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.