lesson 13: linear approximation
DESCRIPTION
New perspectives on an old idea: the derivative measures the slope of the tangent line, which is the line which best fits the graph near a point.TRANSCRIPT
![Page 1: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/1.jpg)
. . . . . .
Section 2.8Linear Approximation and
Differentials
V63.0121, Calculus I
February 26/March 2, 2009
Announcements
I Midterm I Wednesday March 4 in class.I OH this week: MT 1–2pm, W 2–3pmI Get half of additional ALEKS points through March 22, 11:59pm
.
.Image credit: cobalt123
![Page 2: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/2.jpg)
. . . . . .
Outline
The linear approximation of a function near a pointExamples
DifferentialsThe not-so-big idea
![Page 3: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/3.jpg)
. . . . . .
The Big Idea
QuestionLet f be differentiable at a. What linear function best approximates fnear a?
AnswerThe tangent line, of course!
QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x − a)
![Page 4: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/4.jpg)
. . . . . .
The Big Idea
QuestionLet f be differentiable at a. What linear function best approximates fnear a?
AnswerThe tangent line, of course!
QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x − a)
![Page 5: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/5.jpg)
. . . . . .
The Big Idea
QuestionLet f be differentiable at a. What linear function best approximates fnear a?
AnswerThe tangent line, of course!
QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x − a)
![Page 6: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/6.jpg)
. . . . . .
The Big Idea
QuestionLet f be differentiable at a. What linear function best approximates fnear a?
AnswerThe tangent line, of course!
QuestionWhat is the equation for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′(a)(x − a)
![Page 7: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/7.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2
andf′
(π3
)=
12
.
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 8: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/8.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2
andf′
(π3
)=
12
.
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 9: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/9.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2
andf′
(π3
)=
12
.
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 10: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/10.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)=
12
.
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 11: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/11.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 12: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/12.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 13: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/13.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)
I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 14: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/14.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)I Thus
sin(
61π
180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 15: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/15.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)I Thus
sin(
61π
180
)≈ 0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 16: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/16.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)I Thus
sin(
61π
180
)≈ 0.87475
Calculator check: sin(61◦) ≈
0.87462.
![Page 17: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/17.jpg)
. . . . . .
Example
ExampleEstimate sin(61◦) by using a linear approximation(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 So the linear approximationnear 0 is L(x) = 0 + 1 · x = x.
I Thus
sin(
61π
180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(
π3
)=
√3
2 andf′
(π3
)= 1
2 .
I So L(x) =
√3
2+
12
(x − π
3
)I Thus
sin(
61π
180
)≈ 0.87475
Calculator check: sin(61◦) ≈ 0.87462.
![Page 18: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/18.jpg)
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.big difference!
.y = L2(x) =√
32 + 1
2
(x − π
3
)
.
.π/3
.
.very little difference!
![Page 19: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/19.jpg)
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.big difference!.y = L2(x) =
√3
2 + 12
(x − π
3
)
.
.π/3
.
.very little difference!
![Page 20: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/20.jpg)
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.big difference!
.y = L2(x) =√
32 + 1
2
(x − π
3
)
.
.π/3
.
.very little difference!
![Page 21: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/21.jpg)
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.big difference!
.y = L2(x) =√
32 + 1
2
(x − π
3
)
.
.π/3
.
.very little difference!
![Page 22: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/22.jpg)
. . . . . .
Illustration
. .x
.y
.y = sin x
.61◦
.y = L1(x) = x
..0
.
.big difference!
.y = L2(x) =√
32 + 1
2
(x − π
3
)
.
.π/3
. .very little difference!
![Page 23: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/23.jpg)
. . . . . .
Another Example
ExampleEstimate
√10 using the fact that 10 = 9 + 1.
SolutionThe key step is to use a linear approximation to f(x) =
√x near a = 9 to
estimate f(10) =√
10.
√10 ≈
√9 +
ddx
√x
∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(
196
)2
=36136
.
![Page 24: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/24.jpg)
. . . . . .
Another Example
ExampleEstimate
√10 using the fact that 10 = 9 + 1.
SolutionThe key step is to use a linear approximation to f(x) =
√x near a = 9 to
estimate f(10) =√
10.
√10 ≈
√9 +
ddx
√x
∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(
196
)2
=36136
.
![Page 25: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/25.jpg)
. . . . . .
Another Example
ExampleEstimate
√10 using the fact that 10 = 9 + 1.
SolutionThe key step is to use a linear approximation to f(x) =
√x near a = 9 to
estimate f(10) =√
10.
√10 ≈
√9 +
ddx
√x
∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(
196
)2
=36136
.
![Page 26: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/26.jpg)
. . . . . .
Another Example
ExampleEstimate
√10 using the fact that 10 = 9 + 1.
SolutionThe key step is to use a linear approximation to f(x) =
√x near a = 9 to
estimate f(10) =√
10.
