AP Calculus

CLASS OCTOBER 14 2009

WE GOT OUR TESTS BACK
THEY WERE SCALED

QUESTIONS THAT WE WENT OVER:
PART 1: 2,4,6,7
QUESTION 2: REMEMBER TO FACTOR OUT AS MUCH AS POSSIBLE-SYNTHETIC DIVISION ON CUBIC. ANSWER B

QUESTION 4, WHAT DOES THIS FUNCTION LOOK LIKE?
ANSWER: C
THE LIMIT OF THE FUNCTION APPROACHING INFINITY IS ZERO, SO THE X AXIS IS THE VERTICAL ASYMPTOTE.
THE DENOMINATOR IS THE DIFFERENCE OF SQUARES OF ONE, MEANING THERE ARE TWO VERTICAL ASYMPTOTES, AT 1 AND -1.

QUESTION  6: WHAT MUST BE TRUE ABOUT THE FUNCTION?
ANSWER: E
ALL OF THESE OPTIONS CAN BE TRUE, BUT NONE OF THEM HAVE TO BE.

QUESTION 7: SOLVE FOR K
ANSWER: -1/2
MAKE SURE TO WRITE
OR ELSE IT MAKES NO SENSE

PART 2

QUESTION 3: FIND f ‘(2)
ANSWER: f ‘(2)=42
THERE WERE MULTIPLE WAYS TO DO THIS, ALL USING SOME FORM OF THE DIFFERENCE QUOTIENT.

QUESTION 5: FIND DISCONTINOUS POINTS AND LABEL
THIS CAME DOWN TO FACTORING CORRECTLY AND KNOWING YOUR DIFFERENT TYPES OF DISCONTINUITY

THEN WE PLAYED A GAME PLOTTING DERIVATIVE FUNCTIONS
FRONT TEAM WON!!!!!

DISPLACEMENT, VELOCITY, AND ACCELERATION
s(t)= DISPLACEMENT
v(t)=VELOCITY
a(t)=ACCLERATION

THIS LED TO THE DISCOVERY OF SECOND DERIVATIVES:

s’(t)=v(t)
s”(t)=v’(t)=a(t)

AKA: THE DERIVATIVE EQUATION OF A DISPLACEMENT EQUATION IS VELOCITY, AND THE DERIVATIVE EQUATION OF A VELOCITY EQUATION IS ACCELERATION, MAKING THE SECOND DERIVATIVE OF DISPLACEMENT ACCELERATION.

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