Here is the solution again step-by-step:
We are given:
- At 3 hours, there are 10,000 bacteria
- At 5 hours, there are 40,000 bacteria
We know the model for bacterial growth is exponential: y' = ky
The general solution is: y = y0ekt
Setting up the equations:
- At 3 hours: 10,000 = y0ek(3)
- At 5 hours: 40,000 = y0ek(5)
Dividing the equations:
40,000/10,000 = ek(5)-k(3)
4 = e2k
Taking the ln of both sides:
ln
When a quantity changes in proportion to itself (for instance, bacteria reproduction or radioactive decay), the growth or decay is exponential in nature. There are many many examples of this to be found.
Lesson 15: Exponential Growth and Decay (Section 041 slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 15: Exponential Growth and Decay (Section 021 handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 15: Exponential Growth and Decay (Section 041 handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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.
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.
On an Optimal control Problem for Parabolic Equationsijceronline
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.
When a quantity changes in proportion to itself (for instance, bacteria reproduction or radioactive decay), the growth or decay is exponential in nature. There are many many examples of this to be found.
Lesson 15: Exponential Growth and Decay (Section 041 slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 15: Exponential Growth and Decay (Section 021 handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 15: Exponential Growth and Decay (Section 041 handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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.
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.
On an Optimal control Problem for Parabolic Equationsijceronline
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.
Numerical Solution of Nth - Order Fuzzy Initial Value Problems by Fourth Orde...IOSR Journals
In this paper, a numerical method for Nth - order fuzzy initial value problems (FIVP) based on
Seikkala derivative of fuzzy process is studied. The fourth order Runge-Kutta method based on Centroidal Mean
(RKCeM4) is used to find the numerical solution and the convergence and stability of the method is proved. This
method is illustrated by solving second and third order FIVPs. The results show that the proposed method suits
well to find the numerical solution of Nth – order FIVPs.
Molecular Solutions For The Set-Partition Problem On Dna-Based Computingijcsit
Consider that the every element in a finite set S having q elements is a positive integer. The set-partition
problem is to determine whether there is a subset T Í S such that ,
Î Î
=
x T x T
x x where T = {x| x Î S and
x Ï T}. This research demonstrates that molecular operations can be applied to solve the set-partition
problem. In order to perform this goal, we offer two DNA-based algorithms, an unsigned parallel adder
and a parallel Exclusive-OR (XOR) operation, that formally demonstrate our designed molecular solutions
for solving the set-partition problem.
Question bank Engineering Mathematics- ii Mohammad Imran
its a very short Revision of complete syllabus with theoretical as well Numerical problems which are related to AKTU SEMESTER QUESTIONS, UPTU PREVIOUS QUESTIONS,
On the Application of the Fixed Point Theory to the Solution of Systems of Li...BRNSS Publication Hub
In this study, I worked on how to solve the biological and physical problems using systems of linear differential equations. A differential equation is an equation involving an unknown function and one or more of its derivatives. In this work, we consider systems of differential equations and their underlying theories illustrated with some solved examples. Finally, two applications, one to cell biology and the other to physical problems, were considered precisely.
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.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Numerical Solution of Nth - Order Fuzzy Initial Value Problems by Fourth Orde...IOSR Journals
In this paper, a numerical method for Nth - order fuzzy initial value problems (FIVP) based on
Seikkala derivative of fuzzy process is studied. The fourth order Runge-Kutta method based on Centroidal Mean
(RKCeM4) is used to find the numerical solution and the convergence and stability of the method is proved. This
method is illustrated by solving second and third order FIVPs. The results show that the proposed method suits
well to find the numerical solution of Nth – order FIVPs.
Molecular Solutions For The Set-Partition Problem On Dna-Based Computingijcsit
Consider that the every element in a finite set S having q elements is a positive integer. The set-partition
problem is to determine whether there is a subset T Í S such that ,
Î Î
=
x T x T
x x where T = {x| x Î S and
x Ï T}. This research demonstrates that molecular operations can be applied to solve the set-partition
problem. In order to perform this goal, we offer two DNA-based algorithms, an unsigned parallel adder
and a parallel Exclusive-OR (XOR) operation, that formally demonstrate our designed molecular solutions
for solving the set-partition problem.
