Sure, let's tackle this problem step by step.
### Part (a)
We are given the initial mass of the radioactive substance and its half-life. To find the formula relating the remaining mass [tex]\( y \)[/tex] to time [tex]\( t \)[/tex] under the continuous exponential decay model, we start by defining the relationship through the exponential decay formula:
[tex]\[ y = y_0 e^{kt} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial mass of the substance.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in hours.
We need to determine the decay constant [tex]\( k \)[/tex]. Using the fact that the half-life of the substance is 23 hours, we know that:
[tex]\[ y \bigg|_{t=23} = \frac{y_0}{2} \][/tex]
Using the exponential decay formula when [tex]\( t = 23 \)[/tex]:
[tex]\[ \frac{y_0}{2} = y_0 e^{k \cdot 23} \][/tex]
We can cancel [tex]\( y_0 \)[/tex] on both sides of the equation:
[tex]\[ \frac{1}{2} = e^{k \cdot 23} \][/tex]
Taking the natural logarithm on both sides:
[tex]\[ \ln\left(\frac{1}{2}\right) = k \cdot 23 \][/tex]
Simplifying for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{1}{2}\right)}{23} \][/tex]
[tex]\[ k = \frac{\ln(0.5)}{23} \][/tex]
The formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is then:
[tex]\[ y = 769.3 e^{\left(\frac{\ln(0.5)}{23}\right) t} \][/tex]
So in the given formula [tex]\( y = \square e^{Dt} \)[/tex], [tex]\( \square \)[/tex] should be replaced by the initial mass 769.3, and [tex]\( D \)[/tex] should be replaced by [tex]\( \frac{\ln(0.5)}{23} \)[/tex].
Thus, the exact formula is:
[tex]\[ y = 769.3 e^{\left(\frac{\ln(0.5)}{23} \right) t} \][/tex]
### Part (b)
To find out how much substance will be present after 21 hours, we substitute [tex]\( t = 21 \)[/tex] into our formula:
[tex]\[ y = 769.3 e^{\left( \frac{\ln(0.5)}{23} \right) 21} \][/tex]
Evaluating the exponent:
[tex]\[ \frac{\ln(0.5)}{23} \approx -0.0301368 \][/tex]
So:
[tex]\[ y = 769.3 e^{-0.0301368 \cdot 21} \][/tex]
[tex]\[ y \approx 769.3 e^{-0.632873} \][/tex]
[tex]\[ y \approx 769.3 \times 0.5312 \][/tex]
[tex]\[ y \approx 408.5 \][/tex]
Therefore, the amount of the substance present after 21 hours will be approximately 408.5 mg, rounded to the nearest tenth.
