1. A pendulum is connected to a rope 3m long, which is connected to a ceiling 4m high. The angle between its widest swing and vertical hanging position is pi/3. If the pendulum swings out to its widest position for 2 seconds, model the horizontal displacement of the pendulum using a sine function, considering vertical to be x=0.
2. The tides at Cape Capstan change the depth of the water in the harbour. On one day in September, the tides have a high point of approximately 13m at 3pm and 1.6m at 9pm. The sailboat has a draft of 2.5m deep. The captain of the sailboat plans to exit the harbourer at 8:30pm.
a. Write a cosine trigonometric equation that represents the situation described.
b. Plot an accurate graph 1 cycle.
c. Determine whether the sailboat can exit the harbour safely.
3. The average number of customers, c, at a 24-hour sandwich shop per hour is modelled roughly by the equation c(h)=-5cos[pih/12]+12, with h=0 representing midnight.
a. How many hours per day are there 13 customers per hour, to the nearest hour?
b. What is the maximum number of customers?
c. What time of day is peak(max) business?
4. A 3m ladder is leaning against a vertical wall such that the angle between the ground and the ladder is pi/3. What is the exact height that the ladder reaches up the wall?
Please show all the work. Please and thanks.
1. To model the horizontal displacement of the pendulum, we can use the equation of a sine function: x = A*sin(ωt), where x is the horizontal displacement, A is the amplitude, ω is the angular frequency, and t is the time.
Given:
Length of the rope = 3m
Height of the ceiling = 4m
Angle of widest swing = π/3
Time for widest swing = 2 seconds
First, we need to find the amplitude of the pendulum's swing. The amplitude is equal to the length of the rope.
Amplitude (A) = Length of the rope = 3m
Next, we need to find the angular frequency (ω) of the pendulum. The angular frequency can be calculated using the formula ω = √(g/L), where g is the acceleration due to gravity (approximately 9.8 m/s^2) and L is the length of the rope.
Angular frequency (ω) = √(g/L) = √(9.8/3) ≈ 1.785
Now we can plug in the values of A and ω into the equation to model the horizontal displacement of the pendulum:
x = 3*sin(1.785t)
2. a. To write a cosine trigonometric equation that represents the situation described, we need to consider the high point of the tide as the maximum value and the low point of the tide as the minimum value. Let's assume that the time is given in hours and represents the number of hours after 3pm.
Maximum value (high tide) = 13m
Minimum value (low tide) = 1.6m
Time at high tide = 3pm = 0 hours
Time at low tide = 9pm = 6 hours
Since we want a cosine function, we'll use the general form: y = A*cos(Bx + C) + D
Amplitude (A) = (Maximum - Minimum)/2 = (13 - 1.6)/2 = 5.7/2 = 2.85
Period (P) = 6 hours (time between high and low tides)
Vertical shift (D) = (Maximum + Minimum)/2 = (13 + 1.6)/2 = 7.3/2 = 3.65
Phase shift (C) = -B*t, where t is the time at high tide (in this case, 0 hours)
B = 2π/P = 2π/6 = π/3
Therefore, the cosine equation that represents the situation is:
y = 2.85*cos((π/3)*x) + 3.65
b. To plot a graph of 1 cycle, we can choose a suitable range for the values of x. Let's choose x from 0 to 6 (1 cycle = 6 hours).
c. To determine whether the sailboat can exit the harbor safely, we need to check if the depth of the water at 8:30pm (8.5 hours after 3pm) is greater than the draft of the sailboat (2.5m).
Substituting x = 8.5 into the equation:
y = 2.85*cos((π/3)*8.5) + 3.65
y ≈ -1.486 + 3.65
y ≈ 2.164m
The depth of the water at 8:30pm is approximately 2.164m, which is greater than the draft of the sailboat (2.5m). So, the sailboat can safely exit the harbor.
3. a. To find the number of hours per day with 13 customers, we need to solve the equation c(h) = 13.
Given: c(h) = -5*cos(πh/12) + 12
-5*cos(πh/12) + 12 = 13
-5*cos(πh/12) = 1
cos(πh/12) = -1/5
To find the value of h, we can take the inverse cosine (arccos) of -1/5:
πh/12 = arccos(-1/5)
h = (12/arccos(-1/5))*π
Using a calculator, we find that h ≈ 8.342 (approximately)
Therefore, there are approximately 8 hours per day with 13 customers.
b. The maximum number of customers can be found by finding the maximum value of the equation c(h).
Given: c(h) = -5*cos(πh/12) + 12
The maximum value of the cosine function is 1, so we substitute that in:
c(max) = -5*1 + 12
c(max) = 12 - 5
c(max) = 7
Therefore, the maximum number of customers is 7.
c. The peak (maximum) business occurs when the number of customers is at its highest point. From part b, we know that the maximum number of customers is 7.
To find the time of day when this occurs, we need to solve the equation c(h) = 7.
Given: c(h) = -5*cos(πh/12) + 12
-5*cos(πh/12) + 12 = 7
-5*cos(πh/12) = -5
cos(πh/12) = 1
πh/12 = 0
h = 0
Therefore, the peak business occurs at midnight (h = 0), which represents the start of the day.
4. Given that the ladder forms an angle of π/3 with the ground and the length of the ladder is 3m, we need to find the height that the ladder reaches up the wall.
Let h be the height reached by the ladder.
Using trigonometry, we can write the equation:
sin(π/3) = h/3
We know that sin(π/3) = √3/2.
