Derivative as an Instantaneous rate of change

[ve3rfd]

I am having trouble with this problem. Any help how to start it would be appreciated?

http://img297.imageshack.us/img297/1382/untitledsk9.jpg

 

[Galileo]

Mathematically, what is asked is $$V'(d)$$ (prime denotes differentiation), where V(d) is the volume as a function of the depth d. So find V(d) first.

 

[radou]

Yes, mathematically, the result should be a constant, but there's no serious physical sense here.

 

[ve3rfd]

Thats the thing i am having trouble with, starting the equation? I know I need the Volume for a cylinder equation but how do I use it to figure it out?

 

[StatusX]

The volume of a cone is 1/3 b*h, where b is the area of the base (pi r^2) and h is the height.

 

[ve3rfd]

So would the new equation be 1/3 4 - d * 4 - d (pi r^2)?

 

[radou]

The equation is V = 1/3 * d^3 * Pi . Since d can change, it is a variable, which makes the equation a function, so you can write V(d) = 1/3 * d^3 * Pi , because the volume of the fluid V depends of it's depth d.

 

[ve3rfd]

Okay, Thanks a lot i didn't get the d^3.

 

[radou]

The d^3 is implied by radius = height.