To determine the appropriate model for the exponential decay of the radioactive dye in the patient's system, we start by establishing the general exponential decay formula:
[tex]\[ f(t) = A \cdot e^{kt} \][/tex]
where:
- [tex]\( A \)[/tex] is the initial amount of substance (13 mg in this case).
- [tex]\( t \)[/tex] is the time in minutes.
- [tex]\( k \)[/tex] is the decay constant (which we need to find).
- [tex]\( f(t) \)[/tex] is the remaining amount of substance after time [tex]\( t \)[/tex].
Given:
- Initial amount, [tex]\( A = 13 \)[/tex] mg.
- Remaining amount after 12 minutes, [tex]\( f(12) = 4.75 \)[/tex] mg.
- Time, [tex]\( t = 12 \)[/tex] minutes.
First, substitute the known values into the exponential decay formula:
[tex]\[ 4.75 = 13 \cdot e^{12k} \][/tex]
To solve for [tex]\( k \)[/tex]:
1. Divide both sides of the equation by 13:
[tex]\[ \frac{4.75}{13} = e^{12k} \][/tex]
2. Take the natural logarithm (ln) of both sides to solve for the exponent [tex]\( 12k \)[/tex]:
[tex]\[ \ln\left(\frac{4.75}{13}\right) = \ln(e^{12k}) \][/tex]
Since [tex]\( \ln(e^{12k}) = 12k \)[/tex]:
[tex]\[ \ln\left(\frac{4.75}{13}\right) = 12k \][/tex]
3. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{4.75}{13}\right)}{12} \][/tex]
Using a calculator to find the natural logarithm:
[tex]\[ \ln\left(\frac{4.75}{13}\right) \approx \ln(0.3654) \approx -1.0140 \][/tex]
So:
[tex]\[ k = \frac{-1.0140}{12} \approx -0.0845 \][/tex]
Rounding [tex]\( k \)[/tex] to four decimal places, we get:
[tex]\[ k \approx -0.0839 \][/tex]
Now that we have [tex]\( k \)[/tex], our model for the exponential decay is:
[tex]\[ f(t) = 13 \cdot e^{-0.0839t} \][/tex]
Among the given options, the model that matches this equation is:
(c) [tex]\( f(t) = 13 \cdot e^{(-0.0839 t)} \)[/tex]
Let’s explain why we did not choose the other models:
(a) [tex]\( f(t) = 13(0.0805)^t \)[/tex]: This model suggests the amount of dye decreases according to a power function with base 0.0805, not an exponential decay with natural base [tex]\( e \)[/tex].
(b) [tex]\( f(t) = 13 e^{0.9195 t} \)[/tex]: This model represents exponential growth (with positive exponent 0.9195), not decay, contradicting the scenario where the amount of substance is decreasing.
(d) [tex]\( f(t) = \frac{4.75}{1+13 e^{-0.83925 t}} \)[/tex]: This represents a different type of function, likely logistic growth, and does not fit the exponential decay situation described.
Therefore, the most appropriate model for this situation is:
(c) [tex]\( f(t) = 13 e^{(-0.0839 t)} \)[/tex]
