1.08k Separable differential equations: dy/dx = f(x)g(y)

1. The height above ground, \(H\) metres, of a passenger on a roller coaster can be modelled by the differential equation

$$\frac { \mathrm { d } H } { \mathrm {~d} t } = \frac { H \cos ( 0.25 t ) } { 40 }$$ where \(t\) is the time, in seconds, from the start of the ride. Given that the passenger is 5 m above the ground at the start of the ride,

1. show that \(H = 5 \mathrm { e } ^ { 0.1 \sin ( 0.25 t ) }\)
2. State the maximum height of the passenger above the ground. The passenger reaches the maximum height, for the second time, \(T\) seconds after the start of the ride.
3. Find the value of \(T\).

7. \begin{figure}[h] \end{figure} Figure 2 Figure 2 shows a cylindrical tank of height 1.5 m .
Initially the tank is full of water.
The water starts to leak from a small hole, at a point \(L\), in the side of the tank.
While the tank is leaking, the depth, \(H\) metres, of the water in the tank is modelled by the differential equation $$\frac { \mathrm { d } H } { \mathrm {~d} t } = - 0.12 \mathrm { e } ^ { - 0.2 t }$$ where \(t\) hours is the time after the leak starts.
Using the model,

1. show that $$H = A \mathrm { e } ^ { - 0.2 t } + B$$ where \(A\) and \(B\) are constants to be found,
2. find the time taken for the depth of the water to decrease to 1.2 m . Give your answer in hours and minutes, to the nearest minute. In the long term, the water level in the tank falls to the same height as the hole.
3. Find, according to the model, the height of the hole from the bottom of the tank.

1. A balloon is being inflated.

In a simple model,

• the balloon is modelled as a sphere
• the rate of increase of the radius of the balloon is inversely proportional to the square root of the radius of the balloon

At time \(t\) seconds, the radius of the balloon is \(r \mathrm {~cm}\).

1. Write down a differential equation to model this situation. At the instant when \(t = 10\)
   • the radius is 16 cm
   • the radius is increasing at a rate of \(0.9 \mathrm {~cm} \mathrm {~s} ^ { - 1 }\)
   • Solve the differential equation to show that
   $$r ^ { \frac { 3 } { 2 } } = 5.4 t + 10$$
2. Hence find the radius of the balloon when \(t = 20\) Give your answer to the nearest millimetre.
3. Suggest a limitation of the model.

1. A large spherical balloon is deflating.

At time \(t\) seconds the balloon has radius \(r \mathrm {~cm}\) and volume \(V \mathrm {~cm} ^ { 3 }\) The volume of the balloon is modelled as decreasing at a constant rate.

1. Using this model, show that $$\frac { \mathrm { d } r } { \mathrm {~d} t } = - \frac { k } { r ^ { 2 } }$$ where \(k\) is a positive constant. Given that
   • the initial radius of the balloon is 40 cm
   • after 5 seconds the radius of the balloon is 20 cm
   • the volume of the balloon continues to decrease at a constant rate until the balloon is empty
   • solve the differential equation to find a complete equation linking \(r\) and \(t\).
   • Find the limitation on the values of \(t\) for which the equation in part (b) is valid.

1. A spherical mint of radius 5 mm is placed in the mouth and sucked. Four minutes later, the radius of the mint is 3 mm .

In a simple model, the rate of decrease of the radius of the mint is inversely proportional to the square of the radius. Using this model and all the information given,

1. find an equation linking the radius of the mint and the time.
   (You should define the variables that you use.)
2. Hence find the total time taken for the mint to completely dissolve. Give your answer in minutes and seconds to the nearest second.
3. Suggest a limitation of the model.

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.

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.

15 A model for the motion of a small object falling through a thick fluid can be expressed using the differential equation \(\frac { \mathrm { d } v } { \mathrm {~d} t } = 9.8 - k v\),
where \(v \mathrm {~ms} ^ { - 1 }\) is the velocity after \(t \mathrm {~s}\) and \(k\) is a positive constant.

1. Given that \(v = 0\) when \(t = 0\), solve the differential equation to find \(v\) in terms of \(t\) and \(k\).
2. Sketch the graph of \(v\) against \(t\). Experiments show that for large values of \(t\), the velocity tends to \(7 \mathrm {~ms} ^ { - 1 }\).
3. Find the value of \(k\).
4. Find the value of \(t\) for which \(v = 3.5\).