Question bank Engineering Mathematics- ii Mohammad Imran
its a very short Revision of complete syllabus with theoretical as well Numerical problems which are related to AKTU SEMESTER QUESTIONS, UPTU PREVIOUS QUESTIONS,
On the Application of the Fixed Point Theory to the Solution of Systems of Li...BRNSS Publication Hub
In this study, I worked on how to solve the biological and physical problems using systems of linear differential equations. A differential equation is an equation involving an unknown function and one or more of its derivatives. In this work, we consider systems of differential equations and their underlying theories illustrated with some solved examples. Finally, two applications, one to cell biology and the other to physical problems, were considered precisely.
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.
The derivative of a composition of functions is the product of the derivatives of those functions. This rule is important because compositions are so powerful.
Lesson 19: The Mean Value Theorem (Section 021 slides)Matthew Leingang
The Mean Value Theorem is the most important theorem in calculus. It is the first theorem which allows us to infer information about a function from information about its derivative. From the MVT we can derive tests for the monotonicity (increase or decrease) and concavity of a function.
Lesson 18: Maximum and Minimum Values (Section 021 slides)Matthew Leingang
There are various reasons why we would want to find the extreme (maximum and minimum values) of a function. Fermat's Theorem tells us we can find local extreme points by looking at critical points. This process is known as the Closed Interval Method.
Lesson 15: Exponential Growth and Decay (Section 021 slides)Mel Anthony Pepito
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 15: Exponential Growth and Decay (Section 041 slides)Mel Anthony Pepito
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
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.
Lesson 15: Exponential Growth and Decay (handout)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 021 ...Mel Anthony Pepito
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.
An antiderivative of a function is a function whose derivative is the given function. The problem of antidifferentiation is interesting, complicated, and useful, especially when discussing motion.
This is the slideshow version from class.
Lesson 14: Derivatives of Exponential and Logarithmic Functions (Section 041 ...Mel Anthony Pepito
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.
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.
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.
Lesson 15: Exponential Growth and Decay (slides)Matthew Leingang
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Many problems in nature are expressible in terms of a certain differential equation that has a solution in terms of exponential functions. We look at the equation in general and some fun applications, including radioactivity, cooling, and interest.
Streamlining assessment, feedback, and archival with auto-multiple-choiceMatthew Leingang
Auto-multiple-choice (AMC) is an open-source optical mark recognition software package built with Perl, LaTeX, XML, and sqlite. I use it for all my in-class quizzes and exams. Unique papers are created for each student, fixed-response items are scored automatically, and free-response problems, after manual scoring, have marks recorded in the same process. In the first part of the talk I will discuss AMC’s many features and why I feel it’s ideal for a mathematics course. My contributions to the AMC workflow include some scripts designed to automate the process of returning scored papers
back to students electronically. AMC provides an email gateway, but I have written programs to return graded papers via the DAV protocol to student’s dropboxes on our (Sakai) learning management systems. I will also show how graded papers can be archived, with appropriate metadata tags, into an Evernote notebook.
Integration by substitution is the chain rule in reverse.
NOTE: the final location is section specific. Section 1 (morning) is in SILV 703, Section 11 (afternoon) is in CANT 200
Lesson 24: Areas and Distances, The Definite Integral (handout)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
Lesson 24: Areas and Distances, The Definite Integral (slides)Matthew Leingang
We can define the area of a curved region by a process similar to that by which we determined the slope of a curve: approximation by what we know and a limit.
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.
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.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
Uncountably many problems in life and nature can be expressed in terms of an optimization principle. We look at the process and find a few good examples.
JMeter webinar - integration with InfluxDB and GrafanaRTTS
Watch this recorded webinar about real-time monitoring of application performance. See how to integrate Apache JMeter, the open-source leader in performance testing, with InfluxDB, the open-source time-series database, and Grafana, the open-source analytics and visualization application.
In this webinar, we will review the benefits of leveraging InfluxDB and Grafana when executing load tests and demonstrate how these tools are used to visualize performance metrics.
Length: 30 minutes
Session Overview
-------------------------------------------
During this webinar, we will cover the following topics while demonstrating the integrations of JMeter, InfluxDB and Grafana:
- What out-of-the-box solutions are available for real-time monitoring JMeter tests?
