Water is leaking out of an inverted conical tank at a rate of at the same time that water is being pumped into the tank at a constant rate. the tank has height m and the diameter at the top is m. if the water level is rising at a rate of when the height of the water is m, find the rate at which water is being pumped into the tank.

Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/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 which water is being pumped into the tank

The rate at which water is being pumped into the tank is 289,752

According to question,

There is an inverted conical tank, through which water is leaking at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate

The dimension of tank are:

Diameter = 4cm

Radius(r) = = 2cm

Height = 6cm

Clearly we can see that height is 3 times radius so, we can write

h = 3r OR r = h/3    ……………………. (1)

The volume of cone "V" is given as:

  -------- (2)

From (1) and (2)

  -------- (3)

Now we calculate the derivate:-  

 --------- (4)

According to question, when height is 2m = 200cm, the water level is rising at a rate of 20 cm/min

On putting above values in equation(4) and solving we get

Hence, the rate at which water is being pumped is 289,752 which is the sum of water volume increasing at rate of 279,252 and 10,500 leaking out

idk sry

step-by-step explanation: