The rate of change of the surface of a sphere, differentiated with respect to radius, is 8pr, right? So suppose a vacuum is sucking water to a central spot at rate 1. What will be the instantaneous rate of change of the speed of a miniature submarine, relative to the vacuum, 1 radius out?MP

"What will be the instantaneous rate of change of the speed of a miniature submarine ... ?"I'll have a look the next time I take my Phoenix for a drive

And people say innaccurate stock prices don't lead to poor, yet unrecoverable, decisions...Do you suppose Michael Saylor commissioned this thing?MP

Lets see if I've got this right.We have a point-like sink that is sucking in water at a rate of 1 unit/s. What is the speed of a submarine that is beeing sucked in relative to the sink?Well,The volume of a sphere is (4*pi*r^3)/3. Inverting this we get r(V)=(3V/(4*pi))^(1/3).if we differentiate w.r.t. V, we getdr/dV=1/(4*pi)*(3V/(4*pi))^(-2/3)this givesdr/dt = dr/dV * dV/dt = dr/dV * 1 = 1/(4*pi) * (3V/(4*pi))^(-2/3) = 1/(4*pi*r^2)Thus, the radial velocity of the submarine is 1/(4*pi*r^2). Notice that this tends to infinity as we approach the sink. This is due to our assumption that the sink is point-like.If we look at r=1 we get dr/dt = 1/(4*pi)I hope this is what you where looking for.Regards,Niclas

They gave me such a big discount that I didn't feel like asking who originally commissioned it ...