Download Section 3.9 - Differentials

Section 3.9 - Differentials
Local Linearity
If a function is differentiable at a point, it is at
least locally linear.
Differentiable
Local Linearity
If a function is differentiable at a point, it is at
least locally linear.
Not differentiable.
The function is
NOT smooth at
this point.
Local Linearity
If a function is differentiable at a point, it is at
The function is
least locally linear.
differentiable at every point.
The function is
always smooth.
How to Approximate a Function
A line that best
approximates the
graph of a function is
the tangent line (if
the values are near
the point of
tangency).
How to Approximate a Function

x, y
Point of Tangency
( c , f (c) )
Tangent Line
y – f (c) = f '(c) (x – c)

 c , f c 
A point (x,y) on the
tangent line
approximates the value
of the function (must be
near the point of
tangency)
y = f (c) + f '(c) (x – c)
Linear Approximation
The linear approximation or tangent line
approximation of a function f at x = c:
L  x  f c  f 'c   x  c
This is also referred to as the linearization of f
at x = c.
Our Book uses T(x) instead of L(x)
Example 1
Find the local linear approximation of f  x  
at x0 = 1.
Use the linearization formula if c = x0 = 1.
L  x  f c  f ' c   x  c
f  c   f 1  1  1
1
f ' x  2 x
1
f '  c   f ' 1  2 1 
1
2
L  x  1
1
2
 x  1
x
Example 2
Use the linearization of f  x   sin x at π/6 to
approximate sin 0.5.
Use the linearization formula if c = π/6.
f  c   f  6   sin 6 
f '  x   cos x
1
2
L  x  
1
2
f '  c   f '  6   cos 6 
3
2
 x  6 
3
2
Evaluate the linearization formula at x = 0.5.
L  0.5   
1
2
3
2
 0.5  6   0.479563

Error in Local Linear Approximations
As a general rule, the accuracy of the linearization to f (x)
at c deteriorates as x gets further away from c. The size of
the error of the approximation is simply the vertical gap
between the graph and the tangent line:
Error  L  x   f  x 
Under Estimate
Over Estimate
 c , f c 
Error
 c , f c 
Example 2 (Continued)
Use the linearization of f  x   sin x at π/6 to
approximate sin 0.5.
L  0.5     0.5  6   0.479563
Since sin 0.5  0.479426
3
2
1
2
Our approximation is an over estimate and the error is:
0.479563  0.479426  0.000137
The further away
from π/6 the
worse the
approximation:
x
Linearization
Actual
Error
0.5
0.479563
0.479426
0.000137
0.4
0.39296
0.389418
0.003542
0.01
0.05521
.01000
0.045211
Example 3
Use linearization to approximate 4.05 . Is the
approximation an overestimate or underestimate.
Let the function be:
Since the square root of 4 is
well known, let c = 4.
f  c   f  4  4  2
f ' x 
1
2 x
f 'c  f '  4  2 14 
Since
1
4
f  x  x
Evaluate the linearization
formula at x = 0.5.
L  x   2  14  x  4 
L  4.05   2  14  4.05  4 
 2.0125
4.05  2.01246
Our approximation is
an over estimate
Introduction to Differentials
So far we associated the following items to be
equivalent:
y'
f ' x
dy
dx
Now we will consider the differentials “dy” and
“dx” to represent two different real numbers.
Thus dy/dx becomes a real number ratio. Also,
dy
if f is differentiable at x, then dx  f '  x 
becomes
dy  f '  x   dx
Introduction to Differentials
Thus the horizontal differences
are always equal:
x  dx
f  x  x 
y
f  x
x
f ' x 
dy
dx
foristhe
differential
dy:
Error Solve
Error
not
the only
dy  f '  xwe
 dxcan

