The tangent
line to the graph of y = x3 at the point (2, 8) has equation
y = 12(x-2) + 8, as your calculator will tell you.
So it seems
reasonable to expect that when x is close to 2, x3
is close to 12(x-2) + 8.
Check this
on your calculator by comparing the graphs.
Now we'll
introduce some traditional symbols that we can use to say the same thing
over again.
Suppose the
point (a, b) is on the graph of y = x3, near the point
(2, 8).
Then b:::12(a-2)
+ 8, where we use ":::" to mean "is approximately equal to." Another
way to write this is b-8:::12(a-2).
If we think of (a, b) as a point on the graph which moves a small distance away from (2,8), then a-2 is the change in the x-coordinate of the point and b-8 is the change in the y-coordinate. Using traditional symbols, we can write this as Dy:::12Dx, where the Greek capital delta means "the change in." (Note that Dy is supposed to be all one symbol, not some quantity D multiplied by some quantity y.)
The differential
of
y, denoted dy (another indivisible symbol), is the change
in the y-coordinate as we move away from the point (2, 8) along the tangent
line to the curve. dx has the corresponding meaning.
Note that
nothing in the notation tells us which tangent line we're using, or even
which function we're using. You have to be very careful with context
here.
So what we've said so far is
These three
approximations are what you're supposed to use if you're doing differential
approximation.
Error estimation. If x-coordinate of a point on the curve y = x3 is between 1.99 and 2.01, then x:::2 with an error less than 0.01. So you know that y:::8, but what is the maximum error? If Dx is the difference between the actual and the estimated x-coordinates and Dy is the difference between the actual and the estimated y-coordinates, then Dy:::12Dx, so the maximum error in y is approximately 0.12. That is, the actual y-coordinate should be between 7.88 and 8.12.
(In fact,
if 1.99 < x < 2.01, then 7.880599 < y < 8.120601.
There's no real point in using differential approximation when you can
get exact answers.)
Exercises:
1. Expand(2+dx)3 on your calculator. What's the difference between dy and y?
2. Use differentials to approximate the square root of 81.789. Compare with your calculator.
3. If F(2) = -4 and F'(2) = 1/2, use differentials to approximate F(2.05).
4. Suppose G(5) = 14 and G'(5) = -3/4. Suppose that x is measured as 5 with an error of at most 0.1. Estimate G(x). What is the maximum error in this estimate?