The temperature T(t) varies sinusoidally on a certain day in December. The minimum temperature is 35 degrees Fahrenheit at midnight. The maximum temperature is 50 degrees Fahrenheit at noon. Let t be the number of hours since midnight (t=o at midnight).
a.) Sketch and label a graph showing exactly two periods of T(t)beginning at t=0.
b.)Determine a function for T(t) using the cosine function.
c.)Determine a function for T(t) using the sine function.
d.) Use your equation(s) to find the temperature at 1 am.
e.) I want to go on a bike ride, but I prefer to ride when the temperature is at least 45 degrees Fahrenheit. What is the earliest time of day that I can leave for my ride? How long can I stay out before I get cold?
the max-min value is 50-35=15, so the amplitude is 7.5. The center line is (35+50)/2 = 37.5
y = 7.5 sin(x) + 37.5
y has a max at t=12, and the period is 24 hours, so the minimum k is a t t=0.
cos(x) has a max at x=0, so we have -cos(x) and thus
y = -7.5 cos(π/12 x) + 37.5
Now use that to answer the other questions. recall that cos(x) = sin(π/2 - x).
when t=0 , temp = 35
when t = 12, temp = 50
Period = 24 hours or k = π/12
how about
Temp = 7.5cos (π/12)(t + 12) + 42.5
check:
when t = 0, Temp = 7.5cos(π) + 42.5 = 35
when t = 12, Temp = 7.5cos(2π) + 42.5 = 50
You try it with a sine curve, be aware that your phase shift will have to be different.
a 1:00 am , t = 1
temp = 7.5cos (π/12)(13) + 42.5 = 35.26° F
for T ≥ 45
7.5cos (π/12)(t+12) + 42.5 = 45
7.5cos (π/12)(t+12) = 2.5
cos (π/12)(t+12) = .3333...
(π/12)(t+12) = 1.231 or (π/12)(t+12) = 2π - 1.231 = 5.052
t+12 = 4.702 or t+12 = 19.297
t = -7.298 or t = 7.297 because of the symmetry of the curve
and
t = 19.297
t = 7.297 hrs = appr 7:18 am
t = 19.297 = appr 7:18 pm
so the temp is above 45° F from 7:18 am to 7:18 pm
check my arithmetic
Reiny is right: 42.5, not 37.5
but I'm sure you caught my error.
Note how he offset his function by using a phase shift, which has the same effect as changing the sign as I did, since shifting by 1/2 period does that flip:
cos(x+pi) = -cos(x)
Outside temperature over a day can be modelled using a sine or cosine function. Suppose you know the high temperature for the day is 101 degrees and the low temperature of 65 degrees occurs at 4 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.
A ferris wheel is 10 meters in diameter and boarded from a platform that is 5 meters above the ground. The six o'clock position on the ferris wheel is level with the loading platform. The wheel completes 1 full revolution in 2 minutes. The function h = f(t) gives your height in meters above the ground t minutes after the wheel begins to turn.
What is the amplitude?
What is the equation of the Midline?
What is the period?
The equation that models the height of the ferris wheel after
t minutes is:
How high are you off of the ground after 3 minutes? Round your answer to the nearest meters
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 (2π 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((π / 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 = -π/2 (phase shift of -π/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((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
Solving for t, we find:
(π / 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.