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given trig function applied to tides. students need to find max depth, time of low tide and times between which a boat of given depth can use the port.
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\nLow tide will be at the minimum value on the graph. This will be the central line $\\var{a}$ minus the amplitude of $\\var{b}$ which gives $\\simplify{{a}-{b}}$
\n\nPart b
\nThe graph is shown below.
\nIt can be found by applying the following transformations to $y=cosx$
\n{geogebra_applet('https://www.geogebra.org/m/fzfgusaj',defs)}
\n\nFrom the graph you can see that the first high tide after midnight is at $t=12$ which is 12pm.
\n\nYou could also have solved the equation equal to the maximum value as shown below
\n\\[\\simplify{{a}+{b}}=\\var{a}+\\var{b}cos \\left( \\frac{\\pi t}{6} \\right) \\]
\n\nPart c
\nTo do this you need to solve as follows
\n\\[\\var{c}=\\var{a}+\\var{b}cos \\left( \\frac{\\pi t}{6} \\right) \\]
\n
Rearranging this gives
\\begin{align}
cos \\left( \\frac{\\pi t}{6} \\right) & =\\left( \\frac{\\var{c}-\\var{a}}{\\var{b}} \\right) \\\\\\\\
\\left( \\frac{\\pi t}{6}\\right) & = cos^{-1}\\left( \\frac{\\var{c}-\\var{a}}{\\var{b}} \\right)=\\var{pt_6} \\text{ or }2\\pi-\\var{pt_6} \\\\\\\\
t & = \\var{t1} \\text{ or } \\var{t2}
\\end{align}
So the boat cannot use the port between $\\var{t1_min} \\text{ minutes past } \\var{t1_hour}$ and $\\var{t2_min} \\text{ minutes past } \\var{t2_hour}$
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\n\\[x=\\var{a}+\\var{b}cos \\left( \\frac{\\pi t}{6} \\right) \\]
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\n[[0]]
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