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Ex: Exponential Growth Function - Population - YouTube
Channel: Mathispower4u
[1]
- A GROWING CITY HAD A
POPULATION OF 500,000 IN 2005,
[5]
IN 2010 THE POPULATION
WAS 760,000.
[10]
WE WANT TO ASSUME
EXPONENTIAL GROWTH
[12]
AND THEN EXPRESS
THE POPULATION AFTER T YEARS
[15]
AS A FUNCTION OF T,
[17]
PREDICT THE POPULATION
IN 2025,
[20]
AND THEN DETERMINE
IN WHICH YEAR
[22]
THE POPULATION WILL REACH
1,000,000.
[25]
SO WE WANT TO START BY FINDING
THE EXPONENTIAL FUNCTION
[27]
THAT'S GOING TO MODEL
THIS POPULATION.
[29]
SO WE'LL BE USING THIS
EXPONENTIAL FUNCTION HERE
[32]
WHERE P OF T WOULD BE
THE POPULATION
[34]
AFTER A CERTAIN NUMBER
OF YEARS, P SUB 0, OR P NOD,
[38]
WOULD BE THE INITIAL
POPULATION,
[40]
K WOULD BE
THE EXPONENTIAL GROWTH RATE,
[42]
AND T WOULD BE THE TIME
IN YEARS.
[46]
SO LETS START BY FINDING
THE EXPONENTIAL FUNCTION
[48]
FOR THIS POPULATION.
[49]
SO WHEN WE READ THESE
FIRST TWO SENTENCES,
[52]
THE STARTING POPULATION
WOULD BE 500,000,
[55]
SO P SUB 0 OR P NOD IS = TO
500,000, THIS IS IN 2005.
[64]
AND THEN IN 2010
THE POPULATION WAS 760,000,
[68]
SO P OF T WOULD BE
= TO 760,000.
[75]
AND THEN FROM 205 TO 2010,
THAT'S A SPAN OF 5 YEARS,
[79]
SO T IS GOING TO BE = TO 5.
[82]
SO NOW WE'LL PERFORM
SUBSTITUTION
[84]
INTO OUR EXPONENTIAL FUNCTION
[86]
AND THEN SOLVE FOR K,
OUR EXPONENTIAL GROWTH RATE.
[89]
SO WE WANT TO SOLVE
THE EQUATION,
[91]
760,000 = 500,000 x E
RAISED TO THE POWER OF K x T,
[100]
BUT SINCE T IS 5,
WE'LL HAVE 5K.
[104]
NOW, WE WANT TO SOLVE THIS
EXPONENTIAL EQUATION FOR K,
[107]
SO WE'LL FIRST ISOLATE
THE EXPONENTIAL PART.
[109]
SO WE'LL DIVIDE BOTH SIDES BY
500,000, THIS SIMPLIFIES TO 1.
[116]
AND THEN ON THE LEFT SIDE
OF 760,000 DIVIDED BY 500,000,
[126]
WHICH IS = TO 1.52.
[133]
SO WE HAVE 1.52 WOULD = E
TO THE POWER OF 5K.
[138]
AND NOW SINCE WE HAVE BASE E
HERE,
[140]
INSTEAD OF TAKING
THE COMMON LOG OF BOTH SIDES,
[142]
WE'LL TAKE THE NATURAL LOG
OF BOTH SIDES.
[147]
AND NOW ON THE RIGHT SIDE
WE CAN APPLY
[148]
THE POWER PROPERTY
OF LOG RHYTHMS
[150]
TO MOVE THIS 5K TO THE FRONT.
[154]
SO NOW WE'D HAVE NATURAL LOG
1.52 = 5K x NATURAL LOG E,
[164]
BUT NATURAL LOG E IS = 1.
[167]
REMEMBER IF WE HAVE NATURAL
LOG E, THIS IS LOG BASE E,
[171]
AND SINCE E RAISED
TO THE 1ST POWER IS = TO E,
[177]
THIS IS = TO 1.
[179]
SO THIS SIMPLIFIES TO 1.
[181]
SO TO SOLVE THIS FOR K
[183]
WE JUST NEED TO DIVIDE
BOTH SIDES BY 5.
[187]
OF COURSE, WE COULD DIVIDE
BOTH SIDES BY NATURAL LOG E,
[190]
BUT AGAIN THAT'S JUST = TO 1,
[191]
SO IT'S NOT GOING
TO CHANGE ANYTHING.
