Function question

[jsc90]

Hi, I need help on solving this:

The function h(t)=3.9sin0.16pi(t-3)+6.5 gives the depth of water, h metres, at any time, t hours, during a certain day. A cruise ship needs at least 8 metres of water to dock safely. Use the graph of the function to estimate the number of hours in the 24 hour interval starting at t=0 during which the cruise ship can dock safely.

So, how should I approach this question ? (A diagram would help)

 

[Unco]

The equation

. \(\displaystyle \L \mbox{ h(t) = A\sin{(B(t - C))} + D}\)

. . . has:

. . . . amplitude \(\displaystyle \L A\)

. . . . period: \(\displaystyle \L \mbox{\frac{2\pi}{B}}\) (if in radian measure, which we tend to assume unless degrees are stated somewhere, and most certainly when \(\displaystyle \L \pi\)'s involved in the equation.)

. . . . been translated to the right C units (compared to h(t) = sin(t) ).

. . . . been translated up D units (compared to h(t) = sin(t) ).

Your graph has to be pretty good to get a decent estimate!

You should try solving h(t) = 8 afterwards to get a precise interval, provided it isn't outside the scope of what you have covered in class so far.

 

[soroban]

Hello, jsc901

The function $\displaystyle h(t)\:=\:3.9\cdot\sin[0.16\pi(t-3)]\,+\,6.5$
gives the depth of water, $\displaystyle h$ metres, at any time, $\displaystyle t$ hours, during a certain day.

A cruise ship needs at least 8 metres of water to dock safely.

Use the graph of the function to estimate the number of hours in the 24-hour interval
starting at $\displaystyle t=0$ during which the cruise ship can dock safely.

I assume they expect you to graph the function.
$\displaystyle \;\;$Once you do that, locate the times when $\displaystyle h\,\geq\,8$.

I solved it without a graph . . .

When is: $\displaystyle \:3.9\cdot\sin[0.16\pi(t-3)]\,+\,6.5\;\geq\;8$ ?

$\displaystyle \;\;$Do some Algebra: $\displaystyle \;\sin[0.16\pi(t-3)]\:\geq\:\frac{5}{13}$


Now do some Trig . . . carefully!

$\displaystyle \;\;$ If $\displaystyle \,\sin\theta\:\geq\;\frac{5}{13}$, then it would seem that: $\displaystyle \,\theta\:\geq\:\arcsin\left(\frac{5}{13}\right)$

Not quite . . . there is an upper limit, too.
$\displaystyle \;\;\;\arcsin\left(\frac{5}{13}\right)\:\leq\:\theta\:\leq\;\pi\,-\,\arcsin\left(\frac{5}{13}\right)$


So we have: $\displaystyle \:0.3948\:\leq\:0.16\pi(t\,-\,3)\:\leq\:2.7468$

. . . . . . . . . . . . . $\displaystyle 0.7854\:\leq\:t\,-\,3\:\leq\:5.4646$

. . . . . . . . . . . . . . . $\displaystyle 3.7854\:\leq\:t\:\leq\:8.4646$

And there is the "window" for safe docking.