- WELCOME TO PART ONE OF A LESSON ON EXPONENTIAL GROWTH.
IN THIS LESSON WE'LL EXPRESS EXPONENTIAL GROWTH
IN RECURSIVE FORM AS WELL AS EXPLICIT FORM.
SUPPOSE THERE ARE 4,000 STUDENTS
ATTENDING A LOCAL COMMUNITY COLLEGE
AND THE NUMBER OF STUDENTS IS INCREASING BY 5% EACH YEAR.
HOW MANY STUDENTS WILL ATTEND THE COLLEGE IN FOUR YEARS?
SUPPOSE WE LET P SUB N REPRESENT THE NUMBER OF STUDENTS
ATTENDING THE COLLEGE AFTER N YEARS.
SO P SUB ZERO WOULD REPRESENT THE NUMBER OF STUDENTS
NOW AT TIME ZERO, WHICH IN THIS CASE P SUB ZERO
WOULD BE EQUAL TO 4,000 STUDENTS.
P SUB ONE WOULD REPRESENT THE NUMBER OF STUDENTS
AFTER ONE YEAR.
P SUB TWO WOULD REPRESENT THE NUMBER OF STUDENTS
AFTER TWO YEARS AND SO ON.
SO IF P SUB 0 = 4,000,
WE CAN SAY P SUB N, THE STUDENT POPULATION AFTER N YEARS,
WOULD BE EQUAL TO P SUB N - ONE,
THE STUDENT POPULATION FROM THE PREVIOUS YEAR,
+ 0.05 x P SUB N - 1.
THIS PRODUCT HERE REPRESENTS THE 5% GROWTH
FROM THE PREVIOUS YEAR.
TO FIND A PERCENT OF A NUMBER WE CONVERT THE PERCENT TO A DECIMAL
AND MULTIPLY AS WE SEE HERE.
BUT THESE ARE LIKE TERMS,
NOTICE HOW THESE ARE BOTH P SUB N - ONE TERMS.
SO WE CAN COMBINE THESE TWO TERMS BY ADDING THE COEFFICIENTS
WHICH WOULD GIVE US 1.05 x P SUB N - 1.
THIS IS CALLED A RECURSIVE RELATIONSHIP
WHERE A RECURSIVE RELATIONSHIP IS A FORMULA
WHICH RELATES THE NEXT TERM
IN A SEQUENCE TO THE PREVIOUS VALUE OR VALUES.
HERE THE NUMBER OF STUDENTS IN YEAR N
CAN BE FOUND BY MULTIPLYING THE NUMBER OF STUDENTS
IN THE PREVIOUS YEAR, WHICH IS P SUB N - 1 BY 1.05.
SO WHILE 5% IS THE GROWTH RATE, 1.05, AS WE SEE HERE,
IS CALLED THE GROWTH MULTIPLIER.
NOTICE THAT 1.05 CAN BE THOUGHT OF AS THE ORIGINAL 100%
PLUS AN ADDITIONAL 5%.
AGAIN, 100% + 5% = 105% EXPRESSED AS A DECIMAL
WOULD BE 1.05.
SO NOW IF WE WANT TO CALCULATE THE STUDENT POPULATION
OVER THE FIRST 5 YEARS,
IF P SUB 0 = 4,000, P SUB 1 = 1.05 x P SUB 0,
P SUB 2 = 1.05 x P SUB 1 AND SO ON.
NOTICE HOW FOR WHATEVER VALUE OF N WE USE,
N - ONE IS ALWAYS ONE LESS.
SO WE'RE ALWAYS USING THE POPULATION
FROM THE PREVIOUS YEAR TO FIND THE POPULATION AFTER YEAR N.
SO GOING BACK TO OUR QUESTION,
NOTICE AFTER 4 YEARS WE FOUND THE NUMBER OF STUDENTS
WOULD BE 4,862.
NOW LETS LOOK AT THE AMOUNT OF GROWTH IN STUDENTS PER YEAR.
SO TO FIND THE AMOUNT OF GROWTH DURING THE FIRST YEAR,
WE TAKE THE POPULATION AFTER YEAR ONE
AND SUBTRACT THE POPULATION FROM THE PREVIOUS YEAR.
SO DURING THE FIRST YEAR THERE IS A GROWTH OF 200 STUDENTS.
DURING THE SECOND YEAR,
NOTICE HOW THERE IS A GROWTH OF 2010 STUDENTS
AND DURING THE THIRD YEAR THERE IS A GROWTH OF 221 STUDENTS.
