Applied/modelling contexts

4. During an industrial process, the mass of salt, \(S \mathrm {~kg}\), dissolved in a liquid \(t\) minutes after the process begins is modelled by the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } + \frac { 2 S } { 120 - t } = \frac { 1 } { 4 } , \quad 0 \leq t < 120$$ Given that \(S = 6\) when \(t = 0\),

1. find \(S\) in terms of \(t\),
2. calculate the maximum mass of salt that the model predicts will be dissolved in the liquid at any one time during the process.
   (4)(Total 12 marks)

7. A population \(P\) is growing at a rate which is modelled by the differential equation $$\frac { d P } { d t } - 0.1 P = 0.05 t$$ where \(t\) years is the time that has elapsed from the start of observations.
It is given that the population is 10000 at the start of the observations.

1. Solve the differential equation to obtain an expression for \(P\) in terms of \(t\).
2. Show that the population doubles between the sixth and seventh year after the observations began.
   (2)

5. (a) Obtain the general solution of the differential equation $$\frac { \mathrm { d } S } { \mathrm {~d} t } - 0.1 S = t$$ (b) The differential equation in part (a) is used to model the assets, \(\pounds S\) million, of a bank \(t\) years after it was set up. Given that the initial assets of the bank were \(\pounds 200\) million, use your answer to part (a) to estimate, to the nearest \(\pounds\) million, the assets of the bank 10 years after it was set up.

14 In a chemical reaction, the mass \(m\) grams of a chemical at time \(t\) minutes is modelled by the differential equation
\(\frac { \mathrm { d } m } { \mathrm {~d} t } = \frac { m } { t ( 1 + 2 t ) }\).
At time 1 minute, the mass of the chemical is 1 gram.

1. Solve the differential equation to show that \(m = \frac { 3 t } { ( 1 + 2 t ) }\).
2. Hence
   1. find the time when the mass is 1.25 grams,
   2. show what happens to the mass of the chemical as \(t\) becomes large.

10 A particle \(B\), of mass 3 kg , moves in a straight line and has velocity \(v \mathrm {~ms} ^ { - 1 }\).
At time \(t\) seconds, where \(0 \leqslant t < \frac { 1 } { 4 } \pi\), a variable force of \(- ( 15 \sin 4 \mathrm { t } + 6 \mathrm { v } \tan 2 \mathrm { t } )\) Newtons is applied to \(B\). There are no other forces acting on \(B\). Initially, when \(t = 0 , B\) has velocity \(4.5 \mathrm {~ms} ^ { - 1 }\). The motion of \(B\) can be modelled by the differential equation \(\frac { d v } { d t } + P ( t ) v = Q ( t )\) where \(P ( t )\) and \(\mathrm { Q } ( \mathrm { t } )\) are functions of \(t\).

1. Find the functions \(\mathrm { P } ( \mathrm { t } )\) and \(\mathrm { Q } ( \mathrm { t } )\).
2. Using an integrating factor, determine the first time at which \(B\) is stationary according to the model.

10 A particle of mass 0.5 kg is initially at point \(O\). It moves from rest along the \(x\)-axis under the influence of two forces \(F _ { 1 } \mathrm {~N}\) and \(F _ { 2 } \mathrm {~N}\) which act parallel to the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\).
\(F _ { 1 }\) is acting in the direction of motion of the particle and \(F _ { 2 }\) is resisting motion.
In an initial model

• \(F _ { 1 }\) is proportional to \(t\) with constant of proportionality \(\lambda > 0\),
• \(F _ { 2 }\) is proportional to \(v\) with constant of proportionality \(\mu > 0\).
  1. Show that the motion of the particle can be modelled by the following differential equation.

$$\frac { 1 } { 2 } \frac { d v } { d t } = \lambda t - \mu v$$

Solve the differential equation in part (a), giving the particular solution for \(v\) in terms of \(t\), \(\lambda\) and \(\mu\). You are now given that \(\lambda = 2\) and \(\mu = 1\).

