00:01
Hello, in this question we have a tank in the shape of an inverted cone whose height is 6 meters and the top has a circular cross section and the diameter of the top of the tank is 6 .5 meters.
00:24
Now, water is being pumped into the tank at some rate and water is leaking out of the tank at the rate of 8700 centimeter cube per minute and at some point of time the height of the water in the tank is 3 meters.
01:11
At some point of time the height of the water in the tank is 3 meters and at this point of time the water level is rising at 26 centimeter per minute.
01:32
Under this in this situation and from this information we have to find what is the rate at which water is being pumped into the tank.
01:45
Okay, so first let the height of the water in the tank be denoted by h, let us use the variable h to denote it and diameter of the top of the water that is inside the tank let it be d.
02:02
We will have that d by h will be equal to, geometrically this cone will be congruent to this cone, so we will have d by h is equal to 6 .5 over 6 which we can write as, if you multiply by 2 to both sides we will have 13 over 12.
02:39
So, h or sorry d is equal to 13 over 12 times h which means that the radius of the cross section of the water is d by 2 diameter by 2 which is 13 by 24 times h.
02:59
Now, the volume of the water in the tank when the height is h, height of the water in the tank is h is 1 by 3 pi r square h which is 1 by 3 pi r is this fraction times h, so it is 13 by 24 square times h square but h square times h is h cube so we can well yeah times h which is one third times pi times 13 by 24 square times h cube.
03:47
This is some constant times h cube.
03:50
So, we have that dv by dt from this geometry we can see that rate of change of the volume with respect to time is equal to this constant 1 by 3 pi 13 by 24 square times 3 times h square times dh by dt.
04:14
Since h is a function of time we have that derivative of h cube with respect to time is 3 h square times dh by dt.
04:27
We are using chain rule here.
04:29
So, this is dv by dt and it is given that this rate of change of the volume of water in the tank is equal to the rate at which water is being pumped into the tank minus the rate at which water is leaking out from the tank...