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The tangent line to the graph of a function at a point can be thought of as a function itself. As such, it is the best linear function which agrees with the given function at the point. The function and its linear approximation will probably diverge away from the point at which they agree, but this "error" can be measured using the differential notation.
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. . . . . .
Section2.8LinearApproximationand
Differentials
V63.0121.034, CalculusI
October12, 2009
Announcements
I MidtermWednesdayonSections1.1–2.4
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
TheBigIdea
QuestionLet f bedifferentiableat a. Whatlinearfunctionbestapproximates f near a?
AnswerThetangentline, ofcourse!
QuestionWhatistheequationforthelinetangentto y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x− a)
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32
andf′
(π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈
0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈
0.87462.
. . . . . .
Example
ExampleEstimate sin(61◦) byusingalinearapproximation(i) about a = 0 (ii) about a = 60◦ = π/3.
Solution(i)
I If f(x) = sin x, then f(0) = 0and f′(0) = 1.
I Sothelinearapproximationnear 0 isL(x) = 0 + 1 · x = x.
I Thus
sin(61π
180
)≈ 61π
180≈ 1.06465
Solution(ii)
I Wehave f(
π3
)=
√32 and
f′(
π3
)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π
180
)≈ 0.87475
Calculatorcheck: sin(61◦) ≈ 0.87462.
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!.y = L2(x) =
√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
.
.verylittledifference!
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.bigdifference!
.y = L2(x) =√32 + 1
2
(x− π
3
)
..π/3
. .verylittledifference!
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=
36136
.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
SolutionThekeystepistousealinearapproximationto f(x) =
√x near
a = 9 toestimate f(10) =√10.
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write
577408
= 1 + 1691
408= 1 + 169× 1
4× 1
102.
ButstillI havetofind1
102.
SolutionLet f(x) =
1x. Weknow f(100) andwewanttoestimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002(2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculatorcheck:577408
≈ 1.41422.
. . . . . .
Dividingwithoutdividing?ExampleSupposeI haveanirrationalfearofdivisionandneedtoestimate577÷ 408. I write
577408
= 1 + 1691
408= 1 + 169× 1
4× 1
102.
ButstillI havetofind1
102.
SolutionLet f(x) =
1x. Weknow f(100) andwewanttoestimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002(2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculatorcheck:577408
≈ 1.41422.
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
Answer
I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)
. . . . . .
Answers
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
Answer
I 100miI 150miI 600mi(?) (Isitreasonabletoassume12hoursatthesamespeed?)
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
Answer
I $100I $150I $600(?)
. . . . . .
Questions
ExampleSupposewearetravelinginacarandatnoonourspeedis50mi/hr. Howfarwillwehavetraveledby2:00pm? by3:00pm?Bymidnight?
ExampleSupposeourfactorymakesMP3playersandthemarginalcostiscurrently$50/lot. Howmuchwillitcosttomake2morelots? 3morelots? 12morelots?
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
. . . . . .
Answers
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
AnswerTheslopeofthelineis
m =riserun
Wearegivena“run”of dx, sothecorresponding“rise”is mdx.
. . . . . .
Answers
ExampleSupposealinegoesthroughthepoint (x0, y0) andhasslope m. Ifthepointismovedhorizontallyby dx, whilestayingontheline,whatisthecorrespondingverticalmovement?
AnswerTheslopeofthelineis
m =riserun
Wearegivena“run”of dx, sothecorresponding“rise”is mdx.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, sowecanwritethisas
∆y ≈ dy = f′(x)dx.
AndthislooksalotliketheLeibniz-Newtonidentity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, sowecanwritethisas
∆y ≈ dy = f′(x)dx.
AndthislooksalotliketheLeibniz-Newtonidentity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linearapproximationmeans ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Usingdifferentialstoestimateerror
If y = f(x), x0 and ∆x isknown, andanestimateof∆y isdesired:
I Approximate: ∆y ≈ dyI Differentiate:
dy = f′(x)dxI Evaluateat x = x0 and
dx = ∆x.
