Question from Math B Regents (modelling the tide)

[tjkubo]

This is from the Math B Regents August 2004.

The tide at a boat dock can be modeled by the equation:

. . .y = –2 cos((π/6)t) + 8

...where t is the number of hours past noon and y is the height of the tide, in feet. For how many hours between t = 0 and t = 12 is the tide at least 7 feet?

I tried doing this algebraically:

. . .7 ≤ –2 cos((π/6)t) + 8
. . .-1 ≤ –2 cos((π/6)t)
. . .1/2 ≥ cos((π/6)t)
. . .1/2 ≥ cos(30t)
. . .30t ≥ arccos(1/2)
. . .30t ≥ 60....or....30t ≥ 300
. . .t ≥ 2................t ≥ 10

As an answer I got 9 hours.

I know it's wrong. I just don't know where I made a mistake.
*π this is pi by the way

 

[pka]

Just graph the function. You will see the answer.

 

[soroban]

Re: Question from Math B Regents

Hello, tjkubo!

A small error . . .

The tide at a boat dock can be modeled by the equation: \(\displaystyle \,
y \:= \: -2\cos\left(\frac{\pi}{6}t\right)\,+\,8\)
where $\displaystyle t$ is the number of hours past noon and $\displaystyle y$ is the height of the tide in feet.
For how many hours between $\displaystyle t\,=\,0$ and $\displaystyle t\,=\,12$ is the tide at least 7 feet?

-\(\displaystyle \L2\cos\left(\frac{\pi}{6}t\right) + 8\:\geq\:7\)

$\displaystyle \;\;\;$-\(\displaystyle \L2\cos\left(\frac{\pi}{6}t\right)\;\geq\;-1\)

\(\displaystyle \L\;\;\;\;\,\cos\left(\frac{\pi}{6}t\right)\;\leq\;\frac{1}{2}\) . . . We divided by a negative!

\(\displaystyle \L\;\;\;\;\;\;\;\;\;\,\frac{\pi}{6}t\;\leq\;\arccos\left(\frac{1}{2}\right)\)


Hence: \(\displaystyle \L\,\frac{\pi}{3}\;\leq\;\frac{\pi}{6}t\;\leq\;\frac{5\pi}{3}\)


Multiply by \(\displaystyle \L\frac{6}{\pi}:\;\;2\;\leq \; t\; \leq\;10\)


Answer: $\displaystyle \,8\text{ hours}$