A quick note about differentials! We said in class that
Now, this is a little confusing, because normally we think about the linear approximation being centered around a specific point, (x0, y0), not (as we've done above) about some arbitrary point (x, y). But we can get through this without any trouble, as is (hopefully!) explained below.
So, let's start with some function f(x,y). The linear approximation to f at the point (a,b) is just
Check that the second line of that makes sense! So, if we started asking what the change in the function value was between the point (a,b) and another point near it, we could approximate this with the change in the linear approximation, and say that
(where "~" means "is approximately equal to"). Let's work this out:
Now, to quantify the distance we go from the point (a,b), we could take x = a + dx and y = b + dy. Let's also call the difference in L values delta L, so that
Notice that the expression on the right-hand side is the same as the differential that we defined above, with the only exception being that we're considering it at (a,b) instead of (x,y). But we never said what a and b are, so we could happily say that they give any point (x,y), at which point it's clear that the change delta L is in fact what we called the differential:
The volume of a paraboloid (graph to the right) with base radius r and height h is
Suppose we measure that the radius is r = 2 cm and height is h = 4 cm, with a possible measurement error of up to 0.2 cm. How far off could our volume estimate be? We'll estimate this with the differential.
The differential for this is
So, with our measured dimensions of 2 and 4 cm, we have
And an estimate for the maximum possible error is if r and h are off by dr = 0.2 cm and dh = 0.2 cm, respectively:
(Note that the volume itself is V(2,4) = 8 pi cm3, so this is quite a big maximum error! Time to go back and measure more carefully!)