Math Problem Statement
Solution
Let's break down the solution provided for Question 1 based on the image you've uploaded:
Problem Overview:
The temperature of water in a tub at time (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 at different times.
Data:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Differentiation
Integration
Riemann Sums
Fundamental Theorem of Calculus
Formulas
W'(12) ≈ (W(15) - W(9)) / (15 - 9)
∫_0^20 W'(t) dt = W(20) - W(0)
Riemann Sum Approximation: (1/20) ∫_0^20 W(t) dt ≈ (4W(0) + 5W(4) + 6W(9) + 5W(15)) / 20
W(25) = W(20) + ∫_20^25 W'(t) dt
Theorems
Fundamental Theorem of Calculus
Suitable Grade Level
Grade 12 / College
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