# rate of change problem diferential equations

so here i need some help. Here we have a poorly drawn cylinder which is filled with water at a rate of $\frac{5m^3}{s}$ and the cylinder have two holes that are regulated to release water at a rate of $\frac{1m^3}{s}$ each no matter the water volume and all of the water falls in the conic container below as a result it will be filled. the question is to find the rate of change of the height of the cone while is filled by water in any given time. asume that recipients don't overflow in other word h(t) where h is the height of the cone.

so i proceed as follows first identifying the objective function, the volume of a cone is $v= \frac{1}{3}\pi r^2h$ and we know that the cone radius and height relation goes as follows

as a result the equation changes to $v= \frac{1}{3}\pi (\frac{h}{8})^2 h$ and simplified becomes $v= \frac{h^3}{192}\pi$ so the rate of change is the derivative of this function which is $\frac{dv}{dt} = \frac{dh}{dt} \frac{h^2}{64}\pi$ so we solve for $\frac{dh}{dt}$ and get $\frac{dh}{dt} = \frac{dv}{dt} \frac{64}{h^2\pi}$

Now we only need to find the rate of change of the volume and this is where the problem begins for me as the rate of change for the volume is variable with time. once it reaches a height of 1m the rate of change will increase to $\frac{2m^3}{s}$ so the volume for the cylinder is $v= \pi r^2h$ a different h and r of course. so the rate of change of the volume of the cylinder $\frac{5m^3}{s} - v_{out}$ $v_{out}$ is the volume that leaves the cylinder that is a function of the height that is $v_{out} = \frac{1m^3}{s}$ that is the height of the cylinder is lower than 1m but $v_{out} = \frac{2m^3}{s}$ if the height is more than 1m so this is the problem i don't know how to make $v_{out}$ one single expression or how to solve the problem. pictures made with paint

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You won't have a single expression, but that's fine. The flux of water into the cone would depend on time (we can easily find the point in time in which the flux doubles), and the height function would then be defined differently before and after that point (but would remain continuous, naturally). –  Jonathan Y. Sep 4 '13 at 7:02

knowing that the flux of the water changes at $t= \frac {1}{5}\pi$ then the expression for the rate of change of the height of the cone will be from this expression $\frac{dh}{dt} = (\frac{1m^3}{s}) \frac{64}{h^2\pi}$ for $t \le \frac {1}{5}\pi$ and $\frac{dh}{dt} = (\frac{2m^3}{s}) \frac{64}{h^2\pi}$ for $t \gt \frac {1}{5}\pi$ but this is in function of the height and not in function of the time and i con't think this can be solved in function of time