This document summarizes a calculus lecture on linear approximations. It provides examples of using the tangent line to approximate the sine function at different points. Specifically, it estimates sin(61°) by taking linear approximations about 0 and about 60°. The linear approximation about 0 is x, giving a value of 1.06465. The linear approximation about 60° uses the fact that the sine is √3/2 and the derivative is √3/2 at π/3, giving a better approximation than using 0.
International Journal of Computational Engineering Research(IJCER)ijceronline
International Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
International Journal of Computational Engineering Research(IJCER)ijceronline
International Journal of Computational Engineering Research (IJCER) is dedicated to protecting personal information and will make every reasonable effort to handle collected information appropriately. All information collected, as well as related requests, will be handled as carefully and efficiently as possible in accordance with IJCER standards for integrity and objectivity.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 ...Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 ...Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 ...Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 ...Matthew Leingang
The exponential function is pretty much the only function whose derivative is itself. The derivative of the natural logarithm function is also beautiful as it fills in an important gap. Finally, the technique of logarithmic differentiation allows us to find derivatives without the product rule.
Implicit differentiation allows us to find slopes of lines tangent to curves that are not graphs of functions. Almost all of the time (yes, that is a mathematical term!) we can assume the curve comprises the graph of a function and differentiate using the chain rule.
Lesson 12: Linear Approximation and Differentials (Section 41 slides)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Lesson 12: Linear Approximation and Differentials (Section 41 slides)Mel Anthony Pepito
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Lesson 12: Linear Approximation (Section 41 handout)Matthew Leingang
The line tangent to a curve, which is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Lesson 12: Linear Approximations and Differentials (handout)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
Lesson 12: Linear Approximation and Differentials (Section 21 handout)Matthew Leingang
The line tangent to a curve is also the line which best "fits" the curve near that point. So derivatives can be used for approximating complicated functions with simple linear ones. Differentials are another set of notation for the same problem.
The tangent line to the graph of a function at a point 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.
At times it is useful to consider a function whose derivative is a given function. We look at the general idea of reversing the differentiation process and its applications to rectilinear motion.
DevOps and Testing slides at DASA ConnectKari Kakkonen
My and Rik Marselis slides at 30.5.2024 DASA Connect conference. We discuss about what is testing, then what is agile testing and finally what is Testing in DevOps. Finally we had lovely workshop with the participants trying to find out different ways to think about quality and testing in different parts of the DevOps infinity loop.
GDG Cloud Southlake #33: Boule & Rebala: Effective AppSec in SDLC using Deplo...James Anderson
Effective Application Security in Software Delivery lifecycle using Deployment Firewall and DBOM
The modern software delivery process (or the CI/CD process) includes many tools, distributed teams, open-source code, and cloud platforms. Constant focus on speed to release software to market, along with the traditional slow and manual security checks has caused gaps in continuous security as an important piece in the software supply chain. Today organizations feel more susceptible to external and internal cyber threats due to the vast attack surface in their applications supply chain and the lack of end-to-end governance and risk management.
The software team must secure its software delivery process to avoid vulnerability and security breaches. This needs to be achieved with existing tool chains and without extensive rework of the delivery processes. This talk will present strategies and techniques for providing visibility into the true risk of the existing vulnerabilities, preventing the introduction of security issues in the software, resolving vulnerabilities in production environments quickly, and capturing the deployment bill of materials (DBOM).
Speakers:
Bob Boule
Robert Boule is a technology enthusiast with PASSION for technology and making things work along with a knack for helping others understand how things work. He comes with around 20 years of solution engineering experience in application security, software continuous delivery, and SaaS platforms. He is known for his dynamic presentations in CI/CD and application security integrated in software delivery lifecycle.
Gopinath Rebala
Gopinath Rebala is the CTO of OpsMx, where he has overall responsibility for the machine learning and data processing architectures for Secure Software Delivery. Gopi also has a strong connection with our customers, leading design and architecture for strategic implementations. Gopi is a frequent speaker and well-known leader in continuous delivery and integrating security into software delivery.
