The temperature, \(T\) (in °C), in a greenhouse over a 24-hour period is modelled by the function \(T(t) = 15\cos(0.262t + 3.14) + 25\) for \(0 \le t \le 24\), where \(t\) is the number of hours after midnight.
(a) Find the temperature in the greenhouse at 06:00 (\(t = 6\)). [2]
(b) Find the minimum temperature in the greenhouse and the time it occurs. [3]
(c) Find the maximum temperature in the greenhouse. [1]
(d) Find the rate of change of temperature at 10:00 (\(t = 10\)). [3]
(e) Find the time intervals during which the temperature is increasing. [3]
(f) The greenhouse manager wants to maintain the temperature above 22°C. Find the total amount of time during the 24-hour period that the temperature is below 22°C. [3]
Question
Find the temperature in the greenhouse at 06:00 (\(t = 6\)). [2]
Hint
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Substitute \(t=6\) into the model \(T(t)=15\cos(0.262t+3.14)+25\).
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\(T(6) \approx 25.0\) °C
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Question
Find the minimum temperature in the greenhouse and the time it occurs. [3]
Hint
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Determine when \(\cos(0.262t+3.14)=-1\) to find the minimum.
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Minimum temperature = 10 °C, occurring at \(t \approx 0\) (or 24 hours)
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Find the maximum temperature in the greenhouse. [1]
Hint
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Substitute \(\cos(0.262t+3.14)=1\) into \(T(t)=15\cos(0.262t+3.14)+25\).
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Maximum temperature = 40 °C
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Find the rate of change of temperature at 10:00 (\(t = 10\)). [3]
Hint
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Differentiate \(T(t)=15\cos(0.262t+3.14)+25\) to find \(T'(t)\), then substitute \(t=10\).
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\(T'(10) \approx 1.93\) °C/hour
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Find the time intervals during which the temperature is increasing. [3]
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Determine where \(T'(t)>0\). From the graph, the temperature increases from the minimum at \(t=0\) to the maximum near \(t=12\).
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Temperature is increasing for \(0 < t < 12\) hours
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The greenhouse manager wants to maintain the temperature above 22°C. Find the total amount of time during the 24‑hour period that the temperature is below 22°C. [3]
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Set \(T(t)=22\) and solve for \(t\). The intersections give the time interval during which \(T(t)<22\).
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Approximately 10.4 hours (with intersections at \(t \approx 6.78\) and \(t \approx 17.2\))
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