- What are the benefits of integrating InfluxDB and Grafana into the load testing stack?
- Which features are provided by Grafana?
- Demonstration of InfluxDB and Grafana using a practice web application
To view the webinar recording, go to:
https://www.rttsweb.com/jmeter-integration-webinar
LF Energy Webinar: Electrical Grid Modelling and Simulation Through PowSyBl -...DanBrown980551
Do you want to learn how to model and simulate an electrical network from scratch in under an hour?
Then welcome to this PowSyBl workshop, hosted by Rte, the French Transmission System Operator (TSO)!
During the webinar, you will discover the PowSyBl ecosystem as well as handle and study an electrical network through an interactive Python notebook.
PowSyBl is an open source project hosted by LF Energy, which offers a comprehensive set of features for electrical grid modelling and simulation. Among other advanced features, PowSyBl provides:
- A fully editable and extendable library for grid component modelling;
- Visualization tools to display your network;
- Grid simulation tools, such as power flows, security analyses (with or without remedial actions) and sensitivity analyses;
The framework is mostly written in Java, with a Python binding so that Python developers can access PowSyBl functionalities as well.
What you will learn during the webinar:
- For beginners: discover PowSyBl's functionalities through a quick general presentation and the notebook, without needing any expert coding skills;
- For advanced developers: master the skills to efficiently apply PowSyBl functionalities to your real-world scenarios.
PHP Frameworks: I want to break free (IPC Berlin 2024)Ralf Eggert
In this presentation, we examine the challenges and limitations of relying too heavily on PHP frameworks in web development. We discuss the history of PHP and its frameworks to understand how this dependence has evolved. The focus will be on providing concrete tips and strategies to reduce reliance on these frameworks, based on real-world examples and practical considerations. The goal is to equip developers with the skills and knowledge to create more flexible and future-proof web applications. We'll explore the importance of maintaining autonomy in a rapidly changing tech landscape and how to make informed decisions in PHP development.
This talk is aimed at encouraging a more independent approach to using PHP frameworks, moving towards a more flexible and future-proof approach to PHP development.
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.
Accelerate your Kubernetes clusters with Varnish CachingThijs Feryn
A presentation about the usage and availability of Varnish on Kubernetes. This talk explores the capabilities of Varnish caching and shows how to use the Varnish Helm chart to deploy it to Kubernetes.
This presentation was delivered at K8SUG Singapore. See https://feryn.eu/presentations/accelerate-your-kubernetes-clusters-with-varnish-caching-k8sug-singapore-28-2024 for more details.
Slack (or Teams) Automation for Bonterra Impact Management (fka Social Soluti...Jeffrey Haguewood
Sidekick Solutions uses Bonterra Impact Management (fka Social Solutions Apricot) and automation solutions to integrate data for business workflows.
We believe integration and automation are essential to user experience and the promise of efficient work through technology. Automation is the critical ingredient to realizing that full vision. We develop integration products and services for Bonterra Case Management software to support the deployment of automations for a variety of use cases.
This video focuses on the notifications, alerts, and approval requests using Slack for Bonterra Impact Management. The solutions covered in this webinar can also be deployed for Microsoft Teams.
Interested in deploying notification automations for Bonterra Impact Management? Contact us at sales@sidekicksolutionsllc.com to discuss next steps.
Dev Dives: Train smarter, not harder – active learning and UiPath LLMs for do...UiPathCommunity
💥 Speed, accuracy, and scaling – discover the superpowers of GenAI in action with UiPath Document Understanding and Communications Mining™:
See how to accelerate model training and optimize model performance with active learning
Learn about the latest enhancements to out-of-the-box document processing – with little to no training required
Get an exclusive demo of the new family of UiPath LLMs – GenAI models specialized for processing different types of documents and messages
This is a hands-on session specifically designed for automation developers and AI enthusiasts seeking to enhance their knowledge in leveraging the latest intelligent document processing capabilities offered by UiPath.
