Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?

Marcelo Mcclain

Marcelo Mcclain

Answered question

2023-03-08

Water is leaking out of an inverted conical tank at a rate of 10,000 cm3/min at the same time water is being pumped into the tank at a constant rate If the tank has a height of 6m and the diameter at the top is 4 m and if the water level is rising at a rate of 20 cm/min when the height of the water is 2m, how do you find the rate at which the water is being pumped into the tank?

Answer & Explanation

repicotit48p

repicotit48p

Beginner2023-03-09Added 3 answers

Let V be the volume of water in the tank, in c m 3 ; let h be the depth/height of the water, in cm; and let r be the radius of the surface of the water (on top), in cm. Since the tank is an inverted cone, so is the mass of water. Since the tank has a height of 6 m and a radius at the top of 2 m, similar triangles implies that h r = 6 2 = 3 so that h = 3 r
The volume of the inverted cone of water is then V = 1 3 π r 2 h = π r 3
Now differentiate both sides with respect to time t (in minutes) to get d V d t = 3 π r 2 d r d t (the Chain Rule is used in this step).
If V i is the volume of water that has been pumped in, then d V d t = d V i d t - 10000 = 3 π ( 200 3 ) 2 20 (when the height/depth of water is 2 meters, the radius of the water is 200 3 cm).
Hence d V i d t = 800000 π 3 + 10000 847758   cm 3 min

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