Sand is pouring from a pipe at the rate of `12c m^3//sdot` The falling sand forms a cone on the ground in such a way that the height of the cone is always 1/6th of the radius of the base. How fast does the height of the sand cone increase when the height in 4 cm?

Answer

To solve the problem step by step, we will follow the given information and derive the required rate of change of height of the sand cone.

### Step 1: Understand the relationship between height and radius
Given that the height \( h \) of the cone is always \( \frac{1}{6} \) of the radius \( r \), we can express this relationship mathematically:
\[ h = \frac{1}{6} r \quad \Rightarrow \quad r = 6h \]

### Step 2: Write the formula for the volume of the cone
The volume \( V \) of a cone is given by the formula:
\[ V = \frac{1}{3} \pi r^2 h \]
Substituting \( r = 6h \) into the volume formula gives:
\[ V = \frac{1}{3} \pi (6h)^2 h = \frac{1}{3} \pi (36h^2) h = 12\pi h^3 \]

### Step 3: Differentiate the volume with respect to time
We know that sand is pouring out at a rate of \( \frac{dV}{dt} = 12 \, \text{cm}^3/\text{s} \). We can differentiate the volume equation with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}(12\pi h^3) = 12\pi \cdot 3h^2 \frac{dh}{dt} = 36\pi h^2 \frac{dh}{dt} \]

### Step 4: Set the rate of volume change equal to the derivative
Now we set the expression for \( \frac{dV}{dt} \) equal to the rate at which sand is pouring:
\[ 36\pi h^2 \frac{dh}{dt} = 12 \]

### Step 5: Solve for \( \frac{dh}{dt} \)
Rearranging the equation to solve for \( \frac{dh}{dt} \):
\[ \frac{dh}{dt} = \frac{12}{36\pi h^2} = \frac{1}{3\pi h^2} \]

### Step 6: Substitute \( h = 4 \, \text{cm} \)
Now we need to find \( \frac{dh}{dt} \) when \( h = 4 \, \text{cm} \):
\[ \frac{dh}{dt} = \frac{1}{3\pi (4^2)} = \frac{1}{3\pi \cdot 16} = \frac{1}{48\pi} \]

### Final Answer
Thus, the rate at which the height of the sand cone increases when the height is 4 cm is:
\[ \frac{dh}{dt} = \frac{1}{48\pi} \, \text{cm/s} \]