Let Q(t) = the mass (mg) remaining after t hours.
We are told that Q diminishes by 3% every hour.
When t = 0, Q = 4 mg
When t = 1, Q = 4*0.97 mg
When t = 2, Q = 4*(0.97)² mg
By induction,
[tex]Q(t) = 4 (0.97)^{t} \, mg[/tex]
Q'(t) = 0.97Q
Therefore the rate of decrease is 3% per hour.
The person receives an additional dosage when Q falls to 0.50 mg. This happens when
[tex]4(0.97)^{t}=0.5 \\0.97^{t} = 0.125 \\ t \,ln(0.97) = ln(0.125) \\ t = \frac{ln(0.125)}{ln(0.97)} =68.27 \, hrs[/tex]
After the second injection, the mass is now 4.5 mg. Therefore
[tex]Q(t)=4.5(0.97)^{t}[/tex]
When the mass again reaches 0.50 mg, then
[tex]4.5(0.97)^{t}=0.5 \\ 0.97^{t}=0.1111 \\ t \, ln(0.97) = ln(0.1111) \\ t = 72.14 \, hrs[/tex]
Answers:
(a) [tex]Q(t) = 4(0.97)^{t}[/tex]
(b) 3% per hour
(c) 68.3 hours
(d) 72.1 hours
