# small changes in a function

• Jun 7th 2009, 03:03 PM
Gul
small changes in a function
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.
• Jun 8th 2009, 12:04 AM
CaptainBlack
Quote:

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.

Linear approximation of a function at a point (first two terms of Taylor series):

$V(h+\delta h)\approx V(h)+ \delta h ~V'(h)$

That's all there is to it.

CB