Tank/reservoir mixing problems

10 Liquid is flowing into a small tank which has a leak. Initially the tank is empty and, \(t\) minutes later, the volume of liquid in the tank is \(V \mathrm {~cm} ^ { 3 }\). The liquid is flowing into the tank at a constant rate of \(80 \mathrm {~cm} ^ { 3 }\) per minute. Because of the leak, liquid is being lost from the tank at a rate which, at any instant, is equal to \(k V \mathrm {~cm} ^ { 3 }\) per minute where \(k\) is a positive constant.

1. Write down a differential equation describing this situation and solve it to show that $$V = \frac { 1 } { k } \left( 80 - 80 \mathrm { e } ^ { - k t } \right)$$
2. It is observed that \(V = 500\) when \(t = 15\), so that \(k\) satisfies the equation $$k = \frac { 4 - 4 e ^ { - 15 k } } { 25 }$$ Use an iterative formula, based on this equation, to find the value of \(k\) correct to 2 significant figures. Use an initial value of \(k = 0.1\) and show the result of each iteration to 4 significant figures.
3. Determine how much liquid there is in the tank 20 minutes after the liquid started flowing, and state what happens to the volume of liquid in the tank after a long time.

10 A rectangular reservoir has a horizontal base of area \(1000 \mathrm {~m} ^ { 2 }\). At time \(t = 0\), it is empty and water begins to flow into it at a constant rate of \(30 \mathrm {~m} ^ { 3 } \mathrm {~s} ^ { - 1 }\). At the same time, water begins to flow out at a rate proportional to \(\sqrt { } h\), where \(h \mathrm {~m}\) is the depth of the water at time \(t \mathrm {~s}\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 0.02\).

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = 0.01 ( 3 - \sqrt { } h )$$ It is given that, after making the substitution \(x = 3 - \sqrt { } h\), the equation in part (i) becomes $$( x - 3 ) \frac { \mathrm { d } x } { \mathrm {~d} t } = 0.005 x$$
2. Using the fact that \(x = 3\) when \(t = 0\), solve this differential equation, obtaining an expression for \(t\) in terms of \(x\).
3. Find the time at which the depth of water reaches 4 m .

8
\includegraphics[max width=\textwidth, alt={}, center]{c687888e-bef0-4ea9-b5b3-e614028cc07c-3_654_805_274_671} An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time \(t\) hours after filling begins, the volume of liquid is \(V \mathrm {~m} ^ { 3 }\) and the depth of liquid is \(h \mathrm {~m}\). It is given that \(V = \frac { 4 } { 3 } h ^ { 3 }\). The liquid is poured in at a rate of \(20 \mathrm {~m} ^ { 3 }\) per hour, but owing to leakage, liquid is lost at a rate proportional to \(h ^ { 2 }\). When \(h = 1 , \frac { \mathrm {~d} h } { \mathrm {~d} t } = 4.95\).

1. Show that \(h\) satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 5 } { h ^ { 2 } } - \frac { 1 } { 20 } .$$
2. Verify that \(\frac { 20 h ^ { 2 } } { 100 - h ^ { 2 } } \equiv - 20 + \frac { 2000 } { ( 10 - h ) ( 10 + h ) }\).
3. Hence solve the differential equation in part (i), obtaining an expression for \(t\) in terms of \(h\).

10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time \(t\) minutes after filling begins the volume of water in the pool is \(V\) litres. The pool has a small leak and loses water at a rate of \(0.01 V\) litres per minute. The differential equation satisfied by \(V\) and \(t\) is of the form \(\frac { \mathrm { d } V } { \mathrm {~d} t } = a - b V\).

1. Write down the values of the constants \(a\) and \(b\).
2. Solve the differential equation and find the value of \(t\) when \(V = 1000\).
3. Obtain an expression for \(V\) in terms of \(t\) and hence state what happens to \(V\) as \(t\) becomes large.

10 A balloon in the shape of a sphere has volume \(V\) and radius \(r\). Air is pumped into the balloon at a constant rate of \(40 \pi\) starting when time \(t = 0\) and \(r = 0\). At the same time, air begins to flow out of the balloon at a rate of \(0.8 \pi r\). The balloon remains a sphere at all times.

1. Show that \(r\) and \(t\) satisfy the differential equation $$\frac { \mathrm { d } r } { \mathrm {~d} t } = \frac { 50 - r } { 5 r ^ { 2 } }$$
2. Find the quotient and remainder when \(5 r ^ { 2 }\) is divided by \(50 - r\).
3. Solve the differential equation in part (a), obtaining an expression for \(t\) in terms of \(r\).
4. Find the value of \(t\) when the radius of the balloon is 12 .

