A large container has the shape of a frustum of a cone with top radius 5m, bottom radius 3m, height 12m. The container is being filled with water at the constant rate of 3.9 m^3/min. At what rate is the level of water rising at the instant the water 9 m deep ?
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To find the rate at which the water level is rising, we need to find the rate of change of the height of the water with respect to time.
Since the container has the shape of a frustum of a cone, we can use similar triangles to relate the changing height of the water to the changing radius of the water.
The formula for the volume of a frustum of a cone is:
V = (1/3) * π * h * (r₁² + r₂² + r₁ * r₂)
where:
V is the volume of the frustum of the cone,
h is the height of the frustum,
r₁ is the top radius of the frustum,
r₂ is the bottom radius of the frustum.
We want to relate the height of the water (h) to the radius of the water (r). Let's let x be the radius of the water at a given time when the height is h.
We can rewrite the formula for the volume as:
V = (1/3) * π * h * (r₁² + r₂² + r₁ * r₂)
V = (1/3) * π * h * (x² + r₂² + x * r₂)
To find the rate at which the water level is rising, we can take the derivative of the volume equation with respect to time (t):
dV/dt = (1/3) * π * (dh/dt) * (x² + r₂² + x * r₂) + (1/3) * π * h * (2x * dx/dt + r₂ * dx/dt)
Since the water is being filled at a constant rate, dV/dt = 3.9 m³/min.
Now we have an equation with three variables: dh/dt (the rate of change of height), x (the radius of the water), and dx/dt (the rate of change of radius).
We know the values of r₁ (5m) and r₂ (3m), and we want to find dh/dt when h = 9m.
Let's substitute these values into the equation and solve for dh/dt.