The tides at cape Capstan, New Brunswick, change the depth of the water in the harbour. On one day in october, the tides have a high point of approximately 8m at 2 p.m and a low point of approximately 0.6m at 8 p.m. A particular sailboat has a draft of 2m. This means that it can only move in water that is at least 2m deep. when can the captin exit the harbour safely in the sailboat? Use an equation that would model the water depth in the harbour.
The motion of tides can be modelled sinusoidally, that is, by either a sine or a cosine curve
from a high of 8 m to a low of .6 m is 7.4m , so the amplitude of my curve will be 3.7.
From 2:00 pm to 8:00 pm is half a period so the period of the tide is 12 hours
period formula
2π/k = 12 ----> k = π/6
Height(t) = 3.7 cos (πt/6) where t is the time in hours past 2:00 pm
but that does not give us a max height of 8m, lets shift our curve vertically by 4.3 m
Height(t) = 3.7 cos(πt/6) + 4.3
test equation:
let t=0 (2:00 pm)
height(0) = 3.7cos 0 + 4.3 = 3.7(1) + 4.3 = 8
let t = 6 (8:00 pm)
height(8) = 3.7cos(π) + 4.3 = 3.7(-1) + 4.3 = .6
YEAHH!
now ...
2 = 3.7cos(πt/6) + 4.3
-2.3/3.7 = cos(πt/6)
cos(πt/6) = -.62162..
πt/6 = 2.24 or 4.04
t = 4.27 or t = 7.72 or 4:16 and 7:43 in time notation.
The water level will be below 2 m between
6:16 pm and 9:43 pm , so he can before 6:16 or after 9:43
Well, let's sea... Oops, I mean, see what we can do here! To model the water depth in the harbor, we can use a simple linear equation.
Let's assume the time is represented by the variable 't' and the water depth by 'd'. At high tide (2 p.m.), the water depth is approximately 8m, and at low tide (8 p.m.), it's around 0.6m. So, we can say that the water depth follows a linear pattern between these two points.
Using the slope-intercept form of an equation (y = mx + c), we can find the equation for the water depth in the harbor. Let me grab my mathematical compass and calculate it for you!
First, we find the slope (m) by subtracting the water depth at low tide from high tide depth and divide it by the time difference:
m = (8 - 0.6) / (2 - 8) = 7.4 / -6 = -1.2333 (approximately)
Now that we have the slope, let's find the y-intercept (c) by substituting the values of one of the known points into the equation:
8 = -1.2333(2) + c
8 = -2.4666 + c
c ≈ 10.4666
So, the equation that models the water depth in the harbor is:
d = -1.2333t + 10.4666
Now, to answer your question, the captain can safely exit the harbor in the sailboat when the water depth is at least 2m. We can substitute this value into the equation and find the corresponding time (t):
2 = -1.2333t + 10.4666
-1.2333t = -8.4666
t ≈ 6.87
Therefore, the captain can exit the harbor safely at approximately 6:52 p.m. I hope this helps you float away from any depth-related concerns!
To model the water depth in the harbor, we can use a sinusoidal function. A sinusoidal function represents the changing depth of water due to the tides. The general equation for a sinusoidal function is:
D(t) = A*sin(B(t - C)) + D
Where:
D(t) represents the depth of the water as a function of time (t).
A represents the amplitude, which is half the difference between the high and low points of the tide.
B represents the period, which is the time it takes for the tide to complete one full cycle.
C represents the phase shift, which is the horizontal shift of the function.
D represents the vertical shift, which is the average depth of water.
Since the high point is approximately 8m and the low point is approximately 0.6m, the amplitude of the function is:
A = (8 - 0.6)/2 = 3.7m
The period of the tide can be determined by finding the time between two consecutive high points or low points. In this case, the time between the high point at 2 p.m and the next high point is about 12 hours or 720 minutes.
B = 2π/720
The phase shift depends on when the tide reaches its maximum or minimum point. We can consider the time of the high point at 2 p.m as the maximum point.
C = 2
The average depth of the water can be calculated by finding the midpoint between the high and low points:
D = (8 + 0.6)/2 = 4.3m
Putting it all together, the equation that models the water depth in the harbor is:
D(t) = 3.7*sin((2π/720)*(t - 2)) + 4.3
To determine when the captain can safely exit the harbor, we need to find the time at which the water depth is at least 2m. Therefore, we need to solve the following equation:
2 <= 3.7*sin((2π/720)*(t - 2)) + 4.3
Now, we can proceed to solve this equation to find the time when the captain can safely exit the harbor in the sailboat.
To determine when the captain can safely exit the harbor in the sailboat, we need to consider the water depth at different times during the day and compare it to the sailboat's draft of 2m.
We can create an equation to model the water depth in the harbor using the given information about the high and low tides. The equation for the water depth can be represented as:
d(t) = a*sin(b*t + c) + d
Where:
d(t) is the water depth at time t.
a represents the amplitude of the tide (half the difference between the high and low points).
b determines the frequency or number of cycles within a given time period.
c is the phase shift, which represents any horizontal translation of the curve.
d is the average water depth.
Given that the high point of the tide is approximately 8m at 2 p.m., and the low point is approximately 0.6m at 8 p.m., we can determine the values of a and d for our equation.
a = (8 - 0.6) / 2 = 7.4 / 2 = 3.7
d = (8 + 0.6) / 2 = 8.6 / 2 = 4.3
So, our equation becomes:
d(t) = 3.7*sin(b*t + c) + 4.3
To find the specific values of b and c, we need more information or assumptions about the tidal patterns. However, we can work with this generalized equation for now.
Now, to determine when the captain can safely exit the harbor, we need to find t values where the water depth is equal to or greater than the sailboat's draft of 2m.
So, we would set the equation d(t) = 3.7*sin(b*t + c) + 4.3 equal to 2 and solve for t.
3.7*sin(b*t + c) + 4.3 = 2
From here, you would need to use mathematical techniques such as numerical methods, graphing, or trial and error to approximate the values of t when the equation is satisfied.
Once you find the values of t, you can determine the corresponding times and inform the captain when it is safe to exit the harbor in the sailboat.