7. Water is flowing into a large container and is leaking from a hole at the base of the container. At time \(t\) seconds after the water starts to flow, the volume, \(V \mathrm {~cm} ^ { 3 }\), of water in the container is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 300 - k V$$ where \(k\) is a constant.

1. Solve the differential equation to show that, according to the model, $$V = \frac { 300 } { k } + A \mathrm { e } ^ { - k t }$$ where \(A\) is a constant.
   (5) Given that the container is initially empty and that when \(t = 10\), the volume of water is increasing at a rate of \(200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\)
2. find the exact value of \(k\).
3. Hence find, according to the model, the time taken for the volume of water in the container to reach 6 litres. Give your answer to the nearest second.

14. \begin{figure}[h] \end{figure} Water flows at a constant rate into a large tank.
The tank is a cuboid, with all sides of negligible thickness.
The base of the tank measures 8 m by 3 m and the height of the tank is 5 m .
There is a tap at a point \(T\) at the bottom of the tank, as shown in Figure 5.
At time \(t\) minutes after the tap has been opened

• the depth of water in the tank is \(h\) metres
• water is flowing into the tank at a constant rate of \(0.48 \mathrm {~m} ^ { 3 }\) per minute
• water is modelled as leaving the tank through the tap at a rate of \(0.1 h \mathrm {~m} ^ { 3 }\) per minute
  1. Show that, according to the model,

$$1200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 24 - 5 h$$ Given that when the tap was opened, the depth of water in the tank was 2 m ,

show that, according to the model, $$h = A + B \mathrm { e } ^ { - k t }$$ where \(A , B\) and \(k\) are constants to be found. Given that the tap remains open,

determine, according to the model, whether the tank will ever become full, giving a reason for your answer.

17 In an industrial process, a container initially contains 1000 litres of liquid. Liquid is drawn from the bottom of the container at a rate of 5 litres per minute. At the same time, salt is added to the top of the container at a constant rate of 10 grams per minute. After \(t\) minutes the mass of salt in the container is \(x\) grams, and you are given that \(x = 0\) when \(t = 0\). In modelling the situation, it is assumed that the salt dissolves instantly and uniformly in the liquid, and that adding the salt does not change the volume of the liquid.

1. 1. Show that the concentration of salt in the liquid after \(t\) minutes is \(\frac { \mathrm { X } } { 1000 - 5 \mathrm { t } }\) grams per litre.
   2. Hence show that the mass of salt in the container is given by the differential equation $$\frac { d x } { d t } + \frac { x } { 200 - t } = 10$$
2. Show by integration that \(\mathrm { x } = 10 ( 200 - \mathrm { t } ) \ln \left( \frac { 200 } { 200 - \mathrm { t } } \right)\).
3. 1. Hence determine the mass of salt in the container when half the liquid is drawn off.
   2. Determine also the time at which the mass of salt in the container is greatest.
4. When the process is run, it is found that the concentration of salt over time is higher than predicted by the model. Suggest a reason for this.

17 In a chemical process, a vessel contains 1 litre of pure water. A liquid chemical is then passed into the top of the vessel at a constant rate of \(a\) litres per minute and thoroughly mixed with the water. At the same time, the resulting mixture is drawn from the bottom of the vessel at a constant rate of \(b\) litres per minute. You may assume that the chemical mixes instantly and uniformly with the water. After \(t\) minutes, the mixture in the vessel contains \(x\) litres of the chemical.

1. 1. Show that the proportion of chemical present in the vessel after \(t\) minutes is $$\frac { x } { 1 + ( a - b ) t } .$$
   2. Hence show that \(\frac { d x } { d t } + \frac { b x } { 1 + ( a - b ) t } = a\).
2. First, consider the case where \(\mathbf { b } = \mathbf { a }\).
   1. Solve the differential equation to find \(x\) in terms of \(a\) and \(t\).
   2. Given that after 1 minute the vessel contains equal amounts of water and chemical, find the rate of inflow of chemical.
3. Now consider the case where \(\mathrm { b } = 2 \mathrm { a }\).
   1. Explain why the differential equation in part (a)(ii) is now invalid for \(\mathrm { t } \geqslant \frac { 1 } { \mathrm { a } }\).
   2. Find the maximum amount of chemical in the vessel.

