differentials: linear and quadratic approximationshatalov/math 151 project.pdfย ยท differentials:...
TRANSCRIPT
Introduction of Linear
Approximation
๐ฟ ๐ฅ = ๐ ๐ + ๐โ ๐ ๐ฅ โ ๐
โข ๐ : point whose tangent line as an approximation of the function
โข ๐ฅ : point whose value is being approximated
โข If ๐(๐ฅ) is concave up after ๐ฅ = ๐, the approximation will be an underestimate, and if it is concave down, it will be an overestimate
Linear approximation of ๐ ๐ฅ = cos ๐ฅ
at ๐ = 0.
Definition of Linear Approximation
โThe equation of the tangent line to the curve ๐ฆ = ๐(๐ฅ) at
(๐, ๐(๐)) is
๐ฆ = ๐ ๐ + ๐โ ๐ ๐ฅ โ ๐ ,
soโฆthe tangent line at ๐(๐, ๐(๐)) [is] an approximation to the
curve ๐ฆ = ๐(๐ฅ) when ๐ฅ is near ๐.
๐ฟ ๐ฅ = ๐ ๐ + ๐โ ๐ ๐ฅ โ ๐
is called the linear approximation or tangent line
approximation of ๐ at ๐.โ
- Stewart, โCalculus: Early Vectorsโ
Introduction of Quadratic
Approximation
๐ ๐ฅ = ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐ + (1
2)๐โฒโฒ(๐)(๐ฅ โ ๐)2
or
๐ ๐ฅ = ๐ฟ ๐ฅ + (1
2)๐โฒโฒ(๐)(๐ฅ โ ๐)2
Quadratic approximation of ๐ ๐ฅ = ๐๐๐ ๐ฅ
at ๐ = 0.
๐ฆ = ๐ฟ(๐ฅ)
๐ฆ = ๐(๐ฅ)
Definition of Quadratic
Approximation The quadratic approximation also uses the point ๐ฅ = ๐ to
approximate nearby values, but uses a parabola instead of
just a tangent line to do so.
This gives a closer
approximation because the
parabola stays closer to the
actual function.
๐ฟ ๐ฅ for ๐๐ฅ at
๐ = 0.
History: Taylorโs Theorem
Linear and Quadratic approximations are based off of Taylorโs theorem of
polynomials. The theorem is named after 18th century mathematician Brook
Taylor who designed a general formula for approximating the values of
functions after a small change of the x-value. The formula was first published in
1712. His theorem is:
๐ ๐ฅ โ ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐ +๐"(๐)
2!(๐ฅ โ ๐)2+ โฏ +
๐ ๐ ๐
๐!๐ฅ โ ๐ ๐ ,
where a is a reference point, x is the nearby value being approximated, and k
is the maximum amount of times the original function can be derived.
History: Differentials
โข Began in 1680โs
โข Bernouli brothers
โข Gottfried Wilhelm Leibniz
โข Applied to geometry and mechanics
Producing Along Standards
A company requires that the bowling balls that it creates
have a volume in the range of 4170 โ 4200๐๐3. If the radii
of the balls are produced to be 10 ๐๐ and have an error of
.01 ๐๐, will the company be able to produce along these
standards?
Error Calculation with Linear
Approximation
๐ =4
3๐๐3 ๐ฟ(๐ฅ) = ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐
๐๐
๐๐= 4๐๐2 ๐ ๐ฅ = ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐
๐๐
๐๐= 4๐102 ๐ ๐ฅ โ ๐ ๐ = ๐โฒ ๐ ๐ฅ โ ๐
โ๐ฆ = ๐โฒ ๐ โ๐ฅ
โ๐ =๐๐
๐๐โ๐
Error Calculation with Linear
Approximation
Now using the formula โ๐ =๐๐
๐๐โ๐
๐๐
๐๐= 400๐
โ๐ = 0.01
โ๐ = 400๐ .01 = 12.57๐๐3
๐ 10 =4
3๐103 = 4188.79๐๐3
Analysis
We found that the volume of the perfectly created ball
would be 4188.79๐๐3 with a deviation of ยฑ12.57๐๐3. This
creates a range of 4176.22 โ 4201.26๐๐3, which does not
fit with the desired range of 4170 โ 4200๐๐3. The company
will not be able to produce along these standards.
Surface Area Errors
In the design of a waterpark fountain like the one shown
below, the cost of paint is important. With the paint costing
$0.25/๐๐๐โ๐๐ 2, what would the relative error in surface area
be if the radius of the circle has an error of
ยฑ .75 ๐๐๐โ๐๐ ?What would be the variation in price?
Take the inner radius to be ๐
4,
the radius of the circle to be 37 ๐๐๐โ๐๐ ,
and the height to be 2R.
