Question

Hello! Was wondering what are the horizontal and vertical shifts of the problem. Thanks!

Answer 1

We know that a cosine function can model the situation. In general, the cosine function can be expressed by:

`h(t)=A\cos(\omega t+\phi)+k`

Where A is the amplitude, ω is the angular frequency, φ is the phase (related to the initial value), and k is the vertical shift.

From the problem, we know that the period T is 20 seconds. We can find the angular frequency from the period using the following relation:

`\begin{gathered} \omega=(2\pi)/(T)=(2\pi)/(20) \\ \\ \Rightarrow\omega=(\pi)/(10) \end{gathered}`

Additionally, by definition, the amplitude is half the distance from the highest to the lowest point:

`\begin{gathered} A=(4)/(2) \\ \\ \Rightarrow A=2\text{ feet} \end{gathered}`

The average value of the cosine function is the so-called "midline", related to k:

`k=10\text{ feet}`

Part B

If for t = 0, h(t) = 10 feet (the average value; the bottle is moving upwards), we can use this result to find the phase of the function:

`\begin{gathered} h(0)=10 \\ 2\cos((\pi)/(10)\cdot0+\phi)+10=10 \\ \\ \cos\phi=0 \\ \\ \Rightarrow\phi=-(\pi)/(2) \end{gathered}`

Then, the equation of the function is:

`h(t)=2\cos((\pi)/(10)t-(\pi)/(2))+10`

Part C

The graph of the function from t = 0 to when it reaches the lowest value for the first time is:

From the graph, we can see that the bottle reaches the lowest value for the first time in 15 seconds.