4.12 The differential 121
If the radius is increased by amount∆r, the corresponding
change in the area is
∆A 1 = 1 π(r 1 + 1 ∆r)
21 − 1 πr
21 = 12 πr 1 ∆r 1 + 1 π(∆r)
2and this is the area of a circular ring of radius rand width
∆r. When∆ris small enough,
The corresponding ‘differential area’ is
dA 1 = 12 πr 1 dr 1 = 1 circumference 1 × 1 width
0 Exercise 92
EXAMPLE 4.27Differential form of the product rule
Ify 1 = 1 uv, then the change in yaccompanying changes in uand vis
∆y 1 = 1 (u 1 + 1 ∆u)(v 1 + 1 ∆v) 1 − 1 uv 1 = 1 u 1 ∆v 1 + 1 v 1 ∆u 1 + 1 (∆u)(∆v)
and the differential form is
dy 1 = 1 d(uv) 1 = 1 u 1 dv 1 + 1 v 1 du
If y, u, and vare functions of x, this expression is equivalent to the normal form of the
product rule: ‘division by dx’ gives
An important application of the differential is as a formal procedure for changing the
independent variable. Considery 1 = 1 f(x)and its differential
dy 1 = 1 f′(x)dx (4.31)
Let xbe a function of some other variable, t, such thatx 1 = 1 g(t)with differential
dx 1 = 1 g′(t)dt (4.32)
Substitution of this in (4.31) then gives
dy 1 = 1 f′(x)dx 1 = 1 f′(x)g′(t)dt
dy
dx
u
d
dx
du
dx
=+
v
v
∆∆∆Arr
dA
dr
≈= 2 π r
..............................................................................................r ∆ r
...........
....................
.......................................................................................................................................
..........
......
.......
....
.....
....
....
....
...
...
....
...
...
...
...
..
...
...
...
..
...
..
...
...
..
..
...
...
..
...
..
..
...
...
..
...
..
...
...
..
...
...
...
...
...
...
...
....
...
....
....
....
....
......
......
.......
.............
....................................................................................................................................................................................................................................................................................
................
.........
.......
......
.....
.....
.....
....
....
....
...
....
...
...
....
..
...
....
..
...
...
...
..
...
...
..
...
...
..
...
..
...
...
..
...
..
...
..
...
..
...
..
...
..
...
..
...
...
..
...
...
..
...
...
...
..
....
..
...
....
...
...
....
...
...
....
....
....
.....
.....
.....
.......
.......
.........
........................................................................................................................................................................Figure 4.14