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Transcript
Announcements
Topics:
- sections 6.4 (l’Hopital’s rule), 7.1 (differential equations), and
7.2 (antiderivatives)
* 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”)
Extreme Value Theorem
If f (x) is continuous for all x Î [a, b] , then there are
points c1, c 2 Î [a, b] such that f (c1) is the global
minimum and f (c 2 ) is the global maximum
of f (x) on [a, b].
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.
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:
1
2
3
Find the absolute extrema of g(x) = x (x - 2) on [-1, 1].
L’Hopital’s Rule
Another 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).
L’Hopital’s Rule
Suppose that f and g are differentiable functions
such that
is an indeterminate form of type 00 or ¥¥ .
If g¢(x) ¹ 0 near a (could be 0 at a) then
L’Hopital’s Rule
Evaluate the following limits using L’Hopital’s
Rule, if it applies.
(a)
(b)
(c)
(d)
L’Hopital’s Rule
Evaluate the following limits using L’Hopital’s
Rule, if it applies.
(a)
(b)
Differential Equations
A differential equation is an equation that
involves an unknown function and one or
more of its derivatives.
Examples:
y'= 2 + y
y"+2xy = x
2
y'= x + e
2
x
Differential Equations
A solution of a differential equation is a function
that, along with its derivatives, satisfies the DE.
Example:
-x 3
Show that y = 2 + e is a solution of the
differential equation y'+3x 2 y = 6x 2 and initial
condition 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:
ds 2
= t - 3t + 5
dt
Note that the formula for the rate of change
depends purely on the time t.
Example 1: Volume of a Cell
Suppose we observe that 2.0 mm 3 of water
enters a cell each second.
Differential Equation:
General Solution:
V is called the ‘state variable’
Example 1: Volume of a Cell
Suppose we observe that 2.0 mm 3 of water
enters a cell each second.
dV
Differential Equation:
= 2.0 ¬ pure - time DE
dt
General Solution:
V is called the ‘state variable’
Example 1: Volume of a Cell
Suppose we observe that 2.0 mm 3 of water
enters a cell each second.
dV
Differential Equation:
= 2.0 ¬ pure - time DE
dt
General Solution:
V(t) = 2.0t +C
V is called the ‘state variable’
Example 1: Volume of a Cell
Suppose we are told that the initial volume of
the cell is 150 mm 3 .
General Solution:
V(t) = 2.0t +C
Initial Condition:
V(0) =150
Particular Solution:
Example 1: Volume of a Cell
Suppose we are told that the initial volume of
the cell is 150 mm 3 .
General Solution:
V(t) = 2.0t +C
Initial Condition:
V(0) =150
Particular Solution: V(t) = 2.0t +150
Autonomous DEs
An 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.
db
= k × b(t) or simply b'= k × b
dt
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:
db
= 0.5b ¬ autonomous DE
dt
Initial Condition:
b(0) =10
Particular Solution:
b(t) = ?
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:
db
= 0.5b ¬ autonomous DE
dt
Initial Condition:
b(0) =10
Particular Solution:
b(t) =10e
0.5t
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


Geometric Solutions


an explicit formula or algorithm for the solution (often,
impossible to find)
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
A Geometric Solution of a
Pure-Time DEs
Example:
Sketch the graph of the solution to s'(t) = 3- t
given the initial condition s(0) =1.
Euler’s Method
What information does an initial value problem
tell us about the solution?
Example:
dy
DE:
=x+y
dx
IC:
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
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
Euler’s Method
Summary:
An approximate solution to the IVP
dy
= G(t, y), y(t 0 ) = y 0
dt
is generated by choosing a step size
and computing values according to the
algorithm
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
Example
Consider the IVP
y'= t + 3, y(0) = 2
Approximate the value of the solution at t=1 by
applying Euler’s method and using a step size of 0.25.
Example
Calculations:
Table of Approximate Values for the
Solution y(t) of the IVP
tn
yn
t0 = 0
y0 = 2
Example
Graph of Approximate
Solution:
Plot points and connect with
straight line segments.
