Question

I was given a graph and told to find an equation for the temperature T in terms of the time t in the form T(t) = a sin(kt-c)+d where a, k, c and d are constants.

The max was 32.9 and the min 21.1

I determined that

amplitude = 5.9
vertical = 27
period = pi/12

The last thing I have to find is phase shift please help!

Answer 1

I agree with a = 5.9
d = 32.9 - 5.9 = 27 , you had that

so forget the phase shift for the time being.

testing your equation:
t(t) = 5.9 sin (π/12t) + 27
32.9 = 5.9sin(π/12t) + 27
sin(12t/π) = (32.9-27)/5.9 = 1
12t/π = π/2
t = (π/2)(12/π) = 6

21.1 = 5.9sin(12t/π) + 27
sin(12t/π) = -1
12t/π = 3π/2
t = (3π/2)(12/π) = 18

so the max occurs at 6 units of time and the min at 18 units of time
period = 24 units of time

As it stands with no phase shift,
at t = 0 , temp = 27
at t = 6, temp = 32.9
at t = 12 , temp = 27
at t =18, temp = 21.1
at t = 24 , temp = 27 , etc

To have a phase shift, there must have been something given such as,
we know that if t = 4, temp = some value

Answer 2

Well I have a chart so I know that when t = 4 temp = 21.1

So where would I go from there?

Answer 3

Ignore most of the above, I have some of the fractions upside down,
the period is right at 24, making k = π/12

You must have meant:
k = π/12 , not period = π/12

and our basic equation is

T(t) = 5.9 sin ((π/12)t -c ) + 27

Now that you told me that temp = 21.1 when t = 4 , we can say

21.1 = 5.9 sin((π/12)(4) - c) + 27
sin (π/3 - c) = -1
π/3 - c =3π/2
-c = 3π/2 - π/3 = 7π/6

<b>T(t) = 5.9 sin( πt/12 + 7π/6 ) + 27</b>

testing:
if t = 16 , we should get 32.9

T(16) = 5.9 sin(16π/12 + 7π/6 ) + 27
= 5.9 sin (5π/2) + 27
= 5.9(1) + 27 = 32.9 ----- YEAHHHH

Answer 4

Thank you so much you've been a HUGE help, this question has had me stumped for so long.