Water is being pumped into an inverted conical tank at a constant rate. The tank has height 12 meters and the diameter at the top is 4 meters. If the water level is rising at a rate of 26 centimeters per minute when the height of the water is 1 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.
To find the rate at which water is being pumped into the tank, we can use the concept of similar triangles.
Let's denote the radius of the water level as r and the height of the water as h.
Given:
- Height of the tank = 12 meters
- Diameter at the top of the tank = 4 meters
- Rate at which the water level is rising (dh/dt) = 26 centimeters per minute
- Water height = 1 meter
Since the tank is inverted and has a conical shape, we can use the relationship between the height and radius of a cone:
h / r = 12 / (4/2) (height / radius = height of the tank / radius at the top)
Simplifying the equation:
h / r = 12 / 2
h / r = 6
Now, let's differentiate both sides of the equation with respect to time (t):
(dh/dt) / r - h(dr/dt) / r^2 = 0
Substituting the given values:
26 / r - 1(dr/dt) / r^2 = 0
To find the rate at which water is being pumped in cubic centimeters per minute, we need to find the value of (dr/dt).
Now, substitute the given value of h = 1 and solve for r:
1 / r = 6
r = 1 / 6
Substitute this value back into the previous equation:
26 / (1/6) - 1(dr/dt) / (1/6)^2 = 0
26 * (6/1) - 1(dr/dt) / (1/36) = 0
156 - (dr/dt) / (1/36) = 0
Now, solve for (dr/dt):
(dr/dt) / (1/36) = 156
(dr/dt) = (156 * 1/36)
(dr/dt) = 156/36
Therefore, the rate at which water is being pumped into the tank is 13/3 cubic centimeters per minute.
To solve this problem, we can make use of the formula for the volume of a cone. The volume of a cone is given by the formula V = (1/3)πr^2h, where V is the volume, π is a mathematical constant with an approximate value of 3.14, r is the radius of the base of the cone, and h is the height.
In this problem, the height of the cone is fixed at 12 meters, the diameter at the top is 4 meters, and the water level is rising at a rate of 26 centimeters per minute when the height is 1 meter. We need to find the rate at which water is being pumped into the tank in cubic centimeters per minute.
We know that the height of the water is changing with respect to time, so we need to express the rate of change of volume in terms of the rate of change of height.
Let's first find the radius of the base of the cone. The diameter is given as 4 meters, so the radius (r) is half of the diameter, which is 4/2 = 2 meters.
Now let's differentiate the formula for volume with respect to time (t) using the chain rule:
dV/dt = (1/3)π(2r)(dh/dt)
Where dV/dt is the rate of change of volume (which we want to find), π is approximately 3.14, 2r is the radius of the base of the cone, and dh/dt is the rate of change of the height.
Given that the water level is rising at a rate of 26 centimeters per minute when the height is 1 meter, we substitute the values into the equation:
26 = (1/3)π(2)(dh/dt)
Now we solve for dh/dt:
dh/dt = (3/2π)(26)
dh/dt ≈ 39.27 cm/min
Therefore, the rate at which water is being pumped into the tank is approximately 39.27 cubic centimeters per minute.