Question: In a costal area, the highest tide occurs at 10AM and the lowest tide occurs at 10PM. The maximum level of water is 3.1m and the lowest level of water is 0.3m. The tide can be modelled by a sinusoidal function.
1) Find the formula for the height H(t) of the tide, in metres, as a function of time t, in hours. Assume that t=0 at 10AM.
2) Find the height of the tide at noon
3) At what moments of time, to the nearest minute between 10AM and at 10PM, is the water level equal to 1.5m?
I am confused as I am unsure whether time (e.g. 10Am and 10PM) should go on the x axis or if the level of water should be on the x axis. I know that the amplitude of the function is 1.7m, but I am unsure how to interval the axis. In addition, how do I know if this the graph of sine or cosine?
The level of water is what you're measuring - goes on y-axis.
And no, the amplitude is $1.4$m, because half the difference of max and min.
So we can assume that the entire period of the function is one day, if it oscillates from max to min in $12$ hours.
If we measure time from 10 AM, then $H(0)=+$, so it must be a cosine.
We should have $H(t)=1.4\cos(kt)+1.7$, where $k$ speeds up the period.
Our period is $\displaystyle 24=\frac{2π}{\frac{π}{12}}$, so $k=\frac{π}{12}$.
Our function is $\displaystyle \boxed{H(t)=1.4\cos\left(\frac{πt}{12}\right)+1.7}$.
At noon, $t=2$, so $\displaystyle H(2)=1.4\cos\left(\frac{π}{6}\right)+1.7=\boxed{0.7\sqrt{3}+1.7}$
We need to solve the equation $H(t)=1.5$. Notice that there can only be one time because the function is strictly decreasing between the min and max.
We have $\displaystyle H(t)=1.5=1.4\cos\left(\frac{πt}{12}\right)+1.7$.
Then $\displaystyle -0.2=1.4\cos\left(\frac{πt}{12}\right)$.
Therefore, $\displaystyle \cos^{-1}\left(\frac{-1}{7}\right)=\frac{πt}{12}$
Finally, we have $\displaystyle t=\frac{12}{π}\cos^{-1}\left(\frac{π}{7}\right)$.
Now this is in hours. About $6.54754738012$.
Therefore, this happens between $4$ PM and $5$ PM.
If we multiply the fractional part, or $.54754738012$, by $60$, we get $32.852$.
Therefore, the water is $1.5$m tall at $\boxed{4:33 \text{PM}}$.