SO NOTICE WHILE THE PERCENT OF GROWTH IS CONSTANT AT 5%,
BUT NOTICE THE INCREASE IN THE NUMBER OF STUDENTS
IS NOT CONSTANT.
AND THAT'S BECAUSE THIS IS EXPONENTIAL GROWTH,
NOT LINEAR GROWTH AS WE SAW IN OUR PREVIOUS LESSON.
NOW LET'S FIND AN EXPLICIT EQUATION
FOR THE EXPONENTIAL GROWTH.
AN EXPLICIT EQUATION ALLOWS US TO CALCULATE P SUB N DIRECTLY
WITHOUT NEEDING TO KNOW P SUB N - ONE
OR THE STUDENT POPULATION FROM THE PREVIOUS YEAR.
YOU MAY BE ABLE TO GUESS THE EXPLICIT EQUATION,
BUT LETS DERIVE IT USING THE RECURSIVE FORMULA.
SO AGAIN, WE SAW THESE FORMULAS HERE ON THE PREVIOUS SLIDE.
SO WE CAN SAY THAT P SUB 1
WOULD BE EQUAL TO 1 FACTOR OF 1.05 x 4,000.
AND NOW FOR P SUB TWO, WE'LL LEAVE P SUB ONE AS THIS PRODUCT.
SO WE HAVE 1.05 x THIS IS P SUB 1.
NOTICE HOW FOR P SUB 2 WE ACTUALLY HAVE 2 FACTORS OF 1.05.
SO THIS IS P SUB 2 AND SINCE P SUB 3 = 1.05 x P SUB 2,
IF WE LEAVE P SUB 2 AS THIS PRODUCT,
NOTICE HOW TO FIND P SUB 3
WE ACTUALLY HAVE 3 FACTORS OF 1.05 x 4,000.
SO THIS IS P SUB 3, SO P SUB 4 = 1.05 x P SUB 3,
WHICH IS ALSO HERE.
SO NOTICE HOW TO FIND P SUB 4,
WE ACTUALLY HAVE 4 FACTORS OF 1.05 x 4,000
AND NOW YOU CAN PROBABLY SEE THE PATTERN.
WE CAN SAY THAT P SUB N = 1.05
RAISED TO THE POWER OF N x 4,000,
OR IF WE CHANGE THE ORDER OF THIS PRODUCT,
WE CAN WRITE P SUB N AS P SUB N = 4,000 x 1.05
TO THE POWER OF N.
THIS IS AN EXPLICIT EXPONENTIAL GROWTH FUNCTION.
IN OUR EXAMPLE, THE NUMBER OF STUDENTS GREW
BY THE SAME PERCENT EACH YEAR.
THIS CONSTANT PERCENT CHANGE IS A DEFINING CHARACTERISTIC
OF EXPONENTIAL OR GEOMETRIC GROWTH.
IF WE PLOT THE VALUES WE CALCULATED
FOR STUDENT POPULATION,
WE CAN SEE THE VALUES ARE NOT GROWING LINEARLY.
IT FORMS WHAT'S CALLED AN EXPONENTIAL FUNCTION.
AGAIN, WE CANNOT SKETCH A STRAIGHT LINE
PASSING THROUGH THESE FIVE POINTS.
THIS GRAPH DOES LOOK ALMOST LINEAR,
BUT IF WE CHANGE THE SCALING ON THE AXIS
IT BECOMES MUCH MORE OBVIOUS
THIS IS NOT A LINEAR RELATIONSHIP,
WHICH WE SEE HERE FROM THIS GRAPH.
THIS GRAPH ILLUSTRATES EXPONENTIAL GROWTH.
SO TO SUMMARIZE, IF A QUANTITY STARTS AT P SUB ZERO
AND GROWS BY CAPITAL R PERCENT WRITTEN AS A DECIMAL
AS LOWER CASE R EVERY TIME PERIOD,
THEN THE QUANTITY AFTER N TIME PERIODS CAN BE DETERMINED
USING EITHER OF THESE TWO RELATIONS.
THE RECURSIVE RELATIONSHIP WOULD BE P SUB N
EQUALS THE QUANTITY ONE + R x P SUB N - ONE.