Kubernetes & AI - Beauty and the Beast !?! @KCD Istanbul 2024Tobias Schneck
As AI technology is pushing into IT I was wondering myself, as an “infrastructure container kubernetes guy”, how get this fancy AI technology get managed from an infrastructure operational view? Is it possible to apply our lovely cloud native principals as well? What benefit’s both technologies could bring to each other?
Let me take this questions and provide you a short journey through existing deployment models and use cases for AI software. On practical examples, we discuss what cloud/on-premise strategy we may need for applying it to our own infrastructure to get it to work from an enterprise perspective. I want to give an overview about infrastructure requirements and technologies, what could be beneficial or limiting your AI use cases in an enterprise environment. An interactive Demo will give you some insides, what approaches I got already working for real.
State of ICS and IoT Cyber Threat Landscape Report 2024 previewPrayukth K V
The IoT and OT threat landscape report has been prepared by the Threat Research Team at Sectrio using data from Sectrio, cyber threat intelligence farming facilities spread across over 85 cities around the world. In addition, Sectrio also runs AI-based advanced threat and payload engagement facilities that serve as sinks to attract and engage sophisticated threat actors, and newer malware including new variants and latent threats that are at an earlier stage of development.
The latest edition of the OT/ICS and IoT security Threat Landscape Report 2024 also covers:
State of global ICS asset and network exposure
Sectoral targets and attacks as well as the cost of ransom
Global APT activity, AI usage, actor and tactic profiles, and implications
Rise in volumes of AI-powered cyberattacks
Major cyber events in 2024
Malware and malicious payload trends
Cyberattack types and targets
Vulnerability exploit attempts on CVEs
Attacks on counties – USA
Expansion of bot farms – how, where, and why
In-depth analysis of the cyber threat landscape across North America, South America, Europe, APAC, and the Middle East
Why are attacks on smart factories rising?
Cyber risk predictions
Axis of attacks – Europe
Systemic attacks in the Middle East
Download the full report from here:
https://sectrio.com/resources/ot-threat-landscape-reports/sectrio-releases-ot-ics-and-iot-security-threat-landscape-report-2024/
Builder.ai Founder Sachin Dev Duggal's Strategic Approach to Create an Innova...Ramesh Iyer
In today's fast-changing business world, Companies that adapt and embrace new ideas often need help to keep up with the competition. However, fostering a culture of innovation takes much work. It takes vision, leadership and willingness to take risks in the right proportion. Sachin Dev Duggal, co-founder of Builder.ai, has perfected the art of this balance, creating a company culture where creativity and growth are nurtured at each stage.
Neuro-symbolic is not enough, we need neuro-*semantic*Frank van Harmelen
Neuro-symbolic (NeSy) AI is on the rise. However, simply machine learning on just any symbolic structure is not sufficient to really harvest the gains of NeSy. These will only be gained when the symbolic structures have an actual semantics. I give an operational definition of semantics as “predictable inference”.
All of this illustrated with link prediction over knowledge graphs, but the argument is general.
The Art of the Pitch: WordPress Relationships and SalesLaura Byrne
Clients don’t know what they don’t know. What web solutions are right for them? How does WordPress come into the picture? How do you make sure you understand scope and timeline? What do you do if sometime changes?
All these questions and more will be explored as we talk about matching clients’ needs with what your agency offers without pulling teeth or pulling your hair out. Practical tips, and strategies for successful relationship building that leads to closing the deal.
Key Trends Shaping the Future of Infrastructure.pdfCheryl Hung
Keynote at DIGIT West Expo, Glasgow on 29 May 2024.
Cheryl Hung, ochery.com
Sr Director, Infrastructure Ecosystem, Arm.
The key trends across hardware, cloud and open-source; exploring how these areas are likely to mature and develop over the short and long-term, and then considering how organisations can position themselves to adapt and thrive.
UiPath Test Automation using UiPath Test Suite series, part 4DianaGray10
Welcome to UiPath Test Automation using UiPath Test Suite series part 4. In this session, we will cover Test Manager overview along with SAP heatmap.
