Originally Posted by

**Gul** Volume of water in a dam is given by:

V(h) = 750h^3 - 600h^2, where h is the depth of water at the dam wall

If on a particular day the water drops by 0.08m from 22m, what will be the difference in volume. Okay, so apart from actually plugging and chugging into the equation, then subtracting, i can use the small changes formula. But I'd like to know why it works.

dV/dh = DV/Dh

dV = DV/Dh * dh

dV = 2250(22)^2 - 600(22) * 0.08

dV = 85,008

Which is quite close to:

**((750 * (22^3)) - (600 * (22^2))) - ((750 * (21.92^3)) - (600 * (21.92^2))) = 84 695.424**

How do I derive the:

dV/dh = DV/Dh

Thanks.