A trough is 15 ft long and its ends have the shape of isosceles triangles that are 4 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft^3/min, how fast is the water level rising when the water is 8 inches deep?
Does 1.4 ft/min sound right? The computer is saying it's wrong.
The water surface area multiplied by the rate of water level rising equals the volume flow rate. Let's write this as
A*dh/dt = dV/dt = 14 ft^/min.
All you need is the area of water surface, A. At a depth of 8 inches, water is 2/3 of the way to the top and the width of the water channel is (2/3)*4 = 8/3 feet. The total surface area is
(8/3) ft x 15 ft = 40 ft^2
Therefore
dy/dt = 14 ft^3/min/40 ft^2 = 0.35 ft/min
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To determine the rate at which the water level is rising, we can use the concept of related rates. Let's break down the problem into steps:
Step 1: Determine the volume of the water in the trough at a given height.
The trough has the shape of a prism with triangular ends. The triangular ends are isosceles triangles with a base of 4 ft and a height of 1 ft. The volume V of a triangular prism is given by the formula V = (1/2)bh times the length (L) of the trough.
Since the trough is 15 ft long, we can use the equation for volume to find the volume V in terms of the height h:
V = (1/2)(4 ft)(h) (15 ft)
Step 2: Find an equation that relates the height h and the rate of change of the height dh/dt.
The volume of the water is changing with respect to time, so we need to find an equation that relates the rate of change of the volume dV/dt and the rate of change of the height dh/dt.
Taking the derivative of the volume equation with respect to time t, we get:
dV/dt = (1/2)(4 ft)(dh/dt) (15 ft)
Step 3: Substitute the given values and solve for dh/dt.
We're given that the trough is being filled at a rate of 14 ft^3/min (dV/dt = 14 ft^3/min). We're required to find the rate at which the water level is rising when the water is 8 inches deep (h = 8 in).
Converting 8 inches to feet, we have h = 8 in * (1 ft/12 in) = 2/3 ft.
Substituting these known values into the equation, we have:
14 ft^3/min = (1/2)(4 ft)(dh/dt) (15 ft)
Simplifying, we can solve for dh/dt:
dh/dt = (14 ft^3/min) / [(1/2)(4 ft)(15 ft)]
dh/dt = (14 ft^3/min) / (2 ft^3) = 7 ft/min
So the rate at which the water level is rising when the water is 8 inches deep is 7 ft/min.
Therefore, the correct answer is 7 ft/min, not 1.4 ft/min.