# Example 4 the hours of daylight in seattle oscillate

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Example 4The hours of daylight in Seattle oscillate from a low of 8.5 hours in January to a high of16 hours in July4.When should you plant a garden if you want to do it during a monthwhere there are 14 hours of daylight?To model this, we first note that the hours of daylight oscillate with a period of 12months.2126Bππ==corresponds to the horizontal stretch, found by using the ratio ofthe 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, at25.1225.816=+.The amplitude will be half the differencebetween the highest and lowest values:75.325.816=-, or equivalently thedistance from the midline to the high orlow value, 16-12.25=3.75.Letting January bet= 0, the graph startsat the lowest value, so it can be modeledas a flipped cosine graph.Putting thistogether, we get a model:25.126cos75.3)(+-=tthπh(t)is our model for hours of day lighttmonths after January.To find when there will be 14 hours of daylight, we solveh(t)= 14.25.126cos75.314+-=tπIsolating the cosine4
444Chapter 6-=t6cos75.375.1πSubtracting 12.25 and dividing by -3.75=-t6cos75.375.1πUsing the inverse0563.275.375.1cos61-=-tπmultiplying by the reciprocal927.360563.2==πtt=3.927 months past JanuaryThere 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, sowe do not need to find the second solution in this case.Try it Now2. The author’s monthlygas usage (in therms) isshown here.Find afunction to model thedata.Example 6An object is connected to the wall with a spring that has anatural length of 20 cm.The object is pulled back 8 cm pastthe natural length and released.The object oscillates 3 timesper second.Find an equation for the horizontal position of theobject ignoring the effects of friction.How much time during each cycle is the objectmore than 27 cm from the wall?If we use the distance from the wall,x, as the desired output, then the object willoscillate equally on either side of the spring’s natural length of 20, putting the midlineof the function at 20 cm.If we release the object 8 cm past the natural length, the amplitude of the oscillation willbe 8 cm.We are beginning at the largest value and so this function can most easily be modeledusing a cosine function.020406080100120140160JanFebMarAprMayJunJulAugSepOctNovDec
Section 6.5 Modeling with Trigonometric Functions445Since the object oscillates 3 times per second, it has a frequency of 3 and the period ofone oscillation is 1/3 of second. Using this we find the horizontal compression using theratios of the periods:ππ63/12=.

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Inverse function