The UiPath Test Manager overview with SAP heatmap webinar offers a concise yet comprehensive exploration of the role of a Test Manager within SAP environments, coupled with the utilization of heatmaps for effective testing strategies.
Participants will gain insights into the responsibilities, challenges, and best practices associated with test management in SAP projects. Additionally, the webinar delves into the significance of heatmaps as a visual aid for identifying testing priorities, areas of risk, and resource allocation within SAP landscapes. Through this session, attendees can expect to enhance their understanding of test management principles while learning practical approaches to optimize testing processes in SAP environments using heatmap visualization techniques
What will you get from this session?
1. Insights into SAP testing best practices
2. Heatmap utilization for testing
3. Optimization of testing processes
4. Demo
Topics covered:
Execution from the test manager
Orchestrator execution result
Defect reporting
SAP heatmap example with demo
Speaker:
Deepak Rai, Automation Practice Lead, Boundaryless Group and UiPath MVP
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf91mobiles
91mobiles recently conducted a Smart TV Buyer Insights Survey in which we asked over 3,000 respondents about the TV they own, aspects they look at on a new TV, and their TV buying preferences.
Smart TV Buyer Insights Survey 2024 by 91mobiles.pdf
Lesson 12: Linear Approximation
1. Section 2.8
Linear Approximation and Differentials
V63.0121.002.2010Su, Calculus I
New York University
May 26, 2010
Announcements
Quiz 2 Thursday on Sections 1.5–2.5
No class Monday, May 31
Assignment 2 due Tuesday, June 1
. . . . . .
2. Announcements
Quiz 2 Thursday on
Sections 1.5–2.5
No class Monday, May 31
Assignment 2 due
Tuesday, June 1
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 2 / 27
3. Objectives
Use tangent lines to make
linear approximations to a
function.
Given a function and a
point in the domain,
compute the
linearization of the
function at that point.
Use linearization to
approximate values of
functions
Given a function, compute
the differential of that
function
Use the differential
notation to estimate error
in linear approximations. . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 3 / 27
4. Outline
The linear approximation of a function near a point
Examples
Questions
Differentials
Using differentials to estimate error
Advanced Examples
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 4 / 27
5. The Big Idea
Question
Let f be differentiable at a. What linear function best approximates f
near a?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 5 / 27
6. The Big Idea
Question
Let f be differentiable at a. What linear function best approximates f
near a?
Answer
The tangent line, of course!
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 5 / 27
7. The Big Idea
Question
Let f be differentiable at a. What linear function best approximates f
near a?
Answer
The tangent line, of course!
Question
What is the equation for the line tangent to y = f(x) at (a, f(a))?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 5 / 27
8. The Big Idea
Question
Let f be differentiable at a. What linear function best approximates f
near a?
Answer
The tangent line, of course!
Question
What is the equation for the line tangent to y = f(x) at (a, f(a))?
Answer
L(x) = f(a) + f′ (a)(x − a)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 5 / 27
9. The tangent line is a linear approximation
y
.
L(x) = f(a) + f′ (a)(x − a)
is a decent approximation to f L
. (x) .
near a. f
.(x) .
f
.(a) .
.
x−a
. x
.
a
. x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 6 / 27
10. The tangent line is a linear approximation
y
.
L(x) = f(a) + f′ (a)(x − a)
is a decent approximation to f L
. (x) .
near a. f
.(x) .
How decent? The closer x is to
a, the better the approxmation f
.(a) .
.
x−a
L(x) is to f(x)
. x
.
a
. x
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 6 / 27
11. Example
.
Example
Estimate sin(61◦ ) = sin(61π/180) by using a linear approximation
(i) about a = 0 (ii) about a = 60◦ = π/3.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
12. 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)
If f(x) = sin x, then f(0) = 0
and f′ (0) = 1.
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
13. 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) Solution (ii)
(π)
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3
and f′ (0) = 1. f′ π = .
