# An inverted cone has a height of 15 mm and a radius of 16mm. The volum

An inverted cone has a height of 15 mm and a radius of 16mm. The volume of the inverted cone is decreasing a rate of 534 cubic mm per second, with the height begin held constant. What is the rate of change of the radius, in mm per second, when the radius is 6 mm?
Remember that the volume of a cone is $$\displaystyle={V}=\frac{{1}}{{3}}\pi{r}^{{{2}}}{h}$$

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Step 1
The volume of a cone is$$\displaystyle={V}=\frac{{1}}{{3}}\pi{r}^{{{2}}}{h}$$
The volume of the inverted cone is decreasing at a rate of 534 cubic mm per second, with the height begin held constant.
$$\displaystyle={d}\frac{{V}}{{\left.{d}{t}\right.}}=\frac{{1}}{{3}}/\pi{h}.{d}\frac{{{r}^{{{2}}}}}{{\left.{d}{t}\right.}}$$ as h is held constant and hence $$\displaystyle={d}\frac{{h}}{{\left.{d}{t}\right.}}={0}$$
Step 2
Find the formula for the rate of change of the volume, in mm per second.
$$\displaystyle={\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}}={\frac{{{1}}}{{{3}}}}\pi{h}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}^{{{2}}}\right)}$$
$$\displaystyle={\frac{{{1}}}{{{3}}}}\pi{h}{\left({2}{r}\right)}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}$$
$$\displaystyle={\frac{{{2}}}{{{3}}}}\pi{r}{h}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}$$
Step 3
To find the rate of changt of the radius, in mm per second, when the radius is 6 mm.
Here $$\displaystyle={d}\frac{{V}}{{\left.{d}{t}\right.}}={534}$$ cubic mm per second, $$\displaystyle={h}={15}{m}{m}$$ and $$\displaystyle={r}={6}{m}{m}$$.
$$\displaystyle={\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{2}}}{{{3}}}}\pi{r}{h}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}$$
$$\displaystyle=\Rightarrow{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{3}}}{{{2}\pi{r}{h}}}}{\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}$$
$$\displaystyle={\frac{{{3}}}{{{2}\pi\times{6}\times{15}}}}\times{534}$$
$$\displaystyle={2.8329}$$