Speakers:
👨🏫 Andras Palfi, Senior Product Manager, UiPath
👩🏫 Lenka Dulovicova, Product Program Manager, UiPath
Search and Society: Reimagining Information Access for Radical FuturesBhaskar Mitra
The field of Information retrieval (IR) is currently undergoing a transformative shift, at least partly due to the emerging applications of generative AI to information access. In this talk, we will deliberate on the sociotechnical implications of generative AI for information access. We will argue that there is both a critical necessity and an exciting opportunity for the IR community to re-center our research agendas on societal needs while dismantling the artificial separation between the work on fairness, accountability, transparency, and ethics in IR and the rest of IR research. Instead of adopting a reactionary strategy of trying to mitigate potential social harms from emerging technologies, the community should aim to proactively set the research agenda for the kinds of systems we should build inspired by diverse explicitly stated sociotechnical imaginaries. The sociotechnical imaginaries that underpin the design and development of information access technologies needs to be explicitly articulated, and we need to develop theories of change in context of these diverse perspectives. Our guiding future imaginaries must be informed by other academic fields, such as democratic theory and critical theory, and should be co-developed with social science scholars, legal scholars, civil rights and social justice activists, and artists, among others.
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.
Connector Corner: Automate dynamic content and events by pushing a buttonDianaGray10
Here is something new! In our next Connector Corner webinar, we will demonstrate how you can use a single workflow to:
Create a campaign using Mailchimp with merge tags/fields
Send an interactive Slack channel message (using buttons)
Have the message received by managers and peers along with a test email for review
But there’s more:
In a second workflow supporting the same use case, you’ll see:
Your campaign sent to target colleagues for approval
If the “Approve” button is clicked, a Jira/Zendesk ticket is created for the marketing design team
But—if the “Reject” button is pushed, colleagues will be alerted via Slack message
Join us to learn more about this new, human-in-the-loop capability, brought to you by Integration Service connectors.
And...
Speakers:
Akshay Agnihotri, Product Manager
Charlie Greenberg, Host
Epistemic Interaction - tuning interfaces to provide information for AI supportAlan Dix
Paper presented at SYNERGY workshop at AVI 2024, Genoa, Italy. 3rd June 2024
https://alandix.com/academic/papers/synergy2024-epistemic/
As machine learning integrates deeper into human-computer interactions, the concept of epistemic interaction emerges, aiming to refine these interactions to enhance system adaptability. This approach encourages minor, intentional adjustments in user behaviour to enrich the data available for system learning. This paper introduces epistemic interaction within the context of human-system communication, illustrating how deliberate interaction design can improve system understanding and adaptation. Through concrete examples, we demonstrate the potential of epistemic interaction to significantly advance human-computer interaction by leveraging intuitive human communication strategies to inform system design and functionality, offering a novel pathway for enriching user-system engagements.
Epistemic Interaction - tuning interfaces to provide information for AI support
Lesson15 -exponential_growth_and_decay_021_slides
1. Section 3.4
Exponential Growth and Decay
V63.0121.021, Calculus I
New York University
October 28, 2010
Announcements
Quiz 3 next week in recitation on 2.6, 2.8, 3.1, 3.2
. . . . . .
2. . . . . . .
Announcements
Quiz 3 next week in
recitation on 2.6, 2.8, 3.1,
3.2
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 2 / 40
3. . . . . . .
Objectives
Solve the ordinary
differential equation
y′
(t) = ky(t), y(0) = y0
Solve problems involving
exponential growth and
decay
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 3 / 40
4. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 4 / 40
5. . . . . . .
Derivatives of exponential and logarithmic functions
y y′
ex
ex
ax
(ln a) · ax
ln x
1
x
loga x
1
ln a
·
1
x
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 5 / 40
6. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 6 / 40
7. . . . . . .
What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
8. . . . . . .
What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′
(t).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
9. . . . . . .
What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′
(t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
−kx(t) = mx′′
(t) =⇒ x′′
(t) +
k
m
x(t) = 0.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
10. . . . . . .
What is a differential equation?
Definition
A differential equation is an equation for an unknown function which
includes the function and its derivatives.
Example
Newton’s Second Law F = ma is a differential equation, where
a(t) = x′′
(t).
In a spring, F(x) = −kx, where x is displacement from equilibrium
and k is a constant. So
−kx(t) = mx′′
(t) =⇒ x′′
(t) +
k
m
x(t) = 0.