3
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
14. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = .
3
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
15. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
So the linear approximation
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
16. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2
So the linear approximation
So L(x) =
near 0 is L(x) = 0 + 1 · x = x.
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
17. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus
( )
61π 61π
sin ≈ ≈ 1.06465
180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
18. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈
180 180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
19. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
20. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180
Calculator check: sin(61◦ ) ≈
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
21. 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) Solution (ii)
(π) √
3
We have f = and
If f(x) = sin x, then f(0) = 0 ( ) 3 2
and f′ (0) = 1. f′ π = 1 .
3 2 √
So the linear approximation 3 1( π)
So L(x) = + x−
near 0 is L(x) = 0 + 1 · x = x. 2 2 3
Thus Thus
( ) ( )
61π 61π 61π
sin ≈ ≈ 1.06465 sin ≈ 0.87475
180 180 180
Calculator check: sin(61◦ ) ≈ 0.87462.
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 7 / 27
22. Illustration
y
.
y
. = sin x
. x
.
. 1◦
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 8 / 27
23. Illustration
y
.
y
. = L1 (x) = x
y
. = sin x
. x
.
0
. . 1◦
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 8 / 27
24. Illustration
y
.
y
. = L1 (x) = x
b
. ig difference! y
. = sin x
. x
.
0
. . 1◦
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 8 / 27
25. Illustration
y
.
y
. = L1 (x) = x
√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
.
. . x
.
0
. .
π/3 . 1◦
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 8 / 27
26. Illustration
y
.
y
. = L1 (x) = x
√ ( )
y
. = L2 (x) = 2
3
+ 1
2 x− π
3
y
. = sin x
. . ery little difference!
v
. . x
.
0
. .
π/3 . 1◦
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 8 / 27
27. Another Example
Example
√
Estimate 10 using the fact that 10 = 9 + 1.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 9 / 27
28. Another Example
Example
√
Estimate 10 using the fact that 10 = 9 + 1.
Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 9 / 27
29. Another Example
Example
√
Estimate 10 using the fact that 10 = 9 + 1.
Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 9 / 27
30. Another Example
Example
√
Estimate 10 using the fact that 10 = 9 + 1.
Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19
Check: =
6
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 9 / 27
31. Another Example
Example
√
Estimate 10 using the fact that 10 = 9 + 1.
Solution
√
The key step is to use a linear approximation to f(x) =
√ x near a = 9
to estimate f(10) = 10.
√ √ d√
10 ≈ 9 + x (1)
dx x=9
1 19
=3+ (1) = ≈ 3.167
2·3 6
( )2
19 361
Check: = .
6 36
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 9 / 27
32. Dividing without dividing?
Example
Suppose I have an irrational fear of division and need to estimate
577 ÷ 408. I write
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
But still I have to find .
102
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 10 / 27
33. Dividing without dividing?
Example
Suppose I have an irrational fear of division and need to estimate
577 ÷ 408. I write
577 1 1 1
= 1 + 169 = 1 + 169 × × .
408 408 4 102
1
But still I have to find .
102
Solution
1
Let f(x) = . We know f(100) and we want to estimate f(102).
x
1 1
f(102) ≈ f(100) + f′ (100)(2) = − (2) = 0.0098
100 1002
577
=⇒ ≈ 1.41405
408 . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 10 / 27
34. Questions
Example
Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 11 / 27
35. Answers
Example
Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 12 / 27
36. Answers
Example
Suppose we are traveling in a car and at noon our speed is 50 mi/hr.
How far will we have traveled by 2:00pm? by 3:00pm? By midnight?
Answer
100 mi
150 mi
600 mi (?) (Is it reasonable to assume 12 hours at the same
speed?)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 12 / 27
37. Questions
Example
Suppose we are traveling in a car and at noon our speed is 50 mi/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 is
currently $50/lot. How much will it cost to make 2 more lots? 3 more
lots? 12 more lots?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 13 / 27
38. Answers
Example
Suppose our factory makes MP3 players and the marginal cost is
currently $50/lot. How much will it cost to make 2 more lots? 3 more
lots? 12 more lots?
