Answer:
(a)[tex]y=5sin(\frac{\pi}{6}(x+2))+5[/tex]
(b) 7.5ft above the low tide.
Step-by-step explanation:
(a) To find the function that computes the height of the tide, you need to select the form of the sinusoidal function. For example, use the form:
[tex]y=Asin(B(x-C))+D[/tex]
Where A is the amplitude, B the frequency, C the phase shift and D the vertical shift.
The amplitude is half the distance between the highest and the lowest tide:
[tex]A=10/2=5ft[/tex]
The frequency is related to the period T by:
[tex]B=\frac{2\pi}{T}[/tex]
The period is 12 hours, then
[tex]B=\frac{2\pi}{12}=\frac{\pi}{6}[/tex]
The high tide is at 1:00 a.m. and 1:00 p.m. , this is the moment when [tex]sin(B(x-C))=1[/tex], if [tex]sin(\frac{\pi}{2})=1[/tex] then [tex]B(x-C)[/tex] must be equal to [tex]\frac{\pi}{2}[/tex] when [tex]x=1[/tex]:
[tex]B(x-C)=\frac{\pi}{2}\\\frac{\pi}{6}(1-C)=\frac{\pi}{2}\\\frac{1}{6}(1-C)=\frac{1}{2}\\(1-C)=\frac{6}{2}\\-C=3-1\\C=-2[/tex]
The vertical shift is the sum of the lowest value, the height of the low tide ([tex]lt[/tex]) and the amplitude:
[tex]D=5+lt[/tex]
The function is:
[tex]y=5sin(\frac{\pi}{6} (x+2))+5+lt[/tex]
Because the function must be the height above low tide height, subtract this heigh from the function:
[tex]y=5sin(\frac{\pi}{6} (x+2))+5+lt-lt[/tex]
[tex]y=5sin(\frac{\pi}{6} (x+2))+5[/tex]
(b) Use x=11 in the function
[tex]y=5sin(\frac{\pi}{6} (11+2))+5=2.5+5=7.5ft[/tex] above the low tide.
