Question

A cistern in the form of an inverted circular cone is being filled with water at the rate of 95 liters per minute. If the cistern is7 meters deep, and the radius of its opening is 5 meters, find the rate at which the water level is rising in the cistern50 minutes after the filling process began. Round any intermediate calculations, if needed, to no less than six decimal places,and round your final answer to three decimal places. (Hint: 1 m3 = 1000 L.)

Answer 1

`\begin{gathered} _{_{_{_{_{}}}}}V=(1)/(3)\pi r^2h=(\pi)/(3)(r^2h) \\ (dv)/(dt)=(\pi)/(3)(2rh(dr)/(dt)+r^2(dh)/(dt) \\ \\ \text{Volume after 50 minutes =95}*50=4.75m^3 \\ \\ \end{gathered}`
`\begin{gathered} (h)/(r)=\frac{7}{5\text{ }} \\ h=(7)/(7)r \\ \\ (dh)/(dt)((7)/(5)(dr)/(dt)) \\ (dr)/(dt)=(5)/(7)(dh)/(dr) \\ \\ \end{gathered}`

Substitute into the Volume

`\begin{gathered} 4.75=(1)/(3)\pi r^2((7)/(5)r) \\ \text{Simplify} \\ \\ 4.75=(7)/(15)\pi r^3 \\ \\ \text{make r subject the formula} \\ r^3=(4.75*15)/(7\pi) \\ r=1.48m \\ h=2.07 \\ \\ 0.095=(\pi)/(3)(2(1.48)(2.07)(5)/(7)(dh)/(dt)+(1.48^2)(dh)/(dt)) \end{gathered}`

Make dh/dt the subject of the formula

`(dy)/(dt)=0.0138143109\text{ m/min}`