√10 ≈
√9 +
ddx
√x
∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(
196
)2
=
36136
.
![Page 27: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/27.jpg)
. . . . . .
Another Example
ExampleEstimate
√10 using the fact that 10 = 9 + 1.
SolutionThe key step is to use a linear approximation to f(x) =
√x near a = 9 to
estimate f(10) =√
10.
√10 ≈
√9 +
ddx
√x
∣∣∣∣x=9
(1)
= 3 +1
2 · 3(1) =
196
≈ 3.167
Check:(
196
)2
=36136
.
![Page 28: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/28.jpg)
. . . . . .
Dividing without dividing?ExampleSuppose I have an irrational fear of division and need to estimate577 ÷ 408. I write
577408
= 1 + 1691
408= 1 + 169 × 1
4× 1
102.
But still I have to find1
102.
SolutionLet f(x) =
1x
. We know f(100) and we want to estimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002 (2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculator check:577408
≈ 1.41422.
![Page 29: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/29.jpg)
. . . . . .
Dividing without dividing?ExampleSuppose I have an irrational fear of division and need to estimate577 ÷ 408. I write
577408
= 1 + 1691
408= 1 + 169 × 1
4× 1
102.
But still I have to find1
102.
SolutionLet f(x) =
1x
. We know f(100) and we want to estimate f(102).
f(102) ≈ f(100) + f′(100)(2) =1
100− 1
1002 (2) = 0.0098
=⇒ 577408
≈ 1.41405
Calculator check:577408
≈ 1.41422.
![Page 30: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/30.jpg)
. . . . . .
Outline
The linear approximation of a function near a pointExamples
DifferentialsThe not-so-big idea
![Page 31: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/31.jpg)
. . . . . .
Differentials are another way to express derivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, so we canwrite this as
∆y ≈ dy = f′(x)dx.
And this looks a lot like theLeibniz-Newton identity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linear approximation means ∆y ≈ dy = f′(x0) dx near x0.
![Page 32: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/32.jpg)
. . . . . .
Differentials are another way to express derivatives
f(x + ∆x) − f(x)︸ ︷︷ ︸∆y
≈ f′(x)∆x︸ ︷︷ ︸dy
Rename ∆x = dx, so we canwrite this as
∆y ≈ dy = f′(x)dx.
And this looks a lot like theLeibniz-Newton identity
dydx
= f′(x) . .x
.y
.
.
.x .x + ∆x
.dx = ∆x
.∆y.dy
Linear approximation means ∆y ≈ dy = f′(x0) dx near x0.
![Page 33: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/33.jpg)
. . . . . .
ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?
SolutionWrite A(ℓ) =
12ℓ2. We want to know ∆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 = ℓ dℓ, which should be a good estimate for ∆ℓ.
When ℓ = 8 and dℓ = 112 , we have dA = 8
12 = 23 ≈ 0.667. So we
get estimates close to the hundredth of a square foot.
![Page 34: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/34.jpg)
. . . . . .
ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?
SolutionWrite A(ℓ) =
12ℓ2. We want to know ∆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 = ℓ dℓ, which should be a good estimate for ∆ℓ.
When ℓ = 8 and dℓ = 112 , we have dA = 8
12 = 23 ≈ 0.667. So we
get estimates close to the hundredth of a square foot.
![Page 35: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/35.jpg)
. . . . . .
ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?
SolutionWrite A(ℓ) =
12ℓ2. We want to know ∆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 = ℓ dℓ, which should be a good estimate for ∆ℓ.
When ℓ = 8 and dℓ = 112 , we have dA = 8
12 = 23 ≈ 0.667. So we
get estimates close to the hundredth of a square foot.
![Page 36: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/36.jpg)
. . . . . .
ExampleA sheet of plywood measures 8 ft × 4 ft. Suppose ourplywood-cutting machine will cut a rectangle whose width is exactlyhalf its length, but the length is prone to errors. If the length is off by1 in, how bad can the area of the sheet be off by?
SolutionWrite A(ℓ) =
12ℓ2. We want to know ∆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 = ℓ dℓ, which should be a good estimate for ∆ℓ.
When ℓ = 8 and dℓ = 112 , we have dA = 8
12 = 23 ≈ 0.667. So we
get estimates close to the hundredth of a square foot.
![Page 37: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/37.jpg)
. . . . . .
GravitationPencils down!
Example
I Drop a 1 kg ball off the roof of the Science Center (50m high).We usually say that a falling object feels a force F = −mg fromgravity.
I In fact, the force felt is
F(r) = −GMmr2
,
where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.
I At r = re the force really is F(re) =GMm
r2e= −mg.
I What is the maximum error in replacing the actual force felt atthe top of the building F(re + ∆r) by the force felt at groundlevel F(re)? The relative error? The percentage error?
![Page 38: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/38.jpg)
. . . . . .
