Transcript Limits and Continuity - Tidewater Community College

Functions of Several
Variables
Differentials
Written by Richard Gill
Associate Professor of Mathematics
Tidewater Community College, Norfolk Campus, Norfolk, VA
With Assistance from a VCCS LearningWare Grant
To begin this section on differentials in three variables, we will begin with a
review of differentials in two variables. Consider the function y = f(x).
Now consider a generic value of x with a tangent to the curve at (x, f(x)).
Compare the initial value of x to a value of x that is slightly larger.
The slope of the tangent line is:
dy
 f ' ( x).
dx
It was at this point that we first
saw dx defined as  x.
rise dy

Since run dx and since dx is
being defined as “run” then dy
becomes “rise” by definition.
( x  x, f ( x  x))
( x, f ( x))
x


 y
dy 
 
x  x

dx  x
For small values of dx, the rise of the tangent line, dy  f ' ( x) dx was
used as an approximation for y , the change in the function.
Now consider the extension of the differential concept to functions of several
variables. Consider an input (x,y) for this function. Its outputs z = f(x,y)
create a set of points (x,y,z) that form the surface you see below.
The differential dz will have two parts: one part
generated by a change in x and the other part
generated by a change in y.
Consider a new point in the domain
generated by a small change in x.
z = f(x,y)
z
Now consider the functional image
of the new point.
We can use the
function to
calculate  z , the
difference
between the two
z-coordinates.
( x  x, y, f ( x  x, y))
By subtracting the z-coordinates:
f ( x  x, y)  f ( x, y)  z
( x, y, f ( x, y))
x
x
y
( x  x, y )
( x, y )
Now consider the extension of the differential concept to functions of several
variables. Consider an input (x,y) for this function and its output (x,y,z).
The differential dz will have two parts: one part
generated by a change in x and the other part
generated by a change in y.
By subtracting the z-coordinates:
z = f(x,y)
z
f ( x  x, y)  f ( x, y)  z
Since the ycoordinate is
constant, we can
use that cross
( x  x, y, f ( x  x, y))
section to draw the
tangent to ( x, y, f ( x, y))
Remember that, dz is
the change in the height
of the tangent line, and
can be used to estimate
the change in z.
( x, y, f ( x, y))
x
dz
x
y
( x  x, y )
( x, y )
Since there has been no change in y we can express the differential so far in
terms of the change in x:
z
dz  dx
x
z = f(x,y)
Now track the influence on z when a new point is
generated by a change in the y direction.
z
( x  x, y  y, f ( x  x, y  y))
Now that the
change in z is
generated by
changes in x and y,
we can define the
total differential:
( x  x, y, f ( x  x, y))
z
z
dx  dy or
x
y
dz  f x ( x, y)dx  f y ( x, y)dy
dz 
( x, y, f ( x, y))
x
y
x
( x  x, y )
( x, y )
y
( x  x, y  y)
Example1. Suppose thegraph wehavebeen working with is generated
by theequation: z  3x 2  2 y 2 . Find thetotaldifferential for z.
Try this on your own first.
z
z
dx  dy
x
y
dz  6 x dx  4 y dy
z = f(x,y)
z
dz 
( x  x, y  y, f ( x  x, y  y))
( x  x, y, f ( x  x, y))
( x, y, f ( x, y))
x
y
x
( x  x, y )
( x, y )
y
( x  x, y  y)
Example 2.
If f ( x, y )  y cos x, evaluate f (2,1) and f (2.1,1.08),
calculatethe totaldifferential dz and compareit toz.
Round to thenearest hundredth.
Hint: first calculate dx and dy.
Solution: dx = 2.1 – 2 = 0.1 and dy = 1.08 – 1 = 0.08
z  y cos x
z
z
dz  dx  dy
x
y
  y sin x dx  cos x dy
( x, y )  (2,1) , dx  0.1, dy  0.8 
dz  1sin 2(0.1)  cos 2(0.08)
 0.0909.... 0.0332.....  0.124
