1.
Chapter 8: Functions of
Several Variables
Section 8.4
Differentials
Written by Richard Gill
Associate Professor of Mathematics
Tidewater Community College, Norfolk Campus, Norfolk, VA
With Assistance from a VCCS LearningWare Grant
2.
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:
( x + ∆x, f ( x + ∆x))
dy
= f ' ( x).
dx
∆y
dy
It was at this point that we first
( x, f ( x))
saw dx defined as ∆x.
rise dy
= x + ∆x
x
Since run dx and since dx is
being defined as “run” then dy
becomes “rise” by definition.
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.
3.
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
z = f(x,y)
generated by a change in y.
z
Consider a new point in the domain
generated by a small change in x.
Now consider the functional image
of the new point.
We can use the
function to
( x + ∆x, y, f ( x + ∆x, y ))
calculate ∆z , the
difference
( x, y, f ( x, y ))
∆x
between the two
z-coordinates.
x
By subtracting the z-coordinates: y
( x, y )
f ( x + ∆x, y ) − f ( x, y ) = ∆z ( x + ∆x , y )
4.
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
z = f(x,y)
generated by a change in y.
z
By subtracting the z-coordinates:
f ( x + ∆x, y ) − f ( x, y ) = ∆z
Since the y-
coordinate 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 ))
( x, y, f ( x, y ))
∆x
Remember that, dz is
the change in the height dz x y
of the tangent line, and
( x, y )
can be used to estimate ( x + ∆x , y )
the change in z.
5.
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,
( x + ∆x, y, f ( x + ∆x, y ))
we can define the
total differential:
( x, y, f ( x, y ))
∆x
∂z ∂z
dz = dx + dy or ∆y
∂x ∂y x y
dz = f x ( x, y )dx + f y ( x, y )dy ( x + ∆x , y )
( x, y ) ( x + ∆x, y + ∆y )
6.
Example 1. Suppose the graph we have been working with is generated
by the equation : z = 3 x 2 + 2 y 2 . Find the total differential for z.
Try this on your own first.
z = f(x,y)
∂z ∂z z
dz = dx + dy
∂x ∂y ( x + ∆x, y + ∆y, f ( x + ∆x, y + ∆y ))
dz = 6 x dx + 4 y dy
( x + ∆x, y, f ( x + ∆x, y ))
( x, y, f ( x, y ))
∆x
∆y
x y
( x + ∆x , y )
( x, y ) ( x + ∆x, y + ∆y )
7.
If f ( x, y ) = y cos x, evaluate f (2,1) and f (2.1,1.08),
Example 2.
calculate the total differential dz and compare it to ∆z.
Round to the nearest 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)) dz is a good approximation for ∆z
= −0.129
8.
Definition of Differentiability
For z = f ( x, y ), the function is differentiable at (a, b) if ∆z can be
expressed in the form ∆z = f x (a, b)∆x + f y (a, b)∆y + ε 1∆x + ε 2 ∆y
where ε 1 and ε 2 → 0 as (∆x, ∆y ) → (0,0).
Since dz = f x ( x, y )dx + f y ( x, y )dy, you can consider ε 1∆x and ε 2 ∆y to be
the error when we use dz to approximate ∆z.
The following
theorem is presented If the partial derivatives f x and f y exist in
without proof though
a region containing (a, b) and are continuous
you can usually find
the proof in the
at (a, b) then f is differentiable at (a, b).
appendix of a
standard Calculus
textbook.
9.
Theorem: If a function of x and y is differentiable at (a,b) then it
is continuous at (a,b).
f ( x, y ) = f ( a, b)
Solution: the objective is to show that lim
(x, y)→ (a, b)
Let z = f ( x, y ) be differentiable at (a, b).
Then, by definition ∆z = ( f x (a, b) + ε 1 )∆x + ( f y (a, b) + ε 2 )∆y
where ∆x = a − x, ∆y = b − y, and ε 1 → 0 and ε 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) + ε 1 )∆x + ( f 2 (a, b) + ε 2 )∆y
= ( f x (a, b) + ε 1 )(a − x) + ( f 2 (a, b) + ε 2 )(b − y )
Taking the limit as ( x, y ) → (a, b), the above expression goes to 0.
f (a, b) − f ( x, y ) = 0 → f (a, b) = f ( x, y ) as ( x, y ) → (a, b)
which completes the proof.
10.
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.
∆h ∆V
∆r dV
Solution:
0.1 ft. 0.1 ft. 7.54 cu ft
V = f (r , h) = πr h
2
0.01 ft. 0.01 ft.
∂V ∂V .754 cu ft
dV = dr + dh
∂r ∂h
0.001 ft. 0.001 ft. .075 cu ft
dV = 2πrh dr + πr 2 dh
∆r = ∆h = 0.1 → dV = 2π (2)(5)(0.1) + π (2 2 )(0.1)
= 2π + 0.4π = 2.4π ≈ 7.540 ft 3
∆r = ∆h = 0.01 → dV = 2π (2)(5)(0.01) + π (2 2 )(0.01)
= 0.2π + .04π = 0.24π ≈ .754 ft 3
∆r = ∆h = 0.001 → dV = 2π (2)(5)(0.001) + π (2 2 )(0.001)
= 0.024π ≈ 0.075 ft 3
11.
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.
∆h ∆V
∆r dV
Solution:
0.1 ft. 0.1 ft. 7.54 cu ft 7.826 cu ft
V = f (r , h) = πr h
2
0.01 ft. 0.01 ft.
∂V ∂V .754 cu ft 0.757 cu ft
dV = dr + dh
∂r ∂h
0.001 ft. 0.001 ft. .075 cu ft 0.075 cu ft
dV = 2πrh dr + πr 2 dh
∆V = f (r + ∆r , h + ∆h) − f (r , h)
= f (2.1, 5.1) − f (2,5) = π (2.1) 2 (5.1) − π (2) 2 (5) ≈ 7.826
∆V = f (2.01, 5.01) − f (2,5) = π (2.01) 2 (5.01) − π (2) 2 (5) ≈ 0.757
∆V = f (2.001, 5.001) − f (2,5) = π (2.001) 2 (5.001) − π (2) 2 (5) ≈ 0.075
The approximation of dV for ∆V gets better as ∆r and ∆h get smaller.
12.
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 = 2πr 2 + 2πrh The total differential will
generate an estimate for
= 2π (25) 2 + 2π (25)(40)
the possible error.
= 1250π + 800π = 2050π
A = 2πr 2 + 2πrh
∂A ∂A
dh = (4πr + 2πh)dr + 2πr dh
dA = dr +
∂r ∂h
= (4π (25) + 2π (40))(.5) + 2π (25)(.5)
= (100π + 80π )(.5) + 50π (.5) = 115π
dA 115π
= = 0.056
The relative error is
A 2050π
The percent error is 5.6%
13.
This concludes the material for Lesson 8.4. This also concludes the lessons
on Blackboard and the material for the semester.
As was the case in previous Bb lessons, we have
posted three sets of exercises on Bb. Any
exercises worked correctly will add to your
thinkwell exercise totals.
The practice exam should be up and
running on Bb. Good luck preparing for
the exam and congratulations on
getting to the end of one of the
toughest courses in the curriculum.
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