For a boat to float in a tidal bay, the water must be at least 2.5 meters deep. The depth of the water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t)=5+4.6sin(0.5t)
Use three decimal places in all intermediate calculations.
(a) What is the period of the tides in hours?
Round your answer to three decimal places.
(b) If the boat leaves the bay at midday, what is the latest time it can return before the water becomes too shallow?
Round your answer to three decimal places.
To find the period of the tides, we need to determine the length of one complete cycle of the sine function within the given equation. The general form of a sinusoidal function is given by:
y = A + B*sin(C(t - D))
Where A represents the vertical shift, B represents the amplitude, C represents the frequency, and D represents the horizontal shift.
Looking at the equation for the depth of water around the boat in the given problem (d(t) = 5 + 4.6*sin(0.5t)), we can see that the amplitude (B) is 4.6 and the frequency (C) is 0.5.
(a) To find the period of the tides (the time it takes for one complete cycle), we use the formula:
Period = 2π / |C|
In this case, the period is 2π / |0.5| = 2π / 0.5 = 4π = 12.566.
Rounding to three decimal places, the period of the tides is approximately 12.566 hours.
(b) To determine the latest time the boat can return before the water becomes too shallow (less than 2.5 meters), we need to find when the depth of the water equals or falls below 2.5 meters.
Given the equation d(t) = 5 + 4.6*sin(0.5t), we set up the inequality:
5 + 4.6*sin(0.5t) ≤ 2.5
Subtracting 5 from both sides of the inequality gives:
4.6*sin(0.5t) ≤ -2.5
Dividing both sides of the inequality by 4.6 gives:
sin(0.5t) ≤ -2.5 / 4.6
Using a calculator, we find that -2.5 / 4.6 ≈ -0.543.
Now, to solve for t, we need to find the inverse sine function. The inverse sine of -0.543 is approximately -0.566 radians or -32.438 degrees. However, we need to find the value of t in hours since midnight.
Since the general equation for the sine function is:
y = sinθ
We can set up the equation:
-0.543 = sin(0.5t)
To find the value of θ, we can take the arcsine (inverse sine) of both sides:
arcsin(-0.543) ≈ -0.586 radians or -33.593 degrees.
Now, we multiply the radian value by the conversion factor (180° / π radians) to convert from radians to degrees:
-0.586 * (180° / π radians) ≈ -33.593 degrees.
Finally, we need to convert the degree value to hours. We know that the period of the tides is 12.566 hours (from part (a)). We can set up the equation:
Degree value / 360° = Time value / Period
Simplifying the equation:
-33.593° / 360° = t / 12.566 hours.
Now, we solve for t:
t = (-33.593° / 360°) * 12.566 hours.
Calculating this gives:
t ≈ -1.042 hours.
Since we are measuring time since midnight, we disregard negative values and find the positive equivalent:
t ≈ 24 hours - 1.042 hours.
Rounding to three decimal places, the latest time the boat can return before the water becomes too shallow is approximately 22.958 hours, which is approximately 10:57 PM.
(a) To find the period of the tides, we need to find the value of t at which the sine function completes one full cycle. The sine function has a period of 2π, which means one complete cycle occurs over a range of 2π.
From the equation d(t) = 5 + 4.6sin(0.5t), we can see that the coefficient of t is 0.5. To find the period, we can set 0.5t equal to 2π and solve for t.
0.5t = 2π
t = 4π
Therefore, the period of the tides is 4π hours. Rounded to three decimal places, the period is approximately 12.566 hours.
(b) The boat can return before the water becomes too shallow only if the depth of the water, d(t), remains above 2.5 meters. We need to find the latest time at which d(t) = 2.5.
Setting d(t) = 2.5 in the equation d(t) = 5 + 4.6sin(0.5t), we get:
2.5 = 5 + 4.6sin(0.5t)
Subtracting 5 from both sides:
-2.5 = 4.6sin(0.5t)
Dividing by 4.6:
-0.5434782609 = sin(0.5t)
To find the latest time, we need to find the value of t where sin(0.5t) is equal to -0.5434782609. We can use the inverse sine function or arcsin to find this value.
arcsin(-0.5434782609) ≈ -0.595
Since the arcsin function gives the reference angle, we need to find the equivalent angle in the range of 0 to 2π.
0.595 + π ≈ 3.737
So, the latest time the boat can return before the water becomes too shallow is approximately 3.737 hours after midday. Rounded to three decimal places, the boat can return before approximately 3.737 hours.
a) period = 2pi/.5 = 12.663 hrs
so we need the height to be 2.5
5 + 4.6sin(.5t)= 2.5
sin(.5t) = (2.5-5)/4.6 = -.543478
make sure your calculator is set to radians and take
inverse sine(+ .543478)
to get .574575 radians
But, the sine was negative so the "angle" must have been in quadr III or quadrant IV
so .5t = pi + .574575 or .5t = 2pi - .574575
t = 7.432336 or t = 11.41722
or t = 7:26 am or t = 11:25 am
remember that midnight was t=0
so the water was below 2.5 m between
7:25 am and 11:25 am
and the period was 12.663 hrs or 12:40 hrs
so the next time the water will go below 2.5 m
= 7:25 + 12:40 = 17:65 = 18:05 or 6:05 pm