Answer:
The sine equation is given below as
[tex]y=Asin(B(x+C)+D[/tex]
From the graph, The highest point on the crest is
[tex]=5.5[/tex]
The lowest point of the trough is
[tex]=0.5[/tex]
Hence,
The amplitude will be
[tex]\begin{gathered} A=\frac{5.5-0.5}{2} \\ A=\frac{5}{2} \\ A=2.5 \end{gathered}[/tex]
From the image also, the period T, is
[tex]\begin{gathered} T=18 \\ T=\frac{2\pi}{B} \\ 18=\frac{2\pi}{B} \\ B=\frac{2\pi}{18} \\ B=\frac{\pi}{9} \end{gathered}[/tex]
There is no hrizontal shift,
Hence,
The value of C is
[tex]C=0[/tex]
The vertical shift D, will be
[tex]D=3[/tex]
Hence,
The model of the sine function is given below as
[tex]y=2.5\sin\left(\frac{\pi}{9}x\right)+3[/tex]
To figure out the depth at t=4hours, we will have substitue x=4 in the equation above
[tex]\begin{gathered} y=2.5\sin(\frac{\pi}{9}x)+3 \\ y=2.5\sin\mleft(\frac{\pi}{9}\times4\mright)+3 \\ y=5,462m \end{gathered}[/tex]
Hence,
The depth of the water at 4 hours is = 5.462m
To figure out the depth at t= 30 hours, we will have to substitute the value of x=30 in the equation above
[tex]\begin{gathered} y=2.5\sin(\frac{\pi}{9}x)+3 \\ y=2.5\sin\mleft(\frac{\pi}{9}\times30\mright)+3 \\ y=0.835m \end{gathered}[/tex]
Hence,
The depth of the water at 30 hours is = 0.835m
