Suppose you live near a bay where the water level fluctuates due to the tides. Your bay is an inverted cone with a radius of 1 mile, and a depth in the center of 100 feet. (There are 5280 feet in a mile.) Water flows through a channel in and out of the bay with the tides. You want to compute the water flow in that channel, measured in cubic feet per minute.
Suppose the depth of the bay is given by
d(t)=90+10sin((pi/360)t)
where the time t is measured in minutes. (Thus we have 2 tides every day.)
Accordingly, at time t water is flowing through the channel at the rate of __________ cubic feet per minute. Your answer will be an expression in t.
just find the derivative dd/dt
the shape of the bay does not matter for this question.
The shape does matter as the question asks for the rate in cubic ft/ min.
If this were a cylinder, we could express this rate of change of volume in terms of t as:
Pi*r^2 * dd/dt cubic ft/min.
However, an inverted cone shape does not make sense for this question - according to the equation d(t), the water height in the bay rises from 90 ft to 100 ft, in the same time that it drops from 90 feet to 80 feet. These represent two different volumes of water in the inverted cone (draw a picture to understand this).
Would the tide not bring in the same volume as it takes out?
Actually, the shape does not matter for this question because it asked about the rate of flow through the channel. I suspect there were other parts where we'd have to work with the volume or depth in the bay, but not here.
dd/dt has units of ft/min.
The question asks for flow rate in ft^3/min....
Since the bay is an inverted cone, we can express the radius, r, in terms of the depth, d(t), by using similar triangles:
r/d(t) = 5280/100
r = 52.8 d(t) ft
Now express the volume of the cone, V, in terms of d(t):
V(d(t)) = (1/3) π [52.8 d(t)] ² ⋅ d(t)
= 929.28 π ⋅ d(t) ³
Sub in the expression given in the question for d(t) to get volume in terms of t:
V(t) = 929.28 π (90 + 10sin((π/360)t))³ ft ³
Lastly, differentiate V with respect to t, to get the rate of flow in ft ³/min:
V'(t) = 929.28 π ⋅ 3 (90 + 10sin((π/360)t))² ⋅ (π/36)cos((π/360)t)
= 77.4 π ² (90 + 10sin((π/360)t))² ⋅ cos((π/360)t) ft ³/min
To compute the water flow in the channel, we need to determine the cross-sectional area of the channel at any given time and multiply it by the velocity of the water.
The first step is to find the height of the water at time t. The function given for the depth of the bay, d(t), represents the height of the water above the bottom of the bay. Since the depth in the center of the bay is 100 feet, we add 100 to d(t) to obtain the total height of the water.
Therefore, the height of the water at time t is given by:
h(t) = 100 + d(t)
= 100 + (90 + 10sin((π/360)t))
= 190 + 10sin((π/360)t)
The next step is to find the radius of the water surface at time t. Since the bay is an inverted cone, the radius of the water surface is proportional to the height of the water. We can use similar triangles to determine this relationship.
The radius at time t is given by:
r(t) = (1 mile / 100 feet) * h(t)
= (1/5280) * (190 + 10sin((π/360)t))
= (19/528) + (1/528)*sin((π/360)t)
Now, we can find the cross-sectional area of the water channel at time t by using the formula for the area of a circle:
A(t) = π * r(t)^2
= π * ((19/528) + (1/528)*sin((π/360)t))^2
Finally, to compute the water flow in the channel, we multiply the cross-sectional area by the velocity of the water. The velocity of the water can vary depending on factors such as tidal currents and other environmental conditions, so it is not given in the problem.
Therefore, the water flow at time t, expressed in cubic feet per minute, can be represented as:
Flow(t) = A(t) * v
where v represents the velocity of the water in the channel.