Math 106 class summaries

MATH 106 Calculus for Business and Economics Spring, '00
Class Summaries/Outlines

2/7

2/8

2/9

2/10

3/6

3/7

3/8

3/9

4/3

4/4

4/5

5/1

5/2

5/3

5/4

3/13

3/14

3/15

3/16

4/10

4/11

4/12

4/13

2/21

2/22

2/23

2/24

3/20

3/21

3/22

3/23

4/17

4/18

4/19

4/20

2/28

2/29

3/1

3/2

3/27

3/28

3/29

3/30

4/24

4/25

4/26

4/27

1/19 Numbers, Variables, Algebra Review 1.1&1.2

Go over Background Analysis.

Solutions:

B: y=2x+5
D: -9/10
A: (x-1)/x^3
D: x^2/(3y^9)
D: 3x + 2 sqrt(15xy) + 5y
C: $52
D: 4

SECTION 1.1
Real numbers are decimals (possibly infinite and non-repeating)
The number line. Points on the line correspond to real numbers.
Absolute values.

Example: |-2|=2; |sqrt(5)-5| = 5-sqrt(5).
Example*: |2 + - 3| < |2| + |-3|

Equality, inequality, and intervals.

Solving a linear equation.

Example. 3x+5=29:

Solving a linear inequality.

Example: 3x+5<29:

Example: -2x < 10:

Exponents and radicals:

integer exponents:

2^3 = 2*2*2 = 8
2^0 = 1
2^-3 =1/ (2^3) = 1/8

fraction exponents and roots

8^(1/3) = cube root of 8 = 2

Properties:

SAME BASE: Products, quotients, powers
SAME POWER: Products. quotients.

SECTION 1.2

Multiplication and Factoring:

a*(b+c)= a*b + a*c
quadratic expressions (see text p 17)

1/20

Class spent much time on problems from the review sheet related to functions.

5. Boyle's law states that , for a certain gas P*V = 320, where P is pressure and V is volume.
A. Draw a complete graph representing this situation. Label your axes and write an equation for each asymptote.
Make a table of values for P and V using P=320/V and P>0, V>0.

V	P
80	4
160	2
320	1
640	.5

Now use these data points to sketch the graph.

B. If 8 < V < 40, what are the corresponding values of P?
Notice that when V=8, P = 40 and when V=40, P= 8. So as V varies from 8 to 40, P varies from 40 to 8.

We also considered problem 115 on page 13.

1/24 start here:

Summary of Chapter 1 in Calculus Applied to the Real World (on-line)
We did several problems from 1.1 and 1.2... including problems about fractional and negative exponents, and factoring to solve a quadratic equation.
[Also Reviewed interval notations]

Roots and equations.
Solving a quadratic equation.

Example: x^2-5x + 6 = 0 :

Example: x^2-5x + 6 = 0 :

Fractions (Rational Expressions)

Factor and "cancel"
Common Denominators [Still to be reviewed.]
Rationalizing expressions with roots. [Still to be reviewed.]

SECTION 1.3

Cartesian Coordiantes
Distance

Example: eq'n of a circle with center at (1,2) and radius 3.

SECTION 1.4

Slope

m = (y₂- y₁)/ (x₂-x₁)
Example: find slope of line through the points (4, 2) and (1,3):

Interpret: m>0; m<0; m=0
parallel non-vertical lines have the same slope.

1/25

Start with a review of lines, then go on to functions (2.1)
Equations for lines:

Vertical:

Example:X=3

Non-vertical:

Point- slope

Example: m=5 through (2,3):

slope- Y intercept

Example: m=5 Y intercept is (0,7)

General Equation:

Example: find X and Y intercepts for 2X + 3Y - 6 = 0:

SECTION 2.1 FUNCTIONS.

What is a function?

A function f is a correspondence (rule) which assigns to each member of the domain (source) set, x, a member of the co-domain (target) set, denoted f(x) .

How are functions defined?

with a formula.

x²+5. This is expressed by the equation: g(x)=x²+5.