THE EXPLICIT EQUATION FOR EXPONENTIAL GROWTH
WOULD BE P SUB N = P SUB ZERO x THE QUANTITY ONE
+ R RAISED TO THE POWER OF N,
WHERE AGAIN R IS THE GROWTH RATE AND ONE + R
IS CALLED THE GROWTH MULTIPLIER.
THESE TWO FORMULAS AT THE BOTTOM ALLOW US TO CALCULATE R
IN THE QUANTITY R + ONE WHEN NEEDED.
AND WE'LL USE THESE MORE IN PART TWO OF OUR LESSON.
THIS FORMULA HERE IS JUST THE PERCENT CHANGE FORMULA.
BEFORE WE END OUR LESSON THOUGH,
LET'S GO BACK TO OUR FIRST QUESTION.
IF THERE ARE 4,000 STUDENTS
ATTENDING A LOCAL COMMUNITY COLLEGE
AND THE NUMBER OF STUDENTS IS INCREASING BY 5% EACH YEAR,
LET'S USE THE EXPLICIT FORMULA THIS TIME
TO FIND THE NUMBER OF STUDENTS AFTER FOUR YEARS.
AGAIN, THE EXPLICIT FORMULA WOULD BE
P SUB N = P SUB 0 = 4,000 x THE QUANTITY 1 + THE GROWTH RATE
EXPRESSED AS A DECIMAL,
WHICH WOULD BE 0.05 RAISED TO THE POWER OF N.
LET'S GO AHEAD AND SIMPLIFY THIS.
THIS WOULD JUST BE 1.05 RAISED TO THE POWER OF N.
AND THEREFORE, TO FIND THE STUDENT POPULATION AFTER 4 YEARS
WE WANT P SUB 4, WHICH WOULD JUST BE 4,000 x 1.05
RAISED TO THE 4th POWER.
SO WE'D HAVE 4,000 x 1.05 RAISED TO THE 4th.
THIS KEY RIGHT HERE IS THE EXPONENT KEY.
NOTICE HOW THIS IS THE SAME VALUE
THAT WE FOUND USING THE RECURSIVE FORMULA.
THERE WOULD BE 4,862 STUDENTS AFTER 4 YEARS.
LET'S TAKE A LOOK AT ONE MORE QUESTION
RELATING TO THIS PROBLEM.
LET'S DETERMINE WHEN THE STUDENT POPULATION WILL REACH 8,000.
STARTING WITH OUR EXPONENTIAL GROWTH EQUATION,
P SUB N = 4,000 x 1.05 RAISED TO THE POWER OF N.
WE WANT TO KNOW THE VALUE OF N WHEN P SUB N = 8,000,
MEANING WE WANT TO SOLVE THE EQUATION
8,000 = 4,000 x 1.05 RAISED TO THE POWER OF N.
NOW THERE ARE TWO WAYS TO SOLVE THIS EQUATION.
WE CAN SOLVE IT USING LOGARITHMS OR WE CAN SOLVE IT GRAPHICALLY.
IN THIS LESSON, LET'S TAKE A LOOK AT IT GRAPHICALLY.
TO SOLVE IT GRAPHICALLY USING THE INTERSECTION METHOD,
WE WOULD GRAPH Y = 8,000 OR I'LL SAY Y SUB 1 = 8,000
AND Y = 4,000 x 1.05 RAISED TO THE POWER OF N
ON THE SAME COORDINATE PLANE.
SO I'LL CALL THIS Y SUB TWO.
IF WE GRAPH BOTH OF THESE FUNCTIONS
ON THE SAME COORDINATE PLANE,
THE N COORDINATE OR X COORDINATE
WOULD REPRESENT THE SOLUTION TO THIS EQUATION.
LOOKING AT OUR GRAPH, HERE IS WHERE Y = 8,000
AND HERE IS WHERE Y = 4,000 x 1.05
RAISED TO THE POWER OF N.
AND THEREFORE, THIS POINT OF INTERSECTION
REPRESENTS THE SOLUTION TO OUR EQUATION,
WHERE THE FIRST COORDINATE WOULD BE THE VALUE OF N
AND THE SECOND COORDINATE
WOULD BE THE FUNCTION VALUE OR Y VALUE.
SO USING OUR GRAPH WE CAN DETERMINE THE STUDENT POPULATION
WOULD REACH 8,000 IN APPROXIMATELY 14.2 YEARS.
WE'RE GOING TO STOP HERE FOR THIS LESSON
AND WE'LL CONTINUE IN PART TWO.
I HOPE YOU FOUND THIS HELPFUL.