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U2S5V04 Implicit differentiation.txt
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U2S5V04 Implicit differentiation.txt
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#
# File: content-mit-18-01-1x-captions/U2S5V04 Implicit differentiation.txt
#
# Captions for MITx 18.01.1x module [2D_-MzoxfOk]
#
# This file has 93 caption lines.
#
# Do not add or delete any lines. If there is text missing at the end, please add it to the last line.
#
#----------------------------------------
You just found two formulas for slopes of tangent lines
to the circle of radius 5.
And the formulas were a bit messy
because well, there were two of them
and they involved a negative 1/2 power.
So let's look at another way of solving
this same problem by using implicit differentiation.
What does this mean?
So I'm going to think of y as being
a function of x implicitly, but I'm not
going to find an explicit formula.
Instead, I'm going to think of the equation x squared
plus y of x squared being equal to 25
has an equation of functions; that is, I
think of the left-hand side-- x squared
plus y of x squared-- as being a function of x
that is everywhere equal to 25, a constant function of x.
And that's exactly what it means to be on this circle.
If I have an x, then y of x is determined
so that x squared plus y of x squared is equal to 25.
So if the two sides of this equation
are equal as functions of x, then I
can differentiate both sides with respect to x,
and they're still going to be equal.
So let's go ahead and differentiate both sides
with respect to x.
And we get that the derivative of x squared is equal to 2x.
And applying the chain rule, we get
that the derivative of y of x squared is 2y times dydx.
And because 25 is a constant function of x,
the derivative is zero.
So now we have an equation and let's
go ahead and solve it for dydx.
So I'm going to subtract 2x from both sides,
and I get 2y dydx is equal to negative 2x.
Dividing both sides by 2y, I get that dydx
is equal to negative x over y.
You've probably noticed that this formula for dydx
involves both x and y, but I want to leave it in this form
and here's why.
The equation x squared plus y squared equals 25
holds for all points on the circle,
whether you're on the pink part or the green part
of the circle.
That means that this derivative formula is
going to hold in both places.
So what does this mean?
It means that if I take any point, x, y, on my circle,
then the slope of the tangent line through this point
is equal to negative x over y.
Let's check that this make sense.
First of all, our equation for the slope of the tangent line
is undefined when y is equal to zero.
And if we look at our circle, there
are exactly two places where y is equal to zero.
And at both of these points, the circle
has a vertical tangent line.
And notice that slopes of vertical tangent lines
are either infinite or we might say that they're undefined.
So the slope is not well-defined there.
The other thing we can notice right away
is that if x is equal to zero, the slope is equal to zero.
And we see that there are two points on our circle
where x is equal to zero, and at both of these points
the tangent line is horizontal of slope zero.
So now let's just go ahead and do a couple
more quick checks to make sure that this really
does make sense.
So if I'm on the circle in the first quadrant over here,
then the slope of any tangent line to the circle
is going to be negative.
And since both x and y are positive,
this formula-- negative x over y-- is going to be negative.
So the sign agrees there.
Let's just check one more quadrant.
Let's look at the fourth quadrant down here;
then the slope of any tangent line
to the circle in the fourth quadrant
is going to be positive.
And then if we look at our formula,
since x is going to be positive but y
is going to be negative, negative something
positive over something negative, that
is going to be a positive.
So the signs do actually seem to make sense.
And this really is a formula for the slopes
of the tangent lines.
So I'm going to go ahead and let you get some practice using
this formula to find slopes of tangent lines to the circle.
And then we'll go ahead and get some more practice looking
at examples using implicit differentiation.