First High & Low Tide in Harbor: Water Depth at 12am & 6pm

  • Thread starter Thread starter TonyC
  • Start date Start date
  • Tags Tags
    Depth Water
AI Thread Summary
The water depth in the harbor is modeled by the equation y = 21 - 5cos(πt/6), where y represents the depth in feet and t is the time in hours since midnight. To find the first high and low tides, the maximum and minimum values of the cosine function must be determined. The maximum value of cos(πt/6) occurs when t = 0, leading to the first high tide at midnight with a depth of 26 feet. The minimum value occurs when cos(πt/6) is -1, which happens at t = 6 hours, resulting in the first low tide at 6 AM with a depth of 16 feet. This analysis highlights the relationship between the cosine function and tidal patterns in the harbor.
TonyC
Messages
86
Reaction score
0
Suppose the depth of the tide in a certain harbor can be modeled by y=21-5cos pi t/6, whre y is the water depth in feet and t is the time in hours. Consider a day in which t=0 represents 12:00 midnight. For that day, when are the first high tide and the first low tide and what is the water depth at each time?
 
Physics news on Phys.org
You have:
-1 \leq \cos \alpha \leq 1
So what can you say about the y = 21 - 5 \cos \frac{\pi t}{6}
Viet Dao,
 
What are you looking for?
 
Okay, when y is minimum, that means:
5 \cos \frac{\pi t}{6} is maximum.
So \cos \frac{\pi t}{6} is maximum.
So y is minimum means that \cos \frac{\pi t}{6} is maximum.
-1 \leq \cos \frac{\pi t}{6} \leq 1 so what t makes \cos \frac{\pi t}{6} maximum?
Viet Dao,
 
you can take the first derivitive and then find when it is 0 to tell you where your critical points are then use that to find the min and max
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...

Similar threads

Back
Top