Let's address the question step-by-step.
(a) To derive the formula relating the area [tex]\( y \)[/tex] to time [tex]\( t \)[/tex] under an exponential growth model, let's start with the general form of an exponential growth equation:
[tex]\[ y = y_0 \cdot e^{kt} \][/tex]
where:
- [tex]\( y_0 \)[/tex] is the initial area,
- [tex]\( k \)[/tex] is the growth rate,
- [tex]\( t \)[/tex] is the time in hours,
- [tex]\( e \)[/tex] is the base of the natural logarithm.
Given:
- The initial area [tex]\( y_0 = 94.9 \, \text{mm}^2 \)[/tex],
- The doubling time is 18 hours.
To find the growth rate [tex]\( k \)[/tex], we use the fact that the area doubles in 18 hours. This means:
[tex]\[ 2y_0 = y_0 \cdot e^{k \cdot 18} \][/tex]
Dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ 2 = e^{18k} \][/tex]
Taking the natural logarithm of both sides:
[tex]\[ \ln(2) = 18k \][/tex]
Thus, solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{\ln(2)}{18} \][/tex]
Now we substitute [tex]\( y_0 \)[/tex] and [tex]\( k \)[/tex] back into the general exponential growth formula:
[tex]\[ y = 94.9 \cdot e^{\left(\frac{\ln(2)}{18} \cdot t\right)} \][/tex]
So, the exact formula is:
[tex]\[ y = 94.9 \cdot e^{\left(\frac{\ln(2)}{18} \cdot t\right)} \][/tex]
(b) Now, let's determine the area of the sample after 12 hours, [tex]\( t = 12 \)[/tex].
[tex]\[ y = 94.9 \cdot e^{\left(\frac{\ln(2)}{18} \cdot 12\right)} \][/tex]
First, compute the exponent:
[tex]\[ \frac{\ln(2)}{18} \cdot 12 \][/tex]
Then, finding [tex]\( e \)[/tex] raised to that exponent and multiplying by the initial area:
[tex]\[ y = 94.9 \cdot e^{\left(\frac{\ln(2) \cdot 12}{18}\right)} \][/tex]
Using the calculated values from our growth model:
[tex]\[ y = 94.9 \cdot e^{0.4621} \approx 150.6 \][/tex]
So, the area of the sample in 12 hours, rounded to the nearest tenth, is:
[tex]\[ 150.6 \, \text{mm}^2 \][/tex]
