Asked by Katy
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?
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?
Answers
Answered by
Steve
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).
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).
Answered by
Reiny
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
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
Answered by
Steve
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)
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)
Answered by
hola
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.
Answered by
�hola
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
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
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