The document discusses tidal patterns over the month of September. It provides the maximum and minimum tide heights, dates of high and low tides, and uses this information to construct a trigonometric model of the tide levels over time. The model is then used to determine on what specific dates the tide height would reach 53 feet, allowing access to a cavern. The length of time the tide stays above 53 feet is also calculated.

1. Question 1 Solution Trigonometric Modeling Cirular Functions and Transformations

2. In order to do this, you must first sketch yourself a graph of the wave on the cartesian plane. Your max value would be the Spring (the higher tide) tide ( 60 ft ). Your min value would be the Neap (the lower tide) tide ( 23 ft ). To figure out how to scale the period on to the cartesian plane, we can use the dates given, which is Sept. 4, Sept. 11, Sept. 18, and Sept. 25. Sept. 4 - Last Quarter (Neap Tide) Sept. 11 - New Moon (Spring Tide) Sept. 18 - 1 st Quarter (Neap Tide) Sept. 25 - Full Moon (Spring Tide) 23 ft. 60 ft. 23 ft. 60 ft. Since Sept. 4 is during a Neap tide, and a Neap tide is a min, we can locate this point when x is 0. This makes it easier to create our equation for sine and cosine Will be at the point, (O, 23) S o l u t i o n t o ( a ) :

3. (max) 60 y 2 4 6 8 10 12 14 16 18 20 22 24 26 28 23 (min) 4 11 18 25 0 H e i g h t (ft) Days in September a way to calculate the period would be from max to max or min to min, which is how we did it here. Full Period: 14 + 0 = 14 days Half Period: 14 = 7 days 2 Note: What is shown here is not one period, but two because of the moon phases

4. Because you are making 4 at 0, each date ( 11 , 18 , and 25 ) have to be 'changed'. (max) 60 y 2 4 6 8 10 12 14 16 18 20 22 24 26 28 41.5 23 average value (min) 4 11 18 25 0 H e i g h t (ft) Days in September The average value (or sinusoidal axis is an invisible line that the sine or cosine graph wraps around - not included in the graph, so it is represented by a dotted line ) is calculated by taking the max height (60 ft) and adding it to the min height (23 ft), and dividing by 2. 60 + 23 2 sinusoidal axis: 41.5 ft =

5. Another way to find the sinusoidal axis is finding the value that is equally distant from the max to the average value, and min to the average value. Once that value is found, we can add it to the min height to get the average value. Max: 60 Min: 23 60-23 = 37 This number is what determines the value of the max and min from the sinusoidal axis. It is called the amplitude . Now to find the sinusoidal axis by adding the min value. 18.5 + 23 = 41.5 To know for sure that it is the sinusoidal axis, add and subtract the amplitude to the sinusoidal axis to make sure it results in the max and min. 18.5 37 2 =

6. Here is this the graph... (max) 60 y 2 4 6 8 10 12 14 16 18 20 22 24 26 28 41.5 23 (average value) (min) 4 11 18 25 0 H e i g h t (ft) Days in September x

7. Now we are ready to write our equations.... sine f(x) = A sin B (x- C )+ D cosine f(x) = A cos B (x- C )+ D sine cosine A B C D A B C D B determines the period and is not the period. -18.5 18.5 41.5 41.5 0 7 A , B , and D always remain the same for both equations. The only thing you should keep in mind is the sign on A. C (horizontal shift) is usually different for both equations. Your equations are: period = 2 π B = 2π B period 2π 14 = 2π 14 2π 14 sine: h(d) = 18.5 sin 2 π 14 [ (d - 7) [ + 41.5 cosine: h(d) = -18.5 cos 2 π 14 [ [ (d) + 41.5

8. S o l u t i o n t o ( b ) : 2. In order to reach the cavern, Shinobi has to reach a height of 53 feet. At what day would the tide be at this ideal height? Using one of the equations: cosine: h(d) = -18.5 cos [ [ (d) + 41.5 We can solve for the day. 53 = -18.5 cos [ [ (d) + 41.5 let = θ (d) 53 = -18.5 cos ( θ ) + 41.5 This makes solving the equation simpler and neater. From here we solve it algebraically. 2.5 = -18.5 cos ( θ ) = cos ( θ ) -0.1351= cos ( θ ) always have your answer to four decimal places (better accuracy) 2 π 14 2 π 14 2 π 14 2.5 -18.5