Answer
$100
$150
$600 (?)
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 14 / 27
39. Questions
Example
Suppose we are traveling in a car and at noon our speed is 50 mi/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 is
currently $50/lot. How much will it cost to make 2 more lots? 3 more
lots? 12 more lots?
Example
Suppose a line goes through the point (x0 , y0 ) and has slope m. If the
point is moved horizontally by dx, while staying on the line, what is the
corresponding vertical movement?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 15 / 27
40. Answers
Example
Suppose a line goes through the point (x0 , y0 ) and has slope m. If the
point is moved horizontally by dx, while staying on the line, what is the
corresponding vertical movement?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 16 / 27
41. Answers
Example
Suppose a line goes through the point (x0 , y0 ) and has slope m. If the
point is moved horizontally by dx, while staying on the line, what is the
corresponding vertical movement?
Answer
The slope of the line is
rise
m=
run
We are given a “run” of dx, so the corresponding “rise” is m dx.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 16 / 27
42. Outline
The linear approximation of a function near a point
Examples
Questions
Differentials
Using differentials to estimate error
Advanced Examples
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 17 / 27
43. Differentials are another way to express derivatives
f(x + ∆x) − f(x) ≈ f′ (x) ∆x y
.
∆y dy
Rename ∆x = dx, so we can
write this as
.
∆y ≈ dy = f′ (x)dx. .
dy
.
∆y
And this looks a lot like the .
.
dx = ∆x
Leibniz-Newton identity
dy .
= f′ (x) x
.
dx x x
. . + ∆x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 18 / 27
44. Differentials are another way to express derivatives
f(x + ∆x) − f(x) ≈ f′ (x) ∆x y
.
∆y dy
Rename ∆x = dx, so we can
write this as
.
∆y ≈ dy = f′ (x)dx. .
dy
.
∆y
And this looks a lot like the .
.
dx = ∆x
Leibniz-Newton identity
dy .
= f′ (x) x
.
dx x x
. . + ∆x
Linear approximation means ∆y ≈ dy = f′ (x0 ) dx near x0 .
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 18 / 27
45. Using differentials to estimate error
y
.
If y = f(x), x0 and ∆x is known,
and an estimate of ∆y is
desired:
Approximate: ∆y ≈ dy .
Differentiate: dy = f′ (x) dx .
∆y
.
dy
Evaluate at x = x0 and .
.
dx = ∆x
dx = ∆x.
. x
.
x x
. . + ∆x
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 19 / 27
46. Example
A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting
machine will cut a rectangle whose width is exactly half its length, but
the length is prone to errors. If the length is off by 1 in, how bad can the
area of the sheet be off by?
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 20 / 27
47. Example
A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting
machine will cut a rectangle whose width is exactly half its length, but
the length is prone to errors. If the length is off by 1 in, how bad can the
area of the sheet be off by?
Solution
1 2
Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.
2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 20 / 27
48. Example
A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting
machine will cut a rectangle whose width is exactly half its length, but
the length is prone to errors. If the length is off by 1 in, how bad can the
area of the sheet be off by?
Solution
1 2
Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 20 / 27
49. Example
A sheet of plywood measures 8 ft × 4 ft. Suppose our plywood-cutting
machine will cut a rectangle whose width is exactly half its length, but
the length is prone to errors. If the length is off by 1 in, how bad can the
area of the sheet be off by?
Solution
1 2
Write A(ℓ) = ℓ . We want to know ∆A when ℓ = 8 ft and ∆ℓ = 1 in.
2 ( )
97 9409 9409
(I) A(ℓ + ∆ℓ) = A = So ∆A = − 32 ≈ 0.6701.
12 288 288
dA
(II) = ℓ, so dA = ℓ dℓ, which should be a good estimate for ∆ℓ.
dℓ
When ℓ = 8 and dℓ = 12 , we have dA = 12 = 2 ≈ 0.667. So we
1 8
3
get estimates close to the hundredth of a square foot.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 20 / 27
50. Why?
Why use linear approximations dy when the actual difference ∆y is
known?
