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 Q(0) = 238*(0.8)^(0/4) = 238 milligrams.
(b) Q is decreasing every 4 hours by a factor of 0.8, which is a 20% decrease. This can be seen by noticing that Q(t+4) = 238*(0.8)^((t+4)/4) = 238*(0.8)^(t/4)*0.8 = Q(t)*0.8.
(c) Q is decreasing every 2 hours by a factor of (0.8)^(1/2), which is a 10% decrease. This can be seen by noticing that Q(t+2) = 238*(0.8)^((t+2)/4) = 238*(0.8)^(t/4)*(0.8)^(1/2) = Q(t)*(0.8)^(1/2).
(d) ∆Q/∆t from t=2 to t=2+1/6 is [(Q(2+1/6)-Q(2))/((2+1/6)-2)]/1000. Evaluating Q(2+1/6) and Q(2) gives ∆Q/∆t ≈ -2.57 milligrams per hour squared. The units of this number are milligrams per hour squared, and it represents the acceleration of the decrease in the medication quantity over the given time interval.
(e) No, the number from part (d) is not a good estimate of Q'(2). Q'(2) represents the instantaneous rate of change of Q at t=2, and cannot be accurately estimated from the average rate of change over a finite time interval (which is what ∆Q/∆t represents).
(a) To find the initial dose of the medication, we need to evaluate the formula when t = 0. Plugging in t = 0 into the formula Q = 238*(0.8)^(t/4), we have:
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) The percentage by which Q is decreasing every 4 hours can be found by comparing the value of Q at t = 4 to the value of Q at t = 0. Let's calculate Q at these two times:
For t = 0:
Q = 238*(0.8)^(0/4)
Q = 238*1
Q = 238
For t = 4:
Q = 238*(0.8)^(4/4)
Q = 238*(0.8)^1
Q = 238*0.8
Q = 190.4
The decrease in Q is given by:
Decrease = Initial Value - Final Value
Decrease = 238 - 190.4
Decrease = 47.6
The percentage decrease is found by dividing the decrease by the initial value and multiplying by 100:
Percentage Decrease = (Decrease / Initial Value) * 100
Percentage Decrease = (47.6 / 238) * 100
Percentage Decrease = 0.2 * 100
Percentage Decrease = 20%
Therefore, Q is decreasing by 20% every 4 hours.
(c) To find the percentage by which Q is decreasing every 2 hours, we can compare the value of Q at t = 2 to the value of Q at t = 0. Let's calculate Q at these two times:
For t = 0:
Q = 238*(0.8)^(0/4)
Q = 238*1
Q = 238
For t = 2:
Q = 238*(0.8)^(2/4)
Q = 238*(0.8)^0.5
Q ≈ 238*0.8944
Q ≈ 212.89
The decrease in Q is given by:
Decrease = Initial Value - Final Value
Decrease = 238 - 212.89
Decrease ≈ 25.11
The percentage decrease is found by dividing the decrease by the initial value and multiplying by 100:
Percentage Decrease = (Decrease / Initial Value) * 100
Percentage Decrease = (25.11 / 238) * 100
Percentage Decrease ≈ 0.1054 * 100
Percentage Decrease ≈ 10.54%
Therefore, Q is decreasing by approximately 10.54% every 2 hours.
(d) To calculate ∆Q/∆t from t = 2 to t = 2+1/6, we need to find the rate of change of Q with respect to t within this interval. The formula for Q is Q = 238*(0.8)^(t/4). We can differentiate this formula with respect to t to find the rate of change:
Q' = (dQ / dt) = (238/4)*(0.8)^(t/4 - 1)
Let's calculate Q' at t = 2:
Q' = (238/4)*(0.8)^(2/4 - 1)
Q' = (238/4)*(0.8)^(1/2 - 1)
Q' = (238/4)*(0.8)^(-1/2)
Q' ≈ (238/4)*1.4142
Q' ≈ 273.70
∆Q/∆t represents the change in Q per unit change in t. In this case, from t = 2 to t = 2+1/6, the change in t is 1/6. Therefore:
∆Q/∆t = (Q at t = 2 + 1/6) - (Q at t = 2) / (1/6)
∆Q/∆t = (Q at t = 2 + 1/6) - (Q at t = 2) * (6/1)
Note that the units of ∆Q/∆t are milligrams per hour, as Q is in milligrams and t is in hours.
This number tells us the average rate of change of Q over the interval from t = 2 to t = 2+1/6. It represents the average amount by which Q changes per hour within this interval.
(e) Without further calculation, it is difficult to determine if the number from the previous part is a good estimate of Q'(2). This is because the value of ∆Q/∆t obtained represents the average rate of change of Q over the interval from t = 2 to t = 2+1/6, whereas Q'(2) represents the instantaneous rate of change of Q at t = 2. If the function Q is changing at a constant rate over this interval, then the number obtained may be a good estimate of Q'(2). However, if the function Q is not changing at a constant rate, then the estimate may not be accurate.