Expert: Socrates - 12/2/2011

Question
The amount of water evaporating from a conical container varies directly with the exposed surface area (circular region) of the water in the cone. The height of the cone is 10 meters when full. When the height of the water is 5 meters, the radius of the exposed circular region is 3 meters. At this point, water is evaporating at a rate of 0.2 liters per hour. The question is:
    1) How fast is the water evaporating when the container is full?
    2) How fast is the water evaporating when the container is half-full?

Answer
Let E(a) be the rate of evaporation when the surface area of the circular region on top is a.
We are told that the rate of evaporation is directly proportional to the area. This means that E(a)=ca , for some constant c. When the height of the water is 5 meters, the radius of the exposed circular region is 3 meters. At this point, water is evaporating at a rate of 0.2 liters per hour. Since the radius is 3 at this point , the area is a= π 3^2 = 9π . The rate of evaporation is .2 , so E(9π) = (c)(9π) = .2 . Solving for c , we get c = .2/(9π)
Since the radius is 3 when the height is 5 , the ratio of the radius to height will always be 3/5 , no matter how high the water is. To see this ,
you need to draw a picture and use similar right triangles. Unfortunately , I can't do this for you.
We are now ready to answer the questions

1)

When the container is full , the height of the water is 10. The ratio of radius to height is 3/5 , so to find the radius when height is 10 , r/10 = 3/5 and r= 6.
The area when the container is full is then π 6^2 = 36π
E(36π) = (.2/(9π))(36π) =  .8

.8  is the rate of evaporation


2)

The volume of a cone is V = (1/3)π(r^2)h , where r is the radius at the top and h is the height.

The total volume of the container is V = (1/3)π(6^2)10 = 120π

When the container is half full , V = 60π = (1/3)π(r^2)h . We know that r/h = 3/5 , so h = (5/3)r

Then 60π = (1/3)π(r^2)(5/3)r = (5/9)π(r^3)

Solving for r , r^3 = 108 = (27)(4) , so r = (3)(4)^(1/3) , when the container is half full.

a = π(r^2)= 18π(2)^(1/3)

E(a) = E(18π(2)^(1/3)) = (.2/(9π))(18π)(2)^(1/3)) = (.4)(2)^(1/3)

(.4)(2)^(1/3) is approximately .50397 liters per hour.