The quantity Q, in milligrams, of medication in a patient is modelled by the formula Q=238*(0.8)^(t/4), where t is the number of hours since the initial dose.
(a) What was the initial dose of the medication?
(b) By what percentage is Q decreasing every 4 hours?
(c) By what percentage is Q decreasing every 2 hours?
(d) Calculate ∆Q/∆t from t = 2 to t = 2+1/6. What are the units of this number? What does this number tell you?
(e) Without further calculation, is the number from the previous part a good estimate of Q'(2)?
(f) Determine Q'(2) to 3 significant figures by recomputing ∆Q/∆t for shrinking values of h until the rate of change is not changing to 3 significant figures.
(a) The initial dose of the medication is equal to Q when t = 0, so Q(0) = 238*(0.8)^(0/4) = 238. Therefore, the initial dose is 238 milligrams.
(b) To find the percentage decrease in Q every 4 hours, we can use the formula:
% decrease = −100(Q(4) − Q(0))/Q(0)
where Q(0) is the initial dose (238), Q(4) is the dose after 4 hours (when t=4), and the negative sign indicates a decrease. Plugging in these values, we get:
% decrease = −100(238*(0.8)^(4/4) − 238)/238
% decrease = 36.5%
Therefore, Q is decreasing by 36.5% every 4 hours.
(c) To find the percentage decrease in Q every 2 hours, we can use the same formula, but with Q(2) as the final dose after 2 hours:
% decrease = −100(Q(2) − Q(0))/Q(0)
% decrease = −100(238*(0.8)^(2/4) − 238)/238
% decrease = 21.3%
Therefore, Q is decreasing by 21.3% every 2 hours.
(d) To calculate ∆Q/∆t from t = 2 to t = 2+1/6, we can use the formula:
∆Q/∆t = [Q(2+1/6) − Q(2)]/[(2+1/6) − 2]
∆Q/∆t = [238*(0.8)^(2+1/6/4) − 238*(0.8)^(2/4)]/[1/6]
∆Q/∆t ≈ −12.77 mg/hour
The units of this number are milligrams per hour, and it tells us the rate at which Q is decreasing between hours 2 and 2+1/6.
(e) No, the number from part (d) is not a good estimate of Q'(2) because it is an average rate of change over a relatively large interval, whereas Q'(2) is the instantaneous rate of change at a specific point.
(f) To determine Q'(2) to 3 significant figures, we can use the formula:
Q'(2) = lim(h→0) [Q(2+h) − Q(2)]/h
We can compute this limit for smaller and smaller values of h until the rate of change is not changing to 3 significant figures. Using a calculator or spreadsheet, we can find:
h = 0.001: Q'(2) ≈ −12.722 mg/hour
h = 0.0001: Q'(2) ≈ −12.774 mg/hour
h = 0.00001: Q'(2) ≈ −12.778 mg/hour
h = 0.000001: Q'(2) ≈ −12.778 mg/hour
Therefore, Q'(2) ≈ −12.778 mg/hour to 3 significant figures. This means that at t = 2 hours, Q is decreasing at a rate of approximately 12.778 milligrams per hour.
(a) The initial dose of the medication is obtained by substituting t = 0 hours into the formula:
Q = 238 * (0.8)^(t/4)
Q = 238 * (0.8)^(0/4)
Q = 238 * (0.8)^0
Q = 238 * 1
Q = 238 milligrams
Therefore, the initial dose of the medication is 238 milligrams.
(b) To find the percentage by which Q is decreasing every 4 hours, we need to calculate the difference in Q over 4 hours:
ΔQ = Q(t+h) - Q(t)
ΔQ = 238 * (0.8)^((t+h)/4) - 238 * (0.8)^(t/4)
Then, we can find the percentage decrease by dividing ΔQ by Q(t) and multiplying by 100:
Percentage decrease = (ΔQ / Q(t)) * 100
(c) Similarly, to find the percentage by which Q is decreasing every 2 hours, we calculate the difference in Q over 2 hours:
ΔQ = Q(t+h) - Q(t)
ΔQ = 238 * (0.8)^((t+h)/4) - 238 * (0.8)^(t/4)
Then, we find the percentage decrease by dividing ΔQ by Q(t) and multiplying by 100:
Percentage decrease = (ΔQ / Q(t)) * 100
(d) To calculate ∆Q/∆t from t = 2 to t = 2+1/6, we can use the following formula:
∆Q/∆t = [Q(2+1/6) - Q(2)] / [(2+1/6) - 2]
(e) Without further calculation, we cannot determine whether the number from part (d) is a good estimate of Q'(2) since we do not know the value of ∆Q/∆t.
(f) To determine Q'(2) to 3 significant figures, we need to calculate ∆Q/∆t for shrinking values of h until the rate of change is no longer changing. However, since we do not have the specific function values, this cannot be done without further information or calculation.