The depth of water at the end of a pier varies with the tides. The low tides occur on a particular day at 2:00 a.m. and 2:00 p.m. with a depth of 1.1 m. The high tides occur at 8:00 a.m. and 8:00 p.m. with a depth of 5.3 m. A large boat needs at least 3 m of water to be safely secured at the end of the pier.

Develop an exact trigonometric function that models the depth of water in meters t hours after midnight.

 

 Note: An exact function does not use decimal approximations.

 

1)Describe the step-by-step process used to determine the exact function from part B, including a discussion on each of the following:

• amplitude

• period

• horizontal translation

• vertical translation

2)Calculate the time to the nearest minute that the boat can first be safely secured by using the function from part B. Explain step-by-step all mathematical calculations, the use of technology, or both.

1. ?f(t)=2.1sin(6π?t−65π?)+3.2?
2. The boat is safe to dock at 4:49 a.m. to 11:11 a.m. and 4:49 p.m. to 11:11 p.m.

Step-by-step explanation

1. ?f(t)=Asin(Bx−C)+D?

a) Amplitude

To find Amplitude we find the difference of the high tide and low tide, and we divide by 2.

5.3 meters - 1.1 meters = 4.2 meters

4.2 meters / 2 = 2.1 meters

Amplitude = 2.1 meters

b) Period

A simple formula to find the period of the graph involves using the wavelength. The wavelength is the distance of a complete wave. Hence the wavelength of the concerning graph is 12 since the a complete wave is completed every 12 hours (8 a.m. to 8 p.m.).

The formula to is ?B2π?=λ?. Since we know ?λ? = 12 we can plug it in.

?B2π?=12?

?B=122π?=6π??

c) Horizontal Translational

To find the horizontal shift, if we represent it as a sine function, we first notice that the difference of hours from the low tide to the high tide is 8 a.m. - 2 a.m. = 6 hours. The middle point of the graph is found right in between these two hours, hence we can find it at 2+6/2 = 5 a.m. or 5 p.m. Now that we know at what time the middle point of the graph occurs, we can find the horizontal shift using the formula ?BC?=h? where ?h? is the known hours.

?6π?C?? ?=5?

?C=65π??

d) Vertical Translation

To find the vertical translation ?D? simply add the amplitude and the low tide point, or subtract the high tide point with the amplitude.

?LT+A=1.1+2.1=3.2?

?HT−A=5.3−2.1=3.2?

Finally the formula looks like this: ?f(t)=2.1sin(6π?t−65π?)+3.2?

2.To find the the time when the boat is able to dock(3 meters), we solve for t in the following equation

?2.1sin(6π?t−65π?)+3.2=3.0?

Subtract both side by 3.2

?2.1sin(6π?t−65π?)=−0.2?

Divide both side by 2.1

?sin(6π?t−65π?)=−0.0952?

Apply arcsin (inverse sin) to both sides

?6π?t−65π?=arcsin(−0.0952)=−0.0953?

Add both side by ?65π??

?6π?t=−0.0953+2.618=2.523?

Divide bot side by ?6π??

?t=π6?∗2.523=4.819?

This translates to 4 hours 49 minutes. Hence, the boat can dock from 4:49 a.m. to 11:11 a.m. and 4:49 p.m. to 11:11 p.m.