Lesson 12: Linear Approximation and Differentials (Section 41 slides)
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Transcript of Lesson 12: Linear Approximation and Differentials (Section 41 slides)
Section 2.8Linear Approximation and Differentials
V63.0121.041, Calculus I
New York University
October 13, 2010
Announcements
I Quiz 2 in recitation this week on §§1.5, 1.6, 2.1, 2.2I Midterm on §§1.1–2.5
. . . . . .
. . . . . .
Announcements
I Quiz 2 in recitation thisweek on §§1.5, 1.6, 2.1,2.2
I Midterm on §§1.1–2.5
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 2 / 27
. . . . . .
Objectives
I Use tangent lines to makelinear approximations to afunction.
I Given a function and apoint in the domain,compute thelinearization of thefunction at that point.
I Use linearization toapproximate values offunctions
I Given a function, computethe differential of thatfunction
I Use the differentialnotation to estimate errorin linear approximations.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 3 / 27
. . . . . .
Outline
The linear approximation of a function near a pointExamplesQuestions
DifferentialsUsing differentials to estimate error
Advanced Examples
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 4 / 27
. . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
. . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
. . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
. . . . . .
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)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 5 / 27
. . . . . .
The tangent line is a linear approximation
L(x) = f(a) + f′(a)(x− a)
is a decent approximation to fnear a.
How decent? The closer x is toa, the better the approxmationL(x) is to f(x)
. .x
.y
.
.
.
.f(a)
.f(x).L(x)
.a .x
.x − a
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 6 / 27
. . . . . .
The tangent line is a linear approximation
L(x) = f(a) + f′(a)(x− a)
is a decent approximation to fnear a.How decent? The closer x is toa, the better the approxmationL(x) is to f(x)
. .x
.y
.
.
.
.f(a)
.f(x).L(x)
.a .x
.x − a
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 6 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32
andf′(π3)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32
andf′(π3)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32
andf′(π3)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)=
12
.
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =
√32
+12
(x− π
3
)I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =√32
+12
(x− π
3
)
I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =√32
+12
(x− π
3
)I Thus
sin(61π180
)≈
0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =√32
+12
(x− π
3
)I Thus
sin(61π180
)≈ 0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =√32
+12
(x− π
3
)I Thus
sin(61π180
)≈ 0.87475
Calculator check: sin(61◦) ≈
0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
Example.
.
Example
Estimate sin(61◦) = sin(61π/180) 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(61π180
)≈ 61π
180≈ 1.06465
Solution (ii)
I We have f(π3)=
√32 and
f′(π3)= 1
2 .
I So L(x) =√32
+12
(x− π
3
)I Thus
sin(61π180
)≈ 0.87475
Calculator check: sin(61◦) ≈ 0.87462.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 7 / 27
. . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
. . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
. . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
. . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
. . . . . .
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!
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 8 / 27
. . . . . .
Another Example
Example
Estimate√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
.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
. . . . . .
Another Example
Example
Estimate√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
.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
. . . . . .
Another Example
Example
Estimate√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
.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
. . . . . .
Another Example
Example
Estimate√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
.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
. . . . . .
Another Example
Example
Estimate√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
.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 9 / 27
. . . . . .
Dividing without dividing?
Example
Suppose 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.
Solution
Let 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 10 / 27
. . . . . .
Dividing without dividing?
Example
Suppose 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.
Solution
Let 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 10 / 27
. . . . . .
Questions
Example
Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Example
Suppose our factory makes MP3 players and the marginal cost iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?
Example
Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 11 / 27
. . . . . .
Answers
Example
Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Answer
I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same
speed?)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 12 / 27
. . . . . .
Answers
Example
Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Answer
I 100miI 150miI 600mi (?) (Is it reasonable to assume 12 hours at the same
speed?)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 12 / 27
. . . . . .
Questions
Example
Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Example
Suppose our factory makes MP3 players and the marginal cost iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?
Example
Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 13 / 27
. . . . . .
Answers
Example
Suppose our factory makes MP3 players and the marginal cost iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?
Answer
I $100I $150I $600 (?)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 14 / 27
. . . . . .
Questions
Example
Suppose we are traveling in a car and at noon our speed is 50mi/hr.How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Example
Suppose our factory makes MP3 players and the marginal cost iscurrently $50/lot. How much will it cost to make 2 more lots? 3 morelots? 12 more lots?
Example
Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 15 / 27
. . . . . .
