Transcript lecture14

Lecture 14
Review for Test II
Rate = Derivative
Rate, rate of change, rate of growth, rate of decay,
velocity, etc
The amount A(t) to which principal P accumulates
after t years at interest rate r per year (r a decimal) is:
Simple Interest:
Compound Interest:
Continuous Interest
A(t) = P(1+ rt)
r 

A( t ) P  1 
m

A( t ) P e
(r t )
(m t )
Simple Interest
$1000 is borrowed for 5 years at 7% per year.
(a)What amount is due after 5 years and, (b) at what rate
is the amount changing at that time?
A( t ) P( 1 r t )
A( t ) 1000 (1+.07*t)
A( 5 ) 1000 (1+.07*5)
(b)
A '( t ) Pr
A '( 5 ) 1000*.07
Compound Interest
$1000 is borrowed for 10 years at 9% per year compounded quarterly.
(a) To what has the principal accumulated after 10 years and
(b) at what rate is the amount due changing after 10 years
r 

A( t ) P  1 
m

(m t )
1



A( t ) 1000  1 .09 
4


A( 10 ) 2435.188965
(4 t )
mt 

 

r

ln( A( t ) ) ln P  1  
m 
 
r 

 m ln 1 
A(t)
m

A ' (t)
A '( 10 ) 4 A( 10 ) ln( 1000*(1+.09/4) )
A '( 10 ) 216.7377493
Continuous Interest
$1000 is borrowed at 9% compounded continuously for 10 years.
a. To what amount has the $1000 accumulated after 10 years
b. At what rate is the total accumulation changing after 10 years
A( t ) P e
(r t )
A( 10 ) 1000 e
.09*10
A(10) 2459.603111
A ' ( t ) r P e
(r t )
A ' ( 10 ) .09 1000 e
.09*10
A ' (10) 221.3642800
Rates and Distance
If a rock is thrown upward from a height of 10 feet with a velocity
of 73 feet per second then in the absence of air resistance
its height at time t will be
2
h( t )  16 t  73 t 10
What are is its velocity and acceleration when it hits the ground?
Solve h(t) = 0 to get time when height is 0. t = -.133, t = 4.695
Velocity at time t is
h ' (t)  32 t 73
Acceleration at time t is h ‘’ (t) = -32
h '( 4.6956 ) -77.2593036
h ‘ ’ (4.6956) = -32
Rates and Distance II
A particle moves on the x-axis so that its position at time t is
x( t ) 1 4 t
At time t = 3 at what rate is its distance from the point (2,5) changing?
Position at time t is
P( t ) ( 1 4 t, 0 )
Let d(t) = distance from (2,5) at time t
d( t )
2
2
( 2 ( 1 4 t ) )  ( 5 0 )
d '( t )
1
8 32 t
2
26 8 t  16 t
2
d '( 3 )
26
97
194
d '( 3 ) 3.733382424
Geometry of the Derivative
Must be able to infer properties of f(x) from
properties of f ‘ (x) and f ‘ ’(x)
Critical number
Of f. Local max
Of f occurs here
Suppose this is the graph of the derivative of f.
critical number of f
Local min of f occurs here
Critical number of
f ‘ (f ‘ ‘ = 0) a point of
Inflection does NOT
occur here.
f dec
f dec
Critical number of
f ‘ (f ‘’ = 0)
– point of inflect
occurs here
f inc
f concave up
f concave down
Geometry of Derivative II
Suppose f ‘ (x) = (x-3)(x+1) what are: (a) the
intervals on which f is increasing and decreasing,
(b) the local maxima and minima of f,
(c) the intervals on which f is concave up and
down, (d) the points of inflection of the graph of f?
x


f ‘’ (x) = 2-2x
Concavity changes
cp  -1
at x=1, flex point
cp  3


Change from inc to
 -2

dec at x= -1 (local max)
Change from dec to  0
Inc at x= 3 (local min) 
1


2


4
-1
3
f ‘ > 0 (inc)
f ‘ < 0 (dec) 1
f ‘’ < 0 (concave down)
f ‘ >0 (inc)
f ‘ ‘ > 0 (concave up)
f(x) f ' (x) f '' (x)


*
0
-4 


*
0
4 


*
5
-6 


*
-3
-2 


*
-4
0  pfp


*
-3
2 


*
5
6 
Inc: (-inf,-1),(3, inf)
dec: (-1.3)
Concave down: (-inf,1)
Concave up: (1, inf)
Optimization
If f is continuous on a closed interval [a,b] then f assumes
its (absolute) maximum and (absolute) minimum values
on [a,b] at critical numbers of f.
Strategy for finding absolute maxima and minima of f on [a,b]:
i. Find the critical numbers c1, c2, …, of f. Be sure to include the
end points“a” and “b” among them.
ii.
Calculate the numbers f(c1), f(c2), …. The largest is the absolute
maximum and the smallest is the absolute minimum
Note that without end points there may be no absolute
maximum or minimum. Local maxima and minima may
exist and will occur at critical numbers.
Maxima and Minima when there
are no end points
Even if there are no (or only one) end point absolute maxima and
minima can be detected in case there is exactly one interior critical point.
If there is a unique interior critical number then it must be the point at
which the absolute maximum or absolute minimum occurs.
In situations such as this the unique critical number is where the
absolute max or min occurs. It will be a min if f ‘ <0 to the left and
f ‘ >0 to the right, a max if ‘ > 0 to the left and f ‘ >0 to the right.
Alternatively one can check concavity: concave up (f ‘’ <0) at the point
indicates local max, concave down (f ‘’ >0) indicates local min.
Example
A particle travels in the plane so that at time t its x-coordinate is
x(t) = 2t-3 and its y-coordinate is y(t) = 4-t. At which time is it
closest to the origin?
2
2
2
d( t )  (  x )  (  y )
2
2
( 2 t 3 0 )  ( 4 t 0 )
d( t )
d ' (t)
1
2
10 t  20
2
5 t  20 t  25
f ‘ <0 (dec)
2 f ‘ > 0 (inc)
t


cp  2

0


3
d(t) d ' (t) d '' (t)


5
0
5  min


*
-* 


*
+
* 
Min dist =
5
at t = 2