The most general solution is x(t) = A sin ωt + B cos ωt, where
ω =
√
k/m.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 7 / 40
11. . . . . . .
Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
x′′
+
k
m
x = 0, where ω =
√
k/m.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
12. . . . . . .
Showing a function is a solution
Example (Continued)
Show that x(t) = A sin ωt + B cos ωt satisfies the differential equation
x′′
+
k
m
x = 0, where ω =
√
k/m.
Solution
We have
x(t) = A sin ωt + B cos ωt
x′
(t) = Aω cos ωt − Bω sin ωt
x′′
(t) = −Aω2
sin ωt − Bω2
cos ωt
Since ω2
= k/m, the last line plus k/m times the first line result in zero.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 8 / 40
13. . . . . . .
The Equation y′
= 2
Example
Find a solution to y′
(t) = 2.
Find the most general solution to y′
(t) = 2.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
14. . . . . . .
The Equation y′
= 2
Example
Find a solution to y′
(t) = 2.
Find the most general solution to y′
(t) = 2.
Solution
A solution is y(t) = 2t.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
15. . . . . . .
The Equation y′
= 2
Example
Find a solution to y′
(t) = 2.
Find the most general solution to y′
(t) = 2.
Solution
A solution is y(t) = 2t.
The general solution is y = 2t + C.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
16. . . . . . .
The Equation y′
= 2
Example
Find a solution to y′
(t) = 2.
Find the most general solution to y′
(t) = 2.
Solution
A solution is y(t) = 2t.
The general solution is y = 2t + C.
Remark
If a function has a constant rate of growth, it’s linear.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 9 / 40
17. . . . . . .
The Equation y′
= 2t
Example
Find a solution to y′
(t) = 2t.
Find the most general solution to y′
(t) = 2t.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
18. . . . . . .
The Equation y′
= 2t
Example
Find a solution to y′
(t) = 2t.
Find the most general solution to y′
(t) = 2t.
Solution
A solution is y(t) = t2
.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
19. . . . . . .
The Equation y′
= 2t
Example
Find a solution to y′
(t) = 2t.
Find the most general solution to y′
(t) = 2t.
Solution
A solution is y(t) = t2
.
The general solution is y = t2
+ C.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 10 / 40
20. . . . . . .
The Equation y′
= y
Example
Find a solution to y′
(t) = y(t).
Find the most general solution to y′
(t) = y(t).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
21. . . . . . .
The Equation y′
= y
Example
Find a solution to y′
(t) = y(t).
Find the most general solution to y′
(t) = y(t).
Solution
A solution is y(t) = et
.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
22. . . . . . .
The Equation y′
= y
Example
Find a solution to y′
(t) = y(t).
Find the most general solution to y′
(t) = y(t).
Solution
A solution is y(t) = et
.
The general solution is y = Cet
, not y = et
+ C.
(check this)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 11 / 40
23. . . . . . .
Kick it up a notch: y′
= 2y
Example
Find a solution to y′
= 2y.
Find the general solution to y′
= 2y.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
24. . . . . . .
Kick it up a notch: y′
= 2y
Example
Find a solution to y′
= 2y.
Find the general solution to y′
= 2y.
Solution
y = e2t
y = Ce2t
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 12 / 40
25. . . . . . .
In general: y′
= ky
Example
Find a solution to y′
= ky.
Find the general solution to y′
= ky.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
26. . . . . . .
In general: y′
= ky
Example
Find a solution to y′
= ky.
Find the general solution to y′
= ky.
Solution
y = ekt
y = Cekt
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
27. . . . . . .
In general: y′
= ky
Example
Find a solution to y′
= ky.
Find the general solution to y′
= ky.
Solution
y = ekt
y = Cekt
Remark
What is C? Plug in t = 0:
y(0) = Cek·0
= C · 1 = C,
so y(0) = y0, the initial value of y.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 13 / 40
28. . . . . . .
Constant Relative Growth =⇒ Exponential Growth
Theorem
A function with constant relative growth rate k is an exponential
function with parameter k. Explicitly, the solution to the equation
y′
(t) = ky(t) y(0) = y0
is
y(t) = y0ekt
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 14 / 40
29. . . . . . .