With this definition we have
g(0)=5, g(1)=6, g(-1)=6,g(2)=9, g(1/2)=21/4, g(t)=t²+5, g(1+t)=(1+t)²+5, etc.

with a table.

x	f(x)
2	5
1	3
0	-2
-1	4
-2	0

with a graph.

The vertical line test: A graph measuring variables X and Y determines Y as a function of X if each vertical line meets the graph at at most one point.

with a context.

How are functions visualized?

Mapping figures/transformation figures.

Graphs.

1/26

Problem #8 was done !

The wall is 200 meters long. We have 140 meters of fence. This is a constraint. In terms of the lengths measured x and L we have that 2x+L = 140. Thus L = 140 -2x.

The area, A, is found by multiplying x by L. Thus A = x (140-2x) = 2x(70-x). We made a table of some values of this function. We noted that 0<=x<=70.

x	L	A
10	120	1200
20	100	2000
70	0	0
60	20	1200
50	40	2000

We looked at the transformation figure from this data as well as the graph.

In fact, A = -2x²+140x but A also can be expressed by A= -2(x-35)² +2450. It should be clear from thinking about this expression for A that A is at most 2450, since for any x other than 35, A will be the result of subtracting a positive number from 2450.

We then examined this function's graph using WINPLOT.

At the end of class we looked briefly at two problems on functions and transformation figures.

1/27

We looked at problems from today's and the previous day's assignment.

Using these problems we reviewed several aspects of functions as indicated below by &&.

More on functions and Mathematical Modeling. (2.1 and 2.3)

Use Winplot to graph examples of these with parameters.

More examples of functions.

Quadratic functions.

f(x)= Ax² + Bx + C

&& Functions defined in cases.

i. s(x)= 2x+1 when x <3 ; -x+10 when x>=3

ii s(x)= 2x+1 when x <3 ; Ax+B when x>=3

Polynomial functions.

cubic c(x)= Ax³ +Bx ² + Cx + D

quartics q(x)= Ax⁴+ Bx³ + Cx² +Dx +E

&& Rational functions.

R(x)=f(x)/g(x) where f and g are polynomials.

Examples: use polynomials above for f and g.

Example: Average cost= C(x)/x

&& Power/root functions.

examples: f(x)=sqrt(x); g(x)= x^1/3

^1/31

^{Finish a few details on functions-
especially considering equilibrium for supply/demand.}

Demand p =f( x_d) N.B. This reverses the more sensible relationship as x_d being a function of p.
Supply p =f( x_s). N.B. This also reverses the more sensible relationship as x_s being a function of p.

Use Winplot to graph examples.

Begin Rates and slopes and margins.

average velocity (rate), slope of a secant line, marginal cost/unit for an indivisible commodity.

Average Velocity Example: Suppose position of an anvil falling from a 100 foot cliff is a function of time,
S = -16t² +100 feet at time t seconds.
The position at times t=1 and t=2.
Find the average velocity during the time interval [1,2].

Solution: S(1) = -16+100 = 84; S(2) = -16*4 + 100 = 36.
Average velocity = change in position/ change in time = [S(2)-S(1)]/[2-1] = (36-84)/(2-1) = -48 ft/sec.
Note: The negative velocity indicates the anvil is falling down. The average speed of the anvil is 48 ft/sec.

Use Winplot to graph examples.

Instantaneous velocity, slope of a tangent line; marginal cost/unit for a divisible commodity .

Average velocity-Instantaneous velocity
Average rate of price change wrt time- instantaneous rate of price change wrt time
average marginal cost per unit- marginal cost per unit
slope of secant line to curve- slope of tangent line to a curve.

Slopes of tangents and secants, instantaneous rates and average rates. Estimation and limits.

Suppose Y= f(X). Consider X=a and X=a+h. These correspond to 2 points on the the graph of f,

(a, f(a)) and (a+h, f(a+h)). The line connecting these two points ids called the secant line.

