- AT THE BEGINNING OF AN EXPERIMENT,
A SCIENTIST HAS 200 GRAMS OF RADIOACTIVE GOO,
AFTER 40 MINUTES HER SAMPLE HAS DECAYED TO 15.2 GRAMS.
WE WANT TO DETERMINE THE HALF LIFE OF THE GOO IN MINUTES,
WE WANT TO FIND A FORMULA FOR THE AMOUNT OF GOO
REMAINING AT ANY TIME T AND THEN ALSO DETERMINE
HOW MANY GRAMS OF GOO WILL REMAIN AFTER 71 MINUTES.
TO DETERMINE AN EXPONENTIAL DECAY MODEL
WE CAN USE THIS FORM OF AN EXPONENTIAL FUNCTION
OR THIS FORM HERE.
FOR THIS EXAMPLE WE'LL USE THE SECOND FORM
WHERE P OF T IS THE AMOUNT AFTER TIME T,
P SUB 0 IS A STARTING AMOUNT OR INITIAL AMOUNT,
K IS THE CONTINUOUS DECAY RATE EXPRESSED AS A DECIMAL
AND T IS THE TIME.
LOOKING AT THE GIVEN INFORMATION,
WE'RE TOLD THE SCIENTIST STARTS WITH 200 GRAMS
OF RADIOACTIVE GOO WHICH MEANS P OF 0 EQUALS 200
OR THE INITIAL AMOUNT EQUALS 200 THEREFORE P SUB 0 = 200
AND THEN WE'RE ALSO TOLD THAT AFTER 40 MINUTES THE SAMPLE
HAS DECAYED TO 15.2 GRAMS WHICH MEANS P OF 40 MUST EQUAL 15.2.
SO WE'LL USE THIS INFORMATION
TO DETERMINE OUR EXPONENTIAL FUNCTION IN THIS FORM HERE.
AGAIN, WE KNOW THAT P OF T
MUST EQUAL 200 x E RAISED TO THE POWER OF KT
BUT WE ALSO KNOW THAT P OF 40 = 15.2,
WHICH MEANS P OF 40 WOULD HAVE TO = 200 x E,
AGAIN T IS 40 SO THE EXPONENT NOW IS 40K
AND THIS MUST EQUAL 15.2.
SO NOW WE CAN SOLVE THIS EQUATION FOR K
TO DETERMINE OUR CONTINUOUS DECAY RATE
AND THEREFORE DETERMINE THE EXPONENTIAL FUNCTION
THAT MODELS THIS DECAY.
SO THE FIRST STEP HERE IS TO DIVIDE BOTH SIDES BY 200.
NOTICE HOW NOW ON THE LEFT SIDE
WE HAVE E TO THE 40K = THIS FRACTION HERE,
15.2 DIVIDED BY 200 AND NOW TO SOLVE FOR K
WE CAN TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION
AND THEN APPLY THE POWER PROPERTY OF LOGARITHMS HERE
SO WE CAN MOVE THIS 40K TO THE FRONT.
SO NOW WE HAVE 40K x NATURAL LOG E
BUT NATURAL LOG E IS EQUAL TO 1,
SO WE CAN LEAVE THAT OFF.
THIS MUST EQUAL NATURAL LOG OF OUR FRACTION 15.2
DIVIDED BY 200.
SO THE LAST STEP TO SOLVE FOR K IS TO DIVIDE BOTH SIDES BY 40.
AND NOW WE'LL GO THE CALCULATOR
TO GET A DECIMAL APPROXIMATION FOR K.
SO WE HAVE NATURAL LOG 15.2 DIVIDED BY 200,
WE'LL DIVIDE THIS BY 40.
WE DO WANT TO INCLUDE SEVERAL DECIMAL PLACES FOR K.
LET'S GO AHEAD AND ROUND THIS TO 6 DECIMAL PLACES.
SO WE'LL HAVE -.064426.
WHICH MEANS P OF T EQUALS THE INITIAL AMOUNT OR 200
x E RAISED TO THE POWER OF -.064426 x T
CAN BE USED TO DETERMINE THE AMOUNT OF GOO
AFTER TIME T IN MINUTES.
SO THIS FUNCTION HERE IS THE FORMULA
THAT WE'RE ASKED TO FIND FOR THIS SECOND QUESTION.
SO NOW WE'LL GO TO THE NEXT SLIDE
TO DETERMINE THE HALF LIFE OF THE GOO
AND ALSO HOW MUCH WOULD REMAIN AFTER 71 MINUTES.
TO DETERMINE THE HALF LIFE,
IF WE START WITH 200 GRAMS
WE WANT TO DETERMINE THE VALUE OF T
SO THAT THE AMOUNT REMAINING OR P OF T
IS EQUAL TO HALF OF 200 OR 100.
WHICH MEANS TO FIND THE HALF LIFE,
WE WANT TO SOLVE THE EQUATION
100 = 200 x E RAISED TO THE POWER OF -.064426 x T.
SO TO DO THIS, THE FIRST STEP IS TO DIVIDE BOTH SIDES BY 200.
NOTICE HOW THE LEFT SIDE
IS NOW JUST 1/2 = E RAISED TO THE POWER OF -.064426T
AND NOW WE'LL TAKE THE NATURAL LOG OF BOTH SIDES
OF THE EQUATION AND ON THE RIGHT SIDE
WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS
TO MOVE THIS TO THE FRONT.
SO NOW WE HAVE NATURAL LOG 1/2 = -.064426T x NATURAL LOG E
BUT WE CAN LEAVE THAT OFF
BECAUSE NATURAL LOG E IS EQUAL TO 1.
SO NOW TO SOLVE FOR T WE JUST DIVIDE BY THE COEFFICIENT OF T,
WHICH IS OUR CONTINUOUS DECAY RATE.
NOW BEFORE WE GO TO THE CALCULATOR
I DO WANT TO POINT OUT THAT THERE IS A SHORTCUT
FOR DETERMINING HALF LIFE
WHEN WE HAVE THE CONTINUOUS DECAY RATE
FROM THE EXPONENTIAL FUNCTION IN THIS FORM HERE.
NATURAL LOG 1/2 DIVIDED BY K WILL ALWAYS GIVE US
THE HALF LIFE.
SO WE HAVE NATURAL LOG
1/2 DIVIDED BY -.064426
SO THE HALF LIFE IS APPROXIMATELY 10.8 MINUTES.
SO THIS ACTUALLY ANSWERS THE FIRST QUESTION
ABOUT THE HALF LIFE,
NOW WE WANT TO DETERMINE HOW MUCH GOO
REMAINS AFTER 71 MINUTES.
SO WE NEED TO EVALUATE P OF 71
WHICH IS GOING TO BE EQUAL TO 200 x E TO THE -.064426 x 71.
AND NOW BACK TO THE CALCULATOR.
200 x E RAISED TO THE POWER OF -.064426 x 71.
SO IF WE ROUND TO THE NEAREST 10th
THERE WOULD BE APPROXIMATELY 2.1 GRAMS
REMAINING AFTER 71 MINUTES,
WHICH IS THE LAST QUESTION WE WANTED TO ANSWER.
I HOPE YOU FOUND THIS HELPFUL.