For a boat to float in a tidal bay, the water must be at least 2.7 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 + 4.6 sin(0.5t).
(a) What is the period of the tides in hours? (Round your answer to three decimal places.)
I think it is 12.566 hrs.
The period is 2pi/B so 2pi/0.5 which equals 12.566
b) If the boat leaves the bay at midday, what is the earliest time it can return before the water becomes too shallow? (Round your answer to the nearest minute.)
I just don't know how to even start solving this question.
I think maybe this is how you do it? Please correct me!
Is it:
2.7 = 5 + 4.6 sin(0.5t)
(subtract 5 from both sides)
-2.3 = 4.6sin(0.5t)
(divide by 4.6 from both sides)
-0.5=sin(0.5t)
arcsin(-0.5) = 0.5t
Then when I plug the arcsin into my calculator I get -30 which doesn't seem right at all.
Thank you so much for all your help!
This looks almost identical to the one I answered here
http://www.jiskha.com/display.cgi?id=1359417542
Thank you so much!! That really helped!
To find the earliest time the boat can return before the water becomes too shallow, you need to solve the equation -0.5 = sin(0.5t), where t represents the time in hours.
First, let's find the angles between -π/2 and π/2 whose sine is -0.5.
Taking the inverse sine (arcsin) of -0.5 on a calculator will give you the principal value of -π/6 (approximately -0.5236 radians). However, this value only represents one possible angle.
Since sine is a periodic function, we know that sin(θ) = sin(θ + 2πk), where k is an integer. Thus, there are infinitely many values that satisfy sin(0.5t) = -0.5.
For the interval -π/2 to π/2, another angle that satisfies sin(0.5t) = -0.5 is 2π/3 (approximately 2.0944 radians).
Note that this value is obtained by adding 2π (360 degrees) to the principal value. This accounts for the fact that sine has a repeating pattern every 2π radians (360 degrees).
So, we have two possible angles:
θ = -π/6 + 2πk
θ = 2π/3 + 2πk
Now, let's solve for t:
For the first angle (-π/6 + 2πk), solve:
-π/6 + 2πk = 0.5t
t = (-π/6 + 2πk) / 0.5
For the second angle (2π/3 + 2πk), solve:
2π/3 + 2πk = 0.5t
t = (2π/3 + 2πk) / 0.5
Since we need to find the earliest time, we are interested in the smallest positive value for t.
Plugging in values for k, we find the smallest positive t value:
For k = 0:
t = (-π/6) / 0.5 = -π/3 (approximately -1.047)
For k = 1:
t = (2π/3 + 2π) / 0.5 = (8π/3) / 0.5 = (16π)/6 = (8π)/3 (approximately 8.377)
Since -1.047 is negative, we discard it as it represents a time before midnight.
Therefore, the earliest time the boat can return is approximately 8.377 hours, or 8 hours and 22 minutes after midnight.
Please note that if the boat leaves the bay at midday (12:00 PM), it will return around 8:22 PM.
To find the earliest time the boat can return before the water becomes too shallow, we need to solve the equation d(t) = 2.7.
Starting with the equation:
d(t) = 5 + 4.6 sin(0.5t)
First, subtract 5 from both sides:
d(t) - 5 = 4.6 sin(0.5t)
Now, divide both sides by 4.6:
(d(t) - 5) / 4.6 = sin(0.5t)
We can use the arcsine function (sin^(-1)) to find the angle whose sine is equal to the left side of the equation.
Taking the arcsine of both sides:
sin^(-1)((d(t) - 5) / 4.6) = 0.5t
Now, multiply both sides by 2:
2sin^(-1)((d(t) - 5) / 4.6) = t
To find the earliest time, we need to find the smallest positive solution for t. Plug in d(t) = 2.7 into the equation and solve for t:
2sin^(-1)((2.7 - 5) / 4.6) = t
Using a calculator, approximate the value inside the arcsine function:
2sin^(-1)(-0.587) = t
Calculating this, we find that t ≈ -1.207.
Since time cannot be negative, this solution is not meaningful in this context. Therefore, for the boat to return to the bay before the water becomes too shallow, there is no meaningful solution.