– 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.

## 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