Exponential Growth is everywhere
Lots of situations have growth rates proportional to the current
value
This is the same as saying the relative growth rate is constant.
Examples: Natural population growth, compounded interest,
social networks
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 15 / 40
30. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 16 / 40
31. . . . . . .
Bacteria
Since you need bacteria to
make bacteria, the amount
of new bacteria at any
moment is proportional to
the total amount of
bacteria.
This means bacteria
populations grow
exponentially.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 17 / 40
32. . . . . . .
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
33. . . . . . .
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′
= ky for bacteria, we have y = y0ekt
. We have
10, 000 = y0ek·3
40, 000 = y0ek·5
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
34. . . . . . .
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′
= ky for bacteria, we have y = y0ekt
. We have
10, 000 = y0ek·3
40, 000 = y0ek·5
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
35. . . . . . .
Bacteria Example
Example
A colony of bacteria is grown under ideal conditions in a laboratory. At
the end of 3 hours there are 10,000 bacteria. At the end of 5 hours
there are 40,000. How many bacteria were present initially?
Solution
Since y′
= ky for bacteria, we have y = y0ekt
. We have
10, 000 = y0ek·3
40, 000 = y0ek·5
Dividing the first into the second gives
4 = e2k
=⇒ 2k = ln 4 =⇒ k = ln 2. Now we have
10, 000 = y0eln 2·3
= y0 · 8
So y0 =
10, 000
8
= 1250.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 18 / 40
36. . . . . . .
Could you do that again please?
We have
10, 000 = y0ek·3
40, 000 = y0ek·5
Dividing the first into the second gives
40, 000
10, 000
=
y0e5k
y0e3k
=⇒ 4 = e2k
=⇒ ln 4 = ln(e2k
) = 2k
=⇒ k =
ln 4
2
=
ln 22
2
=
2 ln 2
2
= ln 2
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 19 / 40
37. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 20 / 40
38. . . . . . .
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously
give off particles.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
39. . . . . . .
Modeling radioactive decay
Radioactive decay occurs because many large atoms spontaneously
give off particles.
This means that in a sample of
a bunch of atoms, we can
assume a certain percentage of
them will “go off” at any point.
(For instance, if all atom of a
certain radioactive element
have a 20% chance of decaying
at any point, then we can
expect in a sample of 100 that
20 of them will be decaying.)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 21 / 40
40. . . . . . .
Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
y
= k
where k is negative.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
41. . . . . . .
Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
y
= k
where k is negative. So
y′
= ky =⇒ y = y0ekt
again!
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
42. . . . . . .
Radioactive decay as a differential equation
The relative rate of decay is constant:
y′
y
= k
where k is negative. So
y′
= ky =⇒ y = y0ekt
again!
It’s customary to express the relative rate of decay in the units of
half-life: the amount of time it takes a pure sample to decay to one
which is only half pure.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 22 / 40
43. . . . . . .
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
44. . . . . . .
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0ekt
, where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365
=⇒ k = −
365 · ln 2
138
.
Therefore
y(t) = 100e−365·ln 2
138
t
= 100 · 2−365t/138
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
45. . . . . . .
Computing the amount remaining of a decaying
sample
Example
The half-life of polonium-210 is about 138 days. How much of a 100 g
sample remains after t years?
Solution
We have y = y0ekt
, where y0 = y(0) = 100 grams. Then
50 = 100ek·138/365
=⇒ k = −
365 · ln 2
138
.
Therefore
y(t) = 100e−365·ln 2
138
t
= 100 · 2−365t/138
Notice y(t) = y0 · 2−t/t1/2 , where t1/2 is the half-life.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 23 / 40
46. . . . . . .
Carbon-14 Dating
The ratio of carbon-14 to
carbon-12 in an organism
decays exponentially:
p(t) = p0e−kt
.
The half-life of carbon-14 is
about 5700 years. So the
equation for p(t) is
p(t) = p0e− ln2
5700
t
Another way to write this would
be
p(t) = p02−t/5700
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 24 / 40
47. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
48. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
49. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1 From the
equation we have
2−t/5700
= 0.1
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
50. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1 From the
equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
51. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1 From the
equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2
· 5700
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
52. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1 From the
equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2
· 5700 ≈ 18, 940
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
53. . . . . . .