Find a formula for an approximation for \(v\) in terms of \(t\) when \(t\) is large. In a refined model

• \(F _ { 1 }\) is constant, acting in the direction of motion with magnitude 2 N ,
• \(F _ { 2 }\) is as before with \(\mu = 1\).
• Write down a differential equation for the refined model.
• Without solving the differential equation in part (d), write down what will happen to the velocity in the long term according to this refined model.

8 A surge in the current, \(I\) units, through an electrical component at a time, \(t\) seconds, is to be modelled. The surge starts when \(t = 0\) and there is initially no current through the component. When the current has surged for 1 second it is measured as being 5 units. While the surge is occurring, \(I\) is modelled by the following differential equation.
\(\left( 2 t - t ^ { 2 } \right) \frac { d l } { d t } = \left( 2 t - t ^ { 2 } \right) ^ { \frac { 3 } { 2 } } - 2 ( t - 1 ) l\)

1. By using an integrating factor show that, according to the model, while the surge is occurring, \(I\) is given by \(\mathrm { I } = \left( 2 \mathrm { t } - \mathrm { t } ^ { 2 } \right) \left( \sin ^ { - 1 } ( \mathrm { t } - 1 ) + 5 \right)\). The surge lasts until there is again no current through the component.
2. Determine the length of time that the surge lasts according to the model.
3. Determine, according to the model, the rate of increase of the current at the start of the surge. Give your answer in an exact form.

8 A particle \(P\) of mass 2 kg can only move along the straight line segment \(O A\), where \(O A\) is on a rough horizontal surface. The particle is initially at rest at \(O\) and the distance \(O A\) is 0.9 m . When the time is \(t\) seconds the displacement of \(P\) from \(O\) is \(x \mathrm {~m}\) and the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). \(P\) is subject to a force of magnitude \(4 \mathrm { e } ^ { - 2 t } \mathrm {~N}\) in the direction of \(A\) for any \(t \geqslant 0\). The resistance to the motion of \(P\) is modelled as being proportional to \(v\). At the instant when \(t = \ln 2 , v = 0.5\) and the resultant force on \(P\) is 0 N .

1. Show that, according to the model, \(\frac { d v } { d t } + v = 2 e ^ { - 2 t }\).
2. Find an expression for \(v\) in terms of \(t\) for \(t \geqslant 0\).
3. By considering the behaviour of \(v\) as \(t\) becomes large explain why, according to the model, \(P\) 's speed must reach a maximum value for some \(t > 0\).
4. Determine the maximum speed considered in part (c).
5. Determine the greatest value of \(t\) for which the model is valid.

5 A particle \(P\) of mass 4.5 kg is free to move along the \(x\)-axis. In a model of the motion it is assumed that \(P\) is acted on by two forces:

• a constant force of magnitude \(f \mathrm {~N}\) in the positive \(x\) direction;
• a resistance to motion, \(R \mathrm {~N}\), whose magnitude is proportional to the speed of \(P\).

At time \(t\) seconds the velocity of \(P\) is \(v \mathrm {~ms} ^ { - 1 }\). When \(t = 0 , P\) is at the origin \(O\) and is moving in the positive direction with speed \(u \mathrm {~ms} ^ { - 1 }\), and when \(v = 5 , R = 2\).

1. Show that, according to the model, \(\frac { d v } { d t } = \frac { 10 f - 4 v } { 45 }\).
2. 1. By solving the differential equation in part (a), show that \(\mathrm { v } = \frac { 1 } { 2 } \left( 5 \mathrm { f } - ( 5 \mathrm { f } - 2 \mathrm { u } ) \mathrm { e } ^ { - \frac { 4 } { 45 } \mathrm { t } } \right)\).
   2. Describe briefly how, according to the model, the speed of \(P\) varies over time in each of the following cases.
      • \(\mathrm { u } < 2.5 \mathrm { f }\)
3. \(\mathrm { u } = 2.5 \mathrm { f }\)
4. \(u > 2.5 f\)
5. In the case where \(\mathrm { u } = 2 \mathrm { f }\), find in terms of \(f\) the exact displacement of \(P\) from \(O\) when \(t = 9\).

