Separable variables - standard (applied/contextual)

9 A biologist is investigating the spread of a weed in a particular region. At time \(t\) weeks after the start of the investigation, the area covered by the weed is \(A \mathrm {~m} ^ { 2 }\). The biologist claims that the rate of increase of \(A\) is proportional to \(\sqrt { } ( 2 A - 5 )\).

1. Write down a differential equation representing the biologist's claim.
2. At the start of the investigation, the area covered by the weed was \(7 \mathrm {~m} ^ { 2 }\) and, 10 weeks later, the area covered was \(27 \mathrm {~m} ^ { 2 }\). Assuming that the biologist's claim is correct, find the area covered 20 weeks after the start of the investigation.

7. \begin{figure}[h] \end{figure} Figure 3 shows a vertical cylindrical tank of height 200 cm containing water. Water is leaking from a hole \(P\) on the side of the tank. At time \(t\) minutes after the leaking starts, the height of water in the tank is \(h \mathrm {~cm}\). The height \(h \mathrm {~cm}\) of the water in the tank satisfies the differential equation $$\frac { \mathrm { d } h } { \mathrm {~d} t } = k ( h - 9 ) ^ { \frac { 1 } { 2 } } , \quad 9 < h \leqslant 200$$ where \(k\) is a constant. Given that, when \(h = 130\), the height of the water is falling at a rate of 1.1 cm per minute,

1. find the value of \(k\). Given that the tank was full of water when the leaking started,
2. solve the differential equation with your value of \(k\), to find the value of \(t\) when \(h = 50\)

9. Bacteria are growing on the surface of a dish in a laboratory. The area of the dish, \(A \mathrm {~cm} ^ { 2 }\), covered by the bacteria, \(t\) days after the bacteria start to grow, is modelled by the differential equation $$\frac { \mathrm { d } A } { \mathrm {~d} t } = \frac { A ^ { \frac { 3 } { 2 } } } { 5 t ^ { 2 } } \quad t > 0$$ Given that \(A = 2.25\) when \(t = 3\)

1. show that $$A = \left( \frac { p t } { q t + r } \right) ^ { 2 }$$ where \(p , q\) and \(r\) are integers to be found. According to the model, there is a limit to the area that will be covered by the bacteria.
2. Find the value of this limit. \includegraphics[max width=\textwidth, alt={}, center]{79ac81c3-cd05-4f28-8840-3c8a6960e7b7-31_2255_50_314_34}
   

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8 Water flows out of a tank through a hole in the bottom and, at time \(t\) minutes, the depth of water in the tank is \(x\) metres. At any instant, the rate at which the depth of water in the tank is decreasing is proportional to the square root of the depth of water in the tank.

1. Write down a differential equation which models this situation.
2. When \(t = 0 , x = 2\); when \(t = 5 , x = 1\). Find \(t\) when \(x = 0.5\), giving your answer correct to 1 decimal place.

9 \includegraphics[max width=\textwidth, alt={}, center]{798da17d-0af5-4aa6-b731-564642dc28d5-4_572_917_294_607} A cylindrical container has a height of 200 cm . The container was initially full of a chemical but there is a leak from a hole in the base. When the leak is noticed, the container is half-full and the level of the chemical is dropping at a rate of 1 cm per minute. It is required to find for how many minutes the container has been leaking. To model the situation it is assumed that, when the depth of the chemical remaining is \(x \mathrm {~cm}\), the rate at which the level is dropping is proportional to \(\sqrt { } x\). Set up and solve an appropriate differential equation, and hence show that the container has been leaking for about 80 minutes.

5 The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(\pounds V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).

1. Express the model as a differential equation. Verify by differentiation that \(V = \left( \frac { 1 } { 2 } k t + c \right) ^ { 2 }\), where \(c\) is an arbitrary constant, satisfies this differential equation.
2. The value of the company's sales in its first year is \(\pounds 10000\), and the total value of the sales in the first two years is \(\pounds 40000\). Find \(V\) in terms of \(t\).

