Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at thich water is being pumped into the tank.
To solve this problem, we can apply the principles of related rates. Let's break down the problem and identify the given information:
- Water is leaking out of an inverted conical tank at a rate of 10,000 cm^3/min.
- The tank has a height of 6 m and a diameter at the top of 4 m.
- The water level is rising at a rate of 20 cm/min when the height of the water is 2 m.
We need to find the rate at which water is being pumped into the tank.
To solve this problem, we'll use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h,
where V is the volume, π (pi) is a constant approximately equal to 3.14159, r is the radius of the circular base, and h is the height.
Now, let's begin solving the problem step by step:
Step 1: Find the rate at which the height of the water changes with respect to time.
Given: dh/dt = 20 cm/min (the water level is rising at a rate of 20 cm/min)
We want to find dV/dt (the rate at which the volume is changing with respect to time).
Step 2: Relate the variables in the equation.
Using similar triangles, we can relate the radius and height of the cone at any given time t.
We can set up a proportion using the similar triangles: (radius of the bottom of the cone) / (height of the cone) = r / h.
Given: r = (4/6) * h = (2/3) * h.
Step 3: Find the volume of the cone in terms of h.
Substitute the expression for r into the formula for the volume of a cone:
V = (1/3) * π * [(2/3)^2 * h^2] * h,
Simplify: V = (4/27) * π * h^3.
Step 4: Find dV/dt by differentiating the volume equation.
Differentiate both sides of the volume equation with respect to t:
dV/dt = (4/27) * π * 3 * h^2 * dh/dt.
Step 5: Plug in the known values and solve for dV/dt.
At the given height of 2 m, substitute h = 2 into the equation:
10,000 cm^3/min = (4/27) * π * 3 * 2^2 * 20 cm/min.
Simplifying the equation yields:
10,000 cm^3/min = (160/27) * π cm^3/min.
To find the rate at which water is being pumped into the tank, we need to solve for dV/dt:
dV/dt = (10,000 cm^3/min) * (27/160π) ≈ 53.05 cm^3/min.
Therefore, the water is being pumped into the tank at a rate of approximately 53.05 cm^3/min.
To solve this problem, we can use related rates.
Let's label the following dimensions:
h = height of the water in the tank (in meters)
r = radius of the water level (in meters)
We are given the following information:
- The diameter at the top of the tank is 4 m, so the radius of the top of the tank is r = 2 m.
- The water level is rising at a rate of 20 cm/min, which is equivalent to dh/dt = 0.2 m/min.
We want to find the rate at which water is being pumped into the tank, which is equivalent to dV/dt (the rate of change of the volume of water in the tank).
To find dV/dt, we need to find a relationship between V (the volume of water in the tank) and h.
The shape of the tank is an inverted cone, so the volume of the water (V) can be expressed as:
V = (1/3)πr^2h
Differentiating the equation with respect to time t, we get:
dV/dt = (1/3)π (2rh (dr/dt) + r^2 (dh/dt))
We are given that dr/dt = 0 because the diameter of the tank remains constant.
Substituting the given values, we have:
dV/dt = (1/3)π (2(2 m)(6 m/min) (0) + (2 m)^2 (0.2 m/min))
Simplifying further:
dV/dt = (1/3)π (4(6)(0) + 4(0.2))
dV/dt = (1/3)π (0 + 0.8)
dV/dt = (1/3)π (0.8)
dV/dt = (0.8/3)π
Therefore, the rate at which water is being pumped into the tank is approximately (0.8/3)π cubic meters per minute.