An inverted cone with height 10 cm and radius 2 cm is partially filled with a my

Question

An inverted cone with height 10 cm and radius 2 cm is partially filled with a mysterious green liquid, which is oozing through the sides of the cone at a rate proportional to the surface area of the cone in contact with the liquid. (The surface area of a cone is given by (pi r l) , where (r) is the radius of the cone and (l) is the "slant height" of the cone.) The green liquid is being replenished by being poured into the top of the cone by aliens at a rate of 1 cm3/min. At this rate, when the depth of the fluid is 4 cm, the depth is decreasing at the rate of 0.1 cm/min. The Aliens want to keep the depth of the fluid constant at 4 cm. By how much should they change the rate at which they are replenishing the fluid in the cone?

Hint:

Find two expressions for the rate of change of volume. The first one comes from the formula for the volume of a cone, and the second comes from the difference between the fluid flowing in and the fluid oozing out. Use this to solve for your proportionality constant (traditionally called k), then go from there.

Thank you so much for the help! I will rate the answer accordingly =)

Explanation / Answer

I'm posting this here because it won't accept my long comment. Just remember that O is the difference between the rate of addition and the rate of oozing and the difference must be compensated to keep the 4 cm level.

Use dV/dh - O = .1   where O = k A since this is the difference between the change in volume due to filling and leakage. This is the amount to add to the 1 cm^3/min to keep the fluid at the 4 cm level. Probably k need not be calculated since O is the difference between the filling and liquid. One would need O as a function of h to calculate the leakage at some value other than 4 cm, but once you have dV/dh (I should have written dV/dh not dV/dt) you already have O the rate of oozing.

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