1. A tank at a chemical plant has a capacity of 250 litres. The tank initially contains 100 litres of pure water.

Salt water enters the tank at a rate of 3 litres every minute. Each litre of salt water entering the tank contains 1 gram of salt. It is assumed that the salt water mixes instantly with the contents of the tank upon entry.
At the instant when the salt water begins to enter the tank, a valve is opened at the bottom of the tank and the solution in the tank flows out at a rate of 2 litres per minute. Given that there are \(S\) grams of salt in the tank after \(t\) minutes,

1. show that the situation can be modelled by the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } = 3 - \frac { 2 S } { 100 + t }$$
2. Hence find the number of grams of salt in the tank after 10 minutes. When the concentration of salt in the tank reaches 0.9 grams per litre, the valve at the bottom of the tank must be closed.
3. Find, to the nearest minute, when the valve would need to be closed.
4. Evaluate the model.

1. A sample of bacteria in a sealed container is being studied.

The number of bacteria, \(P\), in thousands, is modelled by the differential equation $$( 1 + t ) \frac { \mathrm { d } P } { \mathrm {~d} t } + P = t ^ { \frac { 1 } { 2 } } ( 1 + t )$$ where \(t\) is the time in hours after the start of the study.
Initially, there are exactly 5000 bacteria in the container.

1. Determine, according to the model, the number of bacteria in the container 8 hours after the start of the study.
2. Find, according to the model, the rate of change of the number of bacteria in the container 4 hours after the start of the study.
3. State a limitation of the model.

1. Two different colours of paint are being mixed together in a container.

The paint is stirred continuously so that each colour is instantly dispersed evenly throughout the container. Initially the container holds a mixture of 10 litres of red paint and 20 litres of blue paint.
The colour of the paint mixture is now altered by

• adding red paint to the container at a rate of 2 litres per second
• adding blue paint to the container at a rate of 1 litre per second
• pumping fully mixed paint from the container at a rate of 3 litres per second.

Let \(r\) litres be the amount of red paint in the container at time \(t\) seconds after the colour of the paint mixture starts to be altered.

1. Show that the amount of red paint in the container can be modelled by the differential equation $$\frac { \mathrm { dr } } { \mathrm { dt } } = 2 - \frac { \mathrm { r } } { \alpha }$$ where \(\alpha\) is a positive constant to be determined.
2. By solving the differential equation, determine how long it will take for the mixture of paint in the container to consist of equal amounts of red paint and blue paint, according to the model. Give your answer to the nearest second. It actually takes 9 seconds for the mixture of paint in the container to consist of equal amounts of red paint and blue paint.
3. Use this information to evaluate the model, giving a reason for your answer.

10
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2 \end{array} \right)$$ where \(\mu\) is a scalar parameter. The plane \(\Pi _ { 2 }\) contains the line \(l _ { 1 }\) and the line \(l _ { 2 }\)\\ (c) Determine a vector equation for the line of intersection of \(\Pi _ { 1 }\) and \(\Pi _ { 2 }\) The plane \(\Pi _ { 3 }\) has equation r. \(\left( \begin{array} { l } 1 \\ 1 \\ a \end{array} \right) = b\) where \(a\) and \(b\) are constants.\\ Given that the planes \(\Pi _ { 1 } , \Pi _ { 2 }\) and \(\Pi _ { 3 }\) form a sheaf,\\ (d) determine the value of \(a\) and the value of \(b\).

1. Water is flowing into and out of a large tank.

Initially the tank contains 10 litres of water.\\ The rate of flow of the water is modelled so that

• there are \(V\) litres of water in the tank at time \(t\) minutes after the water begins to flow
• water enters the tank at a rate of \(\left( 3 - \frac { 4 } { 1 + \mathrm { e } ^ { 0.8 t } } \right)\) litres per minute
• water leaves the tank at a rate proportional to the volume of water remaining in the tank

Given that when \(t = 0\) the volume of water in the tank is decreasing at a rate of 3 litres per minute, use the model to\\ (a) show that the volume of water in the tank at time \(t\) satisfies $$\frac { \mathrm { d } V } { \mathrm {~d} t } = 3 - \frac { 4 } { 1 + \mathrm { e } ^ { 0.8 t } } - 0.4 V$$ (b) Determine \(\frac { \mathrm { d } } { \mathrm { d } t } \left( \arctan \mathrm { e } ^ { 0.4 t } \right)\) Hence, by solving the differential equation from part (a),\\ (c) determine an equation for the volume of water in the tank at time \(t\). Give your answer in simplest form as \(V = \mathrm { f } ( t )\) After 10 minutes, the volume of water in the tank was 8 litres.\\ (d) Evaluate the model in light of this information.

1. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.\\ (a) Explain why, for \(n \in \mathbb { N }\)

$$\sum _ { r = 1 } ^ { 2 n } ( - 1 ) ^ { r } \mathrm { f } ( r ) = \sum _ { r = 1 } ^ { n } ( \mathrm { f } ( 2 r ) - \mathrm { f } ( 2 r - 1 ) )$$ for any function \(\mathrm { f } ( r )\).\\ (b) Use the standard summation formulae to show that, for \(n \in \mathbb { N }\) $$\sum _ { r = 1 } ^ { 2 n } r \left( ( - 1 ) ^ { r } + 2 r \right) ^ { 2 } = n ( 2 n + 1 ) \left( 8 n ^ { 2 } + 4 n + 5 \right)$$ (c) Hence evaluate $$\sum _ { r = 14 } ^ { 50 } r \left( ( - 1 ) ^ { r } + 2 r \right) ^ { 2 }$$

1. A colony of small mammals is being studied.

In the study, the mammals are divided into 3 categories (a) State one limitation of the model regarding the division into these categories. A model for the population of the colony is given by the matrix equation $$\left( \begin{array} { l } N _ { n + 1 }
J _ { n + 1 }
B _ { n + 1 } \end{array} \right) = \left( \begin{array} { c c c } 0 & 0 & 2
a & b & 0
0 & 0.48 & 0.96 \end{array} \right) \left( \begin{array} { l } N _ { n }
J _ { n }
B _ { n } \end{array} \right)$$ where \(a\) and \(b\) are constants, and \(N _ { n } , J _ { n }\) and \(B _ { n }\) are the respective numbers of the mammals in each category \(n\) months after the start of the study. At the start of the study the colony has breeders only, with no newborns or juveniles.\\ According to the model, after 2 months the number of newborns is 48 and the number of juveniles is 40\\ (b) (i) Determine the number of mammals in the colony at the start of the study.\\ (ii) Show that \(a = 0.8\)\\ (c) Determine, in terms of \(b\), $$\left( \begin{array} { c c c } 0 & 0 & 2
0.8 & b & 0
0 & 0.48 & 0.96 \end{array} \right) ^ { - 1 }$$ Given that the model predicts approximately 1015 mammals in total at the start of a particular month, and approximately 596 newborns, 464 juveniles and 437 breeders at the start of the next month,
(d) determine the value of \(b\), giving your answer to 2 decimal places. It is decided to monitor the number of newborn males and females as a part of the study. Assuming that \(42 \%\) of newborns are male,
(e) refine the matrix equation for the model to reflect this information, giving a reason for your answer.
(There is no need to estimate any unknown values for the refined model, but any known values should be made clear.)

1. A pond initially contains 1000 litres of unpolluted water.

The pond is leaking at a constant rate of 20 litres per day.
It is suspected that contaminated water flows into the pond at a constant rate of 25 litres per day and that the contaminated water contains 2 grams of pollutant in every litre of water. It is assumed that the pollutant instantly dissolves throughout the pond upon entry.
Given that there are \(x\) grams of the pollutant in the pond after \(t\) days,

1. show that the situation can be modelled by the differential equation, $$\frac { \mathrm { d } x } { \mathrm {~d} t } = 50 - \frac { 4 x } { 200 + t }$$
2. Hence find the number of grams of pollutant in the pond after 8 days.
3. Explain how the model could be refined.

8. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 1 shows the central vertical cross section \(A B C D\) of a paddling pool that has a circular horizontal cross section. Measurements of the diameters of the top and bottom of the paddling pool have been taken in order to estimate the volume of water that the paddling pool can contain. Using these measurements, the curve \(B D\) is modelled by the equation $$y = \ln ( 3.6 x - k ) \quad 1 \leqslant x \leqslant 1.18$$ as shown in Figure 2.

1. Find the value of \(k\).
2. Find the depth of the paddling pool according to this model. The pool is being filled with water from a tap.
3. Find, in terms of \(h\), the volume of water in the pool when the pool is filled to a depth of \(h \mathrm {~m}\). Given that the pool is being filled at a constant rate of 15 litres every minute,
4. find, in \(\mathrm { cmh } ^ { - 1 }\), the rate at which the water level is rising in the pool when the depth of the water is 0.2 m .