11 A balloon is being inflated. The balloon is modelled as a sphere with radius \(x \mathrm {~cm}\) at time \(t \mathrm {~s}\). The volume \(V \mathrm {~cm} ^ { 3 }\) is given by \(\mathrm { V } = \frac { 4 } { 3 } \pi \mathrm { x } ^ { 3 }\). The rate of increase of volume is inversely proportional to the radius of the balloon. Initially, when \(t = 0\), the radius of the balloon is 5 cm and the volume of the balloon is increasing at a rate of \(21 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }\).

1. Show that \(x\) satisfies the differential equation \(\frac { \mathrm { dx } } { \mathrm { dt } } = \frac { 105 } { 4 \pi \mathrm { x } ^ { 3 } }\).
2. Find the radius of the balloon after two minutes.
3. Explain why the model may not be suitable for very large values of \(t\).

17

1. Express \(\frac { \left( x ^ { 2 } - 8 x + 9 \right) } { ( x + 1 ) ( x - 2 ) ^ { 2 } }\) in partial fractions.
2. Express \(y\) in terms of \(x\) given that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y \left( x ^ { 2 } - 8 x + 9 \right) } { ( x + 1 ) ( x - 2 ) ^ { 2 } } \text { and } y = 16 \text { when } x = 3 .$$ \section*{END OF QUESTION PAPER}

6 A hot drink is cooling. The temperature of the drink at time \(t\) minutes is \(T ^ { \circ } \mathrm { C }\).
The rate of decrease in temperature of the drink is proportional to \(( T - 20 )\).

1. Write down a differential equation to describe the temperature of the drink as a function of time.
2. When \(t = 0\), the temperature of the drink is \(90 ^ { \circ } \mathrm { C }\) and the temperature is decreasing at a rate of \(4.9 ^ { \circ } \mathrm { C }\) per minute. Determine how long it takes for the drink to cool from \(90 ^ { \circ } \mathrm { C }\) to \(40 ^ { \circ } \mathrm { C }\).

7

1. Express \(\frac { 1 } { x } + \frac { 1 } { A - x }\) as a single fraction. The population of fish in a lake is modelled by the differential equation \(\frac { d x } { d t } = \frac { x ( 400 - x ) } { 400 }\) where \(x\) is the number of fish and \(t\) is the time in years.
   When \(t = 0 , x = 100\).
2. In this question you must show detailed reasoning. Find the number of fish in the lake when \(t = 10\), as predicted by the model.

10

1. Express \(\frac { 1 } { ( 4 x + 1 ) ( x + 1 ) }\) in partial fractions.
2. A curve passes through the point \(( 0,2 )\) and satisfies the differential equation \(\frac { d y } { d x } = \frac { y } { ( 4 x + 1 ) ( x + 1 ) }\),
   for \(x > - \frac { 1 } { 4 }\).
   Show by integration that \(\mathrm { y } = \mathrm { A } \left( \frac { 4 \mathrm { x } + 1 } { \mathrm { x } + 1 } \right) ^ { \mathrm { B } }\) where \(A\) and \(B\) are constants to be determined.

4 The resistive force, \(F\), on a sphere falling through a viscous fluid is thought to depend on the radius of the sphere, \(r\), the velocity of the sphere, \(v\), and the viscosity of the fluid, \(\eta\). You are given that \(\eta\) is measured in \(\mathrm { N } \mathrm { m } ^ { - 2 } \mathrm {~s}\).

1. By considering its units, find the dimensions of viscosity. A model of the resistive force suggests the following relationship: \(\mathbf { F } = 6 \pi \eta ^ { \alpha } \mathbf { r } ^ { \beta } \mathbf { v } ^ { \gamma }\).
2. Explain whether or not it is possible to use dimensional analysis to verify that the constant \(6 \pi\) is correct.
3. Use dimensional analysis to find the values of \(\alpha , \beta\) and \(\gamma\). A sphere of radius \(r\) and mass \(m\) falls vertically from rest through the fluid. After a time \(t\) its velocity is \(v\).
4. By setting up and solving a differential equation, show that \(\mathrm { e } ^ { - \mathrm { kt } } = \frac { \mathrm { g } - \mathrm { kv } } { \mathrm { g } }\) where \(\mathrm { k } = \frac { 6 \pi \eta \mathrm { r } } { \mathrm { m } }\). As the time increases, the velocity of the sphere tends towards a limit called the terminal velocity.
5. Find, in terms of \(g\) and \(k\), the terminal velocity of the sphere. In a sequence of experiments the sphere is allowed to fall through fluids of different viscosity, ranging from small to very large, with all other conditions being constant. The terminal velocity of the sphere through each fluid is measured.
6. Describe how, according to the model, the terminal velocity of the sphere changes as the viscosity of the fluid through which it falls increases.

