Ap Calculus Ab Semester I 2009

469 / 470 BC - 399 BC

Socratic Ignorance "I know that I know nothing"














PITHAGORAS (600 BC) ZENO (500 BC) EUDOXUS (400 BC) EUCLID (300 BC) ARCHIMIDES (200 BC)*V KEPLER (1500 AC) GALILEO (1500 AC) FERMAT (1600 AC) CAVALIERI (1600 AC) DESCARTES (1600 AC) ISAAC BARROW (1600 AC)













NEWTON (1700 AC)*V LIEBNIZ (1700 AC) BERNOULLI’S (1700 AC) EULER (1700 AC) LAGRANGE (1800 AC) LAPLACE (1800 AC) FOURIER (1830 AC) CAUCHY (1857 AC) RIEMANN (1866 AC) WEIERSTRAUSS (1897 AC)




Was born and worked in Syracuse (Greek city in Sicily) 287 BCE and died in 212 BCE




Friend of King Hieron II




“Eureka!” (discovery of hydrostatic law)


Invented many mechanisms, some of which were used for the defence of Syracuse
Other achievements in mechanics usually attributed to Archimedes (the law of the lever, center of mass, equilibrium, hydrostatic pressure)
Used the method of exhaustions to show that the volume of sphere is 2/3 that of the enveloping cylinder
According to a legend, his last words were “Stay away from my diagram!”, address to a soldier who was about to kill him




was designed to find areas and volumes of complicated objects (circles, pyramids, spheres) using

◦ approximations by simple objects (rectangles, triangles, prisms) having known areas (or volumes)

Approximating the circle

Approximating the pyramid


Let C(R) denote area of the circle of radius R
We show that C(R) is proportional to R2

P2

1)

Inner polygons P1 < P2 < P3 <…

2)

Outer polygons Q1 > Q2 > Q3 >…

3)

P1

Q1

Q2

Qi – Pi can be made arbitrary small

4)

Hence Pi approximate C(R) arbitrarily closely

5)

Elementary geometry shows that Pi is proportional to R2 . Therefore, for two circles with radii R and R' we get: Pi(R) : Ri (R’) = R2:R’2

6)

Suppose that C(R):C(R’) < R2:R’2

7)

Then (since Pi approximates C(R)) we can find i such that Pi (R) : Pi (R’) < R2:R’2 which contradicts 5)

Thus Pi(R) : Ri (R’) = R2:R’2

Y

S




Z




1 R

4

7


3

2

Q

6

5 O

Thus A = ∆1 (1+1/4 + (

P

X

1/4)2+<) = 4/3 ∆

1

Triangles ∆1 , ∆2 , ∆3 , ∆4,… Note that ∆2 + ∆3 = 1/4 ∆1 Similarly ∆4 + ∆5 + ∆6 + ∆7 = 1/16 ∆1 and so on

















Calculus appeared in 17th century as a system of shortcuts to results obtained by the method of exhaustion Calculus derives rules for calculations Problems, solved by calculus include finding areas, volumes (integral calculus), tangents, normals and curvatures (differential calculus) and summing of infinite series This makes calculus applicable in a wide variety of areas inside and outside mathematics In traditional approach (method of exhaustions) areas and volumes were computed using subtle geometric arguments In calculus this was replaced by the set of rules for calculations




Cylon, a Crotoniate and leading citizen by birth, fame and riches, but otherwise a difficult, violent, disturbing and tyrannically disposed man, eagerly desired to participate in the Pythagorean way of life. He approached Pythagoras, then an old man, but was rejected because of the character defects just described. When this happened Cylon and his friends vowed to make a strong attack on Pythagoras and his followers. Thus a powerfully aggressive zeal activated Cylon and his followers to persecute the Pythagoreans to the very last man. Because of this Pythagoras left for Metapontium and there is said to have ended his days.




Los cylons son una civilización cibernética que está en guerra con las Doce Colonias de la humanidad en la película y series de Battlestar Galactica ...

"Imagination is more important than knowledge."

ALBERT EINSTEIN WROTE THIS RIDDLE EARLY DURING THE 19th CENTURY. HE SAID THAT 98% OF THE WORLD POPULATION WOULD NOT BE ABLE TO SOLVE IT. ARE YOU IN THE TOP 2% OF INTELLIGENT PEOPLE IN THE WORLD? SOLVE THE RIDDLE AND FIND OUT.

