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 \(\displaystyle={r}={6}{m}{m}\).

\(\displaystyle={\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{2}}}{{{3}}}}\pi{r}{h}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}\)

\(\displaystyle=\Rightarrow{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{3}}}{{{2}\pi{r}{h}}}}{\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}\)

\(\displaystyle={\frac{{{3}}}{{{2}\pi\times{6}\times{15}}}}\times{534}\)

\(\displaystyle={2.8329}\)

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 \(\displaystyle={r}={6}{m}{m}\).

\(\displaystyle={\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{2}}}{{{3}}}}\pi{r}{h}{\frac{{{d}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}\)

\(\displaystyle=\Rightarrow{\frac{{{d}{r}}}{{{\left.{d}{t}\right.}}}}={\frac{{{3}}}{{{2}\pi{r}{h}}}}{\frac{{{d}{V}}}{{{\left.{d}{t}\right.}}}}{\left({r}\right)}\)

\(\displaystyle={\frac{{{3}}}{{{2}\pi\times{6}\times{15}}}}\times{534}\)

\(\displaystyle={2.8329}\)