7

1. 1. Solve the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\) to find \(x\) in terms of \(t\).
   2. Given that \(x = 1\) when \(t = 0\), show that the solution can be written as $$x = ( a - \cos b t ) ^ { 2 }$$ where \(a\) and \(b\) are constants to be found.
2. The height, \(x\) metres, above the ground of a car in a fairground ride at time \(t\) seconds is modelled by the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = \sqrt { x } \sin \left( \frac { t } { 2 } \right)\). The car is 1 metre above the ground when \(t = 0\).
   1. Find the greatest height above the ground reached by the car during the ride.
   2. Find the value of \(t\) when the car is first 5 metres above the ground, giving your answer to one decimal place.

7 Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y ^ { 2 } x \sin 3 x$$ given that \(y = 1\) when \(x = \frac { \pi } { 6 }\). Give your answer in the form \(y = \frac { 9 } { \mathrm { f } ( x ) }\).

5

1. Find \(\int x \sqrt { x ^ { 2 } + 3 } \mathrm {~d} x\).
   (2 marks)
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x \sqrt { x ^ { 2 } + 3 } } { \mathrm { e } ^ { 2 y } }$$ given that \(y = 0\) when \(x = 1\). Give your answer in the form \(y = \mathrm { f } ( x )\).

8

1. Express \(\frac { 1 } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }\) in the form \(\frac { A } { 3 - 2 x } + \frac { B } { 1 - x } + \frac { C } { ( 1 - x ) ^ { 2 } }\).
   (4 marks)
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 \sqrt { y } } { ( 3 - 2 x ) ( 1 - x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\), expressing your answer in the form $$y ^ { p } = q \ln [ \mathrm { f } ( x ) ] + \frac { x } { 1 - x }$$ where \(p\) and \(q\) are constants.

8

1. A water tank has a height of 2 metres. The depth of the water in the tank is \(h\) metres at time \(t\) minutes after water begins to enter the tank. The rate at which the depth of the water in the tank increases is proportional to the difference between the height of the tank and the depth of the water. Write down a differential equation in the variables \(h\) and \(t\) and a positive constant \(k\).
   (You are not required to solve your differential equation.)
2. 1. Another water tank is filling in such a way that \(t\) minutes after the water is turned on, the depth of the water, \(x\) metres, increases according to the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 1 } { 15 x \sqrt { 2 x - 1 } }$$ The depth of the water is 1 metre when the water is first turned on.
      Solve this differential equation to find \(t\) as a function of \(x\).
   2. Calculate the time taken for the depth of the water in the tank to reach 2 metres, giving your answer to the nearest 0.1 of a minute.
      (l mark)

8

1. \(\quad\) Find \(\int t \cos \left( \frac { \pi } { 4 } t \right) \mathrm { d } t\).
2. The platform of a theme park ride oscillates vertically. For the first 75 seconds of the ride, $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { t \cos \left( \frac { \pi } { 4 } t \right) } { 32 x }$$ where \(x\) metres is the height of the platform above the ground after time \(t\) seconds.
   At \(t = 0\), the height of the platform above the ground is 4 metres.
   Find the height of the platform after 45 seconds, giving your answer to the nearest centimetre.
   (6 marks)

8

1. Express \(\frac { 16 x } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }\) in the form \(\frac { A } { 1 - 3 x } + \frac { B } { 1 + x } + \frac { C } { ( 1 + x ) ^ { 2 } }\).
2. Solve the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 16 x \mathrm { e } ^ { 2 y } } { ( 1 - 3 x ) ( 1 + x ) ^ { 2 } }$$ where \(y = 0\) when \(x = 0\).
   Give your answer in the form \(\mathrm { f } ( y ) = \mathrm { g } ( x )\).
   [0pt] [7 marks]