3 Linear functions

HELM booklet  2 has introduced linear functions of the form y = a x + b . Such functions give rise to straight-line graphs. The coefficient a is the slope. If a is positive the graph of y against x slopes upwards. If a is negative the graph slopes downwards. The coefficient b gives the intercept on the y-axis. The terms a and b may be called the parameters of the line. Note that this is a different use of the term ’parameter’ than in the parametrisation of functions discussed in HELM booklet  2.

3.1 Linear models for falling rocks

In modelling it is wise to use a notation which fits in with the application. When modelling velocity under constant acceleration, we shall replace the dependent variable y by v (for velocity), and the independent variable x by t (for time). The acceleration will be denoted by the symbol a . Consider the motion of a rock dislodged from the top of a cliff (35 m high) by a villain during the filming of a thriller. The film producer might be interested in how long the rock would take to fall to the ground below the cliff and how fast it would be travelling at ground impact. The rock may be assumed to have a constant downward acceleration of 9.8 m s − 2 which the acceleration due to gravity. The velocity ( v m s − 1 ) of a rock, falling from the top of a cliff 35 m high, can be modelled by the equation

v = 9.8 t ( 0 ≤ t ≤ 2.7 )

where t is the time in seconds after the rock starts to fall. This follows from the fact that acceleration is the rate of change of velocity with time. If the acceleration is constant and the object starts from rest, then the velocity is given simply by the product of acceleration and time. The upper limit for t is the time at which the rock hits the ground measured with a stop-watch (about 2.7 s in this case).

Figure 2 shows v as a function of t . Velocity is a linearly increasing function of time and its graph is a straight line passing through t = 0 , v = 0 . Note that various assumptions are needed to obtain the quoted result of a linear variation in speed with time: it is assumed that there is no air resistance, no spinning and no wind.

Figure 2 :

{ Graph of $v=9.8t$ for the falling rock}

In what way should the equation for v be altered if the villain were able to throw the rock downwards at 5 m s − 1 ? Provided we are measuring position or displacement downwards, a downwards velocity is positive. Now we have that v = 5 when t = 0 . So a new model for v is

v = 9.8 t + 5 ( 0 ≤ t ≤ T 1 )

Since they are both downwards, the initial velocity simply adds to the velocity at any time resulting from falling under gravity. Note that T 1 is being used now for the upper limit on t (instead of 2.7) because 2.7 is (approximately) the time taken to fall 35 m from rest rather than with an initial downwards velocity. (Using the symbol T 1 saves us trying to work out its value for the moment!) Note that a general form of the model for motion under constant acceleration of magnitude a m s − 2 given an initial speed b m s − 1 is v = a t + b . In the model just considered a = 9.8 and b = 5 .

Task!

For the above example modelling a falling rock:

  1. Determine whether T 1 is more or less than 2.7.
  2. Sketch a graph of v for 0 ≤ t ≤ T 1 .
  1. T 1 will be less than 2.7 since the rock will be moving faster throughout its descent.
  2. The graph is still a straight line but displaced upwards compared with Figure 2.

{Graph of $v=9.8t+5$ for the falling rock}

Consider now how the function for v will change if the villain is even mightier than we previously thought and throws the rock upwards with an initial speed of 5 m s − 1 instead of simply dislodging it or throwing it downwards. In this circumstance, the initial velocity is directed upwards, and since position is being measured downwards, the initial velocity is negative. We can use the equation v = 9.8 t + b again. This time v = − 5 when t = 0 , leading to b = − 5 and

v = 9.8 t − 5 ( 0 ≤ t ≤ T 2 )

The new time at which the rock hits the ground is denoted by T 2 . The rock will rise before falling to the gound this time so T 2 will be larger than T 1 .

From the modelling point of view, there is one other significant time before the rock hits the ground. Figure 3 shows the new graph of v against t . Notice that there is a time at which v (which starts at − 5 ) is zero. What does this mean?

Figure 3 :

{ Graph of $v=9.8t-5$ for the falling rock}

As time goes by, the fact that gravity is causing the rock to accelerate downwards means that the rock’s upward motion will slow. Its velocity will decrease in magnitude until it reaches zero. At this particular instant the rock will be at its highest point and its velocity will change from upwards (negative) to downwards (positive) passing instantaneously through zero in the process.

We can calculate this time by substituting zero for v and working out the corresponding t .

0 = 9.8 t − 5 , so t = 5 9.8 = 0.51 .

