## Differential Equations

I'm having some trouble with the following questions:

The vertical cross-section of a bucket is shown in this diagram. The sides are arcs of a parabola with the y axis as the central axis and the horizontal cross-sections are circular. The depth is 36 cm, the base radius length is 10 cm and the radius length of the top is 20 cm.

a) From this part, we know that the parabolic sides are arcs of the parabola $y = 0.12x^2 - 12$.
b) Water starts leaking from the bucket, initially full, at the rate given by $\frac{dv}{dt} = \frac{-{\sqrt{h}}}{A}}$ where, at time t seconds, the depth is h cm, the surface area is A cm2 and the volume is v cm3. Prove that $\frac{dv}{dt} = \frac{-3{\sqrt{h}}}{25\pi(h+12)}}$.
I can recognise that $\frac{3}{25}$ represents 0.12, so I know that part (a) ties in somehow, but I don't know where to start.
c) Show that $v = \pi{\int_0^h}({\frac{25y}{3}} + 100)dy$. I don't know how to approach this either, but it will probably become clear once I understand part (b).

This container has an open rectangular horizontal top, PQSR, and parallel vertical ends, PQO and RST. The ends are parabolic in shape. The x axis and y axis intersect at O, with the x axis horizontal and the y axis the line of symmetry of the end PQO. The dimensions are shown on the diagram.

a) The equation for the parabolic arc QOP was found to be $\frac{2}{5}x^2$.
b) If water is poured into the container to a depth of y cm, with a volume of V cm3, find the relationship between V and y.
I initially tried to grapple with the idea that the volume = the area multiplied by 60, but to no avail.

Thanks in advance, I know these can be long winded questions.