Question

. For each question below, model the scenarios and find the quantity of interest. ● Consult with peers to determine that you accurately solved. ● If necessary, revise your work. 1. An account compounding continuously quadruples in value after 25 years. What was its interest rate? 2. A new car devalues exponentially with an annualized rate of 4%, compounded semi-annually. How long does it take the car to halve in value? © Success Academy Charter Schools 1
3. A radioactive substance is decaying continuously at a rate of 18% per year. How long does it take for 60% of the substance to decay? 4. A pot of boiling water (212°F) is left to cool back to room temperature (70°F). How long does it take if the water cools continuously at a rate of 3% per minute? © Success Academy Charter Schools 2
5. After an invasive species was introduced to a national park, the native goldenrod population began declining at a rate of 1% continuously. How many years will it take until the population is 95% gone? 6. Mari withdraws $850 more from her account than she deposited after 9 years. If the account had an annualized interest rate of 5%, compounded quarterly, what was the initial deposit?

Answer 1

To solve these problems, we will use relevant exponential growth and decay formulas, mainly the equations for continuous compounding and continuous decay. Let's go through each question one by one.

### 1. Interest Rate of a Continuously Compounding Account
We know the account quadruples in value, so we can set up the equation for continuous compounding:
\[
A = Pe^{rt}
\]
Where:
- \( A \) = final amount
- \( P \) = initial amount
- \( r \) = interest rate
- \( t \) = time

We know \( A = 4P \) and \( t = 25 \). Thus, the equation becomes:
\[
4P = Pe^{25r}
\]
Dividing both sides by \( P \), we have:
\[
4 = e^{25r}
\]
Taking the natural logarithm of both sides, we get:
\[
\ln(4) = 25r
\]
Solving for \( r \):
\[
r = \frac{\ln(4)}{25} \approx \frac{1.386}{25} \approx 0.05544 \text{ or } 5.54\%
\]

### 2. Time for a Car to Halve in Value
Here, we need to use the formula for exponential decay:
\[
V(t) = V_0 e^{-kt}
\]
Where \( k = 0.04/2 = 0.02 \) (since it's compounded semi-annually, we use half the annual rate).

To find the time it takes for the car's value to halve:
\[
\frac{V_0}{2} = V_0 e^{-0.02t}
\]
Dividing by \( V_0 \):
\[
\frac{1}{2} = e^{-0.02t}
\]
Taking the natural logarithm:
\[
\ln\left(\frac{1}{2}\right) = -0.02t
\]
Solving for \( t \):
\[
t = \frac{-\ln(0.5)}{0.02} \approx \frac{0.693}{0.02} \approx 34.65 \text{ years}
\]

### 3. Decay Time for a Radioactive Substance
For continuous decay of a substance at a rate of 18%, we use:
\[
N(t) = N_0 e^{-kt}
\]
where \( k = 0.18 \). We want to find out how long it takes for 60% to decay (leaving 40%):
\[
0.4N_0 = N_0 e^{-0.18t}
\]
Dividing by \( N_0 \):
\[
0.4 = e^{-0.18t}
\]
Taking the natural logarithm:
\[
\ln(0.4) = -0.18t
\]
Solving for \( t \):
\[
t = \frac{-\ln(0.4)}{0.18} \approx \frac{0.916}{0.18} \approx 5.09 \text{ years}
\]

### 4. Cooling Time for Boiling Water
Using Newton's Law of Cooling:
\[
T(t) = T_a + (T_0 - T_a)e^{-kt}
\]
Where \( T_a = 70°F \), \( T_0 = 212°F \), and \( k = 0.03 \):
\[
T(t) = 70 + (212 - 70)e^{-0.03t}
\]
We want to find \( t \) when \( T(t) = 70 + 0 \):
\[
0 = 142 e^{-0.03t}
\]
This equation does not hold as \( T(t) \) will never reach 70. Instead, we want:
\[
T(t) = 100°F \Rightarrow 100 = 70 + 142 e^{-0.03t}
\]
Solving for \( t \):
\[
30 = 142 e^{-0.03t} \Rightarrow e^{-0.03t} = \frac{30}{142}
\]
Taking the natural logarithm:
\[
-0.03t = \ln\left(\frac{30}{142}\right)
\]
Solving for \( t \):
\[
t = \frac{-\ln(30/142)}{0.03} \approx \frac{3.465}{0.03} \approx 115.5 \text{ minutes}
\]

### 5. Declining Population of Goldenrod
Using continuous decay:
\[
P(t) = P_0 e^{-kt} \quad (k = 0.01)
\]
We want to find the time \( t \) when \( P(t) = 0.05P_0 \):
\[
0.05P_0 = P_0 e^{-0.01t}
\]
Dividing by \( P_0 \):
\[
0.05 = e^{-0.01t}
\]
Taking the natural logarithm:
\[
\ln(0.05) = -0.01t
\]
Solving for \( t \):
\[
t = \frac{-\ln(0.05)}{0.01} \approx \frac{2.996}{0.01} \approx 299.6 \text{ years}
\]

### 6. Initial Deposit Calculation
Using the compound interest formula:
\[
A = P(1 + \frac{r}{n})^{nt}
\]
Where:
- \( A = P + 850 \)
- \( r = 0.05 \), \( n = 4 \), \( t = 9 \)

We set this up as:
\[
P(1 + \frac{0.05}{4})^{4 \cdot 9} = P + 850
\]

Calculating:
\[
P(1 + 0.0125)^{36} = P + 850
\]
Calculating:
\[
P \cdot (1.4908) = P + 850
\]
Now:
\[
1.4908P - P = 850
\]
\[
0.4908P = 850
\]
Solving for \( P \):
\[
P = \frac{850}{0.4908} \approx 1732.50
\]

### Summary of Answers
1. Interest Rate = **5.54%**
2. Time to Halve Value = **34.65 years**
3. Decay Time = **5.09 years**
4. Cooling Time to 100°F = **115.5 minutes**
5. Time for Population to Decline = **299.6 years**
6. Initial Deposit = **$1732.50**