GravitationPencils down!
Example
I Drop a 1 kg ball off the roof of the Science Center (50m high).We usually say that a falling object feels a force F = −mg fromgravity.
I In fact, the force felt is
F(r) = −GMmr2
,
where M is the mass of the earth and r is the distance from thecenter of the earth to the object. G is a constant.
I At r = re the force really is F(re) =GMm
r2e= −mg.
I What is the maximum error in replacing the actual force felt atthe top of the building F(re + ∆r) by the force felt at groundlevel F(re)? The relative error? The percentage error?
![Page 39: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/39.jpg)
. . . . . .
SolutionWe wonder if ∆F = F(re + ∆r) − F(re) is small.
I Using a linear approximation,
∆F ≈ dF =dFdr
∣∣∣∣re
dr = 2GMm
r3edr
=
(GMm
r2e
)drre
= 2mg∆rre
I The relative error is∆FF
≈ −2∆rre
I re = 6378.1 km. If ∆r = 50 m,
∆FF
≈ −2∆rre
= −250
6378100= −1.56 × 10−5 = −0.00156%
![Page 40: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/40.jpg)
. . . . . .
Systematic linear approximation
I√
2 is irrational, but√
9/4 is rational and 9/4 is close to 2.
So
√2 =
√9/4 − 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I This is a better approximation since (17/12)2 = 289/144
I Do it again!
√2 =
√289/144 − 1/144 ≈
√289/144 +
12(17/12)
(−1/144) = 577/408
Now(
577408
)2
=332, 929166, 464
which is1
166, 464away from 2.
![Page 41: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/41.jpg)
. . . . . .
Systematic linear approximation
I√
2 is irrational, but√
9/4 is rational and 9/4 is close to 2. So
√2 =
√9/4 − 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I This is a better approximation since (17/12)2 = 289/144
I Do it again!
√2 =
√289/144 − 1/144 ≈
√289/144 +
12(17/12)
(−1/144) = 577/408
Now(
577408
)2
=332, 929166, 464
which is1
166, 464away from 2.
![Page 42: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/42.jpg)
. . . . . .
Systematic linear approximation
I√
2 is irrational, but√
9/4 is rational and 9/4 is close to 2. So
√2 =
√9/4 − 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I This is a better approximation since (17/12)2 = 289/144
I Do it again!
√2 =
√289/144 − 1/144 ≈
√289/144 +
12(17/12)
(−1/144) = 577/408
Now(
577408
)2
=332, 929166, 464
which is1
166, 464away from 2.
![Page 43: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/43.jpg)
. . . . . .
Systematic linear approximation
I√
2 is irrational, but√
9/4 is rational and 9/4 is close to 2. So
√2 =
√9/4 − 1/4 ≈
√9/4 +
12(3/2)
(−1/4) =1712
I This is a better approximation since (17/12)2 = 289/144
I Do it again!
√2 =
√289/144 − 1/144 ≈
√289/144 +
12(17/12)
(−1/144) = 577/408
Now(
577408
)2
=332, 929166, 464
which is1
166, 464away from 2.
![Page 44: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/44.jpg)
. . . . . .
Illustration of the previous example
.
.2
.
( 94 , 3
2)
..(2, 17
12)
![Page 45: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/45.jpg)
. . . . . .
Illustration of the previous example
.
.2
.
( 94 , 3
2)
..(2, 17
12)
![Page 46: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/46.jpg)
. . . . . .
Illustration of the previous example
..2
.
( 94 , 3
2)
..(2, 17
12)
![Page 47: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/47.jpg)
. . . . . .
Illustration of the previous example
..2
.
( 94 , 3
2)
..(2, 17
12)
![Page 48: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/48.jpg)
. . . . . .
Illustration of the previous example
..2
.
( 94 , 3
2)
..(2, 17
12)
![Page 49: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/49.jpg)
. . . . . .
Illustration of the previous example
..2
.
( 94 , 3
2)
..(2, 17
12)
![Page 50: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/50.jpg)
. . . . . .
Illustration of the previous example
..2
.
( 94 , 3
2)
..(2, 17
12)
![Page 51: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/51.jpg)
. . . . . .
Illustration of the previous example
..2
..( 9
4 , 32).
.(2, 17/12)
..( 289
144 , 1712
)..(2, 577
408
)
![Page 52: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/52.jpg)
. . . . . .
Illustration of the previous example
..2
..( 9
4 , 32).
.(2, 17/12)..( 289
144 , 1712
)
..(2, 577
408
)
![Page 53: Lesson 13: Linear Approximation](https://reader034.vdocument.in/reader034/viewer/2022052400/559b5ec71a28ab524f8b457b/html5/thumbnails/53.jpg)
. . . . . .
Illustration of the previous example
..2
..( 9
4 , 32).
.(2, 17/12)..( 289
144 , 1712
)..(2, 577
408
)