There are no tricks, just pure logic, so good luck and don't give up. 1. In a street there are five houses, painted five different colours. 2. In each house lives a person of different nationality 3. These five homeowners each drink a different kind of beverage, smoke different brand of cigar and keep a different pet. THE QUESTION: WHO OWNS THE FISH?

1. 2. 3. 4.

The Brit lives in a red house. The Swede keeps dogs as pets. The Dane drinks tea. The Green house is next to, and on the left of the White house. 5. The owner of the Green house drinks coffee. 6. The person who smokes Pall Mall rears birds. 7. The owner of the Yellow house smokes Dunhill. 8. The man living in the centre house drinks milk. 9. The Norwegian lives in the first house. 10. The man who smokes Blends lives next to the one who keeps cats. 11. The man who keeps horses lives next to the man who smokes Dunhill. 12. The man who smokes Blue Master drinks beer. 13. The German smokes Prince. 14. The Norwegian lives next to the blue house. 15. The man who smokes Blends has a neighbour who drinks water.

Einstein's Riddle - ANSWER

The German owns the fish.

"Do not worry about your difficulties in Mathematics. I can assure you mine are still greater."

CALCULUS: CALCULAE: STONES








TWO FUNDAMENTAL IDEAS OF CALCULUS DERIVATIVE-INTEGRAL CALCULUS APPLICATIONS BOOK RESOURCES TI 84 PLUS











Calculus is deeply integrated in every branch of the physical sciences, such as physics and biology. It is found in computer science, statistics, and engineering; in economics, business, and medicine. Modern developments such as architecture, aviation, and other technologies all make use of what calculus can offer. Finding the Slope of a Curve Calculating the Area of Any Shape Visualizing Graphs Finding the Average of a Function Calculating Optimal Values

HOW TO FIND:


INSTANTANEOUS RATE OF CHANGE R =




∆D ∆T

∆D ∆T

AREA UNDER A CURVE

A

B




R= D / T RATE = CHANGE IN DISTANCE/ CHANGE IN TIME THE AVERAGE RATE OF CHANGE BETWEEN TWO POINTS = THE SLOPE OF THE SECANT LINE CONNECTING THE TWO POINTS

DISTANCE

THE INSTANTANEOUS RATE OF CHANGE = THE SLOPE OF THE TANGENT LINE TIME R = CHANGE IN D / CHANGE IN T R = O / O = UNDEFINED “BIG PROBLEM”







BLACKBOARD EXAMPLE: From home to school. SKETCHPAD Rate of change

THE INSTANTANEOUS RATE OF CHANGE

∆D R = ∆T

f(x) DISTANCE

THE DEFINITION OF THEDERIVATIVE

f (x + ∆x) − f (x) f ' = lim ∆x

x

∆x → 0

TIME

∆T THE DERIVATIVE OF f(x) AT x REPRESENTS THE SLOPE OF THE TANGENT LINE AT A POINT x

f (x + ∆x) − f (x) f ' = lim ∆x ∆x → 0 THE DERIVATIVE OF f(x) AT x REPRESENTS THE SLOPE OF THE TANGENT LINE AT A POINT x

THE INSTANTANEOUS RATE OF CHANGE

R =

∆D ∆T

Given the graph of below, evaluate the following limits.

(a) (d) (g)

lim f ( x ) x →1

lim− f ( x )

x →−4

lim f ( x ) x →1

(b)

lim f ( x )

(c)

x →3

(e)

lim f ( x )

x →∞

(f)

x → −4+

(h)

lim f ( x )

x →−2+

lim f ( x ) lim f ( x)

x →−4

(i) lim f ( x)

x →−2




1st Direct Substitution ◦ If it fails…




2nd Factoring ◦ If it fails…




3rd The Conjugate Method

Algebraic Limits: (a) lim x − 4 x→2 x 2 + x + 2

2 x −9 (b) lim x →3 x 2 − x − 6

(c) lim x + 3

(d)lim x + 3

(e) lim 3 x − 5

(f) lim 4 x + 7 2

x→2

( x − 2)

2

x →∞

2x + 4

x→2

x →∞

x−2

x − 3x + 5




Workout the MAGIC (Algebra)




