A triangular trench 40ft long is 2ft across the top, and 2ft deep. If water flows in at the rate of 3ft³/min, find how fast the surface is rising when the water is 6 in deep.
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Draw a diagram. It should be clear that when the water has depth y ft, the cross-section of water has area y^2/2 ft^2. So, the volume
v = 20y^2
dv/dt = 40y dy/dt
when y = 6 in = 1/2 ft,
3 = 40(1/2) dy/dt
now just solve for fy/dt, the rate the depth is changing. That is, how fast the surface is rising.
To find how fast the surface of the triangular trench is rising when the water is 6 inches deep, we need to use related rates. The formula for the volume of a triangular trench is:
V = (1/2 * b * h) * l
Where:
V = volume
b = base of the triangle (2 ft)
h = height of the triangle (2 ft)
l = length of the trench (40 ft)
We know that the water flows in at a rate of 3 ft³/min. Let's assign the rate of change of the volume with respect to time as dV/dt and the rate of change of the depth with respect to time as dh/dt.
We are interested in finding the rate of change of the surface level, which is the rate of change of the depth (dh/dt) when the depth is 6 inches (0.5 ft).
To solve the problem, we need to use the chain rule of differentiation:
dV/dt = (∂V/∂b * db/dt) + (∂V/∂h * dh/dt) + (∂V/∂l * dl/dt)
Since the base (b) and length (l) of the trench are constant, db/dt = dl/dt = 0.
Now, let's differentiate the volume formula with respect to each variable:
∂V/∂b = (1/2 * h) * l = hl/2
∂V/∂h = (1/2 * b) * l = bl/2
∂V/∂l = (1/2 * b * h) = bh/2
Substituting these values in the chain rule equation, we get:
dV/dt = (hl/2 * 0) + (bl/2 * dh/dt) + (bh/2 * 0)
dV/dt = (bl/2)(dh/dt)
We can rearrange this equation to solve for dh/dt:
dh/dt = (2 * dV/dt) / bl
Now, let's substitute the known values into the equation:
dh/dt = (2 * 3) / (2 * 2 * 40)
dh/dt = 6 / 160
dh/dt = 0.0375 ft/min
Therefore, the surface level of the triangular trench is rising at a rate of 0.0375 ft/min when the water is 6 inches deep.