. .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
ExampleA sheetofplywoodmeasures 8 ft× 4 ft. Supposeourplywood-cuttingmachinewillcutarectanglewhosewidthisexactlyhalfitslength, butthelengthispronetoerrors. Ifthelengthisoffby 1 in, howbadcantheareaofthesheetbeoffby?
SolutionWrite A(ℓ) =
12ℓ2. Wewanttoknow ∆A when ℓ = 8 ft and
∆ℓ = 1 in.
(I) A(ℓ + ∆ℓ) = A(9712
)=
9409288
So
∆A =9409288
− 32 ≈ 0.6701.
(II)dAdℓ
= ℓ, so dA = ℓdℓ, whichshouldbeagoodestimatefor
∆ℓ. When ℓ = 8 and dℓ = 112 , wehave
dA = 812 = 2
3 ≈ 0.667. Sowegetestimatesclosetothehundredthofasquarefoot.
. . . . . .
Why?
Whyuselinearapproximations dy whentheactualdifference ∆yisknown?
I Linearapproximationisquickandreliable. Finding ∆yexactlydependsonthefunction.
I Theseexamplesareoverlysimple. Seethe“AdvancedExamples”later.
I Inreallife, sometimesonly f(a) and f′(a) areknown, andnotthegeneral f(x).
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
MidtermFacts
I Coverssections1.1–2.4(Limits, Derivatives,DifferentiationuptoQuotientRule)
I CalculatorfreeI Hasabout7problemseachcouldhavemultipleparts
I Somefixed-response,somefree-response
I Tostudy:I outlineI doproblemsI metacognitionI askquestions!
. . . . . .
Outline
ThelinearapproximationofafunctionnearapointExamples
DifferentialsThenot-so-bigideaUsingdifferentialstoestimateerror
MidtermReview
AdvancedExamples
. . . . . .
GravitationPencilsdown!
Example
I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.
I Infact, theforcefeltis
F(r) = −GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
I At r = re theforcereallyis F(re) =GMmr2e
= −mg.
I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re + ∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?
. . . . . .
GravitationPencilsdown!
Example
I Dropa1 kgballofftheroofoftheSilverCenter(50mhigh).Weusuallysaythatafallingobjectfeelsaforce F = −mgfromgravity.
I Infact, theforcefeltis
F(r) = −GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
I At r = re theforcereallyis F(re) =GMmr2e
= −mg.
I Whatisthemaximumerrorinreplacingtheactualforcefeltatthetopofthebuilding F(re + ∆r) bytheforcefeltatgroundlevel F(re)? Therelativeerror? Thepercentageerror?
. . . . . .
SolutionWewonderif ∆F = F(re + ∆r) − F(re) issmall.
I Usingalinearapproximation,
∆F ≈ dF =dFdr
∣∣∣∣re
dr = 2GMmr3e
dr
=
(GMmr2e
)drre
= 2mg∆rre
I Therelativeerroris∆FF
≈ −2∆rre
I re = 6378.1 km. If ∆r = 50m,
∆FF
≈ −2∆rre
= −2 506378100
= −1.56×10−5 = −0.00156%
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2.
So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Systematiclinearapproximation
I√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I Thisisabetterapproximationsince (17/12)2 = 289/144
I Doitagain!
√2 =
√289/144− 1/144 ≈
√289/144+
12(17/12)
(−1/144) = 577/408
Now(577408
)2
=332, 929166, 464
whichis1
166, 464awayfrom 2.
. . . . . .
Illustrationofthepreviousexample
.
.2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
.
.2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
.
(94 ,32)
..(2, 1712)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)
..(289144 ,
1712
)..(2, 577408
)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)..(289144 ,
1712
)
..(2, 577408
)
. . . . . .
Illustrationofthepreviousexample
..2
..(94 ,
32).
.(2, 17/12)..(289144 ,
1712
)..(2, 577408
)