How would you begin this problem?
On a standard summer day in upstate New York, the temperature outside can be modeled using the sinusoidal equation O(t) = 11 cos (pi/12 t) + 71, where t represents the number of hours since the peak temperature for the day.
For 0 less than or equal to t less than or equal to 24, graphically determine all points in time when the outside temperature is equal to 75 degrees. Round your answer to the nearest tenth of an hour.
set 11cos(π/12 t) + 71 = 75
cos(πt/12) = .363636...
set your calculator to radians and take cos^-1 (.363636...) to get 1.19862779
so πt/12 = 1.19862779 or πt/12 = 2π-1.19862779 or 5.0845575.. (the cosine is + in quads I and IV )
t = 4.578 or t = 19.4215
t = 4.6 or t = 19.4 to the nearest tenth of an hour
To begin this problem, we need to set up the equation and solve for the values of t when the outside temperature is equal to 75 degrees.
The given equation is:
O(t) = 11 cos (π/12 t) + 71
We want to find the values of t when O(t) = 75.
So, we set up the equation:
75 = 11 cos (π/12 t) + 71
Now, we can solve this equation to find the values of t.
To begin solving this problem, we need to find the points in time when the outside temperature is equal to 75 degrees. We can do this by setting the equation equal to 75 and solving for t.
The given equation is: O(t) = 11 cos(pi/12 t) + 71
Setting it equal to 75, we have:
11 cos(pi/12 t) + 71 = 75
Subtracting 71 from both sides:
11 cos(pi/12 t) = 4
Now, we need to isolate the cosine term to find the value of t.
Divide both sides by 11:
cos(pi/12 t) = 4/11
To solve for t, we need to take the inverse cosine (or arccos) of both sides:
t = (12/pi) * arccos(4/11)
Using a calculator, we can find the value of arccos(4/11) and then multiply it by (12/pi) to get the value of t. Finally, we round this value to the nearest tenth of an hour.
So, to begin this problem, set up the equation O(t) = 11 cos (pi/12 t) + 71 and solve for t by using the steps mentioned above.