√3/2 = h/3
Solving for h:
h = (√3/2)*3
h = √3*3/2
h = (3/2)√3
h = (3√3)/2
Therefore, the exact height that the ladder reaches up the wall is (3√3)/2 meters.
1. To model the horizontal displacement of the pendulum, we can use the equation for simple harmonic motion:
x(t) = A * sin(ωt + φ)
where:
- x(t) represents the horizontal displacement at time t
- A represents the amplitude of the motion
- ω represents the angular frequency of the motion
- φ represents the phase constant
Given that the angle between the widest swing and vertical hanging position is π/3, we can determine the amplitude of the pendulum by considering the right-angled triangle formed by the pendulum, the ceiling, and the vertical line. The opposite side of the triangle is the amplitude, and the adjacent side is the height of the ceiling.
Using the sine function, we can relate the angle π/3 to the amplitude:
sin(π/3) = opposite/hypotenuse
sin(π/3) = A/4
A = 4 * sin(π/3)
A = 4 * (√3/2)
A = 2√3
The angular frequency can be determined using the relation: ω = 2π/T, where T is the period of the pendulum.
Given that the pendulum swings out to its widest position for 2 seconds, the period (T) would be twice this time:
T = 2s * 2
T = 4s
Therefore, ω = 2π/4 = π/2
The phase constant (φ) can be determined based on the initial conditions. In this case, since we want the pendulum at its maximum displacement when t = 0, we have:
x(0) = A * sin(φ) = A
Solving for φ:
sin(φ) = 1
φ = π/2
Substituting all the values into the equation, we get:
x(t) = A * sin(ωt + φ)
x(t) = (2√3) * sin((π/2)t + π/2)
2. a) To represent the tides with a cosine equation, we can use a similar approach to question 1. Let's assume time t is measured in hours and consider that the high point occurs at 3 pm, which corresponds to t = 3. The low point at 9 pm corresponds to t = 9.
Let H be the amplitude of the water depth and t0 be the time of the high point. The depth of the water can be modeled as:
D(t) = H * cos(ω(t - t0))
Given that the tide has a high point of approximately 13m at 3 pm (t0 = 3) and a low point of approximately 1.6m at 9 pm (t0 = 9), we can calculate the amplitude, H:
H = (13 - 1.6)/2
H = 11.7/2
H = 5.85m
Therefore, the cosine equation that represents the situation is:
D(t) = 5.85 * cos(ω(t - t0))
b) To plot an accurate graph for one cycle, we need to determine the period (T) of the tide. The period is the time it takes for the tide to complete one cycle, which is from the high point to the next high point.
In this case, the high point occurs at 3 pm and then repeats every 12 hours. Therefore, the period T = 12 hours.
Using the amplitude (H) and period (T), we can plot the graph. The x-axis represents time (t) in hours, and the y-axis represents the depth of the water (D) in meters.
c) To determine if the sailboat can safely exit the harbor, we need to consider the draft of the sailboat. The draft represents how deep the boat goes under the water.
In this case, the draft of the sailboat is 2.5m. To ensure safe passage, the water depth (D) should always be greater than or equal to the draft (2.5m).
We need to evaluate the cosine equation for the time the sailboat plans to exit the harbor at 8:30 pm, which corresponds to t = 8.5.
Evaluate D(t) = 5.85 * cos(ω(t - t0)) at t = 8.5.
If D(t) ≥ 2.5, it is safe for the sailboat to exit the harbor. Otherwise, it is not safe.
3. a) To find the number of hours per day with 13 customers per hour, we need to solve the equation c(h) = 13, where c(h) represents the average number of customers at time h.
-5cos(πh/12) + 12 = 13
-5cos(πh/12) = 13 - 12
-5cos(πh/12) = 1
cos(πh/12) = -1/5
Since we want to find the number of hours per day, we consider 0 ≤ h ≤ 24.
To find solutions in this interval, we evaluate the inverse cosine function:
πh/12 = arccos(-1/5)
Now, solve for h:
h = (12/arccos(-1/5)) * π
Rounding to the nearest hour, we find the number of hours per day with 13 customers per hour.
b) The maximum number of customers can be found by evaluating c(h) at its maximum value. In this case, we want to find the maximum value of -5cos(πh/12) + 12.
To find the maximum value of a cosine function, we need to determine its amplitude. The amplitude of -5cos(πh/12) is 5.
Therefore, the maximum number of customers is:
Max customers = 5 + 12
Max customers = 17
c) The peak (maximum) business time corresponds to when the maximum number of customers occurs. We can determine this by finding the h value that results in the maximum value of -5cos(πh/12) + 12.
To find this value, we need to evaluate when the cosine function is at its maximum, which happens when the angle inside the cosine is 0.
πh/12 = 0
h = 0
Therefore, the peak business time is midnight (12:00 am).
4. To find the exact height that the ladder reaches up the wall, we can use trigonometry and the given angle between the ground and the ladder, which is π/3.
Let h be the height the ladder reaches up the wall. Using the sine function, we can relate the angle π/3 to the height and the length of the ladder:
sin(π/3) = h/3
Simplify:
√3/2 = h/3
Cross-multiply:
2h = 3√3
Divide:
h = 3√3/2
Therefore, the exact height that the ladder reaches up the wall is (3√3)/2.
I hate this you guys are supposed to do this on your own
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this is called PLAIGARISM