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Linear Approximation and Differentials
Lesson 3.9
Tangent Line Approximation
• Consider a tangent to a function at a point where x = a
• Close to the point, the tangent line is an approximation for f(x)
•
a
f(a)
y=f(x)
•The equation of the tangent line:y = f(a) + f ‘(a)(x – a)
•The equation of the tangent line:y = f(a) + f ‘(a)(x – a)
Tangent Line Approximation
• We claim that
• This is called linearization of the function at x=a.
• Recall that when we zoom in on an interval of a differentiable function far enough, it looks like a line.
1 1( ) ( ) '( )( )f x f a f a x a
New Look at
• dy = rise of tangent relative to x = dx y = change in y that occurs relative to
x = dx
dy
dx
x
x = dx
• dy y
•• x + x
New Look at
• We know that
then
• Recall that dy/dx is NOT a quotient it is the notation for the derivative
• However … sometimes it is useful to use dy and dx as actual quantities
dy
dx
'( )y
f xx
'( )y f x x
The Differential of y
• Consider
• Then we can say
this is called the differential of y the notation is d(f(x)) = f ’(x) * dx it is an approximation of the actual change of y
for a small change of x
'( )y dy
f xx dx
'( )dy f x dx y
Animated Graphical View
• Note how the "del y" and the dy in the figure get closer and closer
Example: Find the differential of the function:
Remember a differential is dy = f ‘(x)· dx (assuming y=f(x))
DifferentialsDifferentials
2)35(
)43(5)35(3
x
xx
dx
dy2)35(
2015915
x
xx
2)35(
11
xdx
dy
35
43
x
xy
dxx
dy2)35(
11
Try It Out
• Note the examples and rules for differentialson page 238.
• Find the differential of:
1) y = 3 – 5x2
2) f(x) = xe-2x
DifferentialsLet ( ) be a differentiable function. The is an
independent variable. The is '( ) .
y f x
dy f x dx
differential
differential
dx
dy
Example Finding the Differential dyFind the differential dy and evaluate dy for the given value of x and dx.
01.0 ,1 ,25 dxxxxy
dxxdy 25 4
01.0215 4 dy
07.0
Differential Estimate of Change (needed for #34 and #37 on the homework)
Three types of changes that can be found. Absolute, Relative and Percent Change
Absolute
True
Estimated
afxafxf
dxxfdf
Relative and Percent Change
True Estimate
Relative
Percent .
afxf
afdf
100
af
xf 100af
df% %
Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume. To solve this problem, we will use the equation:
∆V ≈ dV = V ‘(r)· dr
DifferentialsDifferentials
3
3
4 sphere a of volume)( rrV
24 )(
rdr
rdV drrrdV 24)(
01.0)2(4 2 dV35027.016.0 indV <- Approximate error in
volume
Example: A company makes ball-bearings with a radius of 2 inches. The measurements are considered to be correct to within 0.01 in. Use differentials to determine the approximate error in volume, approximate relative error in volume, AND approximate relative error percentage in volume.The estimate of relative error is given by:
DifferentialDifferential
afdf
The estimate percentage of relative error is given by: %100
af
df
015. )2(
34
16.
3
%5.1100%* )2(
34
16.
3
<- Approximate relative error in volume
<- Approximate relative error percentage in volume
Differentials for Approximations
• Consider
• Use
• Then with x = 25, dx = .3 obtain approximation
25.3
( )
1( ) '( )
2
f x x
f x x f x f x dx x dxx
Propagated Error
• Consider a rectangular box with a square base Height is 2 times length
of sides of base Given that x = 3.5 You are able to measure with 3% accuracy
• What is the error propagated for the volume?
xx
2x
Propagated Error
• We know that
• Then dy = 6x2 dx = 6 * 3.52 * 0.105 = 7.7175This is the approximate propagated error for the volume
32 2
3% 3.5 0.105
V x x x x
dx
Propagated Error
• The propagated error is the dy sometimes called the df
• The relative error is
• The percentage of error relative error * 100% (in this case, it would be
9%)
7.71750.09
( ) 85.75
dy
f x