Let's go through each part of the question step by step:
### (a) Period of the function [tex]\( y = H(t) \)[/tex]
The period of a periodic function is the time it takes to complete one full cycle. Since the temperature reaches its maximum after two and a half hours and then takes the same amount of time to return to its starting value, one complete cycle includes both the rise to the maximum and the fall back to the initial temperature.
- Time to reach maximum: 2.5 hours
- Time to return to initial temperature: 2.5 hours
Therefore, the period of the function is:
[tex]\[ \text{Period} = 2 \times 2.5 \ \text{hours} = 5 \ \text{hours} \][/tex]
So, the period of the function [tex]\( y = H(t) \)[/tex] is [tex]\( 5 \)[/tex] hours.
### (b) Midline of the function [tex]\( y = H(t) \)[/tex]
The midline of a periodic function is the horizontal line that represents the average value of the maximum and minimum temperatures.
- Minimum temperature: [tex]\( 10^\circ \text{C} \)[/tex]
- Maximum temperature: [tex]\( 135^\circ \text{C} \)[/tex]
The midline is calculated as the average of these temperatures:
[tex]\[ \text{Midline} = \frac{10^\circ \text{C} + 135^\circ \text{C}}{2} = \frac{145^\circ \text{C}}{2} = 72.5^\circ \text{C} \][/tex]
So, the midline of the function [tex]\( y = H(t) \)[/tex] is [tex]\( 72.5^\circ \text{C} \)[/tex].
### (c) Amplitude of the function [tex]\( y = H(t) \)[/tex]
The amplitude of a periodic function is half the range of the function, which is the difference between the maximum and minimum values.
- Minimum temperature: [tex]\( 10^\circ \text{C} \)[/tex]
- Maximum temperature: [tex]\( 135^\circ \text{C} \)[/tex]
The amplitude is:
[tex]\[ \text{Amplitude} = \frac{135^\circ \text{C} - 10^\circ \text{C}}{2} = \frac{125^\circ \text{C}}{2} = 62.5^\circ \text{C} \][/tex]
So, the amplitude of the function [tex]\( y = H(t) \)[/tex] is [tex]\( 62.5^\circ \text{C} \)[/tex].
### (d) Making a Graph of the Function [tex]\( y = H(t) \)[/tex]
To make a graph of the function [tex]\( y = H(t) \)[/tex], we plot the temperature against time.
1. Midline: Draw a horizontal line at [tex]\( y = 72.5^\circ \text{C} \)[/tex].
2. Amplitude and Period: Oscillate this line up and down by [tex]\( 62.5^\circ \text{C} \)[/tex] every [tex]\( 5 \)[/tex] hours.
- At [tex]\( t = 0 \)[/tex]: [tex]\( y = 10^\circ \text{C} \)[/tex] (low point)
- At [tex]\( t = 2.5 \)[/tex]: [tex]\( y = 135^\circ \text{C} \)[/tex] (high point)
- At [tex]\( t = 5 \)[/tex]: [tex]\( y = 10^\circ \text{C} \)[/tex] (low point again)
The resulting graph should show a sinusoidal wave oscillating between [tex]\( 10^\circ \text{C} \)[/tex] and [tex]\( 135^\circ \text{C} \)[/tex] with a period of [tex]\( 5 \)[/tex] hours. The best matching graph will follow these parameters.
Given these criteria, the correct graph is marked as:
[tex]\[ \checkmark \ \text{A} \][/tex]
