What is the answer for question 7) ? http://postimg.org/image/qfwflx4j1/
You will need to plot the data points.
The equation will be of the form y=asin(kt-c)+d. I will start from the form y=asin(b(x-h))+d.
a: a is the amplitude. The amplitude can be found by `a=("max"-"min")/2`
Using the largest data point as an estimate for the maximum and the smallest as an estinate for the minimum we get `a=(32.9-21.1)/2=5.9`
b: Is correlated with the period; `b=(2pi)/p` for the sine function. Since the period is 24 we have `b=(2pi)/24=pi/12` .
d: d is the vertical translation. The midline is `y=("max"+"min")/2` so here we have `y=(32.9+21.1)/2=27` as the midline. This represents a vertical translation up 27 units from the parent function.
h: h is the horizontal translation or phase shift. Since the sine curve starts at the midline and then increases, we look at the graph and estimate that this occurs at approximately 9:30 in the morning. So the parent curve has been shifted right 9.5 units.
So a=5.9, `b=pi/12` , h=9.5, and d=27. We have `y=5.9sin(pi/12(x-9.5))+27`
In the form y=asin(kx-c)+d we have:
** My calculator gives a regression of
y=5.97141sin(.261x-2.4685)+27.016; so we have a good approximation **
(1) What is the temperature at 4:30pm? We substitute x=16.5 into the equation:
`y=5.9sin(.262(16.5)-2.487)+27~~32.6937` so the temperature will be approximately `32.7^@`
(2) When will the temperature be `30^@` ? Substitute 30 for y:
sin(.262x-2.487)=.50847 Take the inverse sin of both sides
.262x-2.487=.5334 or 2.6082 (using radian mode)
.262x-2.487=.5334 ==> .262x=3.0204==> x=11.528
.262x-2.487=2.6082 ==> .262x=5.0952==> x=19.447
So the temperature will be 30 degrees Celsius at approximately 11:30 am and 7:26pm