Announcements
Topics:
- finish section 3.2; work on sections 3.3, 4.1, and 4.2
* 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”)
elimination of chemicals
*** filtration by kidneys (kidneys break down
constant amount per hour … caffeine)
*** breaking down the chemicals using
enzymes from the liver (amount of chemical
broken down depends on the amount present …
alcohol)
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Absorption of Caffeine:
Our bodies eliminate caffeine at a constant rate
of 13% per hour.
DTDS:
ct 1  0.87ct  d
amount of caffeine
(mg) 1 hour later
amount of
caffeine now
amount of
“new” caffeine
consumed at
time t+1

* Similar
to “methadone” example
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Absorption of Caffeine:
Our bodies eliminate caffeine at a constant rate
of 13% per hour.
DTDS:
ct 1  0.87ct  d
amount of caffeine
(mg) 1 hour later
amount of
caffeine now
amount of
“new” caffeine
consumed at
time t+1

* Similar
to “methadone” example
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Absorption of Caffeine:
Our bodies eliminate caffeine at a constant rate
of 13% per hour.
DTDS:
ct 1  0.87ct  d
amount of caffeine
(mg) 1 hour later
amount of
caffeine now
amount of
“new” caffeine
consumed at
time t+1

* Similar
to “methadone” example
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Absorption of Caffeine:
Our bodies eliminate caffeine at a constant rate
of 13% per hour.
DTDS:
ct 1  0.87ct  d
amount of caffeine
(mg) 1 hour later
amount of
caffeine now
amount of
“new” caffeine
consumed at
time t+1

* Similar
to “methadone” example
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
The amount of alcohol that is broken down by
the liver depends on the amount of alcohol
present in the body.
The larger the amount, the smaller the
proportion of alcohol being eliminated.
*Similar to the limited growth population model
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
The amount of alcohol that is broken down by
the liver depends on the amount of alcohol
present in the body.
The larger the amount, the smaller the
proportion of alcohol being eliminated.
*Similar to the limited growth population model
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
The amount of alcohol that is broken down by
the liver depends on the amount of alcohol
present in the body.
The larger the amount, the
smaller the proportion of
alcohol being eliminated.
*Similar to the limited growth population model
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
The amount of alcohol that is broken down by
the liver depends on the amount of alcohol
present in the body.
The larger the amount, the
smaller the proportion of
alcohol being eliminated.
*Similar to the limited growth population model
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
The amount of alcohol that is broken down by
the liver depends on the amount of alcohol
present in the body.
The larger the amount, the
smaller the proportion of
alcohol being eliminated.
*Similar to the limited growth population model
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
DTDS:
rate of elimination
at 1  at  r(at )at  d
amount of alcohol
(g) 1 hour later

amount of
alcohol now
amount of
“new” alcohol
consumed at
time t+1
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
Rate of Elimination:
10.1
r(at ) 
4.2  at
r(at )

 10.1  
DTDS: at 1  at  
at  d
4.2  at 
at
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
r(at )
Rate of Elimination:
10.1
r(at ) 
4.2 at
at

 10.1 
DTDS: at 1  at  
at  d
4.2  at 

Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
A standard drink contains 14g of alcohol.
Compare what happens over time for the following
situations:
(a) You consume two drinks right away and
continue to have half of a drink every hour
(a) You consume one drink every hour
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
A standard drink contains 14g of alcohol.
Compare what happens over time for the following
situations:
(a) You consume two drinks right away and
continue to have half of a drink every hour
(a) You consume one drink every hour
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
A standard drink contains 14g of alcohol.
Compare what happens over time for the following
situations:
(a) You consume two drinks right away and
continue to have half of a drink every hour
(a) You consume one drink every hour
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
Example:
A standard drink contains 14g of alcohol.
Compare what happens over time for the following
situations:
(a) You consume two drinks right away and
continue to have half of a drink every hour
(a) You consume one drink every hour
Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
(a) You consume two drinks right away and
continue to have half of a drink every hour
 10.1 
f (at )  at  
at  7, a0  28
4.2  at 

Substance Absorption (Elimination) and
Replacement (Consumption) Models
Elimination of Alcohol:
(b) You consume one drink every hour
 10.1 
f (at )  at  
at  14, a0  0
4.2

a

t 

Calculus On Continuous Functions
In order to start studying continuous-time
dynamical systems, we need to develop the
usual tools of calculus on continuous functions:




