A water tank has the shape of an inverted right circular cone with its axis vertical and vertex lowermost. Its semi-vertical angle is tan^-1(0.5). Water is poured into it at a constant rate of 5 cubic meters per hour. Find the rate at which the level of the water is rising at the instant when the depth of water in the tank is 4m.

Semi-vertical angle is half of the angle at vertex as shown and let r be the radius of base of cone and h be the height

AC is the radius of base r, AB is the height h and ∠ABC = tan^-1(0.5) which is the semi-vertical angle

The rate at which water is poured is given as 5 cubic meter per hour which means the rate of change of volume with respect to time is 5 m³/h

If V is the volume and t is time then the change is denoted by

Volume of cone is given by

We need to find the change in height with respect to time that is hence we need to eliminate the r we need V in terms of h

From figure

⇒ 0.5h = r

⇒ r = 1/2h

Hence put r = 1/2h in V

Differentiate with respect to t

Now it is given that

We have to find the rate when the depth that is height is 4 m

Hence put h = 4

Hence the rate at which level of water is rising at depth 4m is