The equation given below models the average monthly temperature (ΛC). t denotes the number of months with t=1 representing January. During which months is the average monthly temperature 12.5ΛC?
π(π‘) = 12.5 + 15.8sin (π/6 (π‘) β2π/3)
reason im asking this question is because i got April (only) but i feel as if im missing something. as in the questions it says "monthS"
the period of your function is 2Ο/(Ο/6) or 12.
So it represents 1 year of temperature change.
first look at the graph:
http://www.wolframalpha.com/input/?i=plot+y++%3D+12.5+%2B+15.8sin+(%CF%80%2F6+x+%E2%88%922%CF%80%2F3)+,+y+%3D+12.5+from+0+to+12
notice the sine curve crosses y = 12.5 twice, once at t = 4 and t = 10
now let's do it algebraically.
12.5 + 15.8sin (π/6 (π‘) β2π/3) = 12.5
15.8sin (π/6 (π‘) β2π/3) = 0
sin (π/6 (π‘) β2π/3) = 0
I know sin 0 = 0 and sin Ο = 0
so Ο/6t - 2Ο/3 = 0
Ο/6t = 2Ο/3
t/6 = 2/3
3t = 12
t = 4
use similar steps to show t = 10
if t = 4, that would present April.
What would t = 10 represent?
october
To determine during which months the average monthly temperature is 12.5ΛC, we need to solve the equation T(t) = 12.5.
The equation given is:
T(t) = 12.5 + 15.8sin(Ο/6(t) - 2Ο/3)
To find the values of t for which T(t) = 12.5, we can set up the equation and solve for t.
12.5 = 12.5 + 15.8sin(Ο/6(t) - 2Ο/3)
First, we can subtract 12.5 from both sides of the equation to eliminate the constant term:
0 = 15.8sin(Ο/6(t) - 2Ο/3)
Next, we can divide both sides of the equation by 15.8 to simplify:
0 = sin(Ο/6(t) - 2Ο/3)
Now, we need to solve for the value inside the sine function. To do this, we can set the argument of the sine to different angles and find the corresponding values of t.
Ο/6(t) - 2Ο/3 = Ο/2
Solving for t:
t = (Ο/2 + 2Ο/3) / (Ο/6)
t = (3Ο/6 + 4Ο/6) / (Ο/6)
t = 7Ο/6 / (Ο/6)
t = 7/1
t = 7
So, when the average monthly temperature is 12.5ΛC, it corresponds to the month t = 7, which represents July.