- IF A QUANTITY GROWS AT A RATE PROPORTIONAL TO IT'S SIZE
AND TO THE DISTANCE FROM THE MAXIMUM VALUE M
IT CAN BE MODELED WITH LOGISTIC GROWTH
USING THE DIFFERENTIAL EQUATION Y PRIME = R x Y
x THE QUANTITY 1 - Y DIVIDED BY M
WHERE WHY IS THE GROWTH RATE ABSENT CONSTRAINTS
AND M IS THE MAXIMUM SIZE OF Y.
A DIFFERENTIAL EQUATION IN THIS FORM HAS SOLUTIONS
OF THE FORM Y = M DIVIDED BY THE QUANTITY 1 + A x E
RAISE TO THE POWER OF -RT.
LET'S LOOK AT AN EXAMPLE.
BIOLOGISTS STOCK THE LAKE WITH 400 FISH
AND ESTIMATED THE CARRYING CAPACITY TO BE 8,700.
THE NUMBER OF FISH DOUBLED IN THE FIRST YEAR.
ASSUMING THAT THE SIZE OF THE FISH POPULATION
SATISFIES THE LOGISTIC EQUATION
DETERMINE THE CONSTANT K WHICH IN OUR PREVIOUS NOTES WAS R
AND THEN P OF T TO FIND AN EQUATION
FOR THE SIZE OF THE POPULATION AFTER T YEARS.
SO NOTICE HOW HERE THIS PROBLEM IS USING
SOME DIFFERENT VARIABLES
BUT THE REPRESENT THE SAME INFORMATION
AS THE PREVIOUS SLIDE.
NOTICE HERE INSTEAD OF Y PRIME WE'RE USING DPDT.
INSTEAD OF R WE'RE USING K.
INSTEAD OF Y WE'RE USING P.
BUT BECAUSE WE HAVE LOGISTIC GROWTH
WE KNOW P OF T FITS THIS FORM HERE
WHERE M IS THE MAXIMUM POPULATION OF 8,700
WHICH WE CAN SUBSTITUTE INTO THE EQUATION FOR P OF T.
BUT WE'RE ALSO TOLD THE LAKE WAS STOCKED WITH 400 FISH
WHICH GIVES US THE INITIAL CONDITION P OF 0 = 400.
AND WE'RE ALSO TOLD THAT THE FISH POPULATION DOUBLED
IN THE FIRST YEAR
WHICH MEANS P OF 1 WOULD BE = 400 x 2 OR 800.
SO OUR GOAL HERE IS TO FIND K AND THEN P OF T.
SO FIRST WE'LL FIND "A" USING THE FACT THAT P OF 0 = 400.
AND THEN WE'LL FIND K USING P OF 1 = 800.
BECAUSE THERE'S SO MUCH WORK TO DO I'VE ALREADY SET THIS UP
SO LET'S GO THROUGH IT.
SO HERE WE HAVE P OF T WITH A MAXIMUM POPULATION OF 8,700.
AND WE USE THE INITIAL CONDITION P OF 0 = 400 TO FIND "A".
WHICH MEANS I'LL SUBSTITUTE 400 FOR P OF T AND 0 FOR T.
THIS WOULD GIVE US THE EQUATION 400 = 8,700 DIVIDED BY 1 + "A"
x E RAISE TO THE POWER OF 0 SINCE T IS 0.
E TO THE 0 = 1.
SO TO CLEAR THIS FRACTION WE WOULD MULTIPLY BOTH SIDES
BY THE QUANTITY "A" + 1 WHICH SEE HERE.
NOTICE HOW THE FRACTION IS GONE ON THE RIGHT
AND WE HAVE THIS EXTRA FACTOR OF 1 + "A" ON THE LEFT.
SO TO SOLVE FOR "A" NOTICE HOW WE WOULD FIRST DIVIDE BY 400
AND THEN SUBTRACT 1 WHICH WE SEE HERE.
THIS GIVES US "A" = 20.75
AND WE SUBSTITUTE THAT VALUE IN FOR "A"
GIVING US THE NEWEST FORM OF P OF T.
AND NOW WE'LL USE THE SECOND INITIAL CONDITION TO FIND K.
SO HERE IS P OF T FROM THE PREVIOUS SLIDE
WITH A VALUE "A" IN THE EQUATION.
