3. The volume of a tank used to provide water for animals at an animal enclosure at the local show is given by V(h) = (0.2h^3+3h)m^3 where h metres is the depth of the water in the tank at any time t seconds. Water is being added to the tank at a constant rate of 0.4 m3 per second. Find the rate at which the water level is rising when the depth of water in the tank is 2.1 metres. So i differentiated the equation: v(h) = (0.2h^3+3h) m^3, which is v'(h) = (0.6h^2+3) m^3. But because the limit is 2.1m, and the amount of liquid going into the object is going in at 0.4m^3 a second. so if i use v'(2.1)= (0.6h^2+3) m^3 and solve for h.. i can divide h by 0.4 and find out the time it took to reach 2.1m? Thank you.