Math Problem Statement
Solution
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The temperature in Middletown Park is modeled by the equation:
Where:
- is the rate of temperature change at time hours after 6:00 AM last Sunday.
- The temperature at 6:00 AM last Sunday is 50.2°F.
1. First Question: Rate of Temperature Increase at 10:00 AM
To find the rate at which the temperature is increasing at 10:00 AM, we need to calculate the derivative of , which will give the rate of change of temperature. We will then evaluate this derivative at hours (since 10:00 AM is 4 hours after 6:00 AM).
The derivative of is:
Now, evaluate :
= -3.44 \cdot \frac{\pi}{12} \sin\left( \frac{\pi}{3} \right)$$ Since $$\sin\left( \frac{\pi}{3} \right) = \frac{\sqrt{3}}{2}$$, we get: $$r'(4) = -3.44 \cdot \frac{\pi}{12} \cdot \frac{\sqrt{3}}{2}$$ Now let’s calculate this: $$r'(4) \approx -3.44 \cdot 0.2618 \cdot 0.866$$ $$r'(4) \approx -0.777 \text{ degrees per hour}$$ So, the temperature was decreasing by approximately $$-0.78$$ degrees per hour at 10:00 AM. ### 2. Second Question: Temperature Increase from 6:00 AM to 1:00 PM The total increase in temperature is found by calculating the change in temperature between 6:00 AM (when $$t = 0$$) and 1:00 PM (when $$t = 7$$). We compute $$r(7) - r(0)$$. At $$t = 0$$: $$r(0) = 3.44 \cos(0) + 0.044 = 3.44 \cdot 1 + 0.044 = 3.484$$ At $$t = 7$$: $$r(7) = 3.44 \cos\left( \frac{7\pi}{12} \right) + 0.044$$ Now, let’s calculate the cosine term: $$\cos\left( \frac{7\pi}{12} \right) \approx -0.2588$$ So: $$r(7) = 3.44 \cdot (-0.2588) + 0.044 = -0.890 \, + \, 0.044 = -0.846$$ Now, the total temperature increase from 6:00 AM to 1:00 PM is: $$r(7) - r(0) = -0.846 - 3.484 = -4.33 \text{ degrees}$$ The temperature actually decreased by approximately $$4.33$$ degrees over this time. ### 3. Third Question: Temperature at 1:00 PM Last Sunday We already computed $$r(7) = -0.846$$, so the rate at 1:00 PM is: $$\text{Temperature at 1:00 PM} = 50.2 - 0.846 = 49.35^\circ F$$ ### 4. Fourth Question: Temperature at 4:00 PM Last Friday Friday is 5 days (or 120 hours) before last Sunday. Therefore, 4:00 PM on Friday is: $$t = 6 + 120 + 10 = 136 \text{ hours}$$ We calculate $$r(136)$$: $$r(136) = 3.44 \cos\left( \frac{136\pi}{12} \right) + 0.044$$ Simplifying the angle: $$\frac{136\pi}{12} = \frac{34\pi}{3} = 11\pi + \frac{\pi}{3}$$ Thus: $$r(136) = 3.44 \cos\left( \frac{\pi}{3} \right) + 0.044 = 3.44 \cdot 0.5 + 0.044 = 1.72 + 0.044 = 1.764$$ The temperature at 4:00 PM on Friday would be: $$\text{Temperature at 4:00 PM on Friday} = 50.2 + 1.764 = 51.96^\circ F$$ --- Let me know if you'd like more details or further clarifications. ### Follow-up questions: 1. What is the rate of temperature change at 12:00 PM last Sunday? 2. How would the equation look if we changed the amplitude of the cosine term? 3. What is the maximum temperature change possible according to the given equation? 4. Can we predict when the temperature will be the same as it was at 6:00 AM? 5. How would the temperature change if the cosine term's period was altered? **Tip:** For sinusoidal functions, the amplitude determines how much fluctuation occurs from the mean value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Rates of Change
Cosine Function
Temperature Modeling
Formulas
r(t) = 3.44 cos((πt/12)) + 0.044
Derivative of r(t): r'(t) = -3.44 * (π/12) * sin((πt/12))
Theorems
Sine and Cosine Properties
Rate of Change (Derivative)
Suitable Grade Level
Grades 10-12
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