Sinusoidal function can be defined as follows.
Let A, B, C and D be fixed constants, where A and B are both positive. Then we can form the new function
which is called a sinusoidal-function. The four constants can be interpreted graphically as indicated:
Given a sinusoidal-function in the standard form
once the constants A, B, C, and D are specified, any graphing device can produce an accurate graph. However, it is pretty straight forward to sketch a rough graph by hand and the process will help reinforce the graphical meaning of the constants A, B, C, and D. Here is a “five step procedure” one can follow, assuming we are given A, B, C, and D. It is a good idea to follow the example given below as you read this procedure; that way it will seem a lot less abstract.
Step 1 :
Draw the horizontal line given by the equation y = D; this line will
“split” the graph of
into symmetrical upper and lower halves.
Step 2 :
Draw the two horizontal lines given by the equations y = D±A. These two lines determine a horizontal strip inside which the graph of the sinusoidal-function will oscillate. Notice, the points where the sinusoidal-function has a maximum value lie on the line y = D + A.
Likewise, the points where the sinusoidal function has a minimum value lie on the line y = D− A. Of course, we do not yet have a prescription that tells us where these maxima (peaks) and minima (valleys) are located; that will come out of the next steps.
Step 3 :
Since we are given the period B, we know these important facts:
(i) The period B is the horizontal distance between two successive maxima (peaks) in the graph. Likewise, the period B is the horizontal distance between two successive minima (valleys) in the graph.
(ii) The horizontal distance between a maxima (peak) and the successive minima (valley) is 0.5B.
Step 4 :
Plot the point (C,D). This will be a place where the graph of the sinusoidal function will cross the mean line y = D on its way up from a minima to a maxima.
This is not the only place where the graph crosses the mean line; it will also cross at the points obtained from (C,D) by horizontally shifting by any integer multiple of 0.5B. For example, here are three places the graph crosses the mean line:
(C, D), (C + 0.5B, D), (C+B, D)
Step 5 :
Finally, midway between (C,D) and (C+0.5B, D) there will be a maxima (peak); i.e. at the point (C + 0.25B, D + A). Likewise, midway between (C + 0.5B, D) and (C + B, D) there will be a minima (valley); i.e. at the point (C+ 0.75B, D−A).
It is now possible to roughly sketch the graph on the domain C ≤ x ≤ C + B by connecting the points described. Once this portion of the graph is known, the fact that the function is periodic tells us to simply repeat the picture in the intervals
C+B ≤ x ≤ C+2B, C − B ≤ x ≤ C, etc.
To make sense of this procedure, let’s do an explicit example to see
how these five steps produce a rough sketch
To have better understanding on "How to roughly sketch Sinusoidal function?", let us go through the following example.
The temperature (in ◦C) of Adri-N’s dorm room varies
during the day according to the sinusoidal function
where t represents hours after midnight. Roughly sketch the graph of d(t) over a 24 hour period.. What is the temperature of the room at 2:00 pm? What is the maximum and minimum temperature of the room?
We begin with the rough sketch. Start by taking an inventory of the constants in this sinusoidal function:
Conclude that A = 6, B = 24, C = 11, D = 19. Following the first four steps of the procedure outlined, we can sketch the lines y = D = 19, y = D ± A = 19 ± 6 and three points where the graph crosses the mean line (See the figure given below)
According to the fifth step in the sketching procedure, we can plot the maxima (C+ 0.25B, D+A) = (17, 25) and the minima (C+ 0.75B, D−A) = (29, 13). We then “connect the dots” to get a rough sketch on the domain 11 ≤ t ≤ 35.
Finally, we can use the fact the function has period 24 to sketch the
graph to the right and left by simply repeating the picture every 24 horizontal units.
We restrict the picture to the domain 0 ≤ t ≤ 24 and obtain the computer generated graph (see the figure given below ) as you can see, our rough graph is very accurate. The temperature at 2:00 p.m. is just d(14) = 23.24◦ C. From the graph, the maximum value of the function will be D + A = 25◦ C and the minimum value will be
D − A = 13◦ C.