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 3 cm?

To find out how fast the height of the sand cone is increasing when the height is 3 cm, we can use the relationship between the volume of the cone and its dimensions. The volume \( V \) of a cone is given by the formula:

Volume of the Cone

The formula for the volume of a cone is:

V = (1/3)πr²h

In this case, the height \( h \) is one-sixth of the radius \( r \), so we can express \( h \) as:

h = (1/6)r

Substituting Height into Volume Formula

Now, substituting \( h \) into the volume formula gives:

V = (1/3)πr²(1/6)r = (π/18)r³

Finding the Rate of Change

We know that sand is pouring at a rate of 12 cm³/s, which means:

dV/dt = 12 cm³/s

To find how fast the height is increasing, we need to relate the rates of change. We can differentiate the volume with respect to time:

dV/dt = (π/6)r²(dr/dt)

Relating Radius and Height

Since \( h = (1/6)r \), we can express \( r \) in terms of \( h \):

r = 6h

When \( h = 3 \) cm, then:

r = 6 * 3 = 18 cm

Calculating dr/dt

Now we can substitute \( r \) back into the volume rate equation:

12 = (π/6)(18²)(dr/dt)

Solving for \( dr/dt \):

12 = (π/6)(324)(dr/dt)

dr/dt = 12 / (54π) = 2 / (9π) cm/s

Finding dh/dt

Now, we can find \( dh/dt \) using the relationship between \( r \) and \( h \):

dh/dt = (1/6)(dr/dt)

Substituting \( dr/dt \):

dh/dt = (1/6)(2 / (9π)) = 1 / (27π) cm/s

Thus, the height of the sand cone is increasing at a rate of approximately:

dh/dt ≈ 0.0119 cm/s