Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
The height of the sand cone is increasing at the rate of 1/(48π) cm/s.
Explanation
This is a related rates problem similar to Example 31 in the textbook. The key is to express the volume as a function of height only, using the given relationship between height and radius, then differentiate with respect to time.
Solution Steps
Step 1: Let r be the radius and h be the height of the cone. Given that height is always one-sixth of the radius, we have h = r/6, which gives r = 6h.
Step 2: Volume of a cone is V = (1/3)πr²h. Substituting r = 6h, we get V = (1/3)π(6h)²h = (1/3)π(36h²)h = 12πh³.
Step 3: Differentiating V with respect to t using Chain Rule: dV/dt = dV/dh × dh/dt = 36πh² × dh/dt.
Step 4: Given dV/dt = 12 cm³/s and h = 4 cm. Substituting: 12 = 36π(4)² × dh/dt = 36π(16) × dh/dt = 576π × dh/dt.
Step 5: Solving for dh/dt: dh/dt = 12/(576π) = 1/(48π) cm/s.