6
5
4
tn
yn
t0 = 0
y0 = 2
t1 = 0.25
y1 = 2.75
2
t2 = 0.5
y2 = 3.5625
1
t3 = 0.75
y3 = 4.4375
t4 = 1
y4 = 5.375
3
0
0.25
0.5 0.75
1
Determining Properties of a Solution
Example:
A population P(t) of
caribou is modeled by the
autonomous DE
æ P(t) ö
P'(t) = 2P(t)ç1÷, P(t) > 0.
è 2500 ø
Analyze this equation to
describe the behaviour of
the population of caribou.
Solving Pure-Time DEs
The general form of a pure-time differential
equation is
dF
= f (x)
dx
where F(x)is the unknown state variable and f (x)
is the measured rate of change.
Examples:
(a) dF = 4x 3 + 1
dx
(b)
dy
1
x
= 5e +
2
dx
1+ x
Solving Pure-Time DEs
Solve each by “guess and check”.
Ask yourself:
“What function has this as its derivative?”
Examples:
dF
(a)
= 4x 3 + 1
dx
dy
1
x
(b)
= 5e +
dx
1+ x 2
Antiderivatives/Indefinite Integrals
An antiderivative (or indefinite integral) of a
function f (x) is a function F(x) with derivative
equal to f (x) .
An antiderivative F(x) is a solution to the puretime differential equation
dF
= f (x).
dx
Initial Value Problems
A differential equation has a
whole family of solutions.
Example:
y'= 2x - 4
y = x 2 - 4x + C
Initial Value Problems
An initial value problem
provides an initial condition
so you can find a particular
solution.
Example:
y'= 2x - 4, y(0) = 3
y = x 2 - 4x + 3
Initial Value Problems
An initial value problem
provides an initial condition
so you can find a particular
solution.
Example:
y'= 2x - 4, y(0) = 3
y = x 2 - 4x + 3
Antiderivatives/Indefinite Integrals
Theorem 6.1:
If F(x) is an antiderivative of f(x), then the most
general antiderivative of f(x) is F(x)+C; i.e.,
ò f (x)dx = F(x) + C
where C is a real number.
If an initial value of the solution is given, then we
can solve for C to find a specific or particular
antiderivative of f(x).
Rules for Antiderivatives
The Power Rule for Integrals
ò
n +1
x
x n dx =
+C
n +1
for n ¹ -1
Example:
Integrate each.
(a)
ò x dx
7
(b)
ò
1
dt
4
t
(c)
ò
xdx
Rules for Antiderivatives
The Constant Product Rule for Integrals
Suppose
Then
ò f (x)dx = F(x) + C.
ò af (x)dx = aF(x) + C'. for any constant a.
The Sum Rule for Integrals
Suppose
ò f (x)dx = F(x) + C
Then
ò [ f (x) + g(x)]dx =
and
ò g(x)dx = G(x) + C'.
ò f (x)dx + ò g(x)dx = F(x) + G(x) + C''.
Some More Examples
Example 1:
Integrate.
(a)
3
ò (5x - x )dx
4
(b)
Example 2:
Solve the differential equation
f '(x) = 2 +
x
with initial condition
1
2 x
f (0) = 0.
2
4x
(sec
x
+
e
)dx
ò
Application:
A Differential Equation for AIDS
During the early years of the AIDS epidemic, the
number of new AIDS cases per year was » 523.8t 2,
where t is measured in years since the beginning
of 1981.
rate at which new AIDS cases were reported:
dA
2
= 523.8t
dt
Application:
A Differential Equation for AIDS
Surveys indicated that about 340 people had
been infected at the beginning of 1981.
Initial condition: A(0) = 340
Solution:
A(t) represents the number of AIDS cases t
years after 1981.
A(t) =174.6t 3 + 340
Summary Of Some Basic
Integration Formulas
ò
n +1
x
x n dx =
+C
n +1
for n ¹ -1
ò cos xdx = sin x + C
òx
ò sin xdx = -cos x + C
x
x
e
dx
=
e
+C
ò
ò sec
1
ò 1+ x 2 dx = arctan x + C
2
xdx = tan x + C
-1
dx =
ò
1
dx = ln x + C
x