2 Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x \mathrm {~cm}\), and the volume of water is \(V \mathrm {~cm} ^ { 3 }\), where \(V = \frac { 1 } { 3 } x ^ { 3 }\). The rate at which water is lost is proportional to \(x\), so that \(\frac { \mathrm { d } V } { \mathrm {~d} t } = - k x\), where \(k\) is a constant.

1. Show that \(x \frac { \mathrm {~d} x } { \mathrm {~d} t } = - k\). Initially, the depth of water in the container is 10 cm .
2. Show by integration that \(x = \sqrt { 100 - 2 k t }\).
3. Given that the container empties after 50 seconds, find \(k\). Once the container is empty, water is poured into it at a constant rate of \(1 \mathrm {~cm} ^ { 3 }\) per second. The container continues to lose water as before.
4. Show that, \(t\) seconds after starting to pour the water in, \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 - x } { x ^ { 2 } }\).
5. Show that \(\frac { 1 } { 1 - x } - x - 1 = \frac { x ^ { 2 } } { 1 - x }\). Hence solve the differential equation in part (iv) to show that $$t = \ln \left( \frac { 1 } { 1 - x } \right) - \frac { 1 } { 2 } x ^ { 2 } - x$$
6. Show that the depth cannot reach 1 cm .

6

1. The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating \(x\), the number of bacteria, to the time \(t\).
2. In another colony, the number of bacteria, \(y\), after time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 10000 } { \sqrt { y } }$$ Find \(y\) in terms of \(t\), given that \(y = 900\) when \(t = 0\). Hence find the number of bacteria after 10 minutes.

9 Paraffin is stored in a tank with a horizontal base. At time \(t\) minutes, the depth of paraffin in the tank is \(x \mathrm {~cm}\). When \(t = 0 , x = 72\). There is a tap in the side of the tank through which the paraffin can flow. When the tap is opened, the flow of the paraffin is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 4 ( x - 8 ) ^ { \frac { 1 } { 3 } }$$

1. How long does it take for the level of paraffin to fall from a depth of 72 cm to a depth of 35 cm ?
2. The tank is filled again to its original depth of 72 cm of paraffin and the tap is then opened. The paraffin flows out until it stops. How long does this take?

8 A cylindrical tank is initially full of water. There is a small hole at the base of the tank out of which the water leaks. The height of water in the tank is \(x \mathrm {~m}\) at time \(t\) seconds. The rate of change of the height of water may be modelled by the assumption that it is proportional to the square root of the height of water. When \(t = 100 , x = 0.64\) and, at this instant, the height is decreasing at a rate of \(0.0032 \mathrm {~ms} ^ { - 1 }\).

1. Show that \(\frac { \mathrm { d } x } { \mathrm {~d} t } = - 0.004 \sqrt { x }\).
2. Find an expression for \(x\) in terms of \(t\).
3. Hence determine at what time, according to this model, the tank will be empty.

1. Julie decides to start a business breeding rabbits to sell as pets.

Initially she buys 20 rabbits. After \(t\) years the number of rabbits, \(R\), is modelled by the differential equation $$\frac { \mathrm { d } R } { \mathrm {~d} t } = 2 R + 4 \sin t \quad t > 0$$ Julie needs to have at least 40 rabbits before she can start to sell them.
Use two iterations of the approximation formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$ to find out if, according to the model, Julie will be able to start selling rabbits after 4 months.

1. The value, V hundred pounds, of a particular stock thours after the opening of trading on a given day is modelled by the differential equation

$$\frac { d V } { d t } = \frac { V ^ { 2 } - t } { t ^ { 2 } + t V } \quad 0 < t < 8.5$$ A trader purchases \(\pounds 300\) of the stock one hour after the opening of trading.
Use two iterations of the approximation formula \(\left( \frac { \mathrm { dy } } { \mathrm { dx } } \right) _ { 0 } \approx \frac { \mathrm { y } _ { 1 } - \mathrm { y } _ { 0 } } { \mathrm {~h} }\) to estimate, to the nearest \(\pounds\), the value of the trader's stock half an hour after it was purchased.