“differen”ce
dy
calculate
Find the change in y:
dx
x
Slope of tangent line:
x  x
y  f  x  x   f  x 
Thus the two vertical
differences are almost equal:
y  dy
Differentials
If y = f (x), where f is a differentiable function,
then the differential of x, dx, is an independent
variable; that is, dx can be given the value of
any real number. The differential of y, dy, is
then defined in terms of dx by the equation:
dy  f '  x   dx
So dy is the is a dependent variable; it depends
on the values of x and dx.
Remember:
y  f  x  x   f  x 
dy  y & dx  x
Example 1
Compare the values of Δy and dy if
f (x) = x3 + x2 – 2x + 1 and dx = 0.05.
y  f  2  0.05   f  2   9.718  9  0.718
f '  x   3x 2  2 x  2
dy  f '  2   dx  14   0.05   0.7
Thus, Δy and dy are approximately equal
(with Δy being slightly larger)
Example 2
1  x2
Find the differential of y 
1  x2
Use the differential formula for dy.
dy  f '  x   dx
y' 
2
2
1

x

2
x

1

x
       2x
1  x 
2 2
dy 
4 x
1  x 
2 2

 dx
4 x
1  x 
2 2
Example 3
Find the differential dy of x  y  xy
Since this is an implicit equation, you can either solve for y or use
a method similar to the differentiation process in related
rates/implicit differentiation. The following shows the latter:
d  x  y   d  xy 
dx  dy  x  dy  y  dx
Now solve for dy:
dy  x  dy  y  dx  dx
dy 1  x   y  dx  dx
dy 
y dx  dx
1 x
Example 4
The radius r of a circle increases from 10 m to
10.1 m. Use differentials to estimate the increase in
the circles area A.
The function for area is:
Use the differential formula
to find dA:
A '  2 r
dA  2 r  dr
A r
2
Evaluate the differential
formula at r = 10
dA  2 10  10.1  10 
dA  2
The increase in the circles areas is
about 2π m2
Three Ways to Describe Change
As we move from c to a nearby point c + dx, we can
describe the change in f in three ways:
Actual
Error (change)
Relative Error
(change)
Percentage
Error (change)
Actual
Estimated
y  f  c  x   f  c 
dy  f '  c   dx
Other Vocab
y
f c
y
f c
100
dy
f c
dy
f c
100
Measurement Error: dx = Δx
Propagated Error: dy
Example 1
Inflating a bicycle tire changes its radius from 12 inches
to 13 inches. Use differentials to estimate the actual
change, the relative change, and the percentage
change in the perimeter of the tire. Estimate the changes
Calculate necessary
information for the formulas:
f  c   P 12   2 12  24
f '  x   P '  r   drd  2 r   2
f '  c   P ' 12   2
dx  dr  13  12  1
dy  dP  P ' 12   dr  2 1  2
Actual: ~ 2 in
(Errors):
Actual
dy  2
dy
f (c)
dy
f (c)
Relative
 242 
1
12
Percentage
100  242 100 
25
3
Relative: ~ 0.083 Percentage: ~ 8.333%
Example 2
Suppose that the side of a square is measured with
a ruler to be 10 inches with a measurement error of
at most ± 1/32 in. Estimate the error in the
computed area of the square.
The function for area is:
Use the differential formula
to find dA:
A '  2s
dA  2s  ds
As
2
Evaluate the differential
formula at s = 10
dA  2 10   
5
dA   8
Thus, the propagated error in the area is
estimated to be within ± 5/8 in2
1
32

Example 3
The volume of a sphere is to be computed from a
measured value of its radius. Estimate the maximum
permissible percentage error in the measurement if the
percentage error in the volume must be kept within ± 1.2%.
The function for volume is: V
Use the Differential Formula
to find dV:
V '  4 r
2
dV  4 r  dr
2
Thus, the estimated
percentage error in the
volume is within ± 0.4%
 r
4
3
3
Use the known information
to Find % Error of the Radius:
dV
V
4 r 2 dr
4 r3
3
3dr
r
dr
r
Definition of percentage error
% Error 
1.2 
1.2 
0.4 
100
100
100
100