17 A cyclist accelerates from rest for 5 seconds then brakes for 5 seconds, coming to rest at the end of the 10 seconds. The total mass of the cycle and rider is \(m \mathrm {~kg}\), and at time \(t\) seconds, for \(0 \leqslant t \leqslant 10\), the cyclist's velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). A resistance to motion, modelled by a force of magnitude 0.1 mvN , acts on the cyclist during the whole 10 seconds.

1. Explain why modelling the resistance to motion in this way is likely to be more realistic than assuming this force is constant. During the braking phase of the motion, for \(5 \leqslant t \leqslant 10\), the brakes apply an additional constant resistance force of magnitude \(2 m \mathrm {~N}\) and the cyclist does not provide any driving force.
2. Show that, for \(5 \leqslant t \leqslant 10 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = - 2\).
3. 1. Solve the differential equation in part (b).
   2. Hence find the velocity of the cyclist when \(t = 5\). During the acceleration phase ( \(0 \leqslant t \leqslant 5\) ), the cyclist applies a driving force of magnitude directly proportional to \(t\).
4. Show that, for \(0 \leqslant t \leqslant 5 , \frac { \mathrm {~d} v } { \mathrm {~d} t } + 0.1 v = \lambda t\), where \(\lambda\) is a positive constant.
5. 1. Show by integration that, for \(0 \leqslant t \leqslant 5 , v = 10 \lambda \left( t - 10 + 10 \mathrm { e } ^ { - 0.1 t } \right)\).
   2. Hence find \(\lambda\).
6. Find the total distance, to the nearest metre, travelled by the cyclist during the motion.

16 The population density \(P\), in suitable units, of a certain bacterium at time \(t\) hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is \(A\).

1. One simple model is to assume that the rate of change of population density is directly proportional to \(A - P\).
   1. Formulate a differential equation for this model.
   2. Verify that \(P = A \left( 1 - \mathrm { e } ^ { - k t } \right)\), where \(k\) is a positive constant, satisfies
      • this differential equation,
2. the initial condition,
3. the long-term condition.
An alternative model uses the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } - \frac { P } { t \left( 1 + t ^ { 2 } \right) } = \mathrm { Q } ( t )$$ where \(\mathrm { Q } ( t )\) is a function of \(t\).
4. Find the integrating factor for this differential equation, showing that it can be written in the $$\text { form } \frac { \sqrt { 1 + t ^ { 2 } } } { t } \text {. }$$
5. Suppose that \(\mathrm { Q } ( t ) = 0\). $$\text { (i) Show that } P = \frac { A t } { \sqrt { 1 + t ^ { 2 } } } \text {. }$$ (ii) Find the time predicted by this model for the population density to reach half its longterm value. Give your answer correct to the nearest minute.
6. Now suppose that \(\mathrm { Q } ( t ) = \frac { t \mathrm { e } ^ { - t } } { \sqrt { 1 + t ^ { 2 } } }\). $$\text { Show that } \left. P = \frac { A t - t e ^ { - t } } { \sqrt { 1 + t ^ { 2 } } } \text {. [You may assume that } \lim _ { t \rightarrow \infty } t e ^ { - t } = 0 . \right]$$ It is found that the long-term value of \(P\) is 10, and \(P\) reaches half this value after 37 minutes.
7. Determine which of the models proposed in parts (c) and (d) is more consistent with these data.

1. A raindrop falls from rest from a cloud. The velocity, \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) vertically downwards, of the raindrop, \(t\) seconds after the raindrop starts to fall, is modelled by the differential equation

$$( t + 2 ) \frac { \mathrm { d } v } { \mathrm {~d} t } + 3 v = k ( t + 2 ) - 3 \quad t \geqslant 0$$ where \(k\) is a positive constant.

