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Sand falls from a conveyor belt at a rate of $13{\mathrm{m}}^3/\min$ 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 $◻\mathrm{cm}/\min$ 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 $◻\mathrm{cm}/\min$ 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 $\boxed{{\displaystyle 23.4}}$ cm/min and the radius is changing at a rate of $\boxed{{\displaystyle 31.2}}$ cm/min when the height is 5 m.