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Topics: - sections 6.1 (extreme values), 6.4 (l’Hopital’s rule), and 7.1

(differential equations)* Read these sections and study solved examples in your

textbook!

Work On:- Practice problems from the textbook and assignments

from the coursepack as assigned on the course web page (under the link “SCHEDULE + HOMEWORK”)

Maximum and Minimum Values

is a global (absolute) maximum of if for all in the domain of

is a local (relative) maximum of if for all in some interval around

€

f (c) ≥ f (x)

€

f (c)

€

f

€

f (c) ≥ f (x)

€

x

€

c.

€

f (c)

€

f

€

x

€

f .

Maximum and Minimum Values

is a global (absolute) minimum of if for all in the domain of

is a local (relative) minimum of if for all in some interval around

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f (c) ≤ f (x)

€

f (c)

€

f

€

f (c) ≤ f (x)

€

x

€

c.

€

f (c)

€

f

€

x

€

f .

Extrema

Identify the labeled points as local maxima/minima, global maxima/minima, or none of these.

Extrema

Identify the labeled points as local maxima/minima, global maxima/minima, or none of these.

localmax

localmin

local maxglobal max

global min

nothing special

Extreme Values

Notice:Extreme values occur at either a critical number of

f or at an endpoint of the domain.(However, not all critical numbers and endpoints

correspond to an extreme value.)

Also note:By definition, relative extreme values do not occur

at endpoints.

Finding Local Maxima and Minima(First Derivative Test)

Assume that f is continuous at c, where c is a critical number of f.

If f’ changes from + to - at x=c, then f changes from increasing to decreasing at x=c and f(c) is a local maximum value.

If f’ changes from - to + at x=c, then f changes from decreasing to increasing at x=c and f(c) is a local minimum value.

If f’ does not change sign at x=c, then f doesn’t have an extreme value at x=c.

Finding Local Maxima and Minima(First Derivative Test)

Example:Use the first derivative test to find the local extrema of the following functions.

(a)

(b)

€

g(x) =x

1+ x 2

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f (x) = x ln x

Finding Local Maxima and Minima(Second Derivative Test)

Assume that f’’ is continuous near c and f’(c)=0.

If f’’(c)>0 then the graph of f is concave up at x=c and f(c) is a local minimum value.

If f’’(c)<0 then the graph of f is concave down at x=c and f(c) is a local maximum value.

If f’’(c)=0 or f”(c) D.N.E. then the second derivative test doesn’t apply and you have to use the other method.

Application

Assignment 43, #1 (modified):Consider the function , where(a) Find the critical number of f.

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f (t) = Ate−βt

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A,β > 0.

Application

Assignment 43, #1 (modified):(b) Use the second derivative test to determine if the critical number in part (a) corresponds to a local maximum, local minimum, or neither.

Application

Assignment 43, #1 (modified):(c) Determine the values of such that f describes the graph given below.

€

A and β

Extreme Value Theorem

If is continuous for all , then there are points such that is the global minimum and is the global maximum of on

In words:If a function is continuous on a closed, finite interval, then it has a global maximum and a global minimum on that interval.

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f (x)

€

c1, c2 ∈ [a, b]

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f (c1)

€

x ∈ [a, b]

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[a, b].

€

f (c2)

€

f (x)

Finding Absolute Extreme Values on a Closed Interval [a,b]

1. Find all critical numbers in the interval.2. Make a table of values.

The largest value of f(x) is the absolute maximum and the smallest value is the absolute minimum.

Finding Absolute Extreme Values on a Closed Interval [a,b]

Example:Find the absolute extrema of on

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g(x) = x13 (x − 2)2

€

[−1, 1].

L’Hopital’s RuleAnother application of derivatives is to help evaluate limits of the form

where either or

Idea:Instead of comparing the functions f(x) and g(x), compare their derivatives (rates) f’(x) and g’(x).

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limx →a

f (x)

g(x)

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limx →a

f (x) = 0 and limx →a

g(x) = 0

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limx →a

f (x) = ±∞ and limx →a

g(x) = ±∞ .

L’Hopital’s Rule

Suppose that f and g are differentiable functions such that

is an indeterminate form of type or If near a (could be 0 at a) then

€

limx →a

f (x)

g(x)= lim

x →a

′ f (x)′ g (x)

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limx →a

f (x)

g(x)

€

00

€

∞∞ .

€

′ g (x) ≠ 0

L’Hopital’s Rule

Evaluate the following limits using L’Hopital’s Rule, if it applies.

(a)(b)

(c)(d)

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limx →0

sin x

x

€

limx →0

x

ex −1

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limx →∞

ln x

x3

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limx → ∞

x 2e−3x

L’Hopital’s Rule

Evaluate the following limits using L’Hopital’s Rule, if it applies.

(a)(b)

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limx → ∞

x sin 1x

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limx → ∞

x1x

Differential Equations

A differential equation is an equation that involves an unknown function and one or more of its derivatives.

Examples:

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y'= 2 + y

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y"+2xy = x 2

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y'= x 2 + ex

Differential Equations

A solution of a differential equation is a function that, along with its derivatives, satisfies the DE.

Example: Show that is a solution of the differential equation and initial condition

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y'+3x 2y = 6x 2

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y = 2 + e−x 3

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y(0) = 3.

Pure-Time DEs

A pure-time differential equation is obtained by measuring the rate of change of the unknown quantity and expressed as a function of time.

Example:

Note that the formula for the rate of change depends purely on the time t.

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ds

dt= t 2 − 3t + 5

Autonomous DEsAn autonomous differential equation is derived

from a rule describing how a quantity changes and is expressed as a function of the unknown quantity.