This means that the rock is stationary about a half second after being thrown upwards. Subsequently the rock will fall until it hits the ground. But there is yet one more time that may be significant in the modelling context chosen here. During its journey to the ground 35 m beneath the cliff-top, the rock will pass the top of the cliff again. Note that we are modelling the motion of a particular point, say the lowest point, on the rock. A real rock, with appreciable size, will only pass the top of the cliff, without landing on it or hitting it, if it is thrown a little forward as well as up. Anyway, in principle we could use the function that we started with, representing the velocity of an object falling from rest under gravity, to work out how long the rock will take to pass the top of the cliff having reached the highest point in its path. A simpler method is to argue that, as long as the rock is thrown from the cliff top level (this requires the villain to be lying down!), the rock should take exactly the same time (approximately 0.5 s) to return to the level of the cliff top as it took to rise above the cliff top to the highest point in its path. So we simply double 0.5 s to deduce that the rock passes the cliff top again about 1 s after being thrown.

Task!

This Task concerns the falling rock model just discussed.

  1. Add lines to a sketch version of Figure 3 to represent velocity as a function of time if the rock is

    (i) dislodged (ii) thrown with velocity 3 m s − 1 downwards (iii) thrown with velocity − 2 m s − 1 :

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  2. What do you deduce about the effect of the initial velocity on the graph of velocity against time?

    The effect of changing the initial velocity (in size or in sign) is simply to displace the straight line upwards or downwards without changing its slope.

  3. Imagine that the filming was on the Moon with roughly one-sixth the gravitational pull of Earth. Find a linear function that would describe the velocity of a dislodged rock:

    v = 9.8 6 t ≈ 1.6 t

  4. What do you deduce about the effect of changing the acceleration due to gravity on the graph of velocity against time?

    The graph of velocity agains time is still linear but the change in the acceleration due to gravity changes the slop

  5. So, in the context of modelling motion under gravity, the initial velocity determines the vertical displacement of the line, its intercept on the v -axis, and the acceleration determines the slope . Again, given the modelling context, both of these influence the range of validity of the model since they alter the time taken for the rock to reach the ground and this fixes the upper limit on time.

    Like velocity, acceleration has direction as well as magnitude. As long as position is being measured downwards, and only gravity is considered to act, falling objects do not provide any examples of negative accelerations - but rocket motion does. Where downwards accelerations are represented as positive, an upwards acceleration will be negative. So a model of the motion of a rocket accelerating away from the Earth could include a constant negative acceleration. Horizontal acceleration, say of a road vehicle, in the same direction as position as being measured, is represented as positive. Deceleration, for example when this vehicle is being braked, implies that velocity is decreasing with time, and is represented as negative. In mathematical modelling, it is usual to refer to acceleration , whether it represents positive acceleration or deceleration.

    Suppose that we are describing the motion of a rocket taking off vertically during its initial booster stage of 10 s. We might model the acceleration as a constant − 20 m s − 2 . The negative sign arises because downwards is being taken as the positive direction but the acceleration is upwards. Since the rocket is starting from rest, an appropriate function is

    v = − 20 t ( 0 ≤ t ≤ 10 )

    This should describe the variation of its velocity with time until the end of the initial booster stage of its flight. Figure 4 shows the corresponding graph of velocity against time. Note the way in which the graph slopes downwards to the right. This function describes an increasingly negative velocity as time passes, consistent with an increasing upwards velocity. The corresponding graph for a positive acceleration of the same magnitude would slope upwards towards the right.

    Figure 4 :

    { Variation of velocity of rocket during the initial booster stage.}

Task!

Imagine that a satellite is falling towards Earth at 5 m s − 1 when a booster rocket is fired for 5 s accelerating it away from the Earth at 10 m s − 2 .

  1. Write down a corresponding linear function that would describe its velocity during the booster stage.

    If position is measured downwards, acceleration away from the Earth may be written as − 10 m s − 2 . The initial velocity towards the Earth may be denoted by ( + ) 5 m s − 1 so v = − 10 t + 5 ( 0 ≤ t ≤ 5 ) . If position is measured upwards v = 10 t − 5 ( 0 ≤ t ≤ 5 ) .

  2. Sketch the corresponding graph of velocity against time if position is measured downwards towards Earth:

    {Satellite velocity (position measured downwards towards Earth)}

  3. Sketch the corresponding graph of velocity against time if position is measured upwards away from the Earth:

    {Satellite velocity (position measured upwards away from Earth)}

  4. At what time would the velocity of the satellite be zero?

    When v is 0, 0 = − 10 t + 5 , so t = 0.5 . The satellite has zero velocity towards the Earth after 0.5 s.

  5. What is the value of the velocity at the end of the booster stage?

    When t = 5 , either v = − 10 × 1 + 5 = − 5 , so the velocity is 5 m s − 1 away from the Earth, or, using the second equation in (a), v = 10 − 5 = 5 , leading to the same conclusion.

