Question

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.

Answer 1

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

Answer:

The rate at which water is being pumped into the tank is 289,752 $cm^3/min$

Solution:

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) = $\frac{diameter}{2}$ = 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:

$\text { volume of cone }(\mathrm{V})=\frac{1}{3} \pi r^{2} h$   -------- (2)

From (1) and (2)

$\text { Volume of cone(V) }=\frac{1}{3} \pi\left(\frac{h}{3}\right)^{2} h$

$\mathrm{V}=\frac{\pi h^{3}}{27}$   -------- (3)

Now we calculate the derivate:-  

$\frac{d V}{d t}=\frac{3 \pi h^{2}}{27} \frac{d h}{d t}$

$\frac{d V}{d t}=\frac{\pi h^{2}}{9} \frac{d h}{d t}$  --------- (4)

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

$\frac{d h}{d t}=20 \mathrm{cm} / \mathrm{min}$

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

$\frac{d V}{d t}=279252$

Hence, the rate at which water is being pumped is 289,752 $cm^3/min$ which is the sum of water volume increasing at rate of 279,252 and 10,500 leaking out