MATHEMATICS

1.

y = sin ( (pi/3 * x)

Just a normal  sine graph with a different  period from normal  

This has the form

y= sin  (Bx)     where  B = pi/3

Period  = 2pi / B  =     2pi / (pi/3)  =  2pi * 3/pi  =  6

Here's the  graph

2.

We have

y= A cos (Bx)  + C

y = 3cos (4x)  +  5

A= the amplitude  = 3    (vertical stretching)

B = 4

The   period = 2pi/ B =  (2pi) / 4  =  pi/2  =  ( horizontal compression from  normal period of 2pi)

5  just shifts  the normal  graph up 5 units

Here's  the normal graph  and our transformed graph 

3. At a dock on the east coast, low tide occurs at 3 p.m. with a water depth of 5ft. The depth at high tide is 83 ft. High tide occurs every 6 hours. Explain how to find the sinusoidal function that models the depth in terms of time, X.

The function will be

y =  A cos (Bx + C)  +  D

High tide will occur at  noon  and  6 PM

Low tide occurs at 3PM and 9PM

The period =    6

So

2pi / B = 6

B = (2/6)pi  =  pi/3

Amplitude = A  = (high point -low point)/ 2 = (83 - 5) / 2  =  39

D =   (high point + low point) / 2 =    (83 + 5)  / 2 =  44 

Let noon be  x  =  0

The  hardest thing to figure is "C"

When x=  0, y= 83   we can  solve for "C"  thusly

y = 39sin (pi/3 *x + C) + D

83   = 39 sin (pi/3 * 0  + C)   + 44

39  = 39 sin (C)

39/39 = sin (C)

1  = sin (C)

arcsin (1)   = C

We are asking....where is the sin   = 1 ???     answer...at pi/2

arcsin (1)  = pi/2  =  C

So...our  function  is

y = 39 sin (pi/3 * x  + pi/2) + 44

Here's the graph