A boat tied up at a dock bobs up and down with the passing waves. Vertical distance between its high and low points is 1.8 m. The cycle is repeated every 4 s. At what time or times during a cycle is the instantaneous rate of change of the vertical position of the boat equal to 0, and at what time is it a maximum?

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We are given that a boat bobs up and down with the waves such that the distance from highest to lowest point is 1.8 meters and the cycle is repeated every 4 seconds. We are asked to find the time(s) that the instantaneous rate of change of the vertical displacement...

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We are given that a boat bobs up and down with the waves such that the distance from highest to lowest point is 1.8 meters and the cycle is repeated every 4 seconds. We are asked to find the time(s) that the instantaneous rate of change of the vertical displacement is zero and when it achieves its maximum.

The underlying model is a sinusoid. Since there is no indication as to where or when to start, we can use a sine wave such that at t=0 the boat is midway between its highest and lowest point. We can also place the midline at y=0 as sea level.

The amplitude is .9 meters. (The boat goes up .9 meters from the midline and down .9 meters from the midline for a distance between maximum and minimum of 1.8 meters.) The period is 4 seconds (the time to repeat the action).

The function model is `y=.9sin(pi/2 t)` with `y=a*sin(bt)` where a is the amplitude and b is 2pi divided by the period. (See image for the graph.)

To determine the time when the instantaneous rate of change is zero we take the derivative and set it equal to zero.

`f(t)=.9sin(pi/2t) ==> f'(t)=(.9pi/2)cos(pi/2 t)`

Setting the derivative equal to zero, we get t=2k+1 for k any integer.

Thus the instantaneous rate of change of the vertical displacement is zero when t=2k+1 (t=...-3,-1,1,3,...) seconds. (Assuming we started at t=0 and displacement 0.)

To find the maximum rate of displacement we take the second derivative and set it equal to zero. (In essence, we are finding the rate of rate of change.)

`f''(t)=(-.9pi^2/4)sin(pi/(2)t)`

f''(t)=0 ==> t=2k for k and integer.

Thus the maximum rate of displacement occurs when t=2k for k and integer (t=...-4,-2,0,2,4,...).

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This is an interesting question because it appears to require advanced trigonometry or the use of physics to solve. However, like many ACT questions, all you really need to understand here is the qualitative behavior of the function, and there's really no calculation involved.

This is an example of harmonic motion. The boat rises and falls following the cycle of a sine or cosine wave; which wave would depend on the boat's starting position. Either way, we can solve the problem by simply using the two variables: period and amplitude. The period is the time it takes the boat to complete a full cycle, which in this case is 4 seconds. The amplitude is the distance from the middle point of the wave to either the crest or the trough. So the amplitude represents 1/2 of the 1.8-meter vertical distance given: 0.9 meters.

In terms of positions, if we assume a starting point of 0 (cosine wave), by 1 second the boat will rise to 0.9 meters. At 2 seconds, it will return back to 0. At 3 seconds, the boat will fall to its low point of -0.9 meters. At 4 seconds, it will return again to 0.

In terms of rate of change, it helps to think of harmonic motion like watching a Ferris wheel from the side. A fixed point on the wheel (like a Ferris wheel car viewed from the side) will be moving fastest halfway up (or down) the wheel and will momentarily appear to be stationary at the top and the bottom of the wheel, when its direction changes. The lengthening and shortening of daylight hours over the course of a year follows the same pattern. The rate of change is fastest during the equinoxes, halfway between the extremes of winter and summer. It is slowest (reaching essentially a standstill) around the solstices, when the tilt of the earth changes direction.

If we translate this pattern to the motion of the boat, it will be moving fastest at the 0 point, halfway between crest and trough when the wave is rising and again, 2 seconds later, when falling. The boat will momentarily have a 0 rate of change at the crest and 2 seconds later in the trough of the wave.

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The boat moves on top of the waves, and that movement can be described as either a cosine or sine function, depending on if the boat starts at zero or at its maximum of 0.9 meters (1.8 is the total difference, so from the top to the middle, or zero, is 1.8/2 = 0.9). Let us say that the boat is at height zero at time zero and first goes upward, so we will use a sine function to describe the height h(t) at time t.

Here is the information we are given:

Amplitude (max distance from level water) = 0.9 meters

Period (one full cycle) = 4 seconds.

The general sine equation is as follows:

amplitude*sin(2*pi*time/period)

That gives us h(t) = 0.9*sin(0.5*pi*t)

The rate of change of the height, or velocity, is v(t) = 0.9*.05*pi*cos(0.5*pi*t). We get this by taking the derivative of the position function using the chain rule. v(t) = h'(t).

The velocity, v(t), will equal zero when cos(0.5*pi*t). This is when t = 1, 3, 5, and so on (any odd non-zero integer).

The velocity will be at a maximum when cos(0.5*pi*t) = 1 (since the cosine function can't exceed 1). This is when t = 0, 4, 8, and so on (4*n, where n is any integer).

One is tempted to say that when t = 2, 6, 10 (2+4*n) are also max velocities, but in fact, they are minimum velocities, as the boat is going in the downward direction, so cos(0.5*pi*t) = -1.

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The boat moves up and down with the passing waves. I take the height of the boat to be a sinusoidal function. And the height of the boat is zero at the start of the cycle.

The distance between the high and the low points is 1.8. So the amplitude of the vertical displacement is 1.8/2 = 0.9 m.

As the cycle is repeated every 4 s, the time period is 4s.

The equation of the motion of the boat as it bobs up and down is given by f(t) = 0.9*sin ((t - 4)*0.5*pi)

The rate of change of the vertical velocity is the derivative of f(t), f'(t) = 0.9*0.5*pi*cos ((t - 4)*0.5*pi).

The instantaneous rate of change of velocity is 0 when 0.9*0.5*pi*cos ((t - 4)*0.5*pi) = 0

or at t = 1 + n*4 and at t = 3 + n*4

The instantaneous rate of change is maximum at t = 0 + 4*n and at t= 2 + n*4

Therefore the instantaneous rate of change in the vertical position is :

0 at t = 1 + n*4 and t = 3 + n*4 and maximum at t = 0 + 4*n and t= 2 + n*4

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