# Thread: Fluid depth and volumetric flow problem

1. ## Fluid depth and volumetric flow problem

Hello there, I was hoping to get some help on an assignment from one or more kind souls. I have completed part a) and am happy with it, but am stuck on part b). I'm not sure how the derivative term fits in or why. If anyone could explain that and part c) (in particular why the power of m for the k variable is 1/2), I'd be very grateful, thanks.

2. ## Re: Fluid depth and volumetric flow problem

Hey omegaman187.

Hint: Volume = Height*Width*Depth (find the area inside the parabola using normal integration). Remember that you have for y = x^2 the area under the parabola as well as the area above the parabola but bounded by y <= sqrt(x) for -h < x < h.

3. ## Re: Fluid depth and volumetric flow problem

Hi chiro,

Thanks for the quick reply, though as I said I managed to figure out the volume question (part a)). Any hints on parts b) or c)?

4. ## Re: Fluid depth and volumetric flow problem

Sorry I missed the statement about part a).

For part b) you need to show dV/dt = -Q (you will have to use differentiation rules to get derivatives in terms of time).

The reason for the above is that the question states that the flow is equal to Q and the negative sign is there because you are losing water from the tank and not gaining water.

5. ## Re: Fluid depth and volumetric flow problem

Thanks a lot for the help chiro, managed to figure it out