Container filling: find depth rate

4 A large industrial water tank is such that, when the depth of the water in the tank is \(x\) metres, the volume \(V \mathrm {~m} ^ { 3 }\) of water in the tank is given by \(V = 243 - \frac { 1 } { 3 } ( 9 - x ) ^ { 3 }\). Water is being pumped into the tank at a constant rate of \(3.6 \mathrm {~m} ^ { 3 }\) per hour. Find the rate of increase of the depth of the water when the depth is 4 m , giving your answer in cm per minute.

6 The volume, \(V \mathrm {~m} ^ { 3 }\), of liquid in a container is given by $$V = \left( 3 h ^ { 2 } + 4 \right) ^ { \frac { 3 } { 2 } } - 8 ,$$ where \(h \mathrm {~m}\) is the depth of the liquid.

1. Find the value of \(\frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 0.6\), giving your answer correct to 2 decimal places.
2. Liquid is leaking from the container. It is observed that, when the depth of the liquid is 0.6 m , the depth is decreasing at a rate of 0.015 m per hour. Find the rate at which the volume of liquid in the container is decreasing at the instant when the depth is 0.6 m .

8. \begin{figure}[h] \end{figure} A bowl is modelled as a hemispherical shell as shown in Figure 3.
Initially the bowl is empty and water begins to flow into the bowl. When the depth of the water is \(h \mathrm {~cm}\), the volume of water, \(V \mathrm {~cm} ^ { 3 }\), according to the model is given by $$V = \frac { 1 } { 3 } \pi h ^ { 2 } ( 75 - h ) , \quad 0 \leqslant h \leqslant 24$$ The flow of water into the bowl is at a constant rate of \(160 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) for \(0 \leqslant h \leqslant 12\)

1. Find the rate of change of the depth of the water, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), when \(h = 10\) Given that the flow of water into the bowl is increased to a constant rate of \(300 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\) for \(12 < h \leqslant 24\)
2. find the rate of change of the depth of the water, in \(\mathrm { cms } ^ { - 1 }\), when \(h = 20\)

5. \begin{figure}[h] \end{figure} Figure 3 shows a container in the shape of a hollow, inverted, right circular cone.
The height of the container is 30 cm and the radius is 12 cm , as shown in Figure 3.
The container is initially empty when water starts flowing into it.
When the height of water is \(h \mathrm {~cm}\), the surface of the water has radius \(r \mathrm {~cm}\) and the volume of water is \(V \mathrm {~cm} ^ { 3 }\)

1. Show that $$V = \frac { 4 \pi h ^ { 3 } } { 75 }$$ [The volume \(V\) of a right circular cone with vertical height \(h\) and base radius \(r\) is given by the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ] Given that water flows into the container at a constant rate of \(2 \pi \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
2. find, in \(\mathrm { cm } \mathrm { s } ^ { - 1 }\), the rate at which \(h\) is changing, exactly 1.5 minutes after water starts flowing into the container.

\includegraphics{figure_2} A vase with a circular cross-section is shown in Figure 2. Water is flowing into the vase. When the depth of the water is \(h\) cm, the volume of water \(V\) cm\(^3\) is given by $$V = 4\pi h(h + 4), \quad 0 \leq h \leq 25$$ Water flows into the vase at a constant rate of \(80\pi\) cm\(^3\)s\(^{-1}\) Find the rate of change of the depth of the water, in cm s\(^{-1}\), when \(h = 6\) \hfill [5]