Review: ◦ ALGEBRA ◦ ECUATIONS, RELATIONS, AND FUNCTIONS ◦ TRIGONOMETRY

Let f be a function with f(1) = 4 such that for all points (x, y) on the graph of f . The slope is given by 3x2 + 1 2y

(a)Find the slope of the graph of f at the point where x = 1. (b)Write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f(1.2)

The graph of the velocity v(t), in ft/sec, of a car traveling on a straight road, for 0 ≤ t ≤ 50, is shown above. A table of values for v(t), at 5 second intervals of time t, is shown to the right of the graph. (a)During what intervals of time is the acceleration of the car positive? Give a reason for your answer (b)Find the average acceleration of the car, in ft/sec2, over the interval 0 ≤ t ≤ 50, (c)Find one approximation for the acceleration of the car, in ft/sec2, t = 40. Show the computations you used to arrive at at your answer.




An equation for the line tangent to the graph of y = cos(2x) at x = π 4 is:

(a)

π  y − 1 = − x −  4 

(c)

π  y = 2 x −  4 

(e)

π  y = −2  x −  4 

(b)

π  y − 1 = −2  x −  4 

(d)

π  y = − x −  4 




1 2 y= x 2

At what point on the graph of is the tangent line parallel to the line 2 x − 4 y = 3 ?

a (0.5, -0.5) d (1, 0.5)

b (0.5, 0.125) e (2, 2)

c (1, -0.25)

The following table gives US populations at time t: Estimate and interpret P’(1996). t P (t) 1992

255,002,000

1994

260,292,000

1996

265,253,000

1998

270,002,000

2000

274,634,000

ken s. . . It also helps you to practice and develop your logic/reasoning skills. Calculus throws you challenging problems your way which make you think. Life after school and college will likewise undoubtedly throw you problems which you will have to learn to solve. Although you may never use calculus ever again in your lifetime or career, you will definitely hold on to the lessons that calculus taught you. Things like time management, how to be organized and neat, how to hand in things on time, how to perform under pressure when tested, how to be responsible for your future boss, how to be amongst people in your class (who are analogous to your future clients and co-workers). Calculus on face-value may not seem important to you and may seem useless, but the lessons and skills you are learning will be with you your whole lifetime.







Olivia J: learning advanced math helps you strengthen your mind overall. Think of your mind as a muscle. When you lift heavy things for a while, the lighter things seem really easy. whats my name again: If you want to be a math teacher you can use it to torture a whole other generation of kids.




KillerLi...You may not use Calculus, but much of our society relies on it. The financial operation of our economy relies on forecastings and predictions that only Calculus can provide. Electrical Engineers use Calculus to optimize the processing power of the CPU that runs your computer. City planners and surveyors use Calculus to find the exact areas of irregular regions of land. So calculus is very important in life. As for the meaning of life, Calculus gives no answers, as it is strictly analytical, and not interpretational.

If you multiply two terms with the same base (here it’s x), add the powers and keep the base.

If you divide two terms with the same base, subtract the powers and keep the base.

A negative exponent indicates that a variable is in the wrong spot, and belongs in the opposite part of the fraction, but it only affects the variable it’s touching. Note that the exponent becomes positive when it moves to the right place. ILS AP CALCULUS AB

If an exponential expression is raised to a power, you should multiply the exponents and keep the base.

The numerator of the fractional power remains the exponent. The denominator of the power tells you what sort of radical (square root, cube root, etc.). ILS AP CALCULUS AB




Example 4: Simplify Solution: First raise to the third power.

Then Multiply the x’s and y’s together

ILS AP CALCULUS AB




Problem 4: Simplify the expression

using exponential rules.

