2. (10 marks) 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)?
(a) The initial dose of the medication is 238 milligrams, since Q(0) = 238*(0.8)^(0/4) = 238 * 1 = 238.
(b) To find the percentage by which Q is decreasing every 4 hours, we need to find Q(4) and compare it to Q(0).
Q(4) = 238*(0.8)^(4/4) = 238*0.8 = 190.4
The decrease in Q is therefore 238 - 190.4 = 47.6 milligrams.
To find the percentage by which Q is decreasing, we divide this decrease by the initial dose and multiply by 100:
(47.6/238) * 100% = 20%
Therefore, Q is decreasing by 20% every 4 hours.
(c) To find the percentage by which Q is decreasing every 2 hours, we need to find Q(2) and Q(4) and compare them.
Q(2) = 238*(0.8)^(2/4) = 238*0.6325 = 150.83
Q(4) = 190.4 (from part (b))
The decrease in Q over 2 hours is therefore 190.4 - 150.83 = 39.57 milligrams.
To find the percentage by which Q is decreasing every 2 hours, we divide this decrease by Q(2) and multiply by 100:
(39.57/150.83) * 100% = 26.24%
Therefore, Q is decreasing by 26.24% every 2 hours.
(d) We need to calculate ∆Q/∆t from t = 2 to t = 2+1/6. This means we need to find the change in Q over that time interval, and divide by the length of the interval:
∆Q = Q(2+1/6) - Q(2) = 238*(0.8)^(13/6) - 238*(0.8)^(2/4) ≈ -11.2
(We know the decrease in Q will be negative, since Q is decreasing over time.)
The length of the interval is 1/6 hours, or 10 minutes.
∆t = 1/6 hours = 10 minutes
To find ∆Q/∆t, we divide the change in Q by the length of the interval:
∆Q/∆t = (-11.2) / (1/6) ≈ -67.2
The units of this number are milligrams per hour. This number tells us the rate of change of Q with respect to time at the point t = 2, measured in milligrams per hour. In other words, it tells us how fast Q is decreasing at that particular time.
(e) We know that Q(t) is a decreasing function of t, since Q is decreasing over time (from parts (b) and (c)). Therefore, we expect Q'(t) to be negative for all t. We also know that Q'(t) is the instantaneous rate of change of Q with respect to time, whereas ∆Q/∆t is the average rate of change of Q over an interval. Since Q is a decreasing function, we expect Q'(2) to be more negative (i.e. a larger magnitude) than ∆Q/∆t from t = 2 to t = 2+1/6. Therefore, ∆Q/∆t is not a good estimate of Q'(2).
(a) To find the initial dose of the medication, we need to substitute t = 0 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
Therefore, the initial dose of the medication is 238 milligrams.
(b) To determine the percentage by which Q is decreasing every 4 hours, we need to compare the values of Q at different time intervals.
Let's find Q after 4 hours:
Q = 238*(0.8)^(4/4)
Q = 238*(0.8)^1
Q = 238*0.8
Q = 190.4
Q is decreasing by (238 - 190.4) = 47.6 milligrams every 4 hours.
The percentage by which Q is decreasing can be calculated using the formula:
Percentage decrease = (Amount decreased / Initial amount) * 100
Percentage decrease = (47.6 / 238) * 100
Percentage decrease = 20%
Therefore, Q is decreasing by 20% every 4 hours.
(c) To find the percentage decrease in Q every 2 hours, we can use the same approach as in part (b).
Let's find Q after 2 hours:
Q = 238*(0.8)^(2/4)
Q = 238*(0.8)^0.5
Q = 238*0.89443
Q ≈ 212.89
Q is decreasing by (238 - 212.89) = 25.11 milligrams every 2 hours.
The percentage decrease can be calculated as:
Percentage decrease = (Amount decreased / Initial amount) * 100
Percentage decrease = (25.11 / 238) * 100
Percentage decrease ≈ 10.55%
Therefore, Q is decreasing by approximately 10.55% every 2 hours.
(d) To calculate ∆Q/∆t (the rate of change of Q with respect to t) from t = 2 to t = 2+1/6, we need to find the change in Q and the change in t over that interval.
Let's find Q at t = 2:
Q = 238*(0.8)^(2/4)
Q = 238*(0.8)^0.5
Q = 238*0.89443
Q ≈ 212.89
Now, let's find Q at t = 2 + 1/6:
Q = 238*(0.8)^((2+(1/6))/4)
Q = 238*(0.8)^(25/12)
Q ≈ 208.86
The change in Q (∆Q) over this interval is (208.86 - 212.89) ≈ -4.03 milligrams.
The change in t (∆t) is 1/6 hours.
∆Q/∆t = (∆Q) / (∆t)
∆Q/∆t = (-4.03) / (1/6)
∆Q/∆t ≈ -24.18
The units of ∆Q/∆t are milligrams per hour. This number represents the rate of change of Q (the quantity of medication) per hour over the specified interval.
(e) Without further calculation, it is difficult to determine if the number from part (d) is a good estimate of Q'(2) (the derivative of Q with respect to t at t = 2). To accurately estimate Q'(2), we need to calculate the derivative of the Q function and evaluate it at t = 2.