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Example 1
At the shore of Liisa-Petter's summer cottage, the tide changed the depth of water in 12-hour periods. Liisa-Petter measured the maximum depth one night at 0.30 and it was 1,5 metres. At 6.30 in the morning, the water depth was only 0,3 metres.
According to Liisa-Petter's research, the tide at her summer cottage followed the pattern
where t is the time in hours from midnight. We solve a, b, c and d from the function
Constant a
Constant a
The constant a indicates the amplitude of the function. The values of sine vary between [-1,1] and multiplying the sine with a number changes the minimum and maximum value by the same amount. The constant a is thus half the difference between the largest and smallest values.
The tidal period is 12 hours. Then we get
Constant b
Constant b
The constant b changes the base period of the function. The base period of sine is 2𝜋, so
Constant d
Constant d
The constant d moves the graph vertically. The amplitude of the function is 0,6 the maximum value is 1,5 and the minimum is 0,3, the equilibrium position of the graph should be at 0,9
1,5 - 0,6 = 0,9
0,3 + 0,6 = 0,9
Then the constant d = 0,9
Constant c
Constant c
The constant c moves the graph horizontally.
We know that half an hour after midnight the height of the water is 1,5 metres, f(0,5) = 1,5 and our function is now
Let us limit the solution to the base period, the constant c falls in the range [0,12]
From this we get the value of the constant c, with a calculator c is approximately 9,5.
The function that tells the water level, t hours after midnight is
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