A 0.2 ml dose of a drug is injected into a patient steadily for 0.75 seconds. 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 percent per second. Using formulas, express Q as a continuous function of time, t, in seconds
Q(t) = ______ if 0 is less than or equal to t less than or equal to ______
and Q(t) = ______ if _____ less than or equal to t less than or equal to Infinity
Q(t) = 0.2(1 - e^(-0.002t)) for 0 ≤ t ≤ 0.75
Q(t) = 0.2e^(-0.002(t - 0.75)) for t > 0.75
To express Q as a continuous function of time, we can use the exponential decay formula:
Q(t) = Q₀ * e^(-kt)
Where:
- Q(t) is the quantity of the drug in the body at time t
- Q₀ is the initial quantity of the drug (0.2 ml)
- k is the decay constant
First, we need to find the decay constant, k. Given that the drug decays at a continuous rate of 0.2 percent per second, we can convert this to a decimal value:
Decay rate = 0.2 percent = 0.002 (decimal)
Since the decay rate is equal to k, we have k = 0.002.
Now, let's find the time intervals for Q(t):
- For 0 ≤ t ≤ 0.75 seconds: The drug is being injected, so the quantity remains constant at 0.2 ml. Therefore:
Q(t) = 0.2 ml, for 0 ≤ t ≤ 0.75
- For t > 0.75 seconds: The drug starts decaying exponentially. Therefore:
Q(t) = Q₀ * e^(-0.002t), for t > 0.75
Combining both intervals, we have:
Q(t) = 0.2 ml, for 0 ≤ t ≤ 0.75
Q(t) = 0.2 * e^(-0.002t), for t > 0.75
To express Q as a continuous function of time, t, you need to consider two different time intervals: from 0 to 0.75 seconds and from 0.75 seconds to infinity.
First, let's consider the interval from 0 to 0.75 seconds. During this time, the drug is being injected into the patient steadily. The rate of injection is given to be 0.2 ml/0.75 s. So, for this interval, the quantity of the drug in the body is increasing linearly with time.
Q(t) = 0.2 mL / (0.75 s) * t, if 0 ≤ t ≤ 0.75 seconds
Now, let's consider the interval from 0.75 seconds to infinity. After the injection is done, the quantity of the drug in the body starts decaying exponentially at a continuous rate of 0.2 percent per second. To model this decay, we can use the concept of exponential decay.
The general formula for exponential decay is given by:
Q(t) = Q0 * e^(-kt),
where Q0 is the initial quantity of the drug, e is the base of the natural logarithm (approximately 2.71828), k is the decay constant, and t represents time.
In this case, the decay rate is given as 0.2 percent per second. To convert it into a decimal value, we divide by 100, giving us a decay constant of 0.002.
Therefore, for the interval from 0.75 seconds to infinity, the expression becomes:
Q(t) = Q0 * e^(-0.002t), if t > 0.75 seconds
However, since we do not have the initial quantity, Q0, defined in the problem, we cannot give a specific expression for Q(t) for the interval from 0.75 seconds to infinity without sufficient information.
So, to summarize:
Q(t) = 0.2 mL / (0.75 s) * t, if 0 ≤ t ≤ 0.75 seconds
Q(t) = Q0 * e^(-0.002t), if t > 0.75 seconds (assuming Q0 is known)
Q(t+1 second ) = Q(t) -.002Q(t)
so
dQ/dt = -.002 Q
dQ/Q = -.002 dt
lnQ = -.002 t + c'
Q = e^-.002 t + e^c'
Q = e^c^ e^-.002 t
or
Q = C e^-.002 t in ml
where C is Q at t = 0(start of decay period)
now
first period, t = 0 to t = .75
rate = .2ml/.75 s = .267 ml/s
so
Q(t) = _.267 * t ml for 0<t<.75
and second period
Q(t) = .2 e^-.002(t-.75) ml
for .75<t<oo