I really do not understand these questions, so does my tutor and need help with them. Please solve with all steps.
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?
Sure, let's solve each question step by step:
1. To model the horizontal displacement of the pendulum using a sine function, we can start by finding the amplitude and period of the function. The amplitude (A) is the distance from the vertical hanging position to the widest swing, which is 3m in this case. The period (T) is the time taken for one complete swing, and it can be calculated using the formula T = 2π√(L/g), where L is the length of the pendulum (3m) and g is the acceleration due to gravity (approximately 9.8 m/s²).
T = 2π√(3/9.8) ≈ 1.77 seconds
The equation for the horizontal displacement (x) of the pendulum as a function of time (t) can be written as:
x(t) = Asin(2πt/T)
Plugging in the values, we get:
x(t) = 3sin((2π/1.77)t)
2. Let's solve part (a) first. The cosine function will represent the depth of the water in the harbor. Since the high point of the tides is approximately 13m, and the lowest point is 1.6m, we can use these values to determine the amplitude (A) and vertical shift (D) of the cosine function.
Amplitude (A) = (high point - low point)/2 = (13 - 1.6)/2 = 11.7/2 ≈ 5.85m
Vertical shift (D) = (high point + low point)/2 = (13 + 1.6)/2 = 14.6/2 ≈ 7.3m
The equation for the cosine function that represents the situation can be written as:
h(t) = A*cos(ωt) + D
where ω represents the angular frequency. The angular frequency can be calculated using the formula ω = 2π/T, where T is the time taken for one cycle. In this case, the time between the high point and the high point from a previous tidal cycle is 6 hours, so the time for one cycle (T) is 6 hours.
Now we have:
ω = 2π/6
Finally, the cosine equation that represents the situation is:
h(t) = 5.85*cos((2π/6)t) + 7.3
For part (b), you need to plot the graph of this equation for one cycle. Use a time span of 6 hours (from 3pm to 9pm) on the x-axis and the corresponding depths on the y-axis.
For part (c), to determine whether the sailboat can safely exit the harbor, we need to check if the depth at 8:30pm is greater than the draft of the sailboat, which is 2.5m. Evaluate the equation h(t) at t = 8.5 hours (8:30pm) and check if it is greater than 2.5m.
3. For part (a), we need to solve the equation c(h) = 13 for h, using the given equation c(h) = -5cos(πh/12) + 12. Set c(h) to 13 and solve for h:
13 = -5cos(πh/12) + 12
Solve for cos(πh/12):
cos(πh/12) = -1/5
Take the inverse cosine (cos⁻¹) of both sides and solve for h:
πh/12 = cos⁻¹(-1/5)
h = (12/π)cos⁻¹(-1/5)
Approximately, h ≈ 5.7 hours.
For part (b), the maximum number of customers can be found by evaluating c(h) at the maximum value of the cos function, which is 1. So, substitute cos(0) = 1 into the equation c(h) = -5cos(πh/12) + 12:
c(h) = -5(1) + 12
c(h) = 7
Therefore, the maximum number of customers is 7.
For part (c), the peak (maximum) business occurs at the time when c(h) is at its maximum value. Since we found that c(h) is maximum at 7, we can find the corresponding value for h by solving c(h) = 7:
7 = -5cos(πh/12) + 12
Simplify and solve for h:
-5cos(πh/12) = -5
cos(πh/12) = 1
Since cos(0) = 1, we have:
πh/12 = 0
h = 0
Therefore, the peak business occurs at midnight (h = 0), or starting from the beginning of the day.
4. To find the exact height that the ladder reaches up the wall, we can use the trigonometric relationship of a right-angled triangle. In this case, the ladder is the hypotenuse of the triangle, and the angle between the ground and the ladder is π/3.
Using the trigonometric function cosine (cos), which relates the adjacent side of a right triangle to its hypotenuse, we have:
cos(π/3) = adjacent side / hypotenuse
cos(π/3) = h / 3
Since cos(π/3) = 1/2, we can solve for h:
1/2 = h / 3
Multiply both sides by 3:
3/2 = h
Therefore, the exact height that the ladder reaches up the wall is 3/2 meters or 1.5 meters.