ILS AP CALCULUS AB

ss worksheet 3

Exponential Form

103 = 1,000

Logarithmic Form

log101,000 = 3

log416 = 2

4-2 = 1/16

Exponential Form

Logarithmic Form

33 = 27

70 = 1

log55 = 1

102 = 100

3-2 = 1/9

log255 = 1/2

log31 =0

Exponential Form

Logarithmic Form

log28+=

Log41=

log1010,000=

log101/100=

Log327=

Exponential Form

103 = 1,000

42 = 16

33 = 27

51 = 5

70 = 1

Logarithmic Form

log101,000 = 3

log416 = 2

log327 = 3

log55 = 1

log71 = 0

Exponential Form

251/2 = 5

4-2 = 1/16

102 = 100

3-2 = 1/9

30 = 1

Logarithmic Form

log255 = 1/2

log4(1/16) =-2

log10100 = 2

Log3 1/9 = -2

Exponential Form

Logarithmic Form

23 = 8

log28+= 3

40 = 1

Log41= 0

104 = 10000

log1010,000 = 4

10-2 = 1/100

log101/100= -2

log31 =0

33 = 27

Log327= 3




Greatest Common Factors Factoring using the greatest common factor is the easiest method of factoring and is used whenever you see terms that have pieces in common. Take, for example, the expression 4x + 8. Notice that both

terms can be divided by 4, making 4 a common factor. Therefore, you can write the expression in the factored form of 4(x + 2). In effect, I have “pulled out” the common factor of 4, and what’s left behind are the terms once 4 has been divided out of each. In these type of problems, you should ask yourself, “What do each of the terms have in common?” and then pull that greatest common factor out of each to write your answer in factored form. Problem 5: Factor the expression




You should feel comfortable factoring trinomials such as x² + 5x + 4 = 0

using whatever method suits you. Most people play with binomial pairs until they stumble across some-thing that works, in this case (x + 4)(x + 1)

There are some patterns that you should have memorized:
◆ Difference of perfect squares: a² – b² = (a + b)(a – b)








Explanation: A perfect square is a number like 16, which can be created by multiplying something times itself. In the case of 16, that something is 4, since 4 times itself is 16. If you see one perfect square being subtracted from another, you can automatically factor it using the pattern above. For example, x² – 25 is a difference of x² and 25, and both are perfect squares. Thus, it can be factored as (x + 5)(x – 5). You cannot factor the sum of perfect squares so whereas x² – 4 is factorable, x² + 4 is not!




◆ Sum of perfect cubes:

a³ + b³ = (a + b)(a² – ab + b²) Explanation: Perfect cubes are similar to perfect squares. The number 125 is a perfect cube because 5 · 5 · 5 = 125. This formula can be altered just slightly to factor the difference of perfect cubes, as illustrated in the next bullet. Other than a couple of sign changes, the process is the same.




◆ Difference of perfect cubes:

a³ – b³ = (a – b)(a2 + ab + b2) Example 5: Factor x³ – 27 using the difference of perfect cubes factoring pattern. Solution: Note that x is a perfect cube since x · x · x = x³, and 27 is also, since 3 · 3 · 3 = 27. Therefore, x³ – 27 corresponds to a³ – b³ in the formula, making a = x and b = 3. Now, all that’s left to do is plug a and b into the formula:

You cannot factor (x² + 3x + 9) any further,

so you are finished.




Problem 6: Factor the expression

8x³ + 343




Method One: One: Factoring




Method Two: Completing the Square




Method Three: The Quadratic Formula




To begin, set your quadratic equation equal to 0; If the resulting equation is factorable, factor it and set each individual term equal to 0. These equations will give you the solutions to the equation. That’s all there is to it.




Example 6: Solve the equation




3x² + 4x = –1

by factoring

Solution: Always start the factoring method by setting the equation equal to 0. 3x² + 4x + 1 = 0. Now, factor the equation and set each factor equal to 0. (3x + 1)(x + 1) = 0 3x + 1 = 0 x+1=0 x = - 1/3 x=-1 This equation has two solutions: x = -1/3 or x = –1

You can check them by plugging each separately into the original equation, and you’ll find that the result is true.

Factoring: 3x² + 4x +1 Solution 3x x

1=x 1 = 3x 4x

(3x + 1) (x + 1)

(3x + 1)( x + 1) = 0 (3x + 1) = 0 , (x + 1) =0 x = - 1/3 , x = -1




Example 7: Solve the equation 2x² + 12x – 18 = 0

by completing the square. Solution: Move the constant to the right side of the equation: 2x² + 12x = 18

This is important: For completing the square to work, the coefficient of x2 must be 1. Divide every term in the equation by 2: x² + 6x = 9 Here’s the key to completing the square: Take half of the coefficient of the x term, square it, and add it to both sides. In this problem, the x coefficient is 6, so take half of it (3) and square that (3² = 9). Add the result (9) to both sides of the equation: x² + 6x + 9 = 9 + 9 x² + 6x + 9 = 18