[194]
SO WE HAVE K
IS = TO THIS QUOTIENT,
[197]
WHICH WE'LL HAVE TO GET
A DECIMAL APPROXIMATION FOR.
[203]
SO WE HAVE NATURAL LOG OF 1.52
DIVIDED BY 5
[211]
AND THE MORE DECIMAL PLACES
THAT WE USE FOR K,
[213]
THE MORE ACCURATE OUR ANSWER
IS GOING TO BE.
[215]
LET'S GO AHEAD AND TAKE THIS
OUT TO 6 DECIMAL PLACES,
[219]
SO WE'LL HAVE 0.083742.
[226]
AND NOW WE HAVE
OUR EXPONENTIAL FUNCTION
[228]
THAT'S GOING TO MODEL
THIS POPULATION.
[229]
WE'LL HAVE P OF T IS = TO THE
INITIAL POPULATION OF 500,000
[236]
x E RAISED TO THE POWER
OF 0.083742 x T,
[245]
WHERE T IS GOING TO BE
THE NUMBER OF YEARS
[246]
AFTER THE YEAR 2005.
[249]
SO LETS TAKE THIS FUNCTION
BACK TO THE PREVIOUS SCREEN.
[253]
JUST KEEP IN MIND THAT T
IS THE NUMBER OF YEARS
[255]
AFTER OUR BASE YEAR OF 2005.
[258]
SO TO PREDICT THE POPULATION
IN 2025,
[261]
WELL 2025 - THE BASE YEAR
OF 2005 WOULD BE 20.
[269]
SO WE WANT TO FIND P OF 20
TO APPROXIMATE THE POPULATION
[273]
IN THE YEAR 2025.
[289]
SO NOW WE'LL GO BACK
TO THE CALCULATOR
[292]
AND APPROXIMATE THIS VALUE.
[297]
SECOND NATURAL LOG BRINGS UP E
[299]
RAISED TO THE POWER
OF .083742 x 20,
[309]
THIS WOULD BE OUR EXPONENT
ON E,
[312]
AND SO THE POPULATION IS GOING
TO BE APPROXIMATELY 2,668,971
[319]
IF WE ROUND
TO THE NEAREST PERSON.
[326]
NOW, FOR THE LAST QUESTION
THEY WANT TO KNOW
[327]
WHAT YEAR WILL THE POPULATION
REACH 1,000,000,
[330]
SO WE WANT TO SUBSTITUTE
1,000,000 FOR P OF T
[333]
AND SOLVE FOR T.
[335]
SO WE'LL HAVE
1,000,000 = 500,000,
[343]
E RAISED TO THE POWER
OF 0.083742T.
[350]
SO WE'RE GOING TO ISOLATE
THE EXPONENTIAL PART.
[352]
SO WE'LL DIVIDE BOTH SIDES
BY 500,000,
[355]
THIS SIMPLIFIES TO ONE,
THIS WOULD BE 2,
[358]
SO WE HAVE 2 = E
RAISED TO THIS EXPONENT.
[363]
AND JUST AS WE DID BEFORE,
[364]
WE'LL NOW TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION
[368]
AND THEN APPLY THE POWER
OF PROPERTY OF LOG RHYTHMS.
[370]
SO WE'LL MOVE THIS EXPONENT TO
THE FRONT OF THE LOG RHYTHM,
[377]
SO WE'LL HAVE NATURAL LOG 2
= 0.083472T x NATURAL LOG E,
[387]
BUT AS WE SHOWED BEFORE
NATURAL LOG E SIMPLIFIES TO 1.
[390]
SO SOLVE THIS FOR T,
[392]
WE JUST NEED TO DIVIDE
BY THIS DECIMAL COEFFICIENT.
[398]
AGAIN, THIS SIMPLIFIES TO 1
[401]
SO LET'S GO BACK
TO THE CALCULATOR.
[403]
AND THIS QUOTIENT WOULD BE
THE VALUE OF T,
[405]
WHICH WOULD BE THE NUMBER
OF YEARS AFTER 2005.
[421]
SO WE CAN SEE THAT T
IS APPROXIMATELY 8.28 YEARS.
[436]
SO THIS WOULD BE 8.28 YEARS
AFTER THE YEAR 2005,
[441]
AND SINCE THE BASE YEAR
IS THE YEAR 2005,
[444]
2005 + 8.28 MEANS
[447]
THAT THE POPULATION
WOULD REACH 1,000,000
[450]
DURING THE YEAR 2013.
[460]
OKAY, I HOPE YOU FOUND
THIS HELPFUL.
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