Limits
Continuity
Derivatives
Integrals

these are all defined
in terms of limits
Rates of Change
The rate of change of a function tells us how the
dependent variable changes when there is a
change in the independent variable.
Geometrically, the rate of change of a function
corresponds to the slope of it’s graph.
Secant Lines and Tangent Lines
A secant line is a line that intersects two points on a
curve.
A tangent line is a line that just touches a curve at a point
and most closely resembles the curve at that point.
Average Rate of Change
= Slope of Secant Line
The average rate of change
of f(t) from t=t1 to t=t2
corresponds to the slope of
the secant line PQ.
f
f (t 2 )  f (t1 )
mPQ 

t
t2  t1
Average Rate of Change
= Slope of Secant Line
Alternative Notation:
The average rate of change
of f(t) from the base point
t=t0 to t=t0+Δt is
f
f (t 0  t)  f (t 0 )
mPQ 

t
t
Average Rate of Change
= Slope of Secant Line
Example #2 (modified):
Find the average rate of change of the function
h(t)  t 2 1
starting from time t=1 and lasting 1, 0.1, and 0.01 units of
time.
 t0
.
Δt
t0+Δt
1
1
2
1
0.1
1.1
1
0.01
1.01
f(t0+Δt) - f(t0)
Δt
Estimating the Slope of the Tangent
Steps:
1. Approximate the tangent at P
using secants intersecting P and a
nearby point Q.
2. Obtain a better approximation
to the tangent at P by moving Q
closer to P, but Q  P.
3. Define the slope of the tangent
at P to be the limit of the slopes
of secants
PQ as Q approaches P
(if the limit exists).
Instantaneous Rate of Change =
Slope of Tangent Line
The instantaneous rate of
change of f(t) at t=t0
corresponds to the slope of
the tangent line at t=t0.
f
f '(t0 )  lim
t 0 t
f (t 0  t)  f (t0 )
 lim
t 0
t
Note:
The slope of the curve y=f(t)
 at P is the slope of its
tangent line at P.
Instantaneous Rate of Change =
Slope of Tangent Line
This special limit is called
the derivative of f at t0
and is denoted by f’(t0)
(read “f prime of t0”).
Alternative notation:
Instantaneous Rate of Change =
Slope of Tangent Line
Example #10 (modified):
h(t)  t 1
2
(a) Guess the limit of the slopes
of the secants as the second
point approaches the base point.

(b) Use this to find the equation
of the tangent line to h(t) at t=1.
.
t0
Δt
t0+Δt
f(t0+Δt) - f(t0)
Δt
1
0.1
1.1
2.1
1
0.01
1.01
2.01
1
0.001 1.001
2.001
The Limit of a Function
Notations:
f (2)  5
means that the y-value of
the function AT x=2 is 5

x 2  4

if x  2 
f (x)   x  2

if x  2
5
lim f (x)  4
x 2
means that the y-values of
the function APPROACH 4
as x APPROACHES 2
The Limit of a Function
Definition:
lim f (x)  L
x a
“the limit of f(x), as x approaches a, equals L”
means that the values of f(x) (y-values)
approach the number L more and more closely
as x approaches a more and more closely (from
either side of a), but xa.
Limit of a Function
Some examples:
Note: f may or may not be defined at x=a. Limits are only asking how f is defined NEAR a.
Limit of a Function
Some examples:
lim f (x)  3
x 4
f (4) is undefined
Note: f may or may not be defined at x=a. Limits are only asking how f is defined NEAR a.
Limit of a Function
Some examples:
lim f (x)  3
lim g(x)  5
f (4) is undefined
g(2)  5
x 4
x 2
Note: f may or may not be defined at x=a. Limits are only asking how f is defined NEAR a.

Limit of a Function
Some examples:
lim f (x)  3
lim g(x)  5
lim h(x)  
f (4) is undefined
g(2)  5
h(3) is undefined
x 4
x 2
x 3
Note: f may or may not be defined at x=a. Limits are only asking how f is defined NEAR a.


Left-Hand and Right-Hand Limits
lim f (x)  L
x a
means f (x)  L as x a from the left (x  a).
lim f (x)  L
x a

 means f (x) 
 L as x a from the right (x  a).
** The full limit
exists
if
and
only
if
the
left
and
right
 and are the
 (equal a real number)
 limits both exist
same value.
Left-Hand and Right-Hand Limits
For each function below, determine the value of
the limit or state that it does not exist.
Left-Hand and Right-Hand Limits
For each function below, determine the value of
the limit or state that it does not exist.
lim f (x)  3
x 4 
lim f (x)  5
x 4
 lim f (x) D.N.E.
x 4
Left-Hand and Right-Hand Limits
For each function below, determine the value of
the limit or state that it does not exist.
lim f (x)  3
g(x) undefined when x  0
lim f (x)  5
x 4
lim g(x)  0
x 0
 lim f (x) D.N.E.
but lim g(x) D.N.E.
x 4 
x 4
x 0
Left-Hand and Right-Hand Limits
For each function below, determine the value of
the limit or state that it does not exist.
lim f (x)  3
g(x) undefined when x  0
lim f (x)  5
x 4
lim g(x)  0
x 0
 lim f (x) D.N.E.
but lim g(x) D.N.E.
x 4 
x 4
x 0
lim h(x)  3
x 3 
lim h(x)  3
x 3
 lim h(x) = 3
x 3
Evaluating Limits
We can evaluate the limit of a function in 3
ways:
1. Graphically
2. Numerically
3. Algebraically
Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
x
3.5
3.9
3.99
4
4.01
4.1
4.5

f(x)
undefined
Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
x
f(x)
3.5
7.5
3.9
7.9
3.99
7.99
4
undefined
4.01
4.1
4.5

Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
x
f(x)
3.5
7.5
3.9
7.9
3.99
7.99
4
undefined
4.01
4.1
It appears that y  8 as x  4 

4.5
Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
It appears that y  8 as x  4 

x
f(x)
3.5
7.5
3.9
7.9
3.99
7.99
4
undefined
4.01
8.01
4.1
8.1
4.5
8.5
Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
It appears that y  8 as x  4 
 and that y  8 as x  4 .
x
f(x)
3.5
7.5
3.9
7.9
3.99
7.99
4
undefined
4.01
8.01
4.1
8.1
4.5
8.5
Evaluating Limits
Example:
Use a table of values to
estimate the value of
x 2 16
lim
x 4 x  4
It appears that y  8 as x  4 
 and that y  8 as x  4 .
x 2 16
So we guess that lim
 8.
x 4 x  4
x
f(x)
3.5
7.5
3.9
7.9
3.99
7.99
4
undefined
4.01
8.01
4.1
8.1
4.5
8.5
Evaluating Limits Algebraically
BASIC LIMITS
Limit of a Constant
Function
Limit of the Identity
Function
lim c  c, where c  R
x a
Example:
lim x  a
x a
Example:
lim 2  2
x3
 lim
x 3
x 3



LIMIT LAWS
[used to evaluate limits algebraically]
Suppose that c is a constant and the limits
lim f (x) and lim g(x)
exist. Then
x a
x a
[ f (x)  g(x)]  lim f (x)  lim g(x)
1. lim
x a
x a
x a

[ f (x)  g(x)]  lim f (x)  lim g(x)
2. lim
x a
x a
x a
c f (x)  c lim f (x)
3. lim

x a
x a


LIMIT LAWS
[used to evaluate limits algebraically]
Continued…
4.
lim [ f (x)  g(x)]  lim f (x)  lim g(x)
5.
lim [ f (x)  g(x)]  lim f (x)  lim g(x), if lim g(x)  0
x a
x a
x a
x a
x a
x a
x a
Evaluating Limits Algebraically
Example:
Evaluate the limit and justify each step by
indicating the appropriate Limit Laws.
lim (x 2  5x  6)
x 1
 lim x 2  lim 5x  lim 6
x 1
x 1
x 1
 lim x  lim x  5lim x  lim 6
x 1
x 1
 (1)(1)  5(1)  6
2
x 1
x 1
Evaluating Limits Algebraically
Example:
Evaluate the limit and justify each step by
indicating the appropriate Limit Laws.
lim (x 2  5x  6)
x 1
 lim x 2  lim 5x  lim 6
x 1
x 1
x 1
 lim x  lim x  5lim x  lim 6
x 1
x 1
 (1)(1)  5(1)  6
2
x 1
x 1
Evaluating Limits Algebraically
Example:
Evaluate the limit and justify each step by
indicating the appropriate Limit Laws.
lim (x 2  5x  6)
x 1
 lim x 2  lim 5x  lim 6
x 1
x 1
x 1
 lim x  lim x  5lim x  lim 6
x 1
x 1
 (1)(1)  5(1)  6
2
x 1
x 1
Evaluating Limits Algebraically
Example:
Evaluate the limit and justify each step by
indicating the appropriate Limit Laws.
lim (x 2  5x  6)
x 1
 lim x 2  lim 5x  lim 6
x 1
x 1
x 1
 lim x  lim x  5lim x  lim 6
x 1
x 1
 (1)(1)  5(1)  6
2
x 1
x 1
Direct Substitution Property
From the previous slide, we have
lim (x 2  5x  6)  2
x1
f (1)
f (x)
Notice that we could have simply
found the


valueof the limit
by plugging in x=1 into the
function.
Direct Substitution Property
Direct Substitution Property:
If f(x) is an algebraic, exponential, logarithmic,
trigonometric, or inverse trigonometric
function, and a is in the domain of f(x), then
lim f (x)  f (a)
x a


Equal Limits Property
Consider the functions:
x2  4
f (x) 
x 2

g(x)  x  2.

* Note: f(x)=g(x) everywhere except at x=2
Equal Limits Property
Example:
2
x
Calculate lim  4 .
x 2
Note: direct substitution does not work
x 2

FACT:
If f (x)  g(x) when x  a , then lim f (x)  lim g(x)
x a
x a
provided the limits exist.

Strategy for Evaluating Limits
#
 
0

real #

0
0


Evaluating Limits Algebraically
Evaluate each limit or state that it does not exist.
x2
(a) xlim
1 x  1
 1x
(b) lim
x 2 x  2
x 2
(c) lim
x 4 4  x
1 x
(d) lim
x 1 x 1
1
2

