Anyways, this is our solution.
SLIDE 17:
This equation gives the depth of the water, h meters, at an ocean port at any time, t hours during a certain day
h(t) = 2.5 sin[2pi(t - 1.5)/12.4] + 4.3
A) Explain the significance of each number in the equation
II) 12.4 is the Period of the function, which is obtained from parameter B - 2pi/12.4
III) 1.5 - This is parameter C, which is the phase shift, in which this function, the graph shifts to the right 1.5 units/hours.
IV)4.3 - This is parameter D, which is the sinusoidal axis, which shifts the sinusoidal axis up 4.3 metres..
B) What is the minimum depth of the water? When does it occur?
Now, We can go backwards and use that point, but since we can use whichever point, we decided to use the next one, which minimum depth, 1.8 metres, occurs at 10.8 hours.
C) Determine the depth of the water at 9:30 am.
So then we just plug it in.
h(9.5) = 2.5sin[2pi(9.5 - 1.5)/12.4] + 4.3
*t is 9.5 because time ~ 9:30 AM ~ is converted to 9.5 because :30 minutes is .5 hours.*
in which we get... 2.323 metres.
D) Determine one time when the water is 4.0 metres deep.
So then we go something like...
4 = 2.5sin[2pi(t-1.5)/12.4] + 4.3
First we subtracted 4.3 from both sides like so.
-0.3 = 2.5sin[2pi(t-1.5)/12.4]
Then we added the phase shift to both sides to get...
1.2 = 2.5 sin[2pi(t)/12.4]
Then we divided parameter A out.
0.48 = sin[2pi(t)/12.4)
We then divided 2pi/12.4 in which .48 would be multiplied by the reciprocal of 2pi/12.4, and moved sine to the other side to change it to ARCSine to isolate t.
ARCSine(0.9473) = t
Which would equal to --> 1.2447 hours.
To make it efficient, we'll multiply the .2247 by 60 so
Depth of water of 4 metres occurs at 1:13:48 am.
Again, sorry for posting so late. Original personnel that was to post, did not post. Feel free to comment, as we are supposed to.
