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### Problem Overview:
The temperature \( W(t) \) of water in a tub at time \( t \) (measured in minutes) is modeled by a differentiable function. The goal is to perform various operations and estimations based on the given data points of \( W(t) \) at different times.

#### Data:
\[
\begin{array}{|c|c|}
\hline
t \, (\text{minutes}) & W(t) \, (\text{degrees Fahrenheit}) \\
\hline
0 & 55.0 \\
4 & 57.1 \\
9 & 61.8 \\
15 & 67.9 \\
20 & 71.0 \\
\hline
\end{array}
\]

### Solution Outline:

#### Part (a):
Estimate \( W'(12) \), the derivative at \( t = 12 \).

- **Approach**: Use the average rate of change between \( t = 9 \) and \( t = 15 \), where 12 is approximately the midpoint.
  
  \[
  W'(12) \approx \frac{W(15) - W(9)}{15 - 9} = \frac{67.9 - 61.8}{6} = 1.017 \, \text{(or 1.016)}
  \]
  
- **Interpretation**: The temperature is increasing at a rate of approximately \( 1.017 \, ^\circ \text{F/min} \) at \( t = 12 \) minutes.

#### Part (b):
Evaluate \( \int_0^{20} W'(t) \, dt \).

- **Approach**: Use the Fundamental Theorem of Calculus. This integral gives the change in the water's temperature over the interval from \( t = 0 \) to \( t = 20 \).
  
  \[
  \int_0^{20} W'(t) \, dt = W(20) - W(0) = 71.0 - 55.0 = 16.0
  \]

- **Interpretation**: The water temperature increased by \( 16^\circ \text{F} \) from \( t = 0 \) to \( t = 20 \).

#### Part (c):
Find the average temperature over the interval \( [0, 20] \) using a left Riemann sum.

- **Formula**:
  \[
  \frac{1}{20} \int_0^{20} W(t) \, dt \approx \frac{1}{20} \left( 4W(0) + 5W(4) + 6W(9) + 5W(15) \right)
  \]

- **Calculation**:
  \[
  = \frac{1}{20} \left( 4(55.0) + 5(57.1) + 6(61.8) + 5(67.9) \right)
  \]
  \[
  = \frac{1}{20} (220.0 + 285.5 + 370.8 + 339.5) = \frac{1}{20} (1215.8) = 60.79
  \]

- **Interpretation**: This Riemann sum underestimates the average temperature because the function \( W(t) \) is increasing, and left sums tend to underestimate for increasing functions.

#### Part (d):
Model the temperature with the function \( W'(t) = 0.4 \, t \cos(0.06 t) \) for \( t \in [20, 25] \), and estimate \( W(25) \).

- **Formula**:
  \[
  W(25) = W(20) + \int_{20}^{25} W'(t) \, dt
  \]
  
  The integral approximation gives:
  \[
  \int_{20}^{25} W'(t) \, dt = 2.043155
  \]

- **Final calculation**:
  \[
  W(25) = 71.0 + 2.043155 = 73.043^\circ \text{F}
  \]

### Do you need any additional details or further explanation?

---

Here are 5 related questions to deepen your understanding:
1. How would the result in part (c) change if a right Riemann sum was used instead of a left Riemann sum?
2. Why does the increasing nature of \( W(t) \) cause the left Riemann sum to underestimate the average temperature in part (c)?
3. Can you explain the relationship between the derivative \( W'(t) \) and the behavior of the temperature function \( W(t) \)?
4. How would you approximate \( W'(t) \) at points where data isn't directly given (e.g., \( W'(12) \)) using methods like forward and backward differences?
5. How would the integral in part (d) change if the function for \( W'(t) \) was more complex or non-linear?

### Tip:
When solving for a derivative using a table of values, it's important to choose intervals where the data points are close to the value of interest. This leads to more accurate estimations.

The temperature of water in a tub at time t is modeled by a strictly increasing, twice-differentiable function W(t), where W(t) is measured in degrees Fahrenheit and t is measured in minutes. At time t = 0, the temperature of the water is 55°F. The water is heated for 30 minutes, beginning at t = 0. Use the table to estimate W'(12), evaluate the integral from 0 to 20, approximate the average temperature using Riemann sums, and estimate the temperature at t = 25 based on a given model.

Learn how to estimate the rate of temperature change using derivatives, evaluate integrals for total change, and approximate average values with Riemann sums. This example involves differentiable functions of temperature over time, integrals, and numerical approximation methods. Suitable for Grade 12 and College students.

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