Question

Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 13 meters and the diameter at the top is 3 meters. If the water level is rising at a rate of 17 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. Round to the nearest whole number.

Answer 1

Answer:

The rate at which water is being pumped into the tank is 13,913.27 cubic centimetre per minute

Step-by-step explanation:

Given that, the height of the conical tank is 13 meters and diameter is 3 meters.

Radius = $\frac32$ meters

$\therefore \frac{radius }{height}=\frac {\frac{3}{2}}{13}$

$\Rightarrow \frac{radius }{height}=\frac {3}{26}$

$\Rightarrow radius=\frac {3}{26}\times height$   ......(1)

The volume of the cone is   $V=\frac13 \pi r^2 h$

Now putting $r=\frac{3}{26} h$

$V=\frac13 \pi (\frac3{26}h)^2 h$

$\Rightarrow V=\frac{3}{676} \pi h^3$

Differentiating with respect to t

$\frac{dV}{dt}=\frac{3}{676}\pi .3h^2 \frac{dh}{dt}$

$\Rightarrow \frac{dV}{dt}=\frac{9}{676}\pi h^2 \frac{dh}{dt}$

Given that, the water level is rising at a rate of 17 cm per minute when the height 1 meter= 100 cm .i.e $\frac{dh}{dt}=17$ cm/min

Putting $\frac{dh}{dt}=17$  

$\frac{dV}{dt}=\frac{9}{676}\pi h^2 \times 17$

$\frac{dV}{dt}|_{h=100}=\frac{9}{676}\pi (100)^2 \times 17$

             ≈7,113 cubic centimetre per minute

The volume of water increases 7,113 cubic centimetre per minute while 6,800 cubic centimetre per minute is leaking out.

It means the required rate at which water is being pumped into the tank is

=(7,113+6,800)cubic centimetre per minute

=13,913 cubic centimetre per minute