Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 4 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

Question

Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 4 feet, the depth is increasing at  0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

wolf1728 · Answer

I guess the cylindrical tank is oriented this way?

anonymous · Answer

I think it is with the base on the bottom, not on the sides

wolf1728 · Answer

Well that makes the calculations a LOT simpler

anonymous · Answer

So I derived the equation in terms of time and I got $$dv/dt=\pi[2rh dr/dt + r^2 dh/dt]$$ and because the radius is a constant, the derivative of r is 0 so I emelimated the 2rh in the formula and was left with $$dV/dt=\pi[9 (0.2) dh/dt]$$ so is the answer $$dV/dt = 1.8\pi $$?

wolf1728 · Answer

I updated the graphic :

wolf1728 · Answer

I still have the feeling that this problem is for a horizontal cylindrical tank.  Is this for a calculus class?  If it is then I'd say it's almost certainly horizontal.

anonymous · Answer

It's vertical

wolf1728 · Answer

Well okay.  (That does make the calculations much easier).