just find the derivative dd/dt
the shape of the bay does not matter for this question.
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.
5 answers
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?
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....
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
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
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