semicircle.
This one on the other hand has a locus
that looks like this.
And so this is more difficult.
you know, maybe there's some hope that on
the left side of the y axis, so
over here you can solve for, you know,
at least here it passes the vertical line
test.
So, maybe you could solve this and get an
expression y equals something.
Well not maybe you can, but it's not very
much fun.
on the right side, until you get to this
point over here
where this little loop ends, it doesn't
pass the vertical line test.
So here for instance, a vertical line
would intersect the function three times.
But we might still want to be able to take
a derivative of either of these functions.
So suppose I want to find the slope of a
tangent line to the unit circle, so, that
circle I
just had a picture of that has radius one,
that's
called the unit circle, because the radius
is one unit.
So I have to split this up, if I want to
do this using Regular functions.
I have to split this up into two cases.
I have to consider the top semicircle and
the bottom semicircle independently.
But I can use the chain rule to sort of
help me out here.
So what I'm going to do is take the
derivative.
I have y equals.
The square root of 1 minus x squared.
And I'll take the derivative of both sides
of that equation with respect to x.
And by the chain rule, when I'm taking the
derivative
of y with respect to x, I have to
multiply.
So I'm going to take the derivative of y.
But that's just going to be
1, so there's kind of an invisible one
over here that I didn't bother to write
down.
Times the derivative of y with respect to
x
and remember this is what I'm trying to
find.
This is the, the derivative and on the
right hand side I'm now going to take the
derivative with respect to x and this
should be
pretty easy because it's just a function
of x.
So I'll rewrite the square root as
quantity to the one
half power, so it's easier to use the
power rule.
And then, cheat a little bit with the
chain rule again.
So here, the derivative of the inside of
this quantity.
The derivative of one is zero because it's
a constant.
The derivative of minus x squared is going
to be minus 2 x.
So the chain rule pops that minus 2x out
here
and if I simplify so the two is going to
cancel
out the one half that popped out here
because of the power rule.
So I end up with minus x divided by this
square root of 1 minus x squared.
And then case two is going to be very
similar
except there's going to be a minus sign
here.
So I'll just skip down to the bottom.
And we get exactly the same thing except
now I have the minus sign down here.
And when I don't bother cancelling out a
minus
sign, it's because I have something clever
for you coming in the future.
So just keep this bit in mind.
That this is minus x divided by the square
root of 1 minus x squared.
And this is minus x, divided by minus the
square root of 1 minus x squared.
[COUGH]
And
so, now I'm going to try doing this using
the idea of an implicit function.
So.
If I take the, derivative of a function of
y squared, by the chain rule, I can take
the derivative of this with respect to y
and then multiply by dy, dy.
So, this is the same thing I did on
the previous slide, except now, I have a y
squared
here instead of a y.
So, it doesn't just go away.
So instead of having my invisible one, I
end up
with the derivative of this, 2 y times dy
dx.
let's see what I'm going to get down here.
Oh, okay.
So this is my little result.
And now I'm going to try and take the
derivative of.
The equation of a circle.
The implicit equation of a circle, with
respect to x.
So, I still have the linearity property.
So here I want to do
the derivative of a sum, so that's just
going to be
the sum of the derivatives.
And on the right hand side I'm just
going to take the derivative
of one, one is a constant so hopefully we
get zero for there.
And we do.
And when I split this sum up into two
terms I
end up with the derivative with respect to
x of x squared
so that's going to be easy to do because I
have, I've
got my function already in the right
units, in the right variable.
And then when I
want to take the derivative of y squared
with res-, with respect to x, that's when
I'm going to have to, oops, use this
result that I, I went through up here.
So I'm going just end up with 2y, dy dx.
So I end up with this expression here, 2x
plus 2y, times dy dx.
But remember I'm trying to find the
derivative for this tangent line touching
the circle.
And d y d x is going to be the slope of
that tangent line.
So d y d x is what I'm trying to get.
