Answer to Question #93254 in Trigonometry for Joann

Question

May someone help me with this HW problem?

The depth (D metres) of water in a harbour at a time (t hours) after midnight on a particular day can be modelled by the function
D = 2 sin(0.51t - 0.4) +5, t <=15,
where radians have been used.

Select the two options which are correct statements about the predictions based on this model.

Select one or more:
a) The smallest depth is 5 metres. 
b) At midday, the depth is approximately 7 metres. 
c) The model can be used to predict the tide for up to 15 days. 
d) The largest depth is 7 metres.
e) At midnight the depth is approximately 4.2 metres.
f) The time between the two high tides is exactly 12 hours. 
g) The depth of water in the harbour falls after midnight.

Answer 1

Solution.a)

Let us find the maximum and the minimum of the function,which describes the depth in a harbour regarding time.

1.To do this we need to find the derative of the function:

$D'=2\cos(0.51t-0.4)×0.51+5$

$D'=1.02\cos(0.51t-0.4)$

2.Then to find the critical points:

$1.02\cos(0.51t-0.4)=0$

$\cos(0.51t-0.4)=0$

$0.51t-0.4=\frac{\pi}{2}+\pi k$

$0.51t=\frac{\pi}{2}+\pi k+0.4$

$t=\frac{50\pi}{51}+\frac{100\pi k}{51}+\frac{40}{51}$

$t=\frac{50\pi+100\pi k+40}{51}$ ,$k\isin Z$

$t$ is an amount of hours after mignight,so it cannot be negative. Consequently,$t\isin(0;15]$ .

The next step is to consider only those values of $t$ which match the aboveestablished interval by changing the significance of $k$ .

If $k=0$ ,$t\approx4.$

If $k=1$ , $t\approx10.$

By using the interval method regarding the derative of the function we can make out that $t(max)=4$ and $t(min)=10$ (These are the only critical points that can be found on set interval.)

The smallest depth is the minimum of the function. And we can define it :

$D(min)=D(10)$

$D(min)=2\sin(0.51×10-0.4)+5$

$D(min)\approx3(m)$.

So that,the $a)$ statement is false.

b)

At midday means at $12:00$ am. According to this, $t=12$ .

Having this knowledge we can find a depth at midday.

$D(midday)=D(12)$

$D(midday)=2\sin(0.51×12-0.4)+5$

$D(midday)\approx4(m)$ .

That is why the $b)$ statement is wrong.

c)

As long as $t\isin(0;15]$ and $t$ means a number of hours after midnight, the model can be used to predict the tide for up to 15 hours.

The $c)$ statement is wrong either.

d)

We define the largest depth by the following way:

$t(max)=4$ , so $D(max)=D(4)$

$D(max)=2\sin(0.51×4-0.4)+5$

$D(max)\approx7$ .

The $d)$ statement is true.

e)

At midnight means at 0:00 am.

$t=0$ .

$D(midnight)=D(0)$

$D(midnight)=2\sin(0.51×0-0.4)+5$

$D(midnight)\approx4.2$ .

The $e)$ is true.

f)

As long as this exact model can predict only up for 15 hours after a mignight,it cannot define the time between hight tides. If we had a warrant to predict the state of depth of water after 15-st hour we would claim that $f)$ is true because the following $t(max)_2\approx16$ (if we consider a function,where $t\isin(-\infin;+\infin)$ ). And $t(max)_2-t(max)=12$ indeed($t(max)=4)$

But since $t\isin(0;15]$ we don't know what the sutiation will be after 15-st hours, so cannot approve or refute $f)$ statement.

g)

If we draw a graphic of the function,we can see a growth in the significance of depth of water after a midnight(sketch1.0)

sketch1.0

Also we can deduce the pattern by comparing the depth of water at $0:00$ am and $1:00$ am($t=1)$

$D(1)=2\sin(0.51×1-0.4)+5$

$D(1)\approx5$

$5>4$ , so that $D(1)>D(midnight)$ .

We can see,that the depth of water increase after midnight.

The $g)$ statement is false.

Answer: a),b),c),g) are wrong;d),e) are true.