Lecture 14

Review for Test II

Rate = DerivativeRate, 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: A(t) = P(1+ rt)

Compound Interest:

Continuous Interest

( )A t P

1

r

m

( )m t

( )A t P e( )r 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 rateis the amount changing at that time?

( )A t ( )P 1 r t

( )A 5 1000 (1+.07*5)

( )A t 1000 (1+.07*t)

(b) ( )A ' t Pr

( )A ' 5 1000*.07

Compound Interest

( )A t P

1

r

m

( )m t

$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

( )A t 1000

1

1

4.09

( )4 t

( )A 10 2435.188965( )A ' 10 4 ( )A 10 ( )ln 1000*(1+.09/4)

( )A ' 10 216.7377493

( )ln ( )A t

ln P

1

r

m

mt

A ' (t)

A(t)m

ln 1

r

m

Continuous Interest

$1000 is borrowed at 9% compounded continuously for 10 years. a. To what amount has the $1000 accumulated after 10 yearsb. 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 velocityof 73 feet per second then in the absence of air resistance its height at time t will be

( )h t 16 t2

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

At time t = 3 at what rate is its distance from the point (2,5) changing?

Position at time t is

Let d(t) = distance from (2,5) at time t

( )x t 1 4 t

( )P t ( ),1 4 t 0

( )d t ( )2 ( )1 4 t2

( )5 02

( )d ' t1

2

8 32 t

26 8 t 16 t2

( )d ' 326

97194

( )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)

Suppose this is the graph of the derivative of f.

f dec f inc

critical number of fLocal min of f occurs here

Critical number Of f. Local maxOf f occurs here

Critical number off ‘ (f ‘’ = 0) – point of inflectoccurs here Critical number of

f ‘ (f ‘ ‘ = 0) a point ofInflection does NOToccur here.

f dec

f concave up f concave down

Geometry of Derivative IISuppose 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?

-1 3

f ‘’ (x) = 2-2x

1

cpcp

x f(x) f ' (x) f '' (x)

-1 * 0 -4

3 * 0 4

-2 * 5 -6

0 * -3 -2

1 * -4 0

2 * -3 2

4 * 5 6

pfp

f ‘’ < 0 (concave down) f ‘ ‘ > 0 (concave up)

f ‘ > 0 (inc) f ‘ < 0 (dec) f ‘ >0 (inc)

Concavity changesat x=1, flex point

Change from inc todec at x= -1 (local max) Change from dec to

Inc at x= 3 (local min)

Inc: (-inf,-1),(3, inf)dec: (-1.3)Concave down: (-inf,1)Concave up: (1, inf)

OptimizationIf f is continuous on a closed interval [a,b] then f assumes its (absolute) maximum and (absolute) minimum valueson [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 absolutemaximum and the smallest is the absolute minimum

Note that without end points there may be no absolutemaximum or minimum. Local maxima and minima mayexist 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 pointindicates 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?

( )d t2 ( ) x

2( ) y

2

( )d t ( ) 2 t 3 02

( ) 4 t 02

d ' (t)1

2

10 t 20

5 t2

20 t 25

t d(t) d ' (t) d '' (t)

2 5 0 5

0 * -- *

3 * + *

cp

2f ‘ <0 (dec) f ‘ > 0 (inc)

Min dist = at t = 2

min

5