The balance *A* (in dollars) in a savings account is given by 

*A* = 8000*e*0.06*t*,

 where *t* is measured in years. Find the rate at which the balance is changing when *t* = 1 year, *t* = 10 years, and *t* = 50 years. (Round your answers to two decimal places.)

To find the rate at which the balance is changing at different times \( t \), we need to determine the derivative of the balance function \( A(t) \) with respect to time \( t \). This derivative will give us the rate of change of the balance at any given time \( t \).

### Step 1: Given equation
The balance function is:
\[
A(t) = 8000 \cdot e^{0.06t}
\]

### Step 2: Find the derivative of \( A(t) \)
We take the derivative of \( A(t) \) with respect to \( t \):
\[
\frac{dA}{dt} = 8000 \cdot \frac{d}{dt} \left( e^{0.06t} \right)
\]
Using the chain rule, the derivative of \( e^{0.06t} \) is \( 0.06 \cdot e^{0.06t} \), so:
\[
\frac{dA}{dt} = 8000 \cdot 0.06 \cdot e^{0.06t}
\]
\[
\frac{dA}{dt} = 480 \cdot e^{0.06t}
\]

### Step 3: Evaluate the rate of change at different times \( t \)

1. **For \( t = 1 \):**
\[
\frac{dA}{dt} = 480 \cdot e^{0.06 \cdot 1} = 480 \cdot e^{0.06}
\]
Using \( e^{0.06} \approx 1.06184 \), we get:
\[
\frac{dA}{dt} \approx 480 \cdot 1.06184 \approx 509.68
\]
So, the rate of change at \( t = 1 \) year is approximately **509.68 dollars/year**.

2. **For \( t = 10 \):**
\[
\frac{dA}{dt} = 480 \cdot e^{0.06 \cdot 10} = 480 \cdot e^{0.6}
\]
Using \( e^{0.6} \approx 1.82212 \), we get:
\[
\frac{dA}{dt} \approx 480 \cdot 1.82212 \approx 874.62
\]
So, the rate of change at \( t = 10 \) years is approximately **874.62 dollars/year**.

3. **For \( t = 50 \):**
\[
\frac{dA}{dt} = 480 \cdot e^{0.06 \cdot 50} = 480 \cdot e^{3}
\]
Using \( e^{3} \approx 20.0855 \), we get:
\[
\frac{dA}{dt} \approx 480 \cdot 20.0855 \approx 9641.04
\]
So, the rate of change at \( t = 50 \) years is approximately **9641.04 dollars/year**.

### Summary of results:
- Rate of change at \( t = 1 \) year: **509.68 dollars/year**
- Rate of change at \( t = 10 \) years: **874.62 dollars/year**
- Rate of change at \( t = 50 \) years: **9641.04 dollars/year**

Would you like further details on this solution or have any questions?

#### 5 Related Questions:
1. What would the balance be at \( t = 1 \), \( t = 10 \), and \( t = 50 \) years?
2. How does the rate of change compare with the initial deposit?
3. How would the rate of change differ if the interest rate were 5% instead of 6%?
4. Can we approximate the time it takes for the balance to double?
5. What would the derivative look like if the interest compounded quarterly?

**Tip**: When dealing with exponential growth, small changes in time can lead to increasingly larger rates of change, especially as time increases.

The balance A (in dollars) in a savings account is given by A = 8000e^(0.06t), where t is measured in years. Find the rate at which the balance is changing when t = 1 year, t = 10 years, and t = 50 years. (Round your answers to two decimal places.)

Learn how to calculate the rate of change of a savings account balance modeled by exponential growth. The equation A(t) = 8000e^(0.06t) is differentiated to find the rates of change at t = 1, 10, and 50 years, using the chain rule. Detailed step-by-step solution provided, suitable for students in Grades 11-12.

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