1. Solve the differential equation to show that $$v = \frac { k } { 4 } ( t + 2 ) - 1 + \frac { 4 ( 2 - k ) } { ( t + 2 ) ^ { 3 } }$$ Given that \(v = 4\) when \(t = 2\)
2. determine, according to the model, the velocity of the raindrop 5 seconds after it starts to fall.
3. Comment on the validity of the model for very large values of \(t\)

11 A particular radioactive substance decays over time.
A scientist models the amount of substance, \(x\) grams, at time \(t\) hours by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } + \frac { 1 } { 10 } x = \mathrm { e } ^ { - 0.1 t } \cos t .$$

1. Solve the differential equation to find the general solution for \(x\) in terms of \(t\). Initially there was 10 g of the substance.
2. Find the particular solution of the differential equation.
3. Find to 6 significant figures the amount of substance that would be predicted by the model at
   (a) 6 hours,
   (b) 6.25 hours.
4. Comment on the appropriateness of the model for predicting the amount of substance over time. \section*{END OF QUESTION PAPER}

2. (a) Find the general solution of the differential equation $$t \frac { \mathrm {~d} v } { \mathrm {~d} t } - v = t , \quad t > 0$$ and hence show that the solution can be written in the form \(v = t ( \ln t + c )\), where \(c\) is an arbitrary constant.
(b) This differential equation is used to model the motion of a particle which has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t \mathrm {~s}\). When \(t = 2\) the speed of the particle is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find, to 3 significant figures, the speed of the particle when \(t = 4\).
[0pt] [P4 January 2002 Qn 6]

4 A particle of mass 0.5 kg is initially at point \(O\). It moves from rest along the \(x\)-axis under the influence of two forces \(F _ { 1 } \mathrm {~N}\) and \(F _ { 2 } \mathrm {~N}\) which act parallel to the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\).
\(F _ { 1 }\) is acting in the direction of motion of the particle and \(F _ { 2 }\) is resisting motion.
In an initial model

• \(F _ { 1 }\) is proportional to \(t\) with constant of proportionality \(\lambda > 0\),
• \(F _ { 2 }\) is proportional to \(v\) with constant of proportionality \(\mu > 0\).
  1. Show that the motion of the particle can be modelled by the following differential equation.

$$\frac { 1 \mathrm {~d} v } { 2 \mathrm {~d} t } = \lambda t - \mu v$$

Solve the differential equation in part (a), giving the particular solution for \(v\) in terms of \(t\), \(\lambda\) and \(\mu\). You are now given that \(\lambda = 2\) and \(\mu = 1\).

Find a formula for an approximation for \(v\) in terms of \(t\) when \(t\) is large. In a refined model

• \(F _ { 1 }\) is constant, acting in the direction of motion with magnitude 2 N ,
• \(F _ { 2 }\) is as before with \(\mu = 1\).
• Write down a differential equation for the refined model.
• Without solving the differential equation in part (d), write down what will happen to the velocity in the long term according to this refined model.

3 A particle of mass 2 kg moves along the \(x\)-axis. At time \(t\) seconds the velocity of the particle is \(v \mathrm {~ms} ^ { - 1 }\). The particle is subject to two forces.

• One acts in the positive \(x\)-direction with magnitude \(\frac { 1 } { 2 } t \mathrm {~N}\).
• One acts in the negative \(x\)-direction with magnitude \(v \mathrm {~N}\).
  1. Show that the motion of the particle can be modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } + \frac { 1 } { 2 } v = \frac { 1 } { 4 } t$$ The particle is at rest when \(t = 0\).

Find \(v\) in terms of \(t\).

Find the velocity of the particle when \(t = 2\). When \(t = 2\) the force acting in the positive \(x\)-direction is replaced by a constant force of magnitude \(\frac { 1 } { 2 } \mathrm {~N}\) in the same direction.

Refine the differential equation given in part (a) to model the motion for \(t \geqslant 2\).

Use the refined model from part (d) to find an exact expression for \(v\) in terms of \(t\) for \(t \geqslant 2\).