Part 4: Differentiability and Differentials

The definition of the limit can also be used to restate Definition 5.1 by saying that for any e > 0, there is a neighborhood of 0 such that
   | Df-Ñf( p) · Dx |
||Dx||
 < e
for all Dx in that neighborhood. If we multiply both sides by ||Dx||, then we can further modify definition 5.1 by saying that e( Dx) is a continuous function for which e( 0) = 0.    

Definition 5.1b: A function f( x) of n-variables is differentiable at a point p with total derivative Ñf( p) if there is a continuous function e( Dx) with e( 0) = 0 such that


| Df-Ñf( p) ·Dx| < e( Dx) ||Dx||
on some neighborhood of 0.

   

If we let dz = Ñf( p) ·Dx and let Dx = á dx,dy ñ , then Theorem 5.2 implies that
dz = fx( p,q) dx+fy( p,q) dy
Definition 5.1b then implies that
 Dz » dz
whenever dx and dy are sufficiently close to 0.      

 

EXAMPLE 6    Compute Dz and dz for f( x,y) = x2+xy when ( p,q) = ( 1,2) , dx = Dx = 0.01, and dy = Dy = 0.03.

Solution: To begin with, f( 1,2) = 12+1·2 =  3 and f( 1.01,2.03) = ( 1.01)2+1.01·2.03 = 3. 0704, so that
Dz = f( 1.01,2.03) -f( 1,2) = 3. 0704-3 = 0.0704
Moreover, fx( x,y) = 2x+y and fy( x,y) = x, so that
fx( 1,2) = 2·1+2 = 4        and        fy(1,2) = 1
As a result, we have
dz = fx( 1,2) dx+fy( 1,2) dy = 4·0.01+1·0.03 = 0.07
which is within 0.0004 of Dz.        

Let's look at an application of the differential, one in which dx and dy are both interpreted to be tolerances due to the inaccuracy of measuring devices.      

EXAMPLE 7    A large coffee can has a height of 6.5¢¢ to within an accuracy of 1/16¢¢ and a base with a radius of 3.25¢¢ to within an accuracy of 1/16¢¢. Find the volume V and the approximate error dV in the volume of the can.

Solution: If h denotes the height and r denotes the radius of the base of the can, then
V = pr2h
When h = 6.5 and r = 3.25, then V = p( 3.25) 2·6.5 = 215. 69 cubic inches. Moreover,
dV = Vh( 6.5,3.25) dh+Vr( 6.5,3.25) dr
where Vh = pr2 and Vr = 2prh. Since dh = dr = 1/16¢¢, we thus have
dV = p( 3.25) 2·  1
16
 +2p( 3.25)( 6.25) ·  1
16
 = 10.05  in3
Thus, the volume of the coffee can is 215.7 in3, give or take approximately 10 in3.