Question

calculator allowed

1. a cylindrical can of radius 10 millimeters is used to measure rainfall in stormville. the can is initially empty, and rain enters the can during a 60-day period. the height of water in the can is modeled by the function s, where s(t) is measured in millimeters and t is measured in days for 0 ≤ t ≤ 60. the rate at which the height of the water is rising in the can is given by s’(t) = 2 sin(0.03t) + 1.5.

how is the function (or functions) presented?
check all that apply:
☐ graphically ☐ numerically ☑️ algebraically ☐ words (verbally)

based on the problem stem above, write 3 questions that could be asked.

without actually answering the questions, explain how 2 of your questions could be answered. what method should be used?

answer 1 of the questions you made up.

next week we will see some of the questions that actually appeared... stay tuned!

Explanation:

Part 1: Function Presentation

• Algebraically: The rate function \( S'(t) = 2\sin(0.03t)+1.5 \) is given as a mathematical formula.
• Words (Verbally): The problem describes the context (cylindrical can, rainfall over 60 days, meaning of \( S(t) \) and \( t \)) in words.
• Graphically: No graph is provided.
• Numerically: No table of values is given.

So the applicable ones are Algebraically and Words (Verbally).

1. What is the height of the water in the can after 10 days? (Requires integrating the rate function to find \( S(t) \) and then evaluating at \( t = 10 \).)
2. At what time \( t \) (in days) is the height of the water rising the fastest? (Requires finding the maximum of \( S'(t) \) by analyzing the sine function or taking its derivative and finding critical points.)
3. What is the total change in the height of the water over the 60 - day period? (Requires integrating \( S'(t) \) from \( t = 0 \) to \( t = 60 \).)

• Question 1 (Height after 10 days):
• Method: Integration. Since \( S'(t) \) is the rate of change of \( S(t) \), we know that \( S(t)=S(0)+\int_{0}^{t}S'(x)dx \). Given the can is initially empty, \( S(0) = 0 \). So we need to compute \( \int_{0}^{10}(2\sin(0.03x)+1.5)dx \). We can use the rules of integration (antiderivative of \( \sin(ax) \) is \( -\frac{1}{a}\cos(ax) \), antiderivative of a constant \( c \) is \( cx \)) and then evaluate from 0 to 10.
• Question 3 (Total change over 60 days):
• Method: Integration. The total change in height is \( \int_{0}^{60}S'(t)dt \). Using the same integration rules as above, we find the antiderivative of \( 2\sin(0.03t)+1.5 \) and evaluate it from 0 to 60.

Answer:

• Algebraically (checked)
• Words (Verbally) (checked)

Part 2: Three Questions