Example 4 the hours of daylight in seattle oscillate

Example 4
The hours of daylight in Seattle oscillate from a low of 8.5 hours in January to a high of
16 hours in July
4
. When should you plant a garden if you want to do it during a month
where there are 14 hours of daylight?
To model this, we first note that the hours of daylight oscillate with a period of 12
months.
2
12 6
B
π π
= =
corresponds to the horizontal stretch, found by using the ratio of
the original period to the new period.
With a low of 8.5 and a high of 16, the midline will be halfway between these values, at
25 . 12
2
5 . 8 16
=
+
.
The amplitude will be half the difference
between the highest and lowest values:
75 . 3
2
5 . 8 16
=
-
, or equivalently the
distance from the midline to the high or
low value, 16-12.25=3.75.
Letting January be t = 0, the graph starts
at the lowest value, so it can be modeled
as a flipped cosine graph. Putting this
together, we get a model:
25 . 12
6
cos 75 . 3 ) ( +






- = t t h
π
h(t) is our model for hours of day light t months after January.
To find when there will be 14 hours of daylight, we solve h(t) = 14.
25 . 12
6
cos 75 . 3 14 +






- = t
π
Isolating the cosine
4

444 Chapter 6






- = t
6
cos 75 . 3 75 . 1
π
Subtracting 12.25 and dividing by -3.75






= - t
6
cos
75 . 3
75 . 1 π
Using the inverse
0563 . 2
75 . 3
75 . 1
cos
6
1
≈






- =
-
t
π
multiplying by the reciprocal
927 . 3
6
0563 . 2 = ⋅ =
π
t
t =3.927 months past January
There will be 14 hours of daylight 3.927 months into the year, or near the end of April.
While there would be a second time in the year when there are 14 hours of daylight,
since we are planting a garden, we would want to know the first solution, in spring, so
we do not need to find the second solution in this case.
Try it Now
2. The author’s monthly
gas usage (in therms) is
shown here. Find a
function to model the
data.
Example 6
An object is connected to the wall with a spring that has a
natural length of 20 cm. The object is pulled back 8 cm past
the natural length and released. The object oscillates 3 times
per second. Find an equation for the horizontal position of the
object ignoring the effects of friction. How much time during each cycle is the object
more than 27 cm from the wall?
If we use the distance from the wall, x , as the desired output, then the object will
oscillate equally on either side of the spring’s natural length of 20, putting the midline
of the function at 20 cm.
If we release the object 8 cm past the natural length, the amplitude of the oscillation will
be 8 cm.
We are beginning at the largest value and so this function can most easily be modeled
using a cosine function.
0
20
40
60
80
100
120
140
160
Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec

Section 6.5 Modeling with Trigonometric Functions 445
Since the object oscillates 3 times per second, it has a frequency of 3 and the period of
one oscillation is 1/3 of second. Using this we find the horizontal compression using the
ratios of the periods:
π
π
6
3 / 1
2
=
.