"A 0.6ml dose of a drug is injected into a patient steadily forhalf a second. At the end of this time, the quantity, Q, ofthe drug in the body starts to decay exponentially at a continuousrate of 0.2% per second. Using formulas, express Q as acontinuous function of time, t in seconds."

"A 0.6ml dose of a drug is injected into a patient steadily forhalf a second. At the end of this time, the quantity, Q, ofthe drug in the body starts to decay exponentially at a continuousrate of 0.2% per second. Using formulas, express Q as acontinuous function of time, t in seconds."

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asked 2021-02-10
"A 0.6ml dose of a drug is injected into a patient steadily forhalf a second. At the end of this time, the quantity, Q, ofthe drug in the body starts to decay exponentially at a continuousrate of 0.2% per second. Using formulas, express Q as acontinuous function of time, t in seconds."

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2021-02-12
Question:
"A 0.6ml dose of a drug is injected into a patient steadily for half a second. At the end of this time, the quantity, Q, of the drug in the body starts to decay exponentially at a continuous rate of 0.2% per second. Using formulas, express Q as a continuous function of time, t inseconds."
1.PIECEWISE FUNCTION IS NEEDED BECAUSE WHAT HAPPENS DURINGINJECTION OF DRUG IS TOTALLY DIFFERENT FROM WHAT HAPPENS WHEN ITDECAYS.
FIRST MAKE A TABLE LIKE THIS
T..............Q..................RATE
0...............0..............
0.5...........0.6...........DURING INJECTION =0.6/0.5=1.2ML/SEC...HENCE
---------------------Q=1.2T DURING T=0 TO 0.5 SEC.
------------- SUBSEQUENT TIME...THAT IS FORT>0.5
.............=AE^B(T-0.5).....SINCE T IS TIME FROM BEGINING ANDDECAY STARTS AFTER INJECTION ..THAT IS AFTER 0.5 SECS ANDEXPONENTIAL FUNCION HOLDS IN THAT PERIOD ,WE PUT (T-0.5)
NOW FOR A START AT T=0.5...Q=0.6...HENCE
\(\displaystyle{Q}={A}{E}^{{B}}{\left({0.5}-{0.5}\right)}={A}={0.6}\)...GOT IT....
NOW USE THE OTHER DATA OF 0.2%/SEC TO FIND B..HOPE YOU CAN CONTINUEWITH THIS APPROACH
DQ/DT IS RATE OF DECAY....100*(DQ/DT)/Q IS % RATE OF DECAY.
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