Water is leaking out of an inverted conical tank at a rate of 1.5 cm3 /min at the same time that water is being pumped into the
tank at a constant rate. The tank has height 10 cm and the diameter at the top is 6 cm. If the water level is rising at a rate of 1.0 cm/min when the height of the water is 2 cm, find the rate at which water is being pumped into the tank. (a) Draw a picture of the situation for any time t. (b) What quantities are given in the problem
1 answer:
Answer:
a) Check Explanation
b) Check Explanation
c) The rate at which water is being pumped into the tank = 2.631 cm³/min
Explanation:
Let the rate of flow of water into the tank be k cm³/min
a) The image of the conical tank is presented in the attached image
Note, the radius and height of a cone are related through the similar triangles principle.
As shown in the attached image, it is evident that
r/h = 3/10
r = 3h/10 = 0.3 h
b) The quantities given in the problem.
- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3
- total height of the tank, H = 10 cm
- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm
- rate at which water is leaking from the tank = 1.5 cm³/min
- water is being pumped into the tank at constant rate of k cm³/min
- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min
c) volume of the tank at any time = πr²h/3
Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)
dV/dt = k - 1.5
V = πr²h/3 and r = 0.3 h, r² = 0.09 h²
V = 0.03πh³
dV/dt = (dV/dh) × (dh/dt)
dV/dh = 0.09π h²
dV/dt = 0.09π h² (dh/dt)
dV/dt = k - 1.5
0.09π h² (dh/dt) = k - 1.5
But at h = 2 cm, (dh/dt) = 1.0 cm/min
0.09π h² (dh/dt) = k - 1.5
0.09π 2² (1) = k - 1.5
k - 1.5 = 1.131
k = 1.5 + 1.131 = 2.631 cm³/min
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