- YOU GO TO THE DOCTOR AND HE GIVES YOU 12 MILLIGRAMS
OF RADIOACTIVE DYE.
AFTER 10 MINUTES, 5.5 MILLIGRAMS OF DYE REMAIN IN YOUR SYSTEM.
TO LEAVE THE DOCTOR'S OFFICE
YOU MUST PASS THROUGH A RADIATION DETECTOR
WITHOUT SOUNDING THE ALARM.
IF THE DETECTOR WILL SOUND THE ALARM
IF MORE THAN 2 MILLIGRAMS OF THE DYE ARE IN YOUR SYSTEM,
HOW LONG WILL YOUR VISIT TO THE DOCTOR
TAKE ASSUMING YOU WERE GIVEN THE DYE AS SOON AS YOU ARRIVED.
SO TO MODEL THIS EXPONENTIAL DECAY WE HAVE A CHOICE
OF USING THIS EXPONENTIAL FUNCTION
OR THIS EXPONENTIAL FUNCTION.
THIS FIRST EXPONENTIAL FUNCTION IS ACTUALLY EASIER TO USE
IF WE'RE GIVEN INFORMATION ABOUT THE HALF LIFE OR DOUBLING TIME
BUT SINCE WE'RE NOT WE'RE GOING TO GO AHEAD
AND USE THIS FUNCTION HERE WHERE P OF T
IS THE AMOUNT AFTER TIME T, P SUB 0 IS THE INITIAL AMOUNT,
K IS THE CONTINUOUS EXPONENTIAL DECAY RATE AND T IS THE TIME.
SO GOING BACK UP TO OUR PROBLEM,
IF YOU RECEIVED 12 MILLIGRAMS OF RADIOACTIVE DYE
WHEN YOU GO TO THE DOCTOR THAT MEANS YOU HAVE 12 MILLIGRAMS
OF DYE AT TIME T = 0, SO P OF 0 MUST EQUAL 12.
SO 12 IS GOING TO BE THE INITIAL AMOUNT OR THE VALUE OF P SUB 0
AND AFTER 10 MINUTES 5.5 MILLIGRAMS OF DYE REMAIN.
SO THIS MEANS P OF 10 MUST EQUAL 5.5.
AND SINCE 12 WAS THE INITIAL AMOUNT OR THE AMOUNT
WHEN T = 0 WE KNOW THAT P OF T MUST = 12 x E
RAISED TO THE POWER OF KT.
NOTICE HOW WHEN T IS EQUAL TO ZERO,
WE WOULD HAVE 12 x E TO THE 0 AND E TO THE 0 IS EQUAL TO 1,
SO THE FUNCTION VALUE IS 12.
AND NOW USING P OF 10 = 5.5
WE'RE GOING TO DETERMINE THE VALUE OF K
OR A CONTINUOUS EXPONENTIAL DECAY RATE.
SO P OF 10 IS GOING TO BE EQUAL TO 12 x E TO THE--
SINCE T = 10 THE EXPONENT IS GOING TO BE 10K
AND THIS MUST EQUAL 5.5.
AND NOW WE'LL SOLVE THIS EQUATION FOR K.
SO THE FIRST STEP IS TO ISOLATE THE EXPONENTIAL PART
SO WE'LL DIVIDE BOTH SIDES BY 12.
SO NOW WE HAVE E TO THE 10K = 5.5 DIVIDED BY 12.
AND NOW TO SOLVE FOR K
WE'LL TAKE THE NATURAL LOG OF BOTH SIDES OF THE EQUATION
APPLY THE POWER OF PROPERTY OF LOGARITHMS AND THEN SOLVE FOR K.
SO WE'RE GOING TO HAVE NATURAL LOG E TO THE 10K = NATURAL LOG
OF THIS FRACTION.
NOW ON THE LEFT SIDE OF THIS EQUATION
WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS,
WHICH MEANS WE CAN TAKE THE EXPONENT OF 10K
AND MOVE IT TO THE FRONT.
SO WE WOULD HAVE 10K x NATURAL LOG E EQUALS NATURAL LOG OF 5.5
DIVIDED BY 12.
BUT NATURAL LOG E IS EQUAL TO 1.
SO THIS SIMPLIFIES OUT SO TO SOLVE FOR K
WE DIVIDE BOTH SIDES BY 10.
SO NOW WE'LL GO TO THE CALCULATOR AND DETERMINE
THIS APPROXIMATE VALUE.
SO WE'LL HAVE NATURAL LOG OF 5.5 DIVIDED BY 12.
I'M GOING TO DIVIDE THIS BY 10.
NOTICE HOW THE VALUE IS NEGATIVE WHICH IS A GOOD SIGN
BECAUSE WE DO HAVE EXPONENTIAL DECAY.
IT'S ALSO IMPORTANT THAT SINCE THIS IS GOING TO EFFECT
THE EXPONENT WE WANT TO INCLUDE SEVERAL DECIMAL PLACES
I'M GOING TO ROUND THIS TO 6 DECIMAL PLACES.
SO K IS GOING TO BE APPROXIMATELY -.078016,
WHICH MEANS THE EXPONENTIAL FUNCTION
THAT WILL MODEL THE AMOUNT OF DYE IN YOUR SYSTEM
WILL BE P OF T = 12 x E TO THE POWER OF -.078016 x T.
AND NOW TO FINALLY ANSWER THIS QUESTION,
WE WANT TO KNOW WHEN THE AMOUNT OF DYE IN YOUR SYSTEM
WILL BE EQUAL TO 2 MILLIGRAMS
SO YOU CAN LEAVE THE DOCTOR'S OFFICE.
SO WE WANT TO SET P OF T WITH A 2 AND SOLVE FOR T.
SO WE WANT TO SOLVE THE EQUATION 2 = 12 x E
RAISED TO THE POWER OF -.078016 x T.
SO AGAIN, THE FIRST STEP IS TO ISOLATE THE EXPONENTIAL PART
SO WE'LL DIVIDE BOTH SIDES BY 12.
SO WE HAVE 1/6 = E RAISED TO THE POWER OF -.078016T
AND NOW WE'LL TAKE THE NATURAL LOG OF BOTH SIDES
OF THE EQUATION SO THAT WE CAN SOLVE FOR T.
SO WE'LL HAVE NATURAL LOG 1/6 = E TO THE -.078016T
AND NOW WE CAN APPLY THE POWER PROPERTY OF LOGARITHMS,
MEANING WE CAN TAKE THE EXPONENT HERE AND MOVE IT TO THE FRONT.
SO WE HAVE NATURAL LOG 1/6 = -.078016T x NATURAL LOG E.
AGAIN, MOST TEXTBOOKS DON'T SHOW THIS NATURAL LOG E
BECAUSE IT IS EQUAL TO 1.
SO TO SOLVE FOR T WE'LL JUST DIVIDE BY THE COEFFICIENT OF T.
SO NOW WE'LL GO BACK TO THE CALCULATOR ONE MORE TIME.
SO WE HAVE NATURAL LOG 1/6
DIVIDED BY THE CONTINUOUS EXPONENTIAL DECAY RATE.
REMEMBER THE TIME IS IN MINUTES
SO IF WE ROUND THIS TO THE NEAREST TENTH
THIS WOULD BE 23.0 MINUTES.
SO THE DOCTOR'S APPOINTMENT WILL TAKE APPROXIMATELY 23 MINUTES,
WHICH AS FAR AS DOCTOR'S APPOINTMENTS GO
THIS WOULD BE PRETTY AMAZING.