At this point, if you’ve done everything correctly, the left side of the equation will be factorable. In fact, it will be a perfect square! (x + 3)(x + 3) = 18 (x + 3)² = 18 To solve the equation, take the square root of both sides. That will cancel out the exponent. Whenever you do this, you have to add a ± sign in front of the right side of the equation. This is always done when square rooting both sides of any equation: √(x + 3) ² = ± √18 x + 3 = ± √18 Solve for x, and that’s it. It would also be good form to simplify into : x = -3 ± √18 x = -3 ± 3 √2 x = - 3 + 3 √2 x = - 3 - 3 √2




The quadratic formula Set the equation equal to 0, and you’re halfway there. Your equation will then look like this: ax² + bx + c = 0

where a, b, and c are the coefficients as indicated. Take those numbers and plug them straight into this formula : 2 − b ± b − 4ac x= 2a

You’ll get the same answer you would achieve by completing the square.




Solve the equation 2x² + 12x – 18 = 0

using the quadratic formula. Solution: The equation is already set equal to 0, in form ax² + bx + c = 0, and a = 2, b = 12, and c = –18 Plug these values into the quadratic formula and simplify:




Problem 7: Solve the equation 3x² + 12x = 0

three times, using all the methods you have learned for solving quadratic equations.













◆ Basic equation solving is an important skill in calculus. ◆ Reviewing the five exponential rules will prevent arithmetic mistakes in the long run. ◆ You can create the equation of a line with just a little information using point-slope form. ◆ There are three major ways to solve quadratic equations, each important for different reasons.







WHEN IS AN ECUATION A FUNCTION? IMPORTANT FUNCTION PROPERTIES FUNCTION SKILLS THE BASIC PARAMETRIC ECUATIONS

GO TO THE TEXTBOOK




The Rule of four: Tables, Graphs, Formulas, and Words. Cricket Chirp Rate versus Temperature Temperature °F

40

75

100

136

Chirps per minute

0

140

240

384

C=4T - 160

C (Chirps per minute)

500 400 300

The Chirp Rate is a Function of Temperature C(T)=4T-160

200 100 0

T (°F)

40

75

100

136

C=4T - 160

C (Chirps per minute)

500 400

Cricket Chirp Rate versus Temperature Temperature °F

40

75

100

136

300 200

Chirps per minute

0

140

240

384

100 0

T (°F)

40

75

100

136

◦ Domain (inputs) ◦ =All T values between 40°F and 136°F ◦ =All T values with 40≤x≥136 ◦ =[40,136] ◦ Range (outputs) ◦ =All C values from 0 to 384 ◦ =All C value with 0≤C≥384 ◦ =[0,384]




This function, called g, accepts any real

number input. To find out the output g gives, you plug the input into the x slot.




Real life examples… ◦ A person’s height is a function of time ◦ Other examples (by ss)…




Sometimes you’ll plug more than a number into a function—you can also plug a function into

another function. This is called composition of functions.


Example 1: If f(x) = and g(x) = x + 6,

evaluate g( f (25)).

Solution: In this case, 25 is plugged into f, and that output is in turn plugged into g. Evaluate f(25). Now, plug this result into g: g(5) = 5 + 6 = 11 Therefore, g(f (25)) = 11.




Piecewise-defined function




Evaluate ◦ ◦ ◦ ◦ ◦

f(1)= f(2)= f(3)= f(10)= f(0)=




The last important thing you should know test. about functions is the vertical line test

This test is a way to tell whether or not a given graph is the graph of a function or not.




Linear functions y=f(x)=b +mx y-y₁=m(x-x₁)




Número de habitantes ◦ En el II Conteo de Población y Vivienda 2005, realizado por el INEGI, se contaron 103 263 388 habitantes en México.

◦ Por ello, México está entre los once países más poblados del mundo, después de: China, India, Estados Unidos de América, Indonesia, Brasil, Pakistán, Rusia, Bangladesh, Nigeria y Japón.





