Tank/container - constant cross-section (cuboid/cylinder)

10 A water tank is in the shape of a cuboid with base area \(40000 \mathrm {~cm} ^ { 2 }\). At time \(t\) minutes the depth of water in the tank is \(h \mathrm {~cm}\). Water is pumped into the tank at a rate of \(50000 \mathrm {~cm} ^ { 3 }\) per minute. Water is leaking out of the tank through a hole in the bottom at a rate of \(600 \mathrm {~cm} ^ { 3 }\) per minute.

1. Show that \(200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 250 - 3 h\).
   
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2. It is given that when \(t = 0 , h = 50\). Find the time taken for the depth of water in the tank to reach 80 cm . Give your answer correct to 2 significant figures.

12. \begin{figure}[h] \end{figure} Figure 4 shows a right cylindrical water tank. The diameter of the circular cross section of the tank is 4 m and the height is 2.25 m . Water is flowing into the tank at a constant rate of \(0.4 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). There is a tap at a point \(T\) at the base of the tank. When the tap is open, water leaves the tank at a rate of \(0.2 \pi \sqrt { h } \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\), where \(h\) is the height of the water in metres.

1. Show that at time \(t\) minutes after the tap has been opened, the height \(h \mathrm {~m}\) of the water in the tank satisfies the differential equation $$20 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 2 - \sqrt { h }$$ At the instant when the tap is opened, \(t = 0\) and \(h = 0.16\)
2. Use the differential equation to show that the time taken to fill the tank to a height of 2.25 m is given by $$\int _ { 0.16 } ^ { 2.25 } \frac { 20 } { 2 - \sqrt { h } } \mathrm {~d} h$$ Using the substitution \(h = ( 2 - x ) ^ { 2 }\), or otherwise,
3. find the time taken to fill the tank to a height of 2.25 m . Give your answer in minutes to the nearest minute.
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

8. \begin{figure}[h] \end{figure} Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m . Water is flowing into the tank at a constant rate of \(0.48 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\). At time \(t\) minutes, the depth of the water in the tank is \(h\) metres. There is a tap at a point \(T\) at the bottom of the tank. When the tap is open, water leaves the tank at a rate of \(0.6 \pi h \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }\).

1. Show that \(t\) minutes after the tap has been opened $$75 \frac { \mathrm {~d} h } { \mathrm {~d} t } = ( 4 - 5 h )$$ When \(t = 0 , h = 0.2\)
2. Find the value of \(t\) when \(h = 0.5\)

9. \begin{figure}[h] \end{figure} Figure 4 shows a cylindrical tank that contains some water. The tank has an internal diameter of 8 m and an internal height of 4.2 m .
Water is flowing into the tank at a constant rate of \(( 0.6 \pi ) \mathrm { m } ^ { 3 }\) per minute. There is a tap at point \(T\) at the bottom of the tank. At time \(t\) minutes after the tap has been opened,

• the depth of the water is \(h\) metres
• the water is leaving the tank at a rate of \(( 0.15 \pi h ) \mathrm { m } ^ { 3 }\) per minute
  1. Show that

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 12 - 3 h } { 320 }$$ Given that the depth of the water in the tank is 0.5 m when the tap is opened,

find the time taken for the depth of water in the tank to reach 3.5 m .

11. \begin{figure}[h] \end{figure} A tank in the shape of a cuboid is being filled with water.
The base of the tank measures 20 m by 10 m and the height of the tank is 5 m , as shown in Figure 1. At time \(t\) minutes after water started flowing into the tank the height of the water was \(h \mathrm {~m}\) and the volume of water in the tank was \(V \mathrm {~m} ^ { 3 }\) In a model of this situation

• the sides of the tank have negligible thickness
• the rate of change of \(V\) is inversely proportional to the square root of \(h\)
  1. Show that

$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { \lambda } { \sqrt { h } }$$ where \(\lambda\) is a constant. Given that

$$h ^ { \frac { 3 } { 2 } } = A t + B$$ where \(A\) and \(B\) are constants to be found.

Hence find the time taken, from when water started flowing into the tank, for the tank to be completely full.

7 A tank is shaped as a cuboid. The base has dimensions 10 cm by 10 cm . Initially the tank is empty. Water flows into the tank at \(25 \mathrm {~cm} ^ { 3 }\) per second. Water also leaks out of the tank at \(4 h ^ { 2 } \mathrm {~cm} ^ { 3 }\) per second, where \(h \mathrm {~cm}\) is the depth of the water after \(t\) seconds. Find the time taken for the water to reach a depth of 2 cm .

10 A tank with vertical sides and rectangular cross-section is initially full of water. The water is leaking out of a hole in the base of the tank at a rate which is proportional to the square root of the depth of the water. \(V \mathrm {~m} ^ { 3 }\) is the volume of water in the tank at time \(t\) hours.

1. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = a \sqrt { V }\), where \(a\) is a constant.
2. Given that the tank is half full after one hour, show that \(V = V _ { 0 } \left( \left( \frac { 1 } { \sqrt { 2 } } - 1 \right) t + 1 \right) ^ { 2 }\), where \(V _ { 0 } \mathrm {~m} ^ { 3 }\) is the initial volume of water in the tank.
3. Hence show that the tank will be empty after approximately 3 hours and 25 minutes.

\includegraphics{figure_3} Figure 3 shows a large vertical cylindrical tank containing a liquid. The radius of the circular cross-section of the tank is 40 cm. At time \(t\) minutes, the depth of liquid in the tank is \(h\) centimetres. The liquid leaks from a hole \(P\) at the bottom of the tank. The liquid leaks from the tank at a rate of \(32\pi \sqrt{h}\) cm\(^3\) min\(^{-1}\).

1. Show that at time \(t\) minutes, the height \(h\) cm of liquid in the tank satisfies the differential equation $$\frac{dh}{dt} = -0.02\sqrt{h}$$ [4]
2. Find the time taken, to the nearest minute, for the depth of liquid in the tank to decrease from 100 cm to 50 cm. [5]

Fluid flows out of a cylindrical tank with constant cross section. At time \(t\) minutes, \(t \geq 0\), the volume of fluid remaining in the tank is \(V\) m\(^3\). The rate at which the fluid flows, in m\(^3\) min\(^{-1}\), is proportional to the square root of \(V\).

1. Show that the depth \(h\) metres of fluid in the tank satisfies the differential equation $$\frac{dh}{dt} = -k\sqrt{h}, \quad \text{where } k \text{ is a positive constant.}$$ [3]
2. Show that the general solution of the differential equation may be written as $$h = (A - Bt)^2, \quad \text{where } A \text{ and } B \text{ are constants.}$$ [4] Given that at time \(t = 0\) the depth of fluid in the tank is 1 m, and that 5 minutes later the depth of fluid has reduced to 0.5 m,
3. find the time, \(T\) minutes, which it takes for the tank to empty. [3]
4. Find the depth of water in the tank at time \(0.5T\) minutes. [2]