Computing age with Carbon-14 content
Example
Suppose a fossil is found where the ratio of carbon-14 to carbon-12 is
10% of that in a living organism. How old is the fossil?
Solution
We are looking for the value of t for which
p(t)
p0
= 0.1 From the
equation we have
2−t/5700
= 0.1 =⇒ −
t
5700
ln 2 = ln 0.1 =⇒ t =
ln 0.1
ln 2
· 5700 ≈ 18, 940
So the fossil is almost 19,000 years old.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 25 / 40
54. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 26 / 40
55. . . . . . .
Newton's Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature difference
between the object and its
surroundings.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
56. . . . . . .
Newton's Law of Cooling
Newton’s Law of Cooling
states that the rate of
cooling of an object is
proportional to the
temperature difference
between the object and its
surroundings.
This gives us a differential
equation of the form
dT
dt
= k(T − Ts)
(where k < 0 again).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 27 / 40
57. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
58. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
59. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
60. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
=⇒ T − Ts = Cekt
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
61. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
=⇒ T − Ts = Cekt
=⇒ T = Cekt
+ Ts
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
62. . . . . . .
General Solution to NLC problems
To solve this, change the variable y(t) = T(t) − Ts. Then y′
= T′
and
k(T − Ts) = ky. The equation now looks like
dT
dt
= k(T − Ts) ⇐⇒
dy
dt
= ky
Now we can solve!
y′
= ky =⇒ y = Cekt
=⇒ T − Ts = Cekt
=⇒ T = Cekt
+ Ts
Plugging in t = 0, we see C = y0 = T0 − Ts. So
Theorem
The solution to the equation T′
(t) = k(T(t) − Ts), T(0) = T0 is
T(t) = (T0 − Ts)ekt
+ Ts
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 28 / 40
63. . . . . . .
Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦
C is put in a sink of 18 ◦
C water. After 5
minutes, the egg’s temperature is 38 ◦
C. Assuming the water has not
warmed appreciably, how much longer will it take the egg to reach
20 ◦
C?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
64. . . . . . .
Computing cooling time with NLC
Example
A hard-boiled egg at 98 ◦
C is put in a sink of 18 ◦
C water. After 5
minutes, the egg’s temperature is 38 ◦
C. Assuming the water has not
warmed appreciably, how much longer will it take the egg to reach
20 ◦
C?
Solution
We know that the temperature function takes the form
T(t) = (T0 − Ts)ekt
+ Ts = 80ekt
+ 18
To find k, plug in t = 5:
38 = T(5) = 80e5k
+ 18
and solve for k.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 29 / 40
65. . . . . . .
Finding k
Solution (Continued)
38 = T(5) = 80e5k
+ 18
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
66. . . . . . .
Finding k
Solution (Continued)
38 = T(5) = 80e5k
+ 18
20 = 80e5k
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
70. . . . . . .
Finding k
Solution (Continued)
38 = T(5) = 80e5k
+ 18
20 = 80e5k
1
4
= e5k
ln
(
1
4
)
= 5k
=⇒ k = −
1
5
ln 4.
Now we need to solve for t:
20 = T(t) = 80e− t
5
ln 4
+ 18
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 30 / 40
71. . . . . . .
Finding t
Solution (Continued)
20 = 80e− t
5
ln 4
+ 18
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
72. . . . . . .
Finding t
Solution (Continued)
20 = 80e− t
5
ln 4
+ 18
2 = 80e− t
5
ln 4
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
73. . . . . . .
Finding t
Solution (Continued)
20 = 80e− t
5
ln 4
+ 18
2 = 80e− t
5
ln 4
1
40
= e− t
5
ln 4
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
74. . . . . . .
Finding t
Solution (Continued)
20 = 80e− t
5
ln 4
+ 18
2 = 80e− t
5
ln 4
1
40
= e− t
5
ln 4
− ln 40 = −
t
5
ln 4
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
75. . . . . . .
Finding t
Solution (Continued)
20 = 80e− t
5
ln 4
+ 18
2 = 80e− t
5
ln 4
1
40
= e− t
5
ln 4
− ln 40 = −
t
5
ln 4
=⇒ t =
ln 40
1
5 ln 4
=
5 ln 40
ln 4
≈ 13 min
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 31 / 40
76. . . . . . .
