Consider the function
[tex]\begin{gathered} f(t)=Acos(Bt+C)+D \\ A\rightarrow\text{ amplitude} \\ B\rightarrow period=\frac{2\pi}{B} \\ C\rightarrow horizontal\text{ shift} \\ D\rightarrow\text{ vertical shift} \end{gathered}[/tex]
The period of the function is 20 seconds; then,
[tex]\begin{gathered} \frac{2\pi}{B}=20 \\ \Rightarrow B=\frac{\pi}{10} \end{gathered}[/tex]
On the other hand, the distance from the highest to the lowest point is 4 feet; thus,
[tex]\begin{gathered} A=\frac{4}{2}=2 \\ \Rightarrow A=2 \end{gathered}[/tex]
The answers to part A) are amplitude=2ft and period=20 seconds.
On the other hand, the average height of the function is equal to its midline which is located at y=10, according to the question. Then,
[tex]D=10[/tex]
So far, the function that models the problem is
[tex]\begin{gathered} f(t)=2cos(\frac{\pi t}{10}+C)+10 \\ \end{gathered}[/tex]
B) Given that the height of the bottle is equal to 10 at t=0,
[tex]\begin{gathered} f(0)=10 \\ \Rightarrow cos(C)=0 \\ \Rightarrow C=\pm\frac{\pi}{2} \end{gathered}[/tex]
Furthermore, after t=0 the bottle moves upwards; then, C=-pi/2.
Then, the function that represents the situation is
[tex]\Rightarrow f(t)=2cos(\frac{\pi t}{10}-\frac{\pi}{2})+10[/tex]
C) The graph of the function is
According to the graph, after 15 seconds, the function will reach its lowest height.
