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Sand falls from a conveyor belt at a rate of $13 \mathrm{~m}^{3} / \mathrm{min}$ onto the top of a conical pile. The height of the pile is always three-eighths of the base diameter. How fast are the height and the radius changing when the pile is 5 $\mathrm{m}$ high?
The height is changing at a rate of $\square \mathrm{cm} / \mathrm{min}$ when the height is $5 \mathrm{~m}$.
(Type an integer or decimal. Round to the nearest hundredth.)
The radius is changing at a rate of $\square \mathrm{cm} / \mathrm{min}$ when the height is $5 \mathrm{~m}$.
(Type an integer or decimal. Round to the nearest hundredth.)

Step 1 ：The volume of a cone is given by the formula \(V = \frac{1}{3}\pi r^2 h\), where \(r\) is the radius and \(h\) is the height.

Step 2 ：Given that the height of the pile is always three-eighths of the base diameter, we can express the radius in terms of the height as \(r = \frac{4}{3}h\).

Step 3 ：Substituting \(r = \frac{4}{3}h\) into the volume formula, we get \(V = \frac{1}{3}\pi\left(\frac{4}{3}h\right)^2 h = \frac{16}{27}\pi h^3\).

Step 4 ：Differentiating both sides with respect to time \(t\), we get \(\frac{dV}{dt} = \frac{16\pi}{9}h^2\frac{dh}{dt}\).

Step 5 ：We know that \(\frac{dV}{dt} = 13\) m³/min, so we can solve for \(\frac{dh}{dt}\) when \(h = 5\) m:

Step 6 ：\(13 = \frac{16\pi}{9}(5)^2\frac{dh}{dt}\)

Step 7 ：Solving for \(\frac{dh}{dt}\), we get \(\frac{dh}{dt} = \frac{13}{\left(\frac{16\pi}{9}\right)(25)}\)

Step 8 ：Simplifying, we find \(\frac{dh}{dt} = 0.234\) m/min or 23.4 cm/min.

Step 9 ：To find \(\frac{dr}{dt}\), we differentiate \(r = \frac{4}{3}h\) with respect to \(t\) to get \(\frac{dr}{dt} = \frac{4}{3}\frac{dh}{dt}\).

Step 10 ：Substituting \(\frac{dh}{dt} = 0.234\) m/min, we get \(\frac{dr}{dt} = \frac{4}{3}(0.234) = 0.312\) m/min or 31.2 cm/min.

Step 11 ：So, the height is changing at a rate of \(\boxed{23.4}\) cm/min and the radius is changing at a rate of \(\boxed{31.2}\) cm/min when the height is 5 m.