Question

a landscaping company is pouring rock chips into a conical pile with a constant ratio of 2:5 between the radius and height.the volume of the rock chips is increasing at a rate of 1.8m3/min.At what rate is the height increasing when the radius is 3

Answer 1

`\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}\impliedby \textit{the r:h ratio is 2:5}\implies \cfrac{r}{h}=\cfrac{2}{5}\implies r=\cfrac{2h}{5} \\\\\\ V=\cfrac{\pi \left((2h)/(5) \right)^2 h}{3}\implies V=\cfrac{(\pi 2^2h^2h)/(5^2)}{3}\implies V=\cfrac{4h^3\pi }{25}\cdot \cfrac{1}{3} \\\\\\ V=\cfrac{4\pi h^3}{75}\\\\ -------------------------------\\\\`


`\bf \cfrac{dv}{dt}=\cfrac{4\pi }{75}\cdot 3h^2\cfrac{dh}{dt}\implies \cfrac{dv}{dt}=\cfrac{4\pi h^2 }{25}\cdot \cfrac{dh}{dt}\implies \cfrac{25(dv)/(dt)}{4\pi h^2}=\cfrac{dh}{dt} \\\\\\ \begin{cases} (dv)/(dt)=1.8\\ r=3\\ h=(5)/(2)r\\ h=(15)/(2) \end{cases}\implies \cfrac{25\cdot 1.8}{4\pi \left( (15)/(2) \right)^2}=\cfrac{dh}{dt}`