So I'm just going to solve for d y d x
now.
So I subtract 2 x from both sides, and
then divide by 2 y.
So I end up with minus x over y.
And now remember when I did my two cases,
I had
to have the top half of the circle, that
was y
equals the square root of 1 minus x, and
so when
y is greater then zero, this is what I
would get.
And when y is less than zero I get the,
the bottom half.
But in both
cases, oops I guess I can't highlight just
the denominator here, in both cases the
denominator is just the value of y
appropriate for where I am on the circle.
So if I don't insist on getting rid of Y
entirely I can actually
express both of these as this one formula,
dy, dx equals minus x divided by y.
And so that the only caveat,
the, so the thing you have to be careful
of, is that this is
only going to hold four points in the
locus So, you know, I can still
evaluate this at the point 1,1 and get
negative one but that has no
interpretation for what I've done because
the point 1,1 is not on the circle.
Okay, and then just for fun I did this, so
this.
Funny shaped thing with the, the loop in
it is called a Folium which I
guess is Latin for leaf and somebody
thought
that that little loop looked like a leaf.
So here it's going to be a little bit
trickier, but we're still going to
use all of the properties exactly the way
that we had been doing it.
So I'm just going to take the derivative
of
both sides Of this equation with respect
to x.
Since here I have a project.
So it's x times y.
I'm going to have to, woops.
Yeah.
So I need to treat x and y.
So I'm kind of imagining y is a function
of x.
Even though I can't actually write it down
that way.
So what I did here I on the left hand side
I just use linearity
to write the sum as a derivative of a sum
as the sum of the derivative
and on the right hand side I had to use
the product rule so I used 6x times y.
And so I get the derivative of the first
function with respect
to x is 6 times y, so that's where this
came from.
And then I'm also going to take the
derivative of the second function,
which is y, so I'm going to take
derivative with respect to x.
Of y times the first function which is 6
x.
So that's where this sec, second term
comes from.
It's just the product rule.
But the derivative of y with respect to x,
so that's, that's what I'm looking for.
Again, that's d y d x.
So technically it's The chain rule, so I
take the derivative of y with respect
to y times dy dx, but really, you can just
think of this as going straight to dy dx.
But now I've got this dy dx on both sides.
But I'm just going to use algebra to
move both of those terms to one side of
the equation, and everything else to the
other side.
So I'm going to end up with, so here I had
a positive 3y squared.
That's where this came from.
And then I had a 6x on this side, but when
I move it to the other side it'll become
minus 6x.
So I have 3y squared minus 6x times dy dx,
and then everything
else, so 6y minus 3x squared on the right
hand side.
And, then, I'm just going to divide both
sides by 3y square minus 6x and
that gives me an expression for my
derivative.
And, again, so I think, okay, let's see,
2y minus x so, you have to be a little bit
careful with this again because, again,
it's only going to be defined when I'm On
the curve, so if I take any other point,
I'll still be able to evaluate this, and
it will give me a number, but that
number has no interpretation for what I've
just done.
This only gives me the derivative for
points that are on this curve.
And then there's still some, one small
problem so if you look at
Where the curves cross I actually should
have two derivatives somehow here.
I have one flat derivative and one
infinitely steep
derivative and so the way you're going to,
capture that
is that's at the origin so that's at x
equals zero, y equals zero.
And if x and y are equal to 0, then this
is zero and this is zero.
So I get one of these numbers that do not
exist.
And so,
in that case you just have to examine the
function
and try and figure out what its behavior
is here.
But since it's crossing the derivative
it's
not going to be differential at that
point.
Is that the whole idea doesn't really make
any sense.
So it's, you know, I couldn't say
you know, if I wanted to add h to this.
You know, would it be h going up this way
or would it be h going out that way?
I, I can't do that with just one h so
you'd have to
do something more complicated if you
wanted to extend the idea of a derivative.
To this one point where the curves cross.