Example:Rule: The growth rate of a population is

proportional to its size.

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db

dt= k ⋅b(t) or simply b'= k ⋅b

Example 1: Volume of a Cell

Suppose we observe that of water enters a cell each second.

Differential Equation:

General Solution:

is called the ‘state variable’

€

2.0 μm3

€

V

Example 1: Volume of a Cell

Suppose we observe that of water enters a cell each second.

Differential Equation:

General Solution:

is called the ‘state variable’

€

2.0 μm3

€

dV

dt= 2.0

€

← pure − time DE

€

V

Example 1: Volume of a Cell

Suppose we observe that of water enters a cell each second.

Differential Equation:

General Solution:

is called the ‘state variable’

€

2.0 μm3

€

dV

dt= 2.0

€

← pure − time DE

€

V (t) = 2.0t + C

€

V

Example 1: Volume of a Cell

Suppose we are told that the initial volume of the cell is

General Solution:

Initial Condition:

Particular Solution: €

V (t) = 2.0t + C

€

150μm3.

€

V (0) =150

Example 1: Volume of a Cell

Suppose we are told that the initial volume of the cell is

General Solution:

Initial Condition:

Particular Solution: €

V (t) = 2.0t + C

€

150μm3.

€

V (0) =150

€

V (t) = 2.0t +150

Example 2: Population Size

Suppose we know that the growth rate of a population is half of its current population and the initial population is 10.

Differential Equation:

Initial Condition:

Particular Solution:

Example 2: Population Size

Suppose we know that the growth rate of a population is half of its current population and the initial population is 10.

Differential Equation:

Initial Condition:

Particular Solution:€

db

dt= 0.5b

€

← autonomous DE

€

b(t) = ?

€

b(0) =10

Example 2: Population Size

Suppose we know that the growth rate of a population is half of its current population and the initial population is 10.

Differential Equation:

Initial Condition:

Particular Solution:€

db

dt= 0.5b

€

← autonomous DE

€

b(t) =10e0.5t

€

b(0) =10

Example 2: Population Size

€

b(t) =10e0.5t

€

db

dt= 0.5b

Modelling: Verbal Descriptions IVPs

Example:Write a differential equation and an initial condition to describe the

following events.

1. The relative rate of change of the population of wild foxes in an ecosystem is 0.75 baby foxes per fox per month. Initially, the population is 74 thousand.

2. The population of an isolated island is 7500. Initially, 13 people are infected with a flu virus. The rate of change of the number of infected people is proportional to the product of the number who are infected and the number of people who are not yet infected.

Solutions for General DEs

Algebraic Solutions an explicit formula or algorithm for the solution (often,

impossible to find) Geometric Solutions

a sketch of the solution obtained from analyzing the DE Numeric Solutions

an approximation of the solution using technology and and some estimation method, such as Euler’s method

Graphical Solutions of Pure-Time DEs

1. Graph the derivative.

2. Create a chart relating information about the derivative to information about the solution.

3. Sketch the solution using the initial condition to ‘anchor’ the graph.

Graphical Solutions of Pure-Time DEs

Example:Sketch the graph of the solution to given the initial condition

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s(0) =1.

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s'(t) = 3 − t

Euler’s Method

What information does an initial value problem tell us about the solution?

Example:

DE:

IC:

€

dy

dx= x + y

€

y(0) =1

slope of the solution curve y(x)

an exact value of the solution

Euler’s Method

Euler’s Idea:

First, using the initial condition as a base point, approximate the solution curve y(x) by its tangent line.

First Euler approximation

Euler’s Method

Next, travel a short distance along this line, determine the slope at the new location (using the DE), and then proceed in that ‘corrected’ direction.

Euler’s approximation with step size

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Δx = 0.5

Euler’s Method

Repeat, correcting your direction midcourse using the DE at regular intervals to obtain an approximate solution of the IVP.

By increasing the number of midcourse corrections, we can improve our estimation of the solution. Euler approximation with step size

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Δx = 0.25

Euler’s Method

Summary:

An approximate solution to the IVP

is generated by choosing a step size and computing values according to the algorithm

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tn +1 = tn + Δt

yn +1 = yn + G(tn ,yn )Δt

€

dy

dt= G(t, y), y(t0) = y0

€

Δt

Euler’s Method

Algorithm:

Algorithm In Words:

next time = current time + step size

next approximation = current approximation + rate of change at current values x step size

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tn +1 = tn + Δt

yn +1 = yn + G(tn ,yn )Δt

Example

Consider the IVP

Approximate the value of the solution at t=1 by applying Euler’s method and using a step size of 0.25.

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y' = t + 3, y(0) = 2

Example

Calculations:

tn yn

t0 = 0 y0 = 2

Table of Approximate Values for theSolution y(t) of the IVP

Example

Graph of Approximate Solution:

Plot points and connect with straight line segments.

tn yn

t0 = 0 y0 = 2

t1 = 0.25 y1 = 2.75

t2 = 0.5 y2 = 3.5625

t3 = 0.75 y3 = 4.4375

t4 = 1 y4 = 5.375

€

0

€

0.25

€

0.5

€

0.75

€

1€

1

€

2

€

3

€

4

€

5

€

6

Determining Properties of a Solution

Example #36, p. 417A population of caribou is modeled by the autonomous DE

Analyze this equation to describe the behaviour of the population of caribou.€

P'(t) = 2P(t) 1−P(t)

2500

⎛

⎝ ⎜

⎞

⎠ ⎟, P(t) > 0.

€

P(t)