. . . . . .
Section3.8LinearApproximationand
Differentials
Math1a
March10, 2008
Announcements
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.
.Image: Flickruser cobalt123
. . . . . .
Announcements
◮ Midtermisgraded. Cometoofficehoursifyoudon’thaveitbackyet.
◮ ProblemSessionsSunday, Thursday, 7pm, SC 310◮ OfficehoursTues, Weds, 2–4pmSC 323
. . . . . .
Outline
Thelinearapproximationofafunctionnearapoint
Examples
. . . . . .
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)
. . . . . .
Outline
Thelinearapproximationofafunctionnearapoint
Examples
. . . . . .
Example
ExampleEstimate ln(1.02).
SolutionWeknow ln(1) = 0, so ln(1.02) shouldnotbetoofarfrom 0. Infact
ln(1.02) ≈ ln(1) +ddx
ln x∣∣∣∣x=0
(0.02)
= 0 + 1(0.02) = 0.02.
Calculatorcheck: . . . ln(1.02) ≈ 0.0198.
. . . . . .
Example
ExampleEstimate ln(1.02).
SolutionWeknow ln(1) = 0, so ln(1.02) shouldnotbetoofarfrom 0.
Infact
ln(1.02) ≈ ln(1) +ddx
ln x∣∣∣∣x=0
(0.02)
= 0 + 1(0.02) = 0.02.
Calculatorcheck: . . . ln(1.02) ≈ 0.0198.
. . . . . .
Example
ExampleEstimate ln(1.02).
SolutionWeknow ln(1) = 0, so ln(1.02) shouldnotbetoofarfrom 0. Infact
ln(1.02) ≈ ln(1) +ddx
ln x∣∣∣∣x=0
(0.02)
= 0 + 1(0.02) = 0.02.
Calculatorcheck: . . . ln(1.02) ≈ 0.0198.
. . . . . .
Example
ExampleEstimate ln(1.02).
SolutionWeknow ln(1) = 0, so ln(1.02) shouldnotbetoofarfrom 0. Infact
ln(1.02) ≈ ln(1) +ddx
ln x∣∣∣∣x=0
(0.02)
= 0 + 1(0.02) = 0.02.
Calculatorcheck: . . . ln(1.02) ≈
0.0198.
. . . . . .
Example
ExampleEstimate ln(1.02).
SolutionWeknow ln(1) = 0, so ln(1.02) shouldnotbetoofarfrom 0. Infact
ln(1.02) ≈ ln(1) +ddx
ln x∣∣∣∣x=0
(0.02)
= 0 + 1(0.02) = 0.02.
Calculatorcheck: . . . ln(1.02) ≈ 0.0198.
. . . . . .
AnotherExample
ExampleEstimate
√10 usingthefactthat 10 = 9 + 1.
Solution
√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.
Solution
√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.
Solution
√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.
Solution
√10 ≈
√9 +
ddx
√x∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(196
)2
=36136
.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
If y = f(x),
thendydx
= f′(x), and dy = f′(x)dx.
. .x
.y
.x .x + ∆x
.dx = ∆x
.∆y.dy
Then ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
If y = f(x), thendydx
= f′(x),
and dy = f′(x)dx.
. .x
.y
.x .x + ∆x
.dx = ∆x
.∆y.dy
Then ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
If y = f(x), thendydx
= f′(x), and dy = f′(x)dx.
. .x
.y
.x .x + ∆x
.dx = ∆x
.∆y.dy
Then ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Differentialsareanotherwaytoexpressderivatives
If y = f(x), thendydx
= f′(x), and dy = f′(x)dx.
. .x
.y
.x .x + ∆x
.dx = ∆x
.∆y.dy
Then ∆y ≈ dy = f′(x0)dx near x0.
. . . . . .
Anotherexample
ExampleDropa1kgballofftheroofoftheScienceCenter(30mhigh). Weusuallysaythatafallingobjectfeelsaforce F = mg fromgravity.
Infact, theforcefeltis
F(r) =GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
At r = re theforcereallyisGMmr2e
= mg. Whatisthemaximum
errorinmakingthisapproximation? Therelativeerror? Thepercentageerror?
Note:re = 6378.1 km, M = 5.9724× 1024 kg, andG = 6.6742× 10−11 N ·m2 · kg−2
. . . . . .
Anotherexample
ExampleDropa1kgballofftheroofoftheScienceCenter(30mhigh). Weusuallysaythatafallingobjectfeelsaforce F = mg fromgravity.Infact, theforcefeltis
F(r) =GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
At r = re theforcereallyisGMmr2e
= mg. Whatisthemaximum
errorinmakingthisapproximation? Therelativeerror? Thepercentageerror?
Note:re = 6378.1 km, M = 5.9724× 1024 kg, andG = 6.6742× 10−11 N ·m2 · kg−2
. . . . . .
Anotherexample
ExampleDropa1kgballofftheroofoftheScienceCenter(30mhigh). Weusuallysaythatafallingobjectfeelsaforce F = mg fromgravity.Infact, theforcefeltis
F(r) =GMmr2
,
where M isthemassoftheearthand r isthedistancefromthecenteroftheearthtotheobject. G isaconstant.
At r = re theforcereallyisGMmr2e
= mg. Whatisthemaximum
errorinmakingthisapproximation? Therelativeerror? Thepercentageerror?
Note:re = 6378.1 km, M = 5.9724× 1024 kg, andG = 6.6742× 10−11 N ·m2 · kg−2
. . . . . .
Systematiclinearapproximation
◮√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2.
So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
◮ Thisisabetterapproximationsince (17/12)2 = 289/144
◮ 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
◮√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
◮ Thisisabetterapproximationsince (17/12)2 = 289/144
◮ 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
◮√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
◮ Thisisabetterapproximationsince (17/12)2 = 289/144
◮ 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
◮√2 isirrational, but
√9/4 isrationaland 9/4 iscloseto 2. So
√2 =
√9/4− 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
◮ Thisisabetterapproximationsince (17/12)2 = 289/144
◮ 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
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
.
.2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2)
.•.(2, 17/12)
. . . . . .
Illustrationofthepreviousexample
..2
.•.(9/4, 3/2).•
.(2, 17/12)