THE GENERAL EXPONENTIAL FUNCTION P is an exponential funtion of t with base a if

P = P0 a

t

◦ Where P₀ is the initial quantity (when t=0) and a is the factor by which P changes when t increases by 1. ◦ If a>1, we have exponential growth ◦ If 0
Examples ◦ Population in Mexico ◦ Elimination of a drug from the body

P (Population in millions)

Population of Mexico (estimated) estimated) 19801980-1986 CHANGE IN POPULATION (MILLIONS)

1000 800

1980

67.38

1981

69.13

1.75

1982

70.93

1.80

400

1983

72.77

1.84

200

1984

74.66

1.89

1985

76.60

1.94

1986

78.59

1.99

Calculate the Exponential Function: What is the initial quantity?

Exponential Growth

600

P = P0 a t

What is the Growth Rate? Evaluate and Interpret P(2005): P(2009): For what year was the Population estimated in 100 million people?

100

50

t (years since 1980)

10

6

5

4

3

0

0 2

POPULATION (MILLIONS)

1

YEAR

Elimination of a Drug from the Body

Q (mg)

Q (mg)

300

0

250

250

1

150

2

90

3

54

4

32.4

5

19.4

t (hours)

Exponential Decay

200 150 100 50

Calculate the Exponential Function:

0 0

Q = Q0 a t

What is the initial quantity? What is the Growth Rate? Evaluate and Interpret Q(10): How many hours does it take for the drug to decrease to 0.001mg?

1

2

3

4

t (hours) 5

Example 1


Suppose that Q=f(t) is an exponential function of t. If f(20)=88.2 and f(23)=91.4 a. Find the base

b. Find the growth rate

c. Evaluate f(25)




Any exponential Growth function can be written, for some a rel="nofollow">1 and k>0, in the form ◦




P = P0 a t

or

P = P0 e kt

And any exponential Decay function can be written, for some 00, as ◦

Q = Q0 a t

or

Q = Q0 e − kt

◦ We say that P and Q are growing or decaying at a continous rate of k. (k=0.02 corresponds to a Continous rate of 2%)


Example. Convert the functions P = e 0.05t and Into the form ◦

P = P0 a t

and

Q = 5e −0.2t

Q = Q0 a t







The graph of a function is concave up if it bends upward as we move from left to right; It is concave down if it bends downward.

Exercises pg. 14: 1,2,3,4,5,6,7,8,9,10,11 12,17,23,24,25,26,27,37,39




Shifts and Stretches

Multiplying a function by a constant, c, stretches the graph vertically (if c>1). Or shrinks the graph vertically (if 0
y=x² y=(x-2)²




Composite Functions “A Function of a Function” ◦ Example 1. If f(x)=x² and g(x)=x+1, find: ◦ a. f(g(2)) ◦ b. g(f(2)) ◦ c. f(g(x)) ◦ d. g(f(x)) ◦ Exmp 2. Express the following function as a composition. h(t)=(1+t³)²⁷




Odd and Even Functions: Symmetry ◦ The graph of any polynomial involving only even powers of x has symmetry about the x-axis. (Even functions. E.g. f(x)=x²) ◦ Polynomials with only odd powers of x are symmetric about the origin. (Odd functions. E.g. g(x)=x³) Even function

f(x)=x²

Odd function

For any function f, f is an Even function if f(-x)=f(x) for all x. f is an Odd function if f(-x)=-f(x) for all x.

g(x)=x³




Inverse Functions

Exercises. Pg 21. [1,8],14,22,25,26,52

f¯¹(y)=x means y=f(x) A function has an inverse if (and only if) its graph intersects any horizontal line at most once. In other words, each y-value correspond to a unique x-value y=f(x)

Find the Inverse function.

y=f(x)

C=f(T)=4T-160 f¯¹(C)

f¯¹(y)=x

y=x³

The logarithm to base 10 of x, written log₁₀ x, is the power of 10 we need to get x. log₁₀ x = c means 10^c = x The natural logarithm of x, written ln x, is the power of e needed to get x. ln x = c means e^c = x Properties of Logarithms 1. 2. 3. 4. 5.

Log (AB)=log A + log B Log (A/B)=log A – log B Log A^p= p log A Log 10^x= x 10 ^ log x= x

1. 2. 3. 4. 5.

