1. Suppose that the outdoor temperature (in Fahrenheit) on a particular day was approximated by the function

T(t) = 50 + 14sin[(pi(t))/12]

Where t is time in hours after 9 AM.

a) find the max (Tmax) and min (Tmin) temperature, And the average temperature:
Tav = 1/12 ∫(0 —> 12) T(t)dt on that day during the period between 9 AM and 9 PM.

B) show that Tav is not equal to (Tmin + Tmax)/2

C) show that T is not given by the above formula, but rather T(t) is a linear function of t, then Tav= (Tmin+Tmax)/2
(use an integral to explain)

To find the maximum (Tmax) and minimum (Tmin) temperatures, as well as the average temperature (Tav), we can follow these steps:

a)

1. To find the maximum and minimum temperatures, we need to determine when the sine function reaches its maximum and minimum values. The maximum value of sin(x) is 1, and the minimum value is -1.

2. In the given function T(t) = 50 + 14sin[(π(t))/12], the maximum temperature occurs when sin[(π(t))/12] equals 1, and the minimum temperature occurs when sin[(π(t))/12] equals -1.

3. Setting sin[(π(t))/12] = 1, we can solve for t:
π(t)/12 = π/2
t = 12/2 = 6

Therefore, Tmax occurs at 6 hours after 9 AM.

4. Setting sin[(π(t))/12] = -1, we can solve for t:
π(t)/12 = -π/2
t = -12/2 = -6

However, since time values must be positive in this case, Tmin cannot occur at -6. Thus, we need to find the next occurrence of sin[(π(t))/12] = -1 that is positive.

5. Setting sin[(π(t))/12] = -1, we solve for t again:
π(t)/12 = π + π/2
t = 12 + 6 = 18

Therefore, Tmin occurs at 18 hours after 9 AM.

6. Now, substitute these values of t into the original function T(t) to find the corresponding temperatures:
T(6) = 50 + 14sin[(π(6))/12] = 50 + 14sin[π/2] = 50 + 14(1) = 50 + 14 = 64°F
T(18) = 50 + 14sin[(π(18))/12] = 50 + 14sin[3π/2] = 50 + 14(-1) = 50 - 14 = 36°F

Therefore, Tmax = 64°F and Tmin = 36°F.

7. To find the average temperature (Tav), we need to calculate the definite integral of T(t) from 0 to 12 (the period between 9 AM and 9 PM):
Tav = (1/12) * ∫(0 —> 12) T(t) dt

Plugging in the values into the integral:
Tav = (1/12) * ∫(0 —> 12) (50 + 14sin[(π(t))/12]) dt

We can integrate each term separately:
Tav = (1/12) * [50t - 14(12/π)cos[(π(t))/12]] (evaluated from 0 to 12)

Simplifying and substituting the values:
Tav = (1/12) * [50(12) - 14(12/π)cos[π] - (50(0) - 14(12/π)cos[0])]
= (1/12) * (600 - 14(12/π) + 14(12/π))
= (1/12) * 600
= 50°F

Therefore, Tav = 50°F.

b)

Now, we need to show that Tav is not equal to (Tmin + Tmax)/2.

Tav = 50°F (calculated above)
(Tmin + Tmax)/2 = (36°F + 64°F)/2 = 100°F/2 = 50°F

Since Tav = 50°F and (Tmin + Tmax)/2 = 50°F, we can conclude that Tav is equal to (Tmin + Tmax)/2.

c)

To show that T(t) is not given by the given formula, but rather as a linear function of t, then Tav = (Tmin + Tmax)/2, we need to reason why we cannot integrate T(t) to get Tav in the given case.

Considering the given T(t) = 50 + 14sin[(π(t))/12], we can see that the function is not linear because it contains the sine function. Integrating the sine function with respect to t would yield a cosine function, which changes the linearity of the function T(t).

If T(t) were a linear function of t, it would be of the form T(t) = at + b, where a and b are constants. In this case, we could integrate T(t) over the interval from 0 to 12 (9 AM to 9 PM) to find Tav, and it would indeed be equal to (Tmin + Tmax)/2.

However, since T(t) is not represented by a linear function, the integral cannot be used to compute Tav as (Tmin + Tmax)/2.