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

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

The constant b changes the base period of the function. The base period of sine is 2πœ‹, so

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

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

Example 2

The side s of a tent forming the shape of a straight circular cone is 5 metres. Find the magnitude of the angle between the ground plane and side s so that the volume of the tent is as large as possible.

Let x be the requested angle. We call the height h and the radius r of the base by the angle x. The altitude line and the radius of the base are perpendicular to each other. The angle should be between] 0, πœ‹/2 [

The volume of straight circular cone

The volume gets its maximum value at the same point as the function

The factor 125πœ‹ / 3 does not affect the angle at which the maximum value is reached.



Zero product property

These solutions are invalid because they are not between ]0, πœ‹/2[

Another factor

Since we are between ]0, πœ‹/2[, only a positive value for the tangent is valid

The function has an extreme point when x = 0,615

This is the maximum point, because

In this case, the maximum volume of the tent is with an angle of

The problem could also have been solved using angles instead of radians.

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