Before starting these examples you might want to refresh you memory on solving these problems. If so, take a look at the procedure for this in the Modeling Real World Problems section.

Example 1:

If it takes 40 seconds to run around a circular 100 meter track, what is the angular velocity, b, per second?

Solution:
b = 360o/40 = 90o/sec.
Note that this is independent of the size of the track!



Example 2:

Here is a plot of the average weekly temperatures in a city:

Find a curve which best fits the shape of the graph above.

Solution:
From the shape of the graph, it sort of looks like a sine curve. But if we plot the sine curve on the same graph we get:

Which is not even similar to the original plot. However, if we stretch the sine graph and change the amplitude, it just might work. Lets use the general formula asin(bx+c) + d

The VERTICAL CENTER seems to be around 42oF. Therefore, we will let d.=.42. So we get:

Now we must stretch out the amplitude of the sine graph. The maximum value appears to be 72 and the minimum value is 11. The amplitude is (72-10)/2.=.31. So let a.=.31. So plotting the function 31sin(x).+.42 on the original plot we get:

Looks as though we are almost there. All we must do now is stretch the period of the sine function. Since the period of the sine function is 2pi, and the period of the temperature data is 52 weeks, we set b = 2pi/52. Now we plot the function 31sin(2pix/52).+.42 on the original graph to get:

We have accomplished the task! So the curve which best fits the temperature plot is
31sin(pix/52).+.42.



Example 3:

On Fegruary 10, 1990, high tide in Boston was at midnight. The water level at high tide was 9.9 feet; later, at low tide, it was 0.1 feet. Assuming the next high tide is exactly 12 hours later and that the height of the water is given by a sine or cosine curve, find a formula for water level in Boston as a function of time.

Solution:
Suppose the height of the water level is y feet, and let t be the time measured in hours from midnight. Then the oscillations are to have amplitude 4.9 feet (=(9.9-0.1)/2) and period 12, so 12b = 2pi and b.=.pi/6. Since the water is highest at midnight, when t.=.0, the oscillations are best represented by a cosine function, because the cosine is at its maximum at the beginning of the cycle. Thus we can say

    Height above average = 4.9cos(pit/6)

since the average depth of the water was 5 feet (= (9.9+0.1)/2), we want the cosine curve shifted up by 5. We get this by adding 5:

    y = 5 + 4.9cos(pit/6).

The graph would be as follows:

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