- #1

- 267

- 10

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.