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.

1. How fast is the water evaporating when the container is full?

   Hint (a):

   Determine the radius when the cone is filled.

   Step 1(a):

   $\frac{3}{5}=\frac{r}{10}$

   More Help (a):

   Set up and solve a proportion relating the area with the rate of evaporation.

   Step 2(a):

   $\frac{9\pi}{0.2}=\frac{36\pi }{x}$

2. How fast is the water evaporating when it is half full? (Reminder $V=\frac{1}{3}\pi r^2h$.)

   Hint: (b)

   When the cone is full, it has $120\pi$ m³ of water in it.
   What is the volume of water in the cone when it is half full?

   More Help (b):

   Calculate $r$ when the cone is half full. Then set up a proportion with area and evaporation rate similar to the one in part (a).