Module 5: Lesson 2

 

Self-Check 2
  1. Questions 5, 16, and 18 on pages 275 to 279 of the textbook
    1.  
      1. A graphical solution is shown.

         
        The graph of y equals the sine of pi divided by 4 times x minus 6 is shown. A second graph of y = 0.5 is also shown. The intersection point is at (1.33, 0.5).

        x ≈ 1.33
      2. An algebraic solution is shown.

         


         


        363.590…° is not in the domain 0° < x ≤ 360°. However, the coterminal angle 363.590…° − 360° = 3.590…°.

        So, x ≈ 3.59° and 86.41°.
      3. An algebraic solution is shown.

         


        On the domain 0 < 2x − 5 ≤ 2 π,

         


        x ≈ 3.093 + nπ, n ∈ I, x ∈ R and x ≈ 5.048 + nπ, n ∈ I, x ∈ R.
      4. A graphical solution is shown.

         
        The graph of y equals 5.2 times the sine of 45 times x minus 8 degrees all minus 1 is shown. The graph of y equals –3 is also shown. There are intersection points at (4.5, –3) and (7.5, –3).

        The first two solutions occur at x ≈ 4.5° and 7.5°. Because the period is 8°, the general equation is x ≈ 4.5°, 7.5° ± (8°) n, n ∈ I.
    1.  
      1. This diagram shows a scatter plot of the data provided in the question. The domain is from 0 to 12 (representing all of the months in one year) and the range is approximately from –20 to 20 degrees.
        Adapted from:  Pre-Calculus 12. Whitby, ON:
        McGraw-Hill Ryerson, 2011. Reproduced with permission. 

      2. The temperature that is halfway between the maximum average monthly temperature and the minimum average monthly temperature in Winnipeg can be found by performing the following operation:

         
      3. The a value can be determined by finding the amplitude:

        This means that the amplitude is

        The graph appears to be a sinusoidal graph (of cosine) that has been reflected about the x-axis; this means that you can state that the a value is −18.1. If this is done, there will be a phase shift of 1 unit to the right (because the lowest temperature occurs in January, or the first month). The period is about 12 months. This means that the b value is  The vertical displacement is

        Therefore, an approximate equation for the sinusoidal function is  In this formula, x represents the time, in months, and y represents the average monthly temperature, in degrees Celsius, for Winnipeg.
      4.  This diagram shows a scatter plot of the data provided in the question. The domain is from 0 to 12 (representing all of the months in one year) and the range is approximately from –20 to 20 degrees. The graph of the equation as described in 16.c. has been placed on the graph as well. The graph is a very good approximation of the scatter plot.
        Adapted from: Pre-Calculus 12. Whitby, ON:
        McGraw-Hill Ryerson, 2011. Reproduced with permission. 

      5.  The graph of the equation as described in 16.c. is shown. The domain is from 0 to 12 (representing all of the months in one year) and the range is approximately from –20 to 20 degrees. A horizontal line through y = 16 is also shown as are the two intersection points, which occur at (5.76, 16) and (8.24, 16).

        The two intersection points occur at (5.76, 16) and (8.24, 16). This means that the average temperature will be above 16°C for 8.24 − 5.76, or for approximately 2.5 months.
    1.  
      1. This diagram shows a sketch of the sinusoidal function. The points (0.3, 60) and (1.8, 40) are indicated.
        Adapted from Pre-Calculus 12. Whitby, ON: McGraw-Hill Ryerson, 2011.
        Reproduced with permission.


      2. The a value can be determined by finding the amplitude:

        This means that the amplitude is  you can state that the a value is 10. If this is done, there will be a phase shift of 0.45 units to the left. The period is 3.3 s − 0.3 s or 3 s. Thus, the b value is  The vertical displacement is

        Thus, an approximate equation for the sinusoidal function is  In this formula, t represents the time, in seconds, and y represents the height of the mass, in centimetres, above the floor.
      3. To do this, you can substitute 17.2 in place of t.

         
      4. The graph of the equation as described in 18.c. is shown. The domain is from 0 to 2 seconds and the range is approximately from 0 to 65 cm. A horizontal line through y = 59 is also shown is the first intersection point. It occurs at (0.0847, 59).


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