- WELCOME TO A VIDEO THAT WILL SHOW THE RELATIONSHIP
BETWEEN THE DIFFERENTIAL EQUATION DYDT = K x Y
AND THE EXPONENTIAL FUNCTION Y = C x E TO THE POWER KT.
IF THE RATE OF CHANGE OF Y WITH RESPECTS TO T
IS PROPORTIONAL TO Y
THEN THE GENERAL SOLUTION TO THIS DIFFERENTIAL EQUATION
WILL BE Y = C x E TO THE POWER OF KT.
AGAIN, LET'S TAKE A LOOK AT THAT AGAIN.
IF THE CHANGE IN Y WITH RESPECTS TO TIME
IS EQUAL TO SUM CONSTANT x Y
THEN THE SOLUTION IS Y = C x E TO THE POWER OF KT,
WHERE THE VALUE OF K HERE IS THE RATE OF GROWTH OR DECAY
BASED UPON WHETHER K IS POSITIVE OR NEGATIVE.
C WOULD BE THE INITIAL VALUE OF Y.
K IS THE CONSTANT IN PROPORTIONALITY.
IF K IS POSITIVE EXPONENTIAL GROWTH,
K IS NEGATIVE EXPONENTIAL DECAY.
NOW LETS TAKE A CLOSER LOOK AT THIS RELATIONSHIP.
IF WE WERE GIVEN THIS DIFFERENTIAL EQUATION
WHERE DYDT IS EQUAL TO SUM CONSTANT x Y,
LET'S SEE IF WE CAN SOLVE IT.
THE FIRST THING WE MIGHT TRY TO DO BASED
UPON THE STRATEGY OF SEPARATION OF VARIABLES
IS MULTIPLY BOTH SIDES OF THE EQUATION BY 1/Y.
WELL, IF WE DO THAT WE WOULD HAVE 1/Y x DYDT = 0.12.
NOW LET'S REWRITE THIS IN DIFFERENTIAL FORM.
SO WE HAVE 1/Y DY = 0.12DT.
NOW LET'S INTEGRATE BOTH SIDES OF THE EQUATION.
THE ANTI-DERIVATIVE OF 1/Y WITH RESPECTS TO Y
WOULD BE NATURAL LOG Y EQUALS THE ANTI-DERIVATIVE OF 0.12
WITH RESPECTS TO T WOULD BE 0.12T AND THEN + C.
THIS IS A LOG WOULD MAKE EQUATION.
IF WE WANTED TO REWRITE THIS AS AN EXPONENTIAL EQUATION
WE WOULD HAVE E RAISED TO THIS POWER, IT WOULD HAVE TO EQUAL Y.
AND LET'S GO AHEAD AND DO THAT.
E RAISED TO THE POWER OF 0.12T + C = Y.
NOW LET'S TAKE THIS UP HERE
AND WORK WITH THIS A LITTLE BIT MORE.
NOW LET'S TAKE A LOOK AT THIS RIGHT SIDE OF THE EQUATION.
WE COULD REWRITE THIS AS Y = E TO THE 0.12T x E TO THE C.
REMEMBER WHEN YOU'RE MULTIPLYING AND THE BASES ARE THE SAME,
YOU ADD YOUR EXPONENTS.
LET'S CALL THIS SMALL C.
E TO THE SMALL C IS EQUAL TO SUM CONSTANT.
SO LET'S GO AHEAD AND LET BIG C EQUAL E TO THE SMALL C
BECAUSE THEY'RE BOTH CONSTANTS.
SO WHAT WE HAVE LEFT IS Y = C x E TO THE POWER OF 0.12T.
SO WHAT WE'VE SHOWN IS IF WE TAKE THE TIME
TO SOLVE THIS DIFFERENTIAL EQUATION IN THIS FORM
WE DO GET AN EXPONENTIAL FUNCTION
WHERE THIS VALUE OF K IS THE RATE OF GROWTH, OR DECAY,
IN THIS EXPONENTIAL FUNCTION.
SO LET'S SAY THAT AGAIN.
THIS EQUATION REPRESENTS EXPONENTIAL GROWTH.
FOR A BUSINESS VIEW THIS EQUATION MODELS
CONTINUOUS INTEREST,
WHERE 0.12 REPRESENTS THE RATE OF CONTINUOUS INTEREST,
T IS THE TIME IN YEARS, AND C WOULD BE THE INITIAL INVESTMENT.
SO IF YOU INVEST $2,500 AT 12% CONTINUOUS INTEREST,
THIS EQUATION WOULD MODEL THE BALANCE OF THE ACCOUNT.
SO IF WE WANT TO KNOW HOW MUCH MONEY
WOULD BE IN THE ACCOUNT AFTER FOUR YEARS,
WE WOULD HAVE Y = 2,500 x E TO THE POWER OF 0.12 x T = 4
AND Y WOULD BE THE BALANCE OF THE ACCOUNT.