9. -0.1351= cos ( θ ) θ = 1.7063 These are the values of cos( θ ) in radians π + 1.7063 1.7063 = 4.8479 4.8479 , , 53 1.7063 1.4352 11.1312 7.9895 , Add 2 π to both of these values to get the other so you will get the other values necessary to complete this problem. 11.1312 7.9895 Each of these points do not mean that those are what your answers are, but where your answers should be around. (max) 60 y 2 4 6 8 10 12 14 16 18 20 22 24 26 28 41.5 23 (average value) (min) 4 11 18 25 0 H e i g h t (ft) Days in September x

10. -0.1351= cos ( θ ) θ = 1.7063 1.4352 , , 7.7184 7.9895 , (d) = Remember that θ = 1.7063 ( ) To isolate the variable (d), we multiply by the reciprocal for both sides of the equation. 1.7063 d = 3.8019 d = 53 As you can see on the graph below, the answer we get is pretty close to what we indicated on the graph and so we can conclude that at that point, it is right. However... That is not the date we want. This graph is not accurate, but merely shows what you should look for. Also remember that the 'start value' was at 4. So even though it is approxiamately 3.8 or 4 days, it is actually (rounding up) 8 days. So September 8 is one of the answers. 1.7063 (d) = 2 π 14 2 π 14 14 2 π 14 2 π ( ) 14 2 π ( ) (max) 60 y 2 4 6 8 10 12 14 16 18 20 22 24 26 28 41.5 23 (average value) (min) 4 11 18 25 0 H e i g h t (ft) Days in September x 2 π 14

11. (d) = (d) = (d) = 7.9895 So following the same procedure as we did with one of the values we get this: 4.8479 11.1312 (d) = 4.8479 d = 4.8479 d = 10.8019 days d = 11 days 11 + 4 = 15 September 15 (d) = 7.9895 d = 7.9895 d = 17.8020 days d = 18 days 18 + 4 = 22 days September 22 (d) = 11.1312 d = 11.1312 d = 24.8022 d = 25 days 25 + 4 = 29 September 29 2 π 14 2 π 14 2 π 14 14 2 π ( ) 14 2 π ( ) 14 2 π ( ) 14 2 π ( ) 2 π 14 14 2 π ( ) 2 π 14 14 2 π ( ) 14 2 π ( ) 2 π 14 14 2 π ( ) 14 2 π ( )

12. Now the easy way... If you had gotten September 8 and September 15 in the beginning in the first part, instead of doing the same thing two times, all you had to do was add the period. September 8 September 15 + 14 + 14 September 22 September 29 1.7063 (d) = ( ) 1.7063 d = 3.8019 d = 1.7063 (d) = September 8 (d) = 4.8479 (d) = 4.8479 d = 4.8479 d = 10.8019 days September 15 2 π 14 14 2 π 14 2 π ( ) 14 2 π ( ) 2 π 14 2 π 14 14 2 π ( ) 14 2 π ( ) 2 π 14 14 2 π ( )

13. So throughout the month of September, Shinobi was able to reach the ideal height of 53 ft at: September 8 September 15 September 22 September 29 Your answers are...

14. S o l u t i o n t o ( c ) : First you take the dates from the previous question. (max) 60 y 53 You want to find the number of days these lines represent September 8 September 15 Then you subtract them to get this length 15 - 8 = 7 days So now you know that the length of the first line is 7 days. and that is how long the tide is above 53. But you're not done. You still have this length to calculate You can either do use the same method you got the 7 days or you can just multiply the number by since the lines are the same length. 2 4 6 8 10 12 14 16 18 20 22 24 26 28 41.5 23 (average value) (min) 4 11 18 25 0 H e i g h t (ft) Days in September x

15. Your answer is... Shinobi is able to stay above 53 ft for... 14 days