Certainly! Let's work step-by-step through the problem involving the depth of the pond increasing each week due to the rainy season.
### Problem Statement
Due to the rainy season, the depth in a pond increases by 3% each week. Before the rainy season started, the pond was 10 feet deep. We are looking for an equation that best represents the depth of the pond each week.
### Given Information
- Initial depth of the pond [tex]\( P = 10 \)[/tex] feet
- Weekly increase rate [tex]\( r = 3\% = 0.03 \)[/tex]
### Formula to Use
The appropriate formula to represent exponential growth is:
[tex]\[ y = P (1 + r)^x \][/tex]
Where:
- [tex]\( y \)[/tex] is the depth of the pond after [tex]\( x \)[/tex] weeks.
- [tex]\( P \)[/tex] is the initial depth of the pond.
- [tex]\( r \)[/tex] is the rate of increase per period (week, in this case).
- [tex]\( x \)[/tex] is the number of weeks.
### Step-by-Step Solution
1. Initial Depth:
The initial depth [tex]\( P \)[/tex] is given to be 10 feet.
2. Weekly Increase Rate:
The depth of the pond increases by 3% each week which translates to [tex]\( r = 0.03 \)[/tex].
3. Equation for Depth After [tex]\( x \)[/tex] Weeks:
By substituting [tex]\( P = 10 \)[/tex] feet and [tex]\( r = 0.03 \)[/tex] into the general formula:
[tex]\[
y = 10 (1 + 0.03)^x
\][/tex]
Simplifying inside the parentheses:
[tex]\[
y = 10 (1.03)^x
\][/tex]
Therefore, the equation that best represents the depth of the pond each week is:
[tex]\[ y = 10 (1.03)^x \][/tex]
### Verification with Given Data
Let's verify by calculating the depths for few different weeks:
1. After 1 Week:
[tex]\[
y = 10 \times (1.03)^1 = 10 \times 1.03 = 10.3 \text{ feet}
\][/tex]
2. After 2 Weeks:
[tex]\[
y = 10 \times (1.03)^2 = 10 \times 1.0609 = 10.609 \text{ feet}
\][/tex]
3. After 4 Weeks:
[tex]\[
y = 10 \times (1.03)^4 = 10 \times 1.12550881 = 11.2550881 \text{ feet}
\][/tex]
These results verify that the provided equation [tex]\( y = 10 (1.03)^x \)[/tex] correctly represents the increase in depth of the pond due to the rainy season.
