Explanation:
[tex]\begin{gathered} h=\frac{3}{8}d \\ d=2r \\ h=\frac{3}{8}\times2r=\frac{3}{4}r \\ \frac{dh}{dt}=\frac{3}{4}\frac{dr}{dt} \\ r=\frac{4h}{3}=\frac{4\times5}{3}=\frac{20}{3}m=\frac{2000}{3}cm \\ \end{gathered}[/tex]
The volume of a cone is given below as
[tex]\begin{gathered} V=\frac{1}{3}\pi r^2h \\ V=\frac{1}{3}\pi r^2(\frac{3}{4}r) \\ V=\frac{1}{4}\pi r^3 \\ \frac{dV}{dr}=\frac{3}{4}\pi r^2 \end{gathered}[/tex]
Given from the question,
[tex]\begin{gathered} \frac{dV}{dt}=15m^3\text{ /min} \\ \frac{dV}{dt}=15000000cm^3\text{ /min} \end{gathered}[/tex][tex]\begin{gathered} \frac{dV}{dt}=\frac{dV}{dr}\times\frac{dr}{dt} \\ 15000000=\frac{3}{4}\pi r^2\times\frac{dr}{dt} \\ \frac{15000000}{0.75\times\pi\times(\frac{2000}{3})^2}=\frac{dr}{dt} \\ \frac{dr}{dt}=\frac{15,000,000}{0.75\pi(\frac{2,000}{3})^{2}} \\ \frac{dr}{dt}=14.32cm\text{ /min} \end{gathered}[/tex]
Hence,
The height is changing at a rate of
[tex]\begin{gathered} \frac{dh}{dt}=\frac{3}{4}\times14.32 \\ \frac{dh}{dt}=10.74cm\text{ /min} \end{gathered}[/tex]
Hence,
The radius is changing at a rate of
[tex]14.32\text{ cm/min}[/tex]