AND NOW WE KNOW THAT P OF 1 = 800.
SO NOW I'LL FORM ANOTHER EQUATION BY SUBSTITUTING 800
FOR P OF T AND NOW 1 FOR T.
THAT WOULD GIVE US 800 = 8,700 DIVIDED BY THE QUANTITY
1 + 20.75 x E RAISE TO THE POWER OF -K SINCE T IS 1.
SO THE NEXT STEP AGAIN WE'LL CLEAR THIS FRACTION
BY MULTIPLYING BOTH SIDES OF THE EQUATION BY THIS QUANTITY HERE.
SO ONCE AGAIN NOTICE HOW THE FRACTION IS GONE FROM THE RIGHT
AND WE HAVE THIS EXTRA FACTOR HERE ON THE LEFT.
OUR GOAL HERE IS TO SOLVE FOR EITHER THE POWER OF -K.
SO FOR THE FIRST STEP WE'LL DIVIDE BOTH SIDES BY 800
AS WE SEE HERE.
AND THEN FROM HERE WE'LL SUBTRACT 1 ON BOTH SIDES
AND THEN DIVIDE BY 20.75
WHICH WOULD GIVE US E TO THE POWER OF -K
EQUALS THIS QUANTITY HERE
WHICH I'VE ALREADY EVALUATED ON THE CALCULATOR.
IT'S APPROXIMATELY 0.475903.
AND NOW TO SOLVE THIS EQUATION FOR K
WE'LL HAVE TO USE LOG RHYTHMS.
SO HERE'S THE SAME EQUATION HERE.
WE'LL TAKE THE NATURAL LOG OF BOTH SIDES.
ONCE WE DO THIS WE CAN APPLY THE POWER PROPERTY OF LOG RHYTHMS
SO WE CAN TAKE THIS EXPONENT HERE AND MOVE IT TO THE FRONT
WHICH WOULD GIVE US THIS EQUATION HERE.
BUT NATURAL LOG E IS EQUAL TO 1.
SO TO SOLVE FOR K WE WOULD DIVIDE BOTH SIDES BY -1
GIVING US K AS APPROXIMATELY 0.742541
WHICH IS THE FIRST PART OF QUESTION "A".
AND THEN FOR P OF T WE SUBSTITUTE THE VALUE OF K
INTO THE EQUATION AS WE SEE HERE.
SO HERE'S K AND HERE'S P OF T.
AND NOW FOR PART B WE CAN USE P OF T
TO DETERMINE WHEN THE FISH POPULATION WILL REACH 4,350.
TO DO THIS WE'LL TAKE P OF T AND SUBSTITUTE 4,350 FOR P OF T.
AND OUR GOAL HERE IS TO SOLVE THIS EQUATION FOR T.
SO, AGAIN, THE FIRST STEP IS TO CLEAR THE FRACTION
BY MULTIPLYING BOTH SIDES BY THIS QUANTITY HERE
WHICH WE SEE IN THIS NEXT LINE.
AND NOW TO SOLVE FOR THE EXPONENTIAL NEXT STEP
DIVIDE BY 4,350 WHICH WE SEE HERE.
THIS IS ACTUALLY JUST 2.
BUT THEN TO SOLVE FOR THE EXPONENTIAL WE WOULD SUBTRACT 1
AND THEN DIVIDED BY 20.75 WHICH WE SEE ON THIS NEXT LINE.
THIS QUANTITY HERE COMES OUT TO APPROXIMATELY 0.048193.
AND, AGAIN, TO SOLVE THIS EXPONENTIAL
THE NEXT STEP WE'LL TAKE THE NATURAL LOG
ON BOTH SIDES OF THE EQUATION WHICH WE SEE HERE.
APPLY THE POWER PROPERTY LOG RHYTHM.
SO WE'LL TAKE THIS EXPONENT MOVE IT TO THE FRONT.
IT GIVES US THIS T TERM x NATURAL LOG 3 WHICH IS 1.
SO IF WE DIVIDE BOTH SIDES BY -0.742541
IT WOULD GIVE US T IS APPROXIMATELY 4.084.
SO WE CAN SAY THE POPULATION WILL INCREASE TO 4,350
IN APPROXIMATELY, LET'S SAY, 4.08 YEARS.
I HOPE YOU FOUND THIS HELPFUL.