Certainly! Let's proceed step by step:
### Part (a)
We start by establishing the formula for the exponential growth of the substance:
1. Exponential Growth Model:
The basic form of the exponential growth model is given by:
[tex]\[
y = y_0 e^{kt}
\][/tex]
where:
- [tex]\( y \)[/tex] is the amount of the substance at time [tex]\( t \)[/tex],
- [tex]\( y_0 \)[/tex] is the initial amount of the substance,
- [tex]\( k \)[/tex] is the growth rate constant,
- [tex]\( t \)[/tex] is the time.
2. Doubling Time:
The doubling time is the time it takes for the substance to double in quantity. Given that the doubling time is 17 minutes, we can use this information to find [tex]\( k \)[/tex]. We set up the equation based on the doubling condition:
[tex]\[
2 = e^{k \cdot 17}
\][/tex]
3. Solving for [tex]\( k \)[/tex]:
To solve for [tex]\( k \)[/tex], take the natural logarithm (ln) on both sides of the equation:
[tex]\[
\ln(2) = k \cdot 17
\][/tex]
[tex]\[
k = \frac{\ln(2)}{17}
\][/tex]
Substitute this back into the exponential growth formula:
[tex]\[
y = y_0 e^{\left(\frac{\ln(2)}{17}\right) t}
\][/tex]
4. Initial Amount:
The initial amount [tex]\( y_0 \)[/tex] is 8.7 mg. Substituting [tex]\( y_0 \)[/tex] into the formula, we get:
[tex]\[
y = 8.7 e^{\left(\frac{\ln(2)}{17}\right) t}
\][/tex]
So, the complete formula is:
[tex]\[
y = 8.7 e^{\left(\frac{\ln(2)}{17}\right) t}
\][/tex]
### Part (b)
Next, we need to find the amount of the substance present at [tex]\( t = 11 \)[/tex] minutes:
1. Using the Formula:
Substitute [tex]\( t = 11 \)[/tex] minutes into the formula:
[tex]\[
y = 8.7 e^{\left(\frac{\ln(2)}{17}\right) \cdot 11}
\][/tex]
2. Calculation:
Compute the exponent:
[tex]\[
\frac{\ln(2)}{17} \cdot 11 \approx \frac{0.6931471805599453}{17} \cdot 11 \approx 0.448210
\][/tex]
Substituting this value back into the formula, we get:
[tex]\[
y = 8.7 \cdot e^{0.448210}
\][/tex]
Using [tex]\( e^{0.448210} \approx 1.565 \)[/tex]:
[tex]\[
y \approx 8.7 \cdot 1.565 \approx 13.6
\][/tex]
So, the amount of the substance present at 11 minutes is approximately:
[tex]\[
13.6 \text{ mg}
\][/tex]
In conclusion:
- (a) The formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is [tex]\( y = 8.7 e^{\left(\frac{\ln(2)}{17}\right) t} \)[/tex].
- (b) The amount present in 11 minutes is approximately [tex]\( 13.6 \)[/tex] mg.
