Problem+Set+3+Solution+Q1

= Problem Set 3: = = Q1 Solution =

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Problem
There is a causeway in St. Andrews By The Sea, New Brunswick that becomes flooded by the sea a few times each day. This “Road Under the Sea” can be used when the tide water is low but as the tide gets higher, the road becomes covered with water. The rising and falling tide models the shape of the cosine wave where the height of the water is given by: h = cos(0.5t) + 10,where h is the height in metres and t is the time in hours. At low tide the depth of the water is 3 m and the road gets covered by water when the depth of the water is 8 metres. a. What is the depth at high tide? b. If 12:00 midnight is represented by x = 0 on the graph, determine the times when the road can be used.

Connection to Curriculum:
__Specific Expectations: **MCR3U** 2.1, 2.5, 2.6, 3.5
 * Grade 11:**__

Solutions
Before answering the parts of the question, we will first model the given equation by looking at the various transformations to cos(t)
 * Algebraically || Verbally || Graphically ||
 * h = cos(t) || The standard cosine function

- amplitude: 1 - period: 2π - y-intercept: 1 || || + Horizontal stretch by a factor of 0.5
 * h = cos(** 0.5 **t) || The standard cosine function

- amplitude: 1 - period: 4π - y-intercept: 1 || || + Horizontal stretch by a factor of 0.5 + Vertical stretch by a factor of 7 - amplitude: 7 - period: 4π - y-intercept: 7 || || + Horizontal stretch by a factor of 0.5 + Vertical stretch by a factor of 7 + Positive vertical shift by 10 - amplitude: 7 - period: 4π - y-intercept: 17 || ||
 * h = ** 7 ** cos(0.5t) || The standard cosine function
 * h = 7 cos(0.5t) ** + 10 ** || The standard cosine function

Now that we have different models of the given equation, we will look at the parts to the question.

a. What is the depth at high tide? Let t = 0 h = 7 cos(0.5(0)) + 10 h = 7 cos(0) + 10 h = 7 (1) + 10 h = 17 || We know low tide is at 3 metres, and since the amplitude is 7, the highest point on the curve will be 2*7 = 14 metres higher. || Looking at the graph, you can see that the highest point is 17 metres. || Therefore, the depth at high tide is 17 metres.
 * Algebraically || Verbally || Graphically ||
 * Since h = 7 cos(0.5t) + 10 is a standard cosine function without any horizontal shift or flip, so one of the highest points on the curve occurs on the y axis, when t = 0.

b. If 12:00 midnight is represented by x = 0 on the graph, determine the times when the road can be used. Let h = 8 8 = 7 cos(0.5t) + 10 8 – 10 = 7 cos (0.5t) -2/7 = cos(0.5t) cos-1(-2/7) = 0.5t t = 2 cos-1(-2/7) t ≈3.721
 * Algebraically || Graphically ||
 * We know the road can be used when the water depth is less than 8 metres. When h < 8.

Since we are considering a whole period (day), we know that h = 8 has two values. We also know that the curve is symmetrical along x = 2π, since there is a horizontal stretch by a factor of 0.5. Therefore, the second value of t when h = 8 will occur at 4π – 3.721 = 8.845.

We can convert these values of t into time values by cross-multiplication:

3.721 / 2π = x hours / 12 hours

//x// = (12 * 3.721) / 2π //x// ≈ 7.1 hours

0.1 hours / 1 hour = x minutes / 60 minutes

//x// = (60 * 0.1) / 1 = 6 minutes Therefore, the first point at which h = 8 is 7:06 AM.

Applying the same principals of symmetry as above, we can determine the second point on the curve when h = 8 to be 12:00 midnight minus 7.1 hours.

12 – 7 hours and 6 minutes = 4 hours and 54 minutes.

Therefore, the second point at which h = 8 is 4:54 PM

Since we have a cosine curve with no flip and no horizontal shift, we know that the curve’s high points are at 0 (midnight) and the end of the next period (4π, midnight the next day).

Therefore, the tide is below 8 metres between 7:06 AM and 4:54 PM. ||

We can change the x-axis from radians to actual time values by assuming that one period (4π) is one full day, midnight to midnight. Looking at the graph (or using graphing calculators or software and tracing the curve), we can estimate the x-values when the depth of the water is at 8 metres to be approximately 1.2π and 2.8π or 7:12 am and 4:48 pm. ||
 * For our first solution, we assumed that a period of 4π constituted a day, and we changed our x axis to reflect that. After presenting the problem to the class, we saw that it may have been incorrect to do that since the problem gives t as time in hours. Working with that, we realize that a full day could also be interpreted as 24 hours ≈ 7.6π hours, so all occurrences of h = 8 up to 7.6π should be considered.

We already have x = 3.721 from our previous calculations. Using the symmetry of the function, we also have x = 4π – 3.721 = 8.845. We can add 4π to each of those values until we surpass 24 hours. 3.721 + 4π = 16.287 8.845 + 4π = 21.411

Using the same conversion methods as above, we can determine that the road is safe between 3:43 AM and 8:51 AM and again between 4: 17 PM and 9:25 PM. || Similarly to above, you can drop down lines to estimate when h = 8 on the curve, you will just have to look at all those points up to 7.6π. ||