f (2.1,1.08)  1.08(cos(2.1))  0.545
f (2,1)  1.08(cos(2.1))  0.416
z  f (2.1,1.08)  f (2,1)
 1.08(cos(2.1))  1.08(cos(2.1))
 0.129
dz is a good approximation for z
Definition of Differentiability
For z  f ( x, y), thefunctionis differentiable at (a, b) if z can be
expressed in theformz  f x (a, b)x  f y (a, b)y  1x   2 y
where 1 and  2  0 as (x, y)  (0,0).
Since dz  f x ( x, y)dx  f y ( x, y)dy, youcan consider1x and  2 y to be
theerror whenwe use dz to approximate z.
The following
theorem is presented
without proof though
you can usually find
the proof in the
appendix of a
standard Calculus
textbook.
If thepartialderivatives f x and f y exist in
a region containing(a, b) and are continuous
at (a, b) then f is differentiable at (a, b).
Theorem: If a function of x and y is differentiable at (a,b) then it
is continuous at (a,b).
Solution: the objective is to show that
lim
(x, y)(a,b)
f ( x, y )  f ( a , b )
Let z  f ( x, y) be differentiable at (a, b).
T hen,by definitionz  ( f x (a, b)  1 )x  ( f y (a, b)   2 )y
where x  a  x, y  b  y, and 1  0 and  2  0 as ( x, y)  (a, b).
We also know thatz  f(a,b)-f(x,y).
From above,
z  f (a, b)  f ( x, y )  ( f x (a, b)   1 )x  ( f 2 (a, b)   2 )y
 ( f x (a, b)   1 )(a  x)  ( f 2 (a, b)   2 )(b  y )
T akingthelimit as ( x, y )  (a, b), theaboveexpressiongoes to 0.
f (a, b)  f ( x, y )  0  f (a, b)  f ( x, y ) as ( x, y )  (a, b)
which completestheproof.
Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2
ft. These measurements have possible errors in accuracy as laid out in the
table below. Complete the table below. Comment on the relationship between
dV and  V for the indicated errors.
Solution:
V  f (r , h)  r h
V
V
dV 
dr 
dh
r
h
dV  2rh dr  r 2 dh
2
r
h
0.1 ft.
0.1 ft.
7.54 cu ft
0.01 ft.
0.01 ft.
.754 cu ft
0.001 ft. 0.001 ft.
r  h  0.1  dV  2 (2)(5)(0.1)   (2 2 )(0.1)
 2  0.4  2.4  7.540 ft 3
r  h  0.01  dV  2 (2)(5)(0.01)   (2 2 )(0.01)
 0.2  .04  0.24  .754 ft 3
r  h  0.001 dV  2 (2)(5)(0.001)   (2 2 )(0.001)
 0.024  0.075ft 3
dV
.075 cu ft
V
Example 3. A right circular cylinder has a height of 5 ft. and a radius of 2
ft. These measurements have possible errors in accuracy as laid out in the
table below. Complete the table below. Comment on the relationship between
dV and  V for the indicated errors.
Solution:
V  f (r , h)  r h
V
V
dV 
dr 
dh
r
h
dV  2rh dr  r 2 dh
2
r
h
0.1 ft.
0.1 ft.
7.54 cu ft
7.826 cu ft
0.01 ft.
0.01 ft.
.754 cu ft
0.757 cu ft
.075 cu ft
0.075 cu ft
0.001 ft. 0.001 ft.
dV
V
V  f (r  r , h  h)  f (r , h)
 f (2.1, 5.1)  f (2,5)   (2.1) 2 (5.1)   (2) 2 (5)  7.826
V  f (2.01, 5.01)  f (2,5)   (2.01) 2 (5.01)   (2) 2 (5)  0.757
V  f (2.001, 5.001)  f (2,5)   (2.001) 2 (5.001)   (2) 2 (5)  0.075
T heapproximation of dV for V gets betteras r and h get smaller.
Example 4. A right circular cylinder is constructed with a height of 40 cm.
and a radius of 25 cm. What is the relative error and the percent error in
the surface area if the possible error in the measurement of each
dimension is ½ cm.
Solution: If the measurements are correct the surface area will be
A  2r 2  2rh
 2 (25) 2  2 (25)(40)
 1250  800  2050
The total differential will
generate an estimate for
the possible error.
A  2r 2  2rh
A
A
dr 
dh  (4r  2h)dr  2r dh
r
h
 (4 (25)  2 (40))(.5)  2 (25)(.5)
dA 
 (100  80 )(.5)  50 (.5)  115
dA 115

 0.056
A 2050
T hepercenterroris 5.6%
T herelativeerroris
For comments on this presentation you may email
the author Professor Richard Gill at
[email protected] or the publisher of the VML, Dr.
Julia Arnold at [email protected].