SOLUTION: In a particular harbour, high tide occurred at midnight. The depth of the water in a harbour is given by the equation
d = 2.8(cos6πt/35)+11.6
where d is the depth in metres
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Question 919656: In a particular harbour, high tide occurred at midnight. The depth of the water in a harbour is given by the equation
d = 2.8(cos6πt/35)+11.6
where d is the depth in metres and t is the time in hours after midnight.
When was the first time that the depth reached 10m?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
easiest way is to graph this.
you can also solve it algebraically.
replace t with x and d with y so you can graph it.
equation becomes:
y = 2.8 * (cos(6*pi*x)/35) + 11.6
the graph of the equation is shown below.
the blue vertical dashed lines are at 0 hours and 24 hours.
the blue horizontal dashed line is at y = 10.
the first time the tide reaches 10 feet is at 4.046 hours in the morning.
that's roughly 4:03 am.
while the tide is at at it's high point at 0 hours, you can see that it's not perfectly synchronized on a 24 hour basis.
there are two high tides during the day and the second high tide occurs near the 24 hour mark but not on it. high tide will therefore shift as the days progress.
the algebraic solution agrees with the graph.
start with:
y = 2.8 * (cos(6*pi*x)/35) + 11.6
let y = 10 to get:
10 = 2.8 * (cos(6*pi*x)/35) + 11.6
subtract 11.6 from both sides of the equation to get:
10 - 11.6 = 2.8 * (cos(6*pi*x)/35)
divide both sides of the equation by 28 to get:
-1.6/2.8 = cos((6*pi*x)/35)
this equation is true if and only if arc-cosine(-1.6/2.8) = 6*pi*x/35
divide both sides of this equation by (6*pi) and multiply both sides of this equation by 35 to get:
x = (arc-cosine(-1.6/2.8) * 35) / (6 * pi)
use your calculator to solve for x to get:
x = 4.046061722 which agrees with the graph.
4.046 hours is your solution which equates to 4:03 in the morning roughly.
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