Computing time of death with NLC
Example
A murder victim is discovered at
midnight and the temperature of
the body is recorded as 31 ◦
C.
One hour later, the temperature
of the body is 29 ◦
C. Assume
that the surrounding air
temperature remains constant
at 21 ◦
C. Calculate the victim’s
time of death. (The “normal”
temperature of a living human
being is approximately 37 ◦
C.)
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 32 / 40
77. . . . . . .
Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
78. . . . . . .
Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
To find k:
29 = 10ek·1
+ 21 =⇒ k = ln 0.8
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
79. . . . . . .
Solution
Let time 0 be midnight. We know T0 = 31, Ts = 21, and
T(1) = 29. We want to know the t for which T(t) = 37.
To find k:
29 = 10ek·1
+ 21 =⇒ k = ln 0.8
To find t:
37 = 10et·ln(0.8)
+ 21
1.6 = et·ln(0.8)
t =
ln(1.6)
ln(0.8)
≈ −2.10 hr
So the time of death was just before 10:00pm.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 33 / 40
80. . . . . . .
Outline
Recall
The differential equation y′
= ky
Modeling simple population growth
Modeling radioactive decay
Carbon-14 Dating
Newton’s Law of Cooling
Continuously Compounded Interest
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 34 / 40
81. . . . . . .
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
A0
(
1 +
r
n
)nt
after t years.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
82. . . . . . .
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
A0
(
1 +
r
n
)nt
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
A(t) = lim
n→∞
A0
(
1 +
r
n
)nt
= A0ert
.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
83. . . . . . .
Interest
If an account has an compound interest rate of r per year
compounded n times, then an initial deposit of A0 dollars becomes
A0
(
1 +
r
n
)nt
after t years.
For different amounts of compounding, this will change. As
n → ∞, we get continously compounded interest
A(t) = lim
n→∞
A0
(
1 +
r
n
)nt
= A0ert
.
Thus dollars are like bacteria.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 35 / 40
84. . . . . . .
Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
85. . . . . . .
Continuous vs. Discrete Compounding of interest
Example
Consider two bank accounts: one with 10% annual interested
compounded quarterly and one with annual interest rate r compunded
continuously. If they produce the same balance after every year, what
is r?
Solution
The balance for the 10% compounded quarterly account after t years is
A1(t) = A0(1.025)4t
= P((1.025)4
)t
The balance for the interest rate r compounded continuously account
after t years is
A2(t) = A0ert
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 36 / 40
86. . . . . . .
Solving
Solution (Continued)
A1(t) = A0((1.025)4
)t
A2(t) = A0(er
)t
For those to be the same, er
= (1.025)4
, so
r = ln((1.025)4
) = 4 ln 1.025 ≈ 0.0988
So 10% annual interest compounded quarterly is basically equivalent
to 9.88% compounded continuously.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 37 / 40
87. . . . . . .
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
88. . . . . . .
Computing doubling time with exponential growth
Example
How long does it take an initial deposit of $100, compounded
continuously, to double?
Solution
We need t such that A(t) = 200. In other words
200 = 100ert
=⇒ 2 = ert
=⇒ ln 2 = rt =⇒ t =
ln 2
r
.
For instance, if r = 6% = 0.06, we have
t =
ln 2
0.06
≈
0.69
0.06
=
69
6
= 11.5 years.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 38 / 40
89. . . . . . .
I-banking interview tip of the day
The fraction
ln 2
r
can also
be approximated as either
70 or 72 divided by the
percentage rate (as a
number between 0 and
100, not a fraction between
0 and 1.)
This is sometimes called
the rule of 70 or rule of 72.
72 has lots of factors so it’s
used more often.
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 39 / 40
90. . . . . . .
Summary
When something grows or decays at a constant relative rate, the
growth or decay is exponential.
Equations with unknowns in an exponent can be solved with
logarithms.
Your friend list is like culture of bacteria (no offense).
V63.0121.021, Calculus I (NYU) Section 3.4 Exponential Growth and Decay October 28, 2010 40 / 40