Linear approximation is quick and reliable. Finding ∆y exactly
depends on the function.
These examples are overly simple. See the “Advanced Examples”
later.
In real life, sometimes only f(a) and f′ (a) are known, and not the
general f(x).
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 21 / 27
51. Outline
The linear approximation of a function near a point
Examples
Questions
Differentials
Using differentials to estimate error
Advanced Examples
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 22 / 27
52. Gravitation
Pencils down!
Example
Drop a 1 kg ball off the roof of the Silver Center (50m high). We
usually say that a falling object feels a force F = −mg from gravity.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 23 / 27
53. Gravitation
Pencils down!
Example
Drop a 1 kg ball off the roof of the Silver Center (50m high). We
usually say that a falling object feels a force F = −mg from gravity.
In fact, the force felt is
GMm
F(r) = − ,
r2
where M is the mass of the earth and r is the distance from the
center of the earth to the object. G is a constant.
GMm
At r = re the force really is F(re ) = = −mg.
r2
e
What is the maximum error in replacing the actual force felt at the
top of the building F(re + ∆r) by the force felt at ground level
F(re )? The relative error? The percentage error? . . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 23 / 27
54. Gravitation Solution
Solution
We wonder if ∆F = F(re + ∆r) − F(re ) is small.
Using a linear approximation,
dF GMm
∆F ≈ dF = dr = 2 3 dr
dr re re
( )
GMm dr ∆r
= 2
= 2mg
re re re
∆F ∆r
The relative error is ≈ −2
F re
re = 6378.1 km. If ∆r = 50 m,
∆F ∆r 50
≈ −2 = −2 = −1.56 × 10−5 = −0.00156%
F re 6378100
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 24 / 27
55. Systematic linear approximation
√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 25 / 27
56. Systematic linear approximation
√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2. So
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 25 / 27
57. Systematic linear approximation
√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2. So
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12
This is a better approximation since (17/12)2 = 289/144
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 25 / 27
58. Systematic linear approximation
√ √
2 is irrational, but 9/4 is rational and 9/4 is close to 2. So
√ √ √ 1 17
2 = 9/4 − 1/4 ≈ 9/4 + (−1/4) =
2(3/2) 12
This is a better approximation since (17/12)2 = 289/144
Do it again!
√ √ √ 1
2 = 289/144 − 1/144 ≈ 289/144 + (−1/144) = 577/408
2(17/12)
( )2
577 332, 929 1
Now = which is away from 2.
408 166, 464 166, 464
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 25 / 27
59. Illustration of the previous example
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
60. Illustration of the previous example
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
61. Illustration of the previous example
.
2
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
62. Illustration of the previous example
.
.
2
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
63. Illustration of the previous example
.
.
2
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
64. Illustration of the previous example
. 2, 17 )
( 12
. .
.
2
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
65. Illustration of the previous example
. 2, 17 )
( 12
. .
.
2
.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
66. Illustration of the previous example
.
. 2, 17/12)
(
. . 4, 3)
(9 2
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
67. Illustration of the previous example
.
. 2, 17/12)
(
.. ( . 9, 3)
(
)4 2
289 17
. 144 , 12
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
68. Illustration of the previous example
.
. 2, 17/12)
(
.. ( . 9, 3)
(
( 577 ) )4 2
. 2, 408 289 17
. 144 , 12
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 26 / 27
69. Summary
Linear approximation: If f is differentiable at a, the best linear
approximation to f near a is given by
Lf,a (x) = f(a) + f′ (a)(x − a)
Differentials: If f is differentiable at x, a good approximation to
∆y = f(x + ∆x) − f(x) is
dy dy
∆y ≈ dy = · dx = · ∆x
dx dx
Don’t buy plywood from me.
. . . . . .
V63.0121.002.2010Su, Calculus I (NYU) Section 2.8 Linear Approximation May 26, 2010 27 / 27