Question
The equation given below models the average monthly temperature (˚C). t denotes the number of months with t=1 representing January. During which months is the average monthly temperature 12.5˚C?
𝑇(𝑡) = 12.5 + 15.8sin (𝜋/6 (𝑡) −2𝜋/3)
𝑇(𝑡) = 12.5 + 15.8sin (𝜋/6 (𝑡) −2𝜋/3)
Answers
Anonymous
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"
Reiny
the period of your function is 2π/(π/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 (𝑡) −2𝜋/3) = 12.5
15.8sin (𝜋/6 (𝑡) −2𝜋/3) = 0
sin (𝜋/6 (𝑡) −2𝜋/3) = 0
I know sin 0 = 0 and sin π = 0
so π/6t - 2π/3 = 0
π/6t = 2π/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?
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 (𝑡) −2𝜋/3) = 12.5
15.8sin (𝜋/6 (𝑡) −2𝜋/3) = 0
sin (𝜋/6 (𝑡) −2𝜋/3) = 0
I know sin 0 = 0 and sin π = 0
so π/6t - 2π/3 = 0
π/6t = 2π/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?
Anonymous
october