Section 2.8 Linear Approximations and Differentials
The idea is that we use a tangent line to approximate values close to some x.
Let x = a, then the point above is a ,f a If I write out the equation of the tangent line through this point:
when x = a:
y y 1 = f ' x x x1 Basic Form
yf a = f ' a xa
Specific Form
Then just solve for y:
y = f af ' a xa
(or)
f x f af ' axa
So, if we are close to x = a, this y might be a good approximation to f.
Traditionally, we rename the y as L(x)
L x = f a f ' axa
and call L(x) a linearization of f(x) near x = a.
Example: Find the linearization L(x) of the function at a:
1
2x
f x =
a=0
1st Find f ' (x):
Simplify:
2nd
1
f x = 2x 2
3
d
1
f ' x = 2 x 2 2 x
2
dx
3
1
f ' x = 2x 2 1
2
1
f ' x =
3
2 2x L x = f a f ' axa and fill in the pieces.
f a = f 0 =
1
1
=
20 2
f ' a = f ' 0 =
1
2 20
3
=
1
1
1
=
=
2 8 22 2
4 2
3rd Substitute:
L x =
1
1
x0
2 4 2
So, if I wanted an approximate linearized value for x = 0.01 (i.e. an estimate)
L0.01 =
1
1
0.01 would be a good guess
2 4 2
Example: Find the linear approximation of the function g x = 3 1 x at a = 0 and use it to
approximate 3 0.95 and 3 1.1
1st Find g' (x):
1
3
g x = 1x 2
1
g ' x = 1x 3
3
1
g ' x = 3
3 1 x2
2nd
or
L x = f a f ' axa
L x = g ag ' a xa
3
3
g 0 = 10 = 1 = 1
g ' 0 =
1
3 10
3
2
=
1
1
=
31 3
3rd Substitute:
1
L x = 1 x0 3
1
L x = 1 x
3
4th Approximate
For our approximation :
3 0.95 compare to 3 1 x
0.95 = 1x
x = 0.05
So:
1
L x = L0.05 = 1 0.05 3
1 5
= 1
3 100
295
5
=
= 1
300 300
59
=
0.9833
60
(Actual:
3 0.95 = 0.9830
Likewise
3 1.1 compare to
Not Bad)
3 1 x
1.1 = 1x
x = 0.1
So:
1
L x = L0.1 = 1 0.1
3
1 1
= 1
3 10
1
31
= 1 =
30 30
1.0 3
(Actual:
3 1.1 = 1.0323
Again, not bad)
Differentials
Key:
dy
= f ' x
dx
dy
= f ' x
dependent
notation
dx
independent
* Small changes in y depend on the value of a derivative and a small change in x.
Think of it like this:
So:
dx =
dy
x
y
This dy is related through the tangent line not the
function.
y = f x x f x
Notice that
But, dy is related to the amount the tangent line deviates from f through a linearization
But notice that dy
y as
x 0 .
We can notationally change linearization to reflect this idea of differentials.
From Before:
f x f af ' axa
dy = f ' x dx Key: x = a:
dy=f ' adx
Let:
dx = xa
x = adx
So:
f adx f a f ' aadxa
f adx f af ' a dx
Concept: f adx f ady
Example:
The radius of a circular disk is given as 24 cm with a maximum error of 0.2 cm. (a) Use
differentials to estimate the maximum error in the calculated area of the disk. (b) What
is the relative error? (c) What is the percentage error?
(a)
2
A= r
dA
d
= r 2
dr
dr
dA
=2 r
dr
dA = 2 r dr
dA = 2 24 cm0.2 cm
2
2
dA = 9.6 cm 50.15 cm (b)
2
dA
9.6 cm
=
= 0.4 (about twice that of the radius)
A
24 cm2
(c)
Just multiply by 100
0.4 * 100 = 40%
More Examples:
1. Approximate
Notice that
99.8
100 99.8
100x where x = 0.02 makes 99.8
1
Choose f x = 100 x = 100x 2
Linearize around a = 0
1
1
f ' x = 100 x 2 1
2
f a = f 0 = 1000 = 10
1
1
1
1
f ' a = f ' 0 = 1000 2 = =
2
210
20
L x = f a f ' axa
1
L x = 10 x
20
x = 0.2
L0.2 = 10
1 2
1
= 10
= 9.99
20 10
100
2
2. Approximate 8.06 3 = 3 8.062
2
3
Notice that:
8 = 8 = 64 = 4 = 4
3
2
2
3
3
3
2
and 8 3 8.063
2
Choose: f x = 8 x 3
x = 0.06
Linearize around a = 0.
1
2
f ' x = 8 x 3 1
3
2
3
2
3
f 0 = 80 = 8 = 4
1
2
2
1
f ' 0 = 80 3 =
=
3
32 3
Then:
L x = f a f ' axa
L x = f 0f ' 0x0
1
L x = 4 x
3
1 6
3 100
2
= 4
100
= 4.02
L0.06 = 4
2
So: 8.06 3 4.02
3. (From Homework) 21. The edge of cube was found to be 30 cm with a possible error in
measurement of 0.1 cm. Use the differentials to estimate the maximum possible error, relative
error, and percentage error when computing: (a) the volume of the cube and (b) the surface area
of the cube.
(a)
The Volume of the Cube:
x = 30 cm ± 0.1 cm
3
V=x
dV
= 3x 2
dx
2
dV = 3x dx
2
dV = 330cm ±0.1 cm
2
dV = 3900 cm ±0.1 cm
2
dV = 2700 cm ±0.1 cm
dV = ±270 cm3
Relative Error: Divide the error by the total volume:
3
3
dV ±270 cm
±270 cm
=
= 0.01
=
3
V
30 cm
27,000 cm3
Percentage Error: multiply by 100: 0.01 * 100 = 1%
(b)
The Surface Area of the Cube:
2
A = 6x
dA
= 62x dx
dA = 12x dx
dA = 1230cm ±0.1 cm
2
dA = ±36cm
Relative Error: Divide the error by the total area:
2
2
dA
±36cm
±36cm
=
= 0.067
=
2
A
6 30 cm
5,400 cm2
Percentage Error: multiply by 100: 0.0067 * 100 = 0.67%
4. The circumference of a sphere was measured to be 84 cm with possible error of 0.5 cm. (a) Use
differentials to estimate maximum error calculated in surface area. (b) Use differentials to
estimate maximum error in calculating volume.
(a) Use differentials to estimate maximum error calculated in surface area.
C=2 r
84 cm = 2 r
42
cm
r=
And
dC
=2
dr
dC = 2 dr
0.5 cm = 2 dr
2
A=4 r
dA
= 4 2r dr
dA = 4r 2 dr
A=4
A=
42
7056
cm
cm
2
2
Maximum Error:
dA = 4
42 cm 0.5 cm
84
dA =
cm
2
Relative Error:
84
dA
84
=
= 0.0119
=
A
7056
7056
(b) Use differentials to estimate maximum error in calculating volume.
4
r3
3
dV
4
= 3 r 2
dr
3
2
dV = 2 r 2 dr
V=
Maximum Error:
dV = 2
42
dV =
2
cm 0.5 cm
1764
Relative Error:
2
cm 3
1764
dV
=
V
4
3
2
cm
1764
3
42 cm 2
3
=
78,784
2
=
1
= 0.0179
56