Ln (AB)=log A + log B Ln (A/B)=log A – log B Ln A^p= p log A Ln e^x= x e ^ ln x= x

Log x and Ln x are not defined when x is negative or 0. Log 1=0 Ln 1=0







EX 1. Find t such that 2t = 7

EX 2. Find when the population of Mexico reaches 200 million by solving P = 67.38(1.026)t




EX 3. What is the half life of ozone?

(Decaying exponentially at a continuous rate of 0.25% per year)

Q = Q0 e − kt




EX 4. The population of Kenya was 19.5 million in 1984, and 21.2 million in 1986. Assuming it increases exponentially, find a formula for the population of Kenya as a function of time.

P = P0 e kt




Give a formula for the inverse of the following function. (Solve for t in terms of P ) P = f (t ) = 67.38(1.026) t

Exercises pg 27: 1,7,8,9,11,17,25,28,29,41

An angle of 1 radian is defined to be the angle at the center of a unit circle which cuts off an arc of length 1. (measured counterclockwise) Arc length= 1 180° = π radians

1 radian = 180° / π

= 1 Radian

Equation of the unit circle: x² + y² =1 Fundamental Identity: cos² t + sin² t = 1

The Unit Circle




Amplitude, Period, and Phase

For any Periodic function of time Amplitude is half the distance between the maximum and the minimum values. (if it exists) Period is the smallest time needed for the function to execute one complete cycle. Phase is the difference a periodic function is shifted with respect to other.

Amplitude =1

Sine and Cosine graphs are shifted horizontally π/2 cos t = sin(t+ π/2) sin t = cos(t – π/2)

Period = 2π Phase = π/2

The phase difference or phase shift between sin t and cos t is π/2

To describe arbitrary amplitudes and periods of Sinusoidal functions:

f(t)=A sin( B t ) and g(t)=A cos( B t ) Where |A| is the amplitude and 2π/|B| is the period The graph of a sinusoidal function is shifted horizontally by a distance |h| when t is replaced by t-h or t+h. Functions of the form f(t)=A sin (Bt) + C and g(t)=A cos( Bt) + C have graphs which are shifted vertically and oscillate about the value C. Ex 1. Find and show on the graph the Amplitude and Period of the functions. a) y=5 sin(2t) b) y=-5 sin(t/2) c) y=1 + 2sin t

EX 2. Find possible formulas for the following sinusoidal functions g(t) 3

-6π

3

2

6π

-3

h(t)

f(t)

t

-1

3

-2

t

π

-5π

-3

7π

t

EX 3. The High tide was 9.9 feet at midnight. Later at Low tide, it was 0.1 feet. the next High tide is at exactly 12 noon and the height of the water is given by a sine or cosine curve. Find a formula for the water level as a function of time.

Ex 4. The interval between high tides actually averages 12 hours 24 minutes. Give a more accurate formula. Ex 5. Using the info from Ex 4. Write a formula for the water level, when the high tide is at 2 pm. Exercises pg 35. 13,14,15,16,17,19,20,24,25,38

The tangent function tan t=sin t / cos t

The inverse trigonometric functions arcsine y=x means sin x=y with -π/2 ≤x≤ π/2 (sin¯¹) arctan y=x means tan x=y with -π/2 <x< π/2 (tan¯¹) arccos y=x means cos x=y with -π/2 ≤x≤ π/2 (cos¯¹)

A power function has the form

f ( x) = kx p Where k and p are constant.

Ex: the volume, V, of a sphere of radius r is given by V= g(r)=4/3 πr³ Ex2: Newton’s Law of Gravitation F=k/r²

or

F=kr¯²

Polynomials are the sums of power functions with nonnegative integer exponents

y = p( x) = an x n + an −1 x n −1 + ... + a1 x + a0 n is a nonnegative integer called the degree of the polynomial.

y = p( x) = 2 x 3 − x 2 − 5 x − 7 degree of the function=_____ The shape of the graph of a polynomial depends on its degree. A leading negative coefficient turns the graph upside down. The quadratic (n=2) turns around once. The cubic (n=3) turns around twice. The quartic (n=4) turns around three times. th An n degree polynomial turns around at most n-1 times. **There may be fewer turns**

n=2

n=3

n=4

n=5

EX1: Find possible formulas for the polynomials. 4

-2

f(x)

g(x)

2

x

-3

1 -12

h(x)

2

x

-3

2

x




Rational functions p( x) are ratios of polynomials, p and q: f ( x) =

Exercises pg 42: 5,7,8,9,10,12,13

q ( x)

y=

1 x2 + 4

y=0 is a Horizontal Asymptote or y→0 as x→∞ and y→0 as x→-∞

y

x

x=K is a Vertical Asymptote if y→∞ or y→-∞ as x →K y

The graphs in Rational functions may have vertical asymptotes where the denominator is zero. Rational functions have horizontal asymptotes if f(x) approaches a finite number as x→∞ or x→-∞.

