Inverse Trig Function word problem

Question

Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 83 degrees and the low temperature of 67 degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

Answer 1

Hi Cole,

Every sine wave follows the following form: F(t) = A*sin(B*t - C) + D.

Where A is the amplitude of Temperature

B is the frequency of the oscillation

C is the phase shift

and D is the vertical shift of the function

We will look at these values one at a time.

A: This is simply the height of the top and bottom portions of the wave. In your case the total height is 83-67 = 16, so the height of the top or bottom is 16/2 or 8 degrees.

B: you know that your temperature sine curve restarts every 24 hours, or it has 1 cycle per day. This also equals 1 cycle/24 hours = .042 cycles/hour. Now lets convert the cycles to radians with the following conversion: 1 cycle = 2*pi radians. So (.042 cycles/hour)*2*pi rad/cycle = .26 radians/hour. We want units of radians per hour, not radians per seconds or days because the variable t, for time, is given in hours.

C: The phase shift. This is how many units, in radians, is the curve shifted to the left or right based on where a sine curve usually starts. We know that the lowest temperature is at 5am and that our clock starts ticking at midnight. So the lowest point is 5 hours into a 24 hour cycle. It then takes a quarter of the period to get back to the start of where a sine curve typically starts at so we are shifted 11 hours to the right. We can set up a correlation and find how many radians this would be.11/24 = x/(2*pi radians) = 2.88 radians. Now we have to ask ourselves the question, does this shift the graph to the left or to the right? Because the temperature is hitting a low point in 5 hours, this is going to shift our normal sine curve to the right. A right shift corresponds to a negative value in C so your value for C would be -2.88 radians

D: Your vertical shift. The question here is how far off is the center of this sine curve, or its average value, off from 0? Well, we know that the average value is (83+67)/2 = 75 degrees. So our vertical offset is 75 degrees. Your value for D would be 75

Now we have all the pieces to write our function: D(t) = 8*sin(.26*t-2.88)+75

Answer 2

the period is 2*pi/24 and the graph is shifted to the right by 5 hours.

so the variable part is (2*pi*(t-5)/24

The amplitude is 8 = (83-67)/2

and the offset for midline is 75 = (83+67)2

the first peak is negative so I use negative cosine

d(t)= (-8*cos(2*pi*(t-5)/24))+75

check, t=5, d=67

t= -8*cos(2pi(1/2))+75=-8*-1+75=83