SO WE'D HAVE 2,500 SECOND NATURAL LOG BRINGS UP
E 0.12 x 4, THE BALANCE OF THE ACCOUNT WOULD BE $4,040
AND ROUGHLY $.19.
LET'S TAKE A LOOK AT THE GRAPH OF THAT.
HERE WE SEE THE GRAPH OF THE SLOPE FIELD
FROM THE DIFFERENTIAL EQUATION
AND THEN IN BLUE WE HAVE THE PARTICULAR SOLUTION
WHEN THE INITIAL INVESTMENT AMOUNT WAS $2,500.
IF WE INVEST FOR FOUR YEARS,
WE GO OUT TO X EQUALS FOUR AND THEN GO UP TO THE FUNCTION,
HERE'S THAT VALUE THAT WE FOUND OF $4,040.
LOOKS LIKE THE POINT AT THREE YEARS IS ALSO A GRAPH,
BUT WE FOUND THE VALUE AT FOUR YEARS.
LET'S TAKE A LOOK AT ONE MORE EXAMPLE.
CARBON 14 HAS A DECAY RATE OF 0.01205% PER YEAR.
THE RATE OF CHANGE OF ANY AMOUNT N OF CARBON 14
IS GIVEN BY THIS DIFFERENTIAL EQUATION.
NOW THAT WE KNOW THIS MATCHES THE FORM WHERE THE SOLUTION
WOULD BE AN EXPONENTIAL EQUATION WHERE THIS IS K,
WE DON'T HAVE TO TAKE THE TIME TO SOLVE
THIS DIFFERENTIAL EQUATION.
WE KNOW THAT THE FUNCTION WOULD HAVE TO BE,
IN THIS CASE, N OF T.
IT'S GOING TO BE EQUAL
TO SUM INITIAL VALUE x E TO THE POWER OF KT.
SO HERE'S THE VALUE OF K x T, WHICH WOULD BE IN YEARS.
SO THAT'S WHAT'S NICE ABOUT RECOGNIZING
WHEN YOU HAVE A DIFFERENTIAL EQUATION
EQUAL TO SUM CONSTANT x THE ORIGINAL FUNCTION,
WE KNOW IT WILL BE AN EXPONENTIAL FUNCTION
AND THIS REPRESENTS A VALUE OF K.
SO NOW WE CAN GO STRAIGHT TO SOLVING A PROBLEM.
SUPPOSE 200 GRAMS OF CARBON 14 IS PRESENT AT TIME EQUALS 0,
HOW MUCH REMAINS AFTER 800 YEARS?
WELL, IT'S QUITE STRAIGHTFORWARD NOW.
SO C WOULD EQUAL 200 AND T WOULD EQUAL 800.
AND WE CAN GO STRAIGHT TO THE CALCULATOR.
SO OF THE 200 GRAMS--SO OF THE INITIAL 200 GRAMS OF CARBON 14
THERE IS ROUGHLY 181.6 GRAMS LEFT.
NOW THE LAST QUESTION, AFTER HOW MANY YEARS
WILL HALF OF 200 GRAMS OF CARBON 14 REMAIN?
AND SO THAT WOULD BE THE HALF LIFE OF CARBON 14.
WELL, THIS IS AN ALGEBRA QUESTION.
IF WE START WITH 200 GRAMS THEN THE AMOUNT AFTER WOULD BE 100.
WE'D HAVE 100 = 200 E TO THE POWER OF -0.0001205 x T
AND WE HAVE TO SOLVE THIS FOR T.
SO WE'D DIVIDE BOTH SIDES BY 200,
SO NOTICE ON THE LEFT SIDE WE HAVE 0.5.
THAT'LL ALWAYS BE TRUE FOR HALF LIFE EQUALS E TO THIS EXPONENT.
AND NOW TO SOLVE FOR T,
WE'LL HAVE TO TAKE THE NATURAL LOG
OF BOTH SIDES OF THE EQUATION
AND NOW WE CAN APPLY THE POWER PROPERTY.
SO IF WE TAKE THIS EXPONENT, WE CAN MOVE IT TO THE FRONT.
SO WE'RE GOING TO HAVE NATURAL LOG 0.5 = -0.0001205T
x NATURAL LOG E, BUT NATURAL LOG = 1.
SO THE LAST STEP HERE IS TO DIVIDE BY THIS COEFFICIENT.
SO LET'S GO BACK TO THE CALCULATOR ONE MORE TIME,
WE'RE GOING TO HAVE NATURAL LOG 0.5 DIVIDED BY -.0001205
AND IT LOOKS LIKE IT'S ROUGHLY $5,752.3 YEARS.
OKAY, THAT'S GOING TO DO IT FOR THIS VIDEO.
HAVE A GOOD DAY.