The slope of the secant line is [f(a+h)-f(a)]/h.

This is also the average velocity of an object moving on a straight line with its position S = f(t) at time t for the time interval between time time t= a and time t=a+h.

When h is very close to 0, h -> 0,

the value of [f(a+h)-f(a)]/h will approximate the slope of the line tangent to the graph of f at the point (a,f(a). It will also approximate the instantaneous velocity of the object moving on the straight line at time t=a.

Example: f(x)= 100-16x². Find the slope of the tangent to the graph at when x=2.

Solution: . First find the slope of the secant line determined by X=2 and X= 2+h.

f(a)=100-16a²; a =2 so f(2) = 100 - 16*4= 36.

f(a+h)= f(2+h) = 100 -16(2+h)² = 100 - 64 - 64h -16h² = 36 -64h -16 h² ;

Thus f(a+h) - f(a) = -64h -16h² = h(-64-16h) and [f(a+h)-f(a)]/h = -64 -16h.

Now: THINK! When h ->0, this slope = -64-16h -> -64. Thus the slope is -64.

Example: Suppose S =f(t) feet where t is measured in seconds with f from the previous example. Then the instantaneous velocity of the object at time 2 seconds is -64 feet per second.

Continued discussion of the tangent line/instantaneous velocity/rate problem.
The equation of the tangent line when x=2 is given by

y=-64(x-2) + 36.

To find the slope of the tangent at a more general contect when x=a:

Consider the quotient [f(a+h)-f(a)]/h using the general choice of a in place of the specific value x=2:

f(a+h) =100 - 16(a+h)² = 100 -16a² -32ah-16h²

f(a) =100 - 16 a²

f(a+h)-f(a)= -32ah - 16h² = (-32a-16h)h

and thus [f(a+h)-f(a)]/h= -32a -16h.

Now considering what happens when h is close to 0, we have the slopes of the secants =-32a-16h -> -32a.

This is consistent with the result when x=2.

Limit notation for expressing this work:....
Do this for x...

2/3 See the text chapter 2 summary and Section 3.1 for examples of what we did today.

We talked more about the definition of the derivative and finding the derivative using algebra and thinking.
We started an examination of the calculus of derivatives using constants. powers, constant multiples, and sums.
We started to use Leibniz notation for the derivative.

2/7 More on the rules of the calculus for derivatives.

Justifications for some rules: constants, powers (integers, square roots), constant multiples, sums.
Start product and quotient rules.

2/8 More on the rules of the calculus for derivatives.

Justifications and interpetations of the derivatives of powers, constant multiples and sums.
Introduced Product Rule.

2/9 More on the rules of the calculus for derivatives.

Demonstration of the product rule's proof.
Quotient rule introduced and justified based on the product rule.

2/10 More on the rules of the calculus for derivatives.

More work with the quotient rule.
If Q(x) = 1/ P(x) then by applying the quotient rule we have

If f(x) = 1/ x ⁿ = x ^-n then f '(x) = -nx ^n-1 / x ²ⁿ = -n x^-n-1.
If C(x) = the total cost for producing x units, C'(x) is the marginal cost for producing one more unit.

Using time, t , position of a car, s , and the consumption of gas, g, during a trip we started a consideration of the chain rule: dg/dt = dg/ds * ds/dt.

2/14 Class cancelled (power failure)
2/15 More on the chain rule. Marginal cost, reveues, and profit.
2/16

Higher order derivatives.
Implicit differention.

2/17

Related Rates
The differential
Elasticity of demand.

Suppose x= f(p). We wish to consider how a small change in price, h, effects the change in quantity demanded on a percentage basis... i.e.

[Percent change in quantity demand] / [ Percent change in unit price]

or [ {f(p+h) - f(p) }/f(p) *100] / [ h/p *100] .

When h is small this is approximately f '(p) /(f(p)/p) = p*f '(p) / f(p).

The elasticity of demand is defined to be E(p) = -p*f '(p) / f(p) = - p / x * dx/dp.