# Ex: Logistic Growth Differential Equation

– 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.

## 2 thoughts on “Ex: Logistic Growth Differential Equation”

• #### Jo Jon says:

Thank you so much for these videos, keep doing this because you are great at it. You really deserve waaaaay more views though….

• #### Mordecai Israel says:

The answer should be 16.3 years