U2S5V07 Finishing the examples.txt

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So we have this crazy looking curve.
And it's implicitly defined by this function.
And we asked two questions.
One was, what is the slope of the tangent line at the point
x, y equals 1 over root 2 1 over root 2?
And the other is, at what points does this function
have a horizontal tangent line?
Answering both of these questions
involves understanding dydx.
And so to find dydx, you guessed it,
we are going to implicitly differentiate.
You've actually already gotten a start on this problem.
And since we're fortunate enough for differentiation
to be linear, we can break this problem up into pieces.
This is the part that you've actually already computed,
which you should have found to be equal to this.
And we know that the derivative of the constant term
is going to be 0.
So all that's left to compute is the derivative
of this 2x squared y squared term.
So applying the product rule, we get
that this is equal to 4x times y squared
plus 2x squared times the quantity-- now, here's
where we apply the chain rule-- 2y dydx.

So now I'm just going to collect the like terms.
So I collect all the terms that involve
a dydx and all the terms that do not involve dydx.
And I get 4x cubed minus 6x plus 4xy squared
plus 4y cubed plus 2y plus 4x squared y, all times dydx.
This is equal to 0.

Now, I just need to do some more algebra to solve for dydx.
So I subtract off this term here.
And then divide by the multiplier of dydx.
And I get this.
Now, I want to do just a tiny bit of algebra
so that this term is easier for me to understand.
First thing I'm going to do is I'm going to pull out a 2
from the top and the bottom.
And those are going to cancel.
The next thing I'm going to do is
I'm actually going to notice that the numerator has
a common factor of x and the denominator
has a common factor of y.
So I'm going to pull out an x over y term.
And now, this is just a little bit easier for me
to think about so I can go about answering the questions
that we asked at the beginning.
First, let's go ahead and find the slope of the tangent line
to the point x, y equals 1 over root 2, 1 over root 2.
Plugging this in for x and y, we do some simple arithmetic
to find that the slope of the tangent line is equal to 1/3.

OK.
So now let's try to figure out where this curve has
horizontal tangent lines.
So if the tangent line is horizontal,
then dydx is going to be equal to 0.
So this equation is equal to 0 if either x is equal to 0
or this term in parentheses is equal to 0.
And notice that if x is equal to 0,
then, necessarily, y is also equal to 0.
That's the only point on our curve where x is equal to 0.
Just by looking at the curve, I can
see that this curve does have two tangent lines,
neither of which will be horizontal.
So we can actually rule out this point
where x and y are equal to 0, because there's
no horizontal tangent line there.
That means that the only possibility
for dydx to be equal to 0 is when 2x squared minus 3
plus 2 y squared is equal to 0.
This is exactly describing a circle of radius
the square root of 3/2.
If I draw this circle on top of my curve,
then these points where the circle
and the lemniscate intersect are exactly
the points I'm trying to identify.
To do that, I'm simply going to say, OK, well,
x squared is equal to 3/2 minus y squared.
Then I'm going to plug that into the equation defining
this curve, and solve for y.
That's exactly going to find where
these two curves intersect.
So doing that, I do some messy algebra.

And I get that y squared is equal to 9/16.
That tells me that y can be equal to plus or minus 3/4.
Plugging this into the formula for the circle of radius
square root 3/2, I get that x is equal to plus or minus
the square root of 15 over 4.
So we've found those points.

We just solved two problems involving
tangent lines of this curve.
However, I hope you noticed that the vast majority
of the problem-solving was not actually doing calculus.
We computed dydx, and then most of the problem-solving
was doing some algebra and using some geometric reasoning.
This is very common in problem-solving.
And I want you to keep this in mind.
We need to identify some small piece of the problem that
involves calculus.
And the rest will involve some other problem-solving
techniques.