Modifying the Linear Approximation
Formula
๐ฟ ๐ฅ = ๐ ๐ + ๐โฒ ๐ ๐ฅ โ ๐ โ โ๐ฆ =๐๐ฆ
๐๐ฅโ๐ฅ
Adding up the surface areas:
Sphere: 2๐๐ 2
Bottom of sphere: ๐๐ 2 โ ๐(๐
4)2
Cylinder: ๐(๐
4)2+2๐
๐
42๐ (top of cylinder doesnโt matter)
Surface Area = 4๐๐ 2
Proposed Area = 4๐(37)2
= 17,203.36 ๐๐๐โ๐๐ 2
Error in Surface Area
Surface Area Error : โ๐ = .75 ๐๐๐โ๐๐
Now take the derivative of the surface area formula, with
respect to the radius of the hemisphere.(๐ด(๐ ) = 4๐๐ 2)
๐๐ด
๐๐ = 8๐๐ ,
Error in Surface Area
โ๐ด๐ =๐๐ด
๐๐ โ๐ โ 8๐ 37 .75 = 697.43 ๐๐๐โ๐๐ 2
๐๐๐๐๐๐๐๐๐๐ก๐๐ฃ๐ =โ๐ด
๐ด=
697.43
17203.36= 0.0405
โ๐๐๐๐๐ = 697.43 0.25 = $174.36
Analysis
The small error in the radius of the sphere can cause a
substantial change in cost.
The cost of painting the structure: Cost = ๐ถ๐๐ ๐ก0 ยฑ โ๐ถ๐๐ ๐ก
๐ถ๐๐ ๐ก0 = 17203.36 .25 = $4300.84 โ๐ถ๐๐ ๐ก = $174.36
Cost = $ 4300.84 ยฑ 174.36
Percent Error of R: .75
37ร 100 = 2.03%
Percent Error of A: 697.43
17203.36ร 100 = 4.05%
Oscillation with Differentials
The water level in an artificial pond, which has both a
constant drainage, due to absorption and evaporation, and
a periodically active source, over a given 24-hour period is
given by the approximation
โ ๐ฅ = 5sin๐ฅ
2+ 25,
where โ is given in ft. This equation was the result of taking
measurement tests every 30 minutes. That leaves the
depth at many other times indefinite, but assumable. Given
that ๐ฅ = 0 is midnight, ๐ฅ = 1 is 1:00 A.M., and so on, what
would the depth have been at 11:03 A.M.?
Oscillation with Differentials
The first thing to do is identify given values.
โ ๐ฅ = 5sin๐ฅ
2+ 25, ๐ = 11 , ๐๐๐
๐ฅ = 11 +3
60 ๐๐ 11.05 .
And we know that the linear and quadratic approximation
formulas are
๐ฟ ๐ฅ = โ ๐ + โโฒ ๐ ๐ฅ โ ๐ ๐๐๐
๐ ๐ฅ = ๐ฟ ๐ฅ +โ"
2(๐ฅ โ ๐)2
Oscillation with Differentials
Now we plug in values and solve where we can. So if
โ ๐ฅ = 5sin๐ฅ
2+ 25, ๐กโ๐๐ โโฒ ๐ฅ = 2.5cos
๐ฅ
2. Now :
๐ฟ 11.05 = โ 11 + โโฒ 11 11.05 โ 11
๐ฟ 11.05 = 21.472 + 1.772 .05
๐ฟ 11.05 = 21.5606
So according to the linear approximation, at 11:03, the
waterโs depth was 21.5606 ft.
Oscillation with Differentials
Now to find the depth of the water according to a quadratic
approximation, we need โ"(๐ฅ). So if
โโ ๐ฅ = 2.5cos ๐ฅ
2,
then
โ"(๐ฅ) = โ1.25sin ๐ฅ
2.
Oscillation with Differentials
Now:
๐ 11.05 = ๐ฟ 11.05 +โ"(11)
2(11.05 โ 11)2
๐ 11.05 = 21.5606 +.882
2.05 2
๐ 11.05 = 21.5617
So now according to the quadratic approximation, the
depth of the water was 21.5617 ft.
Oscillation with Differentials: Analysis
When ๐ฅ = 11.05 is plugged into the original equation, the
depth is 21.5620 ft. To calculate error for the Linear
approximation it is โ 11.05 โ ๐ฟ(11.05)
โ(11.05) ,
which is an error of 0.0065%. For the Quadratic
approximation error it is โ 11.05 โ ๐(11.05)
โ(11.05) ,
which is an error of .0014%. So while both approximations
have very low error, the Quadratic approximation is even
more accurate.
Visual Map Differentials Functions
Linear/Quadratic
Approximation
Method of Finite
Differences
(Create 1st order methods for
solving/approximating
solutions)
Determining
an Accurate
Inverse
Function
Relative
Errors
Common Mistakes
โข Omission of 1
2 in equation for quadratic approximation
โข (๐ฅ โ ๐) is used instead of (๐ฅ โ ๐)2 in the quadratic
approximation
โข Chain rule forgotten when taking derivatives
โข Plug ๐โฒ(๐ฅ) instead of ๐โฒ(๐) into linear or quadratic
approximation
References
"Linear Approximation." Wikipedia.org. 6 November 2012. Web. 3 June 2012.
<http://en.wikipedia.org/wiki/Linear_approximation>.
Stewart, James. "Calculus: Early Vectors." 1st ed. Pacific Grove: Brooks/Cole, 1999.
Print.
"Taylor's Theorem." Wikipedia.org. 6 November 2012. Web. 5 November 2012.
<http://en.wikipedia.org/wiki/Taylor's_theorem>.