1. Notice how the height of the water is always twice the radius.
Part (a)
r = 2.5 cm
Notice how the height of the water is always twice the radius.
Volume = (1/3)pr2h
Volume = (1/3)p(5/2)2(5)
Volume = (125/12)p cm3

2. Remember, the radius of the water is always half the height.
Part (b)
Find dV/dt when h = 5
Remember, the radius of the water is always half the height.
V = (1/3) p (h/2)2 h
Volume = (1/3)pr2h
V = (1/12) p h3
dV/dt = (1/4)ph2(dh/dt)
dV/dt = (1/4)p(5)2(-3/10)
dV/dt = (-15/8)p cm3/hr

3. From Part (b), we know that dV/dt = (1/4)ph2(dh/dt)
Part (c)
From Part (b), we know that dV/dt = (1/4)ph2(dh/dt)
The exposed surface area of the water is a circle. SA = pr2
Recall that r = h/2.
These rates are directly proportional. The constant of proportionality is (dh/dt) or -3/10.
This makes the exposed surface area of the water SA = p(h/2)2 or (1/4)ph2.