I've got one problem left on a math assignment that I can't figure out. I don't usually post stuff like this here, but this one really confuses me.
"A cylindrical drum with a radius of 1.5 metres is being filled with oil"
Part A: "Find the formula to represent the instantaneous rate of change in the volume of oil with respect to the height of water".
So far I just have V(h)=pi*r^2*h. The formula for the volume of a cylinder.
We're doing derivatives and stuff, so I have to (I'm pretty sure), find the derivative of that function to be able to get the IROC.
I thought the derivative would be V(h)'=pi*2r*h, but apparently that's not right since the radius is constant. But the only other way I can see to do it is if V(h)'=pi*r^2. But that doesn't actually include height anywhere in the function.
Is this something I'm just missing in the derivative, or something in the problem itself?
I don't know where the water part came from in your question -- I'm assuming that's also supposed to be oil.
As to the other part, you need to find dV/dh (rate of change in volume with respect to height) -- so of course it's constant. Adding "one meter of oil" (if we're talking about change in height) is going to add the same amount of oil to the drum, whether it's at the bottom or the top. So the amount you add doesn't depend on h.
Ok. So for example, Part B asks what the IROC is when the height is 2 metres...does that mean that it will be the same as if the height was 3 metres?
Originally Posted by psychopath
Ok. The more I think about it, the more it makes sense. I just didn't understand at first how it would be physically possible.