a) To sketch a graph showing two periods of T(t), we need to consider that:
- The minimum temperature of 35 degrees Fahrenheit occurs at midnight (t = 0).
- The maximum temperature of 50 degrees Fahrenheit occurs at noon (t = 12).
Let's start by plotting these two points on the graph. At t = 0, the temperature is 35 degrees Fahrenheit, and at t = 12, the temperature is 50 degrees Fahrenheit.
Now, since temperature varies sinusoidally, we know that the graph will follow a cosine or sine function. We can assume it follows a cosine function since the maximum occurs in the middle.
To create two periods, we need to extend the graph from t = 0 to t = 24. At t = 24, the temperature would have completed two periods. So, we can add another minimum point (35 degrees Fahrenheit) at t = 24.
Your graph should resemble a cosine wave, starting at a minimum and rising to a maximum at t = 12 (noon), then falling back to the minimum at t = 24 (midnight).
b) Function for T(t) using the cosine function:
T(t) = A * cos(Bt) + C
Given:
Minimum temperature = 35 degrees Fahrenheit (C)
Maximum temperature = 50 degrees Fahrenheit (A)
Period = 24 hours (2Ï€ radians)
Let's calculate the value of B:
B = 2π / period = 2π / 24 = π / 12
Therefore, the function for T(t) using the cosine function is:
T(t) = 7.5 * cos((Ï€ / 12)t) + 42.5
c) Function for T(t) using the sine function:
Since the cosine and sine functions are related (sin(t) = cos(t - π/2)), we can use the same equation as above with a phase shift:
T(t) = A * sin(Bt + D) + C
Given:
D = -Ï€/2 (phase shift of -Ï€/2, which shifts the graph to the right by t = 6 hours)
Therefore, the function for T(t) using the sine function is:
T(t) = 7.5 * sin((π / 12)t - π/2) + 42.5
d) To find the temperature at 1 am (t = 1), substitute t = 1 into the equation obtained in part b or c:
Using the cosine function:
T(1) = 7.5 * cos((π / 12)*1) + 42.5 ≈ 48.66 degrees Fahrenheit
Using the sine function:
T(1) = 7.5 * sin((π / 12)*1 - π/2) + 42.5 ≈ 48.66 degrees Fahrenheit
Therefore, the temperature at 1 am is approximately 48.66 degrees Fahrenheit.
e) Since the minimum temperature required is 45 degrees Fahrenheit and the temperature is increasing from midnight to noon, you need to find the first instance where T(t) is equal to or greater than 45 degrees Fahrenheit.
To do this, let's set up the equation T(t) ≥ 45 and solve for t using the cosine function:
7.5 * cos((π / 12)t) + 42.5 ≥ 45
cos((π / 12)t) ≥ (45 - 42.5) / 7.5
cos((π / 12)t) ≥ 2.5/7.5
cos((π / 12)t) ≥ 1/3
To find the earliest time of day, you need to find the smallest positive value of t that satisfies this inequality:
(π / 12)t ≥ arccos(1/3)
t ≥ (12/π) * arccos(1/3) ≈ 4.02 hours
Therefore, the earliest time of day you can leave for your ride is approximately 4:02 am.
To determine how long you can stay out before getting cold, we need to find the time when the temperature drops below 45 degrees Fahrenheit again.
Using the cosine function:
7.5 * cos((Ï€ / 12)t) + 42.5 = 45
cos((Ï€ / 12)t) = (45 - 42.5) / 7.5
cos((Ï€ / 12)t) = 2.5/7.5
cos((Ï€ / 12)t) = 1/3
Solving for t, we find:
(Ï€ / 12)t = arccos(1/3)
t = (12/π) * arccos(1/3) ≈ 4.02 hours
Since t = 4.02 represents the time when the temperature drops below 45 degrees Fahrenheit, you can stay out for approximately 4.02 hours or 4 hours and 1 minute.
So, you can stay out until approximately 8:03 am.