Answers
Example
Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?
AnswerThe slope of the line is
m =riserun
We are given a “run” of dx, so the corresponding “rise” is mdx.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 16 / 27
. . . . . .
Answers
Example
Suppose a line goes through the point (x0, y0) and has slope m. If thepoint is moved horizontally by dx, while staying on the line, what is thecorresponding vertical movement?
AnswerThe slope of the line is
m =riserun
We are given a “run” of dx, so the corresponding “rise” is mdx.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 16 / 27
. . . . . .
Outline
The linear approximation of a function near a pointExamplesQuestions
DifferentialsUsing differentials to estimate error
Advanced Examples
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 17 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 18 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 18 / 27
. . . . . .
Using differentials to estimate error
If y = f(x), x0 and ∆x is known,and an estimate of ∆y isdesired:
I Approximate: ∆y ≈ dyI Differentiate: dy = f′(x)dxI Evaluate at x = x0 and
dx = ∆x.
. .x
.y
.
.
.x .x+∆x
.dx = ∆x
.∆y.dy
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 19 / 27
. . . . . .
Example
A sheet of plywood measures 8 ft× 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?
Solution
Write 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 = ℓdℓ, 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
. . . . . .
Example
A sheet of plywood measures 8 ft× 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?
Solution
Write 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 = ℓdℓ, 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
. . . . . .
Example
A sheet of plywood measures 8 ft× 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?
Solution
Write 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 = ℓdℓ, 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
. . . . . .
Example
A sheet of plywood measures 8 ft× 4 ft. Suppose our plywood-cuttingmachine will cut a rectangle whose width is exactly half its length, butthe length is prone to errors. If the length is off by 1 in, how bad can thearea of the sheet be off by?
Solution
Write 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 = ℓdℓ, 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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 20 / 27
. . . . . .
Why?
Why use linear approximations dy when the actual difference ∆y isknown?
I Linear approximation is quick and reliable. Finding ∆y exactlydepends on the function.
I These examples are overly simple. See the “Advanced Examples”later.
I In real life, sometimes only f(a) and f′(a) are known, and not thegeneral f(x).
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 21 / 27
. . . . . .
Outline
The linear approximation of a function near a pointExamplesQuestions
DifferentialsUsing differentials to estimate error
Advanced Examples
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 22 / 27
. . . . . .
GravitationPencils down!
Example
I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.
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) =GMmr2e
= −mg.
I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 23 / 27
. . . . . .
GravitationPencils down!
Example
I Drop a 1 kg ball off the roof of the Silver Center (50m high). Weusually say that a falling object feels a force F = −mg from gravity.
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) =GMmr2e
= −mg.
I What is the maximum error in replacing the actual force felt at thetop of the building F(re +∆r) by the force felt at ground levelF(re)? The relative error? The percentage error?
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 23 / 27
. . . . . .
Gravitation Solution
SolutionWe wonder if ∆F = F(re +∆r)− F(re) is small.
I Using a linear approximation,
∆F ≈ dF =dFdr
∣∣∣∣redr = 2
GMmr3e
dr
=
(GMmr2e
)drre
= 2mg∆rre
I The relative error is∆FF
≈ −2∆rre
I re = 6378.1 km. If ∆r = 50m,
∆FF
≈ −2∆rre
= −250
6378100= −1.56× 10−5 = −0.00156%
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 24 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
. . . . . .
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.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 25 / 27
. . . . . .
Illustration of the previous example
.
.2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
.
.2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
.
(94 ,32)
..(2, 1712)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
..(94 ,
32)..(2, 17/12)
..(289144 ,
1712
)..(2, 577408
)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
..(94 ,
32)..(2, 17/12) .
.(289144 ,
1712
)
..(2, 577408
)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Illustration of the previous example
..2
..(94 ,
32)..(2, 17/12) .
.(289144 ,
1712
)..(2, 577408
)
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 26 / 27
. . . . . .
Summary
I Linear approximation: If f is differentiable at a, the best linearapproximation to f near a is given by
Lf,a(x) = f(a) + f′(a)(x− a)
I Differentials: If f is differentiable at x, a good approximation to∆y = f(x+∆x)− f(x) is
∆y ≈ dy =dydx
· dx =dydx
·∆x
I Don’t buy plywood from me.
V63.0121.041, Calculus I (NYU) Section 2.8 Linear Approximation and Differentials October 13, 2010 27 / 27