"I wonder if the water depth will be ideal for fishing then," you wonder aloud as you pick up the Thursday paper. On the front of the paper is a headline that reads "Record High Tides Expected." You skim the article and conclude that there will be a high tide that midnight.
1. Draw a graph of the tide and write a trig function that will be ideal for fishing on Friday at terms of time.
I got the real picture of the graph by the information that was given in the problem but a little confusing at first but later after your it more you will understand it all. I began by making a y and x axis graph and began put my information on which the water level on the y axis and my time on the x axis. We know that the high tide max is at 24 feet and our low tide min is at 12 feet, to find our mid(sinusoidal axis) so it will be 18 feet but it also say that the ideal fishing will be between 18 feet to 15 feet. The time that it will take one full wave was every 11 hours 6 minutes, it tells me that will begin at Friday morning till midnight and a high tide will start at zero at the 24 feet mark and a whole day is 24 hours and I knew it has to have two waves on the graph. So here is my final graph that tells the whole story for when(time) the ideal fishing will be good for that whole day on Friday and what water level will it be.....
max = 24 feet (high tide)
sinusoidal axis= 18 feet (mid tide)
min = 12 feet (low tide)
find the period:
B= 2(pie)/period
= 2(pie)/11.1
half the period = 11.1/2
= 5.55
full 2 period = 11.1(2)
= 22.2
quarter period = 1/2(5.55)
= 2.775
sine
A= -6
B= 2(pie)/11.1
C= 2.775
D= 18
cosine
A= 6
B= 2(pie)/11.1
C= 0
D=18
A is our amplitude or the distance from the sinusoidal axis to out max/min values
B is our value that tells us how long does a full wave makes but its not period but determines it
C is our phase shift(horizontal shift) or moves right or left along the x axis
D is another phase shift(vertical shift) or average value between our max and min values
So my final trig function that i have wrote down is how i will figure out my other questions
y=A(f)B(x-C)+D
d(t)= -6sin[2(pie)/11.1 (x-2.775)]+18 (optional)
d(t)= 6cos[2(pie)/11.1 (x)]+18
Period 0, 11.1, 22.2
2.Use your function to determine if the water depth will be ideal for fishing on Friday at 5:00PM
d(5)= 6cos[2(pie)/11.1(x)]+18
[2(pie)/11.1(x)] is all the same angle so we....
Let 0=[2(pie)/11.1(x)]
5= 6cos0 +18
5-18= 6cos0 +18-18
-3= 6cos0
-3/6=6cos0/6
-3/6=cos0
-.5=cos0
So we know that negative cosine is negative in two quadrants so the solution will be in the quadrants of II & III
arc cos of -.5 or (cos-1)-.5= 2.094395102
so we round it off 4 decimal place... 2.0944
cosine is negative in quadrant II & III
quad II 0= 2.0944 quad III 0=(pie)+2.0944= 5.2360
so, [2(pie)/11.1(x)] = 2.0944 and [2(pie)/11.1(x)] = 5.2360
Multiplying by the reciprocal will make it easier for us to solve the equation, it will help us by taking out the 2(pie) out of the numerator, and moving it to the denominator.
first solving 0= 2.0944
[2(pie)/11.1(x)]=2.0944
(11.1/2(pie))[2(pie)/11.1(x)]=2.0944(11.1/2(pie))
the L.H.S reduce each other, so we are left with..
x=2.0944(11.1/2(pie))
=(2.0994)x(11.1/(2(pie)))
=3.7088
second solving 0= 5.2360
[2(pie)/11.1(x)]=5.2360
(11.1/2(pie))[2(pie)/11.1(x)]=5.2360(11.1/2(pie))
the L.H.S reduce each other, so we are left with..
x=5.2360(11.1/2(pie))
=(5.2360)x(11.1/(2(pie)))
=9.2500
both 0=3.7088 and 0=9.2500 are both water depth
11.06 am
between 11.1 + .9 = 12
5 = 5
5.9 till 5 pm
ideal fishing is 18-15 feet
but 5pm is past the 18-15 feet range
5pm is almost at 12 feet
and he wanted to go at 5 pm
it last about 1.5 hours for ideal fishing before 5 pm
3. What are the best times to go fishing on Friday?
4. How long does the "ideal fishing time" last?
