The captain of a shipping vessel must consider the tides when entering a seaport. High tide is at 5am at 10.6 meters and low tide is at 11am at 6.5 meters. The vessel has a draft of 7.5 meters, during what times can the vessel safely enter the port?

Guest Oct 9, 2014

#2**+10 **

Melody's work was impressive......!!! Here's a slightly less sophisticated manner of looking at this * if* we assume that the tide rises and falls in a "linear" fashion.

The following graph illustrates the situation.........

Let x= 0 on the following graph equate to "5AM" and let x = 6 equate to "11AM"

And if we consider only the difference between the heighest and lowest tides to be 4.1 ft., then let the point (0, 4.1) be the "high" point of the tide. And the "low" point would be at y = 0 six hours later, i.e., the point (6,0).

And the ship can only enter the seaport when the tide is 1 ft above its "low point." This is represented on the graph by the function y = 1.

Therefore, the time that the ship can enter the seaport starting at 5AM would be from x = 0 to x = 4.537 hrs. later....in other words...from 5AM to about 9:32AM

And note that, the ship would not be able to enter or leave again until about 7.463 hrs after 5AM = about 12:28PM!! Thus, assuming the linear model, there is about a 3 hour period when the tide is too low for the ship to enter or leave the port....!!!

Melody's model is probably more "real-world," .....mine is probably more simplistic....(Duh!!)

CPhill
Oct 10, 2014

#1**+10 **

(10.6-6.5)/2=2.05

the amplitude (A) of the tide is 2.05 metres.

7.5-6.5=1

For the lowest 1 metre of the tide the water will be too shallow.

Let y=0 represent the middle tide level.

So at high tide y=-2.05

At low tide y=2.05

the water will be too shallow if y<-1.05

11am-5am=6hours The period of the wave is 6*2=12hours

I am going to use the formula y=A*cos[n(x+s)]

A is amplitude, s is phase shift, period is 2pi/n

Now period is 12 so

2pi/n=12

n=pi/6

If there was no phase shift then high tide would be at midnight but it is at 5am so the phase shift is 5 hours to the right n=-5

So we have have our formula

$$y=2.06*sin[\frac{\pi}{6}(x-5)]$$ (remember that it is in radians)

we need to solve this simultaneously with y=-1.05

You can do this by hand without too much trouble but i have done it graphically.

These decimal numbers must be changed to times. I have calculated them to the nearest minute.

x=0 is midnight

x=0.97 0.97*60=58 This is 12:58am

x=9.03 0.03*60=2 This is 9:02am

x=12.97 This is 12.58pm

x=21.02 This is 9:02pm

So the water will be deep enough between 12:58am and 9:02am

and between 12:58pm and 9:02pm

Better leave an extra minute or 2 just to be safe.

Melody
Oct 10, 2014

#2**+10 **

Best Answer

Melody's work was impressive......!!! Here's a slightly less sophisticated manner of looking at this * if* we assume that the tide rises and falls in a "linear" fashion.

The following graph illustrates the situation.........

Let x= 0 on the following graph equate to "5AM" and let x = 6 equate to "11AM"

And if we consider only the difference between the heighest and lowest tides to be 4.1 ft., then let the point (0, 4.1) be the "high" point of the tide. And the "low" point would be at y = 0 six hours later, i.e., the point (6,0).

And the ship can only enter the seaport when the tide is 1 ft above its "low point." This is represented on the graph by the function y = 1.

Therefore, the time that the ship can enter the seaport starting at 5AM would be from x = 0 to x = 4.537 hrs. later....in other words...from 5AM to about 9:32AM

And note that, the ship would not be able to enter or leave again until about 7.463 hrs after 5AM = about 12:28PM!! Thus, assuming the linear model, there is about a 3 hour period when the tide is too low for the ship to enter or leave the port....!!!

Melody's model is probably more "real-world," .....mine is probably more simplistic....(Duh!!)

CPhill
Oct 10, 2014