Let's solve this step by step.
### Part (a): Find the formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex]
We are given that the substance decomposes according to a continuous exponential decay model. The general formula for exponential decay is:
[tex]\[ y = y_0 e^{kt} \][/tex]
where:
- [tex]\( y \)[/tex] is the amount of substance left at time [tex]\( t \)[/tex]
- [tex]\( y_0 \)[/tex] is the initial amount of the substance
- [tex]\( k \)[/tex] is the decay constant
- [tex]\( t \)[/tex] is the time in hours
We need to find the value of [tex]\( k \)[/tex].
We are given:
- The initial amount [tex]\( y_0 = 7600 \)[/tex] kg
- After 13 hours ([tex]\( t = 13 \)[/tex]), the remaining amount [tex]\( y = 1748 \)[/tex] kg
Using the formula [tex]\( y = y_0 e^{kt} \)[/tex]:
[tex]\[ 1748 = 7600 e^{13k} \][/tex]
First, we solve for [tex]\( k \)[/tex]:
1. Divide both sides by 7600:
[tex]\[ \frac{1748}{7600} = e^{13k} \][/tex]
2. Take the natural logarithm (ln) of both sides:
[tex]\[ \ln\left(\frac{1748}{7600}\right) = 13k \][/tex]
3. Solve for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln\left(\frac{1748}{7600}\right)}{13} \][/tex]
This results in:
[tex]\[ k = -0.11305199769684167 \][/tex]
Therefore, the exact formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is:
[tex]\[ y = 7600 e^{-0.11305199769684167 t} \][/tex]
### Part (b): Find the amount of substance left after 23 hours
We need to find the amount of substance left after [tex]\( t = 23 \)[/tex] hours. We use the formula we derived:
[tex]\[ y = 7600 e^{-0.11305199769684167 \times 23} \][/tex]
Let's compute it:
[tex]\[ y = 7600 e^{-0.11305199769684167 \times 23} \][/tex]
Evaluating the exponent:
[tex]\[ -0.11305199769684167 \times 23 = -2.600195947 "*"\][/tex]
So we get:
[tex]\[ y = 7600 e^{-2.600195947} = 564.368597244732 \][/tex]
Rounding to the nearest whole number:
[tex]\[ y \approx 564 \][/tex]
Therefore, the amount of the substance left after 23 hours is:
[tex]\[ 564 \, \text{kg} \][/tex]
In summary:
- The exact formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is [tex]\( y = 7600 e^{-0.11305199769684167 t} \)[/tex].
- The amount of the substance left after 23 hours is approximately 564 kg.