1. The velocity \(v \mathrm {~ms} ^ { - 1 }\), of a raindrop, \(t\) seconds after it falls from a cloud, is modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.1 v ^ { 2 } + 10 \quad t \geqslant 0$$ Initially the raindrop is at rest.

1. Use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }\) to estimate the velocity of the raindrop 1 second after it falls from the cloud. Given that the initial acceleration of the raindrop is found to be smaller than is suggested by the current model,
2. refine the model by changing the value of one constant.

8

1. Solve the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2 ( x - 6 ) ^ { \frac { 1 } { 2 } }$$ to find \(t\) in terms of \(x\), given that \(x = 70\) when \(t = 0\).
2. Liquid fuel is stored in a tank. At time \(t\) minutes, the depth of fuel in the tank is \(x \mathrm {~cm}\). Initially there is a depth of 70 cm of fuel in the tank. There is a tap 6 cm above the bottom of the tank. The flow of fuel out of the tank is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = - 2 ( x - 6 ) ^ { \frac { 1 } { 2 } }$$
   1. Explain what happens when \(x = 6\).
   2. Find how long it will take for the depth of fuel to fall from 70 cm to 22 cm .

4

1. The number of bacteria in a colony is increasing at a rate that is proportional to the square root of the number of bacteria present. Form a differential equation relating \(x\), the number of bacteria, to the time \(t\).
2. In another colony, the number of bacteria, \(y\), after time \(t\) minutes is modelled by the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} t } = \frac { 10000 } { \sqrt { y } } .$$ Find \(y\) in terms of \(t\), given that \(y = 900\) when \(t = 0\). Hence find the number of bacteria after 10 minutes.

12 A patch of disease on a leaf is being chemically treated. At time \(t\) days after treatment starts, its length is \(x \mathrm {~cm}\) and the rate of decrease of its length is observed to be inversely proportional to the square root of its length. At time \(t = 3 , x = 4\) and, at this instant, the length is decreasing at 0.05 cm per day. Write down and solve a differential equation to model this situation. Hence find the time it takes for the length to decrease to 0.01 cm .
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Water is leaking from a container. After \(t\) seconds, the depth of water in the container is \(x\) cm, and the volume of water is \(V\) cm\(^3\), where \(V = \frac{1}{3}x^3\). The rate at which water is lost is proportional to \(x\), so that \(\frac{dV}{dt} = -kx\), where \(k\) is a constant.

1. Show that \(x \frac{dx}{dt} = -k\). [3]

Initially, the depth of water in the container is 10 cm.

1. Show by integration that \(x = \sqrt{100 - 2kt}\). [4]
2. Given that the container empties after 50 seconds, find \(k\). [2]

Once the container is empty, water is poured into it at a constant rate of 1 cm\(^3\) per second. The container continues to lose water as before.

1. Show that, \(t\) seconds after starting to pour the water in, \(\frac{dx}{dt} = \frac{1-x}{x^2}\). [2]
2. Show that \(\frac{1}{1-x} - x - 1 = \frac{x^2}{1-x}\). Hence solve the differential equation in part (iv) to show that $$t = \ln\left(\frac{1}{1-x}\right) - \frac{1}{2}x^2 - x.$$ [6]
3. Show that the depth cannot reach 1 cm. [1]

The total value of the sales made by a new company in the first \(t\) years of its existence is denoted by \(£V\). A model is proposed in which the rate of increase of \(V\) is proportional to the square root of \(V\). The constant of proportionality is \(k\).

1. Express the model as a differential equation. Verify by differentiation that \(V = (\frac{1}{2}kt + c)^2\), where \(c\) is an arbitrary constant, satisfies this differential equation. [4]
2. The value of the company's sales in its first year is £10000, and the total value of the sales in the first two years is £40000. Find \(V\) in terms of \(t\). [4]