in a certain city the temperature (in degrees fahrenheit) to hours after 9am was modeled by T(t)=50+14Sin((pi(t))/12. Find the highest temperature during the period from 9AM to 9PM. Indicate units

Counting the brackets, I will assume you meant

T(t)=50+14Sin((pi(t))/12)
besides that would make the period 24 hours, which makes sense

since t = 0 corresponds with 9:00 am
the temp at 9:00 am was 50°
9:00 am to 9:00 pm is 12 hours.

the min will be 50 - 14 = 36
and the max will be 50+14 = 64° <--- at 3:00 pm , t = 6

What is the derivative of T(t)=50+14Sin((pi(t))/12)?

We are able to use a calculator but I keep getting different answers because I think Im not sure how to get the derivative on the calculator

There is really no need to actually use Calculus for this.

You asked for the max.
sin(??) runs from -1 to +1
so 14Sin((pi(t))/12) has a max of 14(1) = 14
and the max of T(t) = 50+14 = 64

however, if you want to use Calculus
T'(t) = 14cos(((π(t))/12)(π/12) = (14π/12) cos (((π(t))/12) = 0 for a max/min
cos (((π(t))/12) = 0
π(t)/12 = π/2 , 3π/2
t = 6 or t = 18

sub t = 6 into the original to get 64 , the max
sub t = 18 into the original to get 36, the min

To find the highest temperature during the period from 9 am to 9 pm, we need to determine the highest point (peak) of the function T(t) = 50 + 14sin((πt)/12).

The given function represents a sinusoidal function, where T(t) represents the temperature at a specific time t, and t represents the number of hours after 9 am.

In this model, the peak of the sinusoidal function occurs when the sine function reaches its maximum value of 1. Let's find the time corresponding to this peak temperature:

1. Set the sine function equal to 1: sin((πt)/12) = 1

2. Solve for t:
(πt)/12 = π/2 (since sin(π/2) = 1)
t = (12/π) * (π/2) = 6

So, the peak temperature occurs 6 hours after 9 am, i.e., at 3 pm.

Now, let's find the value of T(6) to determine the highest temperature:

T(6) = 50 + 14sin((π(6))/12)
= 50 + 14sin(π/2)
= 50 + 14 * 1
= 50 + 14
= 64

Therefore, the highest temperature during the period from 9 am to 9 pm is 64 degrees Fahrenheit (°F).