K

x

A continuous function has a graph which can be drawn without lifting the pencil from the paper. A function is said to be continuous on an interval if its graph has no breaks, jumps or holes in that interval To be certain that a function has a zero in an interval on which it changes sign, we need to know that the function is defined and continuous in that interval. f(x)=3x²-x²+2x-1

f(x)=1/x

5

-1

1

x

-1 1

-5

Zero for 0≤x≤1 F(0)=-1 and f(1) =3 have opposite signs

No zero for -1≤x≤1 although f(-1) and f(1) have opposite signs

x

A continuous function cannot skip values

1

f(x)=cos x -2x²

x 0.4 0.6 0.8 1

The function f(x)=cos x -2x² must have a zero because its graph cannot skip over the x-axis. f(x) has at least one zero in the interval 0.6≤x≤0.8 since f(x) changes from positive to negative on that interval.

-1

The Intermediate Value Theorem Suppose f is a continuous function on a closed interval [a, b]. If k is any number between f(a) and f(b), then there is at least One number c in [a, b] such that f(c)=k.

EX: Investigate the continuity of f(x)=x² at x=2 x

1.9

1.99

1.999 2.001 2.01

2.1

x²

3.61

3.96

3.996 4.004 4.04

4.41

The values of f(x)=x² approach f(2)=4 as x approaches 2. Thus f appears to be continuous at x=2

Continuity The function f is continuous at x=c if f is defined at x=c and if

lim f ( x ) = x → c

f (c )

Exercises pg 47: 15, 17, 15. An electrical circuit switches instantaneously from a 6 volt battery to a 12 volt battery 7 seconds after being turned on. Graph the battery voltage against time. Give formulas for the function represented by your graph. What can you say about the continuity of this function? f(t)

t

17. Find k so that the following function is continuous on any interval:

 kx, 0 ≤ x < 2 f ( x) =  2 3 x , 2 ≤ x

Notation:

lim f ( x) = L x→c

Exercises pg 55: 1,2

if the values of f(x) approach L as x approaches c.

lim f ( x) x→2

lim f ( x) x → 2+

lim f ( x)

general limit

right-hand limit

left-hand limit

x → 2−

When Limits Do Not Exist Whenever there is no number L such that lim f ( x) = L

x→c

lim f ( x) x → 2+

≠

lim f ( x) x → 2−

0 0 ∞ -∞

Given the graph of below, evaluate the following limits.

(a) (d) (g)

lim f ( x ) x →1

lim− f ( x )

x →−4

lim f ( x ) x →1

(b)

lim f ( x )

(c)

x →3

(e)

lim f ( x )

x →∞

(f)

x → −4+

(h)

lim f ( x )

x →−2+

lim f ( x ) lim f ( x)

x →−4

(i) lim f ( x)

x →−2




1st Direct Substitution ◦ If it fails… (0/0 Indeterminate form)




2nd Factoring ◦ If it fails…




3rd The Conjugate Method

Algebraic Limits: (a) lim x − 4 x→2 x 2 + x + 2

2 x −9 (b) lim x →3 x 2 − x − 6

(c) lim x + 3

(d)lim x + 3

(e) lim 3 x − 5

(f) lim 4 x + 7 2

x→2

( x − 2)

2

x →∞

2x + 4

(g) lim x − 2 x→4 x−4

x→2

x →∞

x−2

x − 3x + 5




Concepts are key to AP Exams slope of the tangent line  •A derivative is limit of slopes of secant lines instantaneous rate of change limit of average rates of change




•Continuity

Functions ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦

Linear functions Exponential functions New from old functions Logarithmic functions Trigonometric functions Powers, Polynomials, and Rational functions Continuity Limits

lim f(x)=f(x)