3.2 Other contexts for linear models

Linear functions may arise in other contexts. In each of these situations, the slope and intercept values will have some modelling significance. Indeed the behaviour and hence the suitability of a linear function, of the form y = a x + b , when modelling any given situation will be determined by the values of a and b .

Task!

During 20 minutes of rain, a cylindrical rain barrel that is initially empty is filled to a depth of 1.5 cm.

  1. Choose variables to represent the level of water in the barrel and time. Sketch a graph representing the level of water in the barrel if the intensity of rainfall remains constant over the 20 minute period.

    In this answer h cm is used for the level of water measured from the bottom of the barrel and t minutes for time.

    {Height (depth) of rainwater in a barrel.}

  2. Write down a linear function that represents the level of water in the vessel together with its range of validity.

    The intensity of rainfall is stated to be constant, so the rate at which the barrel fills may be taken as constant. The gradient of an appropriate linear function relating level of water ( h  cm) measured from the bottom of the vessel and time ( t , minutes) measured would be 1.5 20 = 0.075 and an appropriate linear function would be h = 0.075 t + c . Since the barrel is empty to start with, h = 0 when t = 0, implying that c = 0. So the appropriate linear function and its range of validity are expressed by h = 0.075 t , ( 0 ≤ t ≤ 20 ) .

  3. State any assumptions that you have made:

    It is assumed that the barrel has a uniformly cylindrical cross section, that no water is removed during the rainfall and there are no holes or leaks up to 1.5 cm depth.

  4. Write down the amended form of your answer to (b), if the vessel contains 2 cm of water initially.

    h = 0.075 t + 2 ( 0 ≤ t ≤ 20 )

Task!

Suppose that you travel often from Nottingham to Milton Keynes which is a distance of 87 miles almost all of which is along the M1 motorway. Usually it takes 1.5 hours. Suppose also that, on one occasion, you have agreed to pick someone up at the Leicester junction (21) of the M1. This is 25 miles from the start of your journey in Nottingham. If you start your journey at 8 a.m., what time should you advise for the pick-up?

(Graphical method)

Assume a constant speed for the whole journey. This means that if 87 miles is covered in 1.5 hours, then half the distance (43.5 miles) is covered in 0.75 hours and so on.

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(Symbolic method)

Let d miles be the distance travelled in time t hours. Then d t = 58 t . This is valid only for the duration of the journey ( 0 ≤ t ≤ 1.5 ) . The equation can be used to find the time at which d = 25. Now 25 = 58 t and so t = 25 58 = 0.43103448 = 0.43 (to two decimal places). Either way, given that 0.43 h is about 26 minutes, a possible suggestion is that the passenger should be advised 8.26 a.m. for the pick-up. But the assumption of constant speed has its limitations. It would be safer to say “be there by 8.20 a.m. but be prepared to wait perhaps until 8.30 a.m.”

Task!

A local authority has flood control plans in which the emergency and rescue services are alerted when the river level rises to critical values. A linear model is used to estimate the variation of height with time. After a period of continuous heavy rain the level one day was 1.5 m at 8 a.m. and 1.8 m at 2 p.m.

  1. Use a linear model to write down an equation for estimating the level of the river at different times of the day:

    If the level of water is represented by L m and time by t hours after 8 a.m. then a linear model for the level as a function of time may be written

    L = a t + b

    where a and b are constants to be found from the other information in the problem. Specifically, it is stated that L = 1.5 when t = 0 and L = 1.8 when t = 6 . The first statement implies that

    1.5 = 0 + b or b = 1.5

    The second statement implies that

    1.8 = 6 a + b

    or, after substituting for b ,

    1.8 = 6 a + 1.5 or 0.3 = 6 a or a = 0.05

    So the equation for estimating the level of the river at different times is

    L = 0.05 t + 1.5

  2. Suggest a suitable range of values of time for which the model could be used:

    The model is valid between 8 a.m. and 2 p.m. and, subsequently, only as long as the river level rises steadily.

  3. What time does the model predict that the level of the river will reach 2 m?

    The model will predict a level of 2 m at time t given by

    2 = 0.05 t + 1.5 or t = 0.5 0.05 = 10

    i.e. 10 hours after 8 a.m. which is 6 p.m.

Task!

During one winter, the roads in a rural area were completely free from snow when it started snowing at midnight. It snowed heavily all night and day. By 10 a.m. it was 19 cm deep.

To save money the local authorities wait until the snow is 30 cm deep before ploughing the snow away from the roads. Forecast when ploughing should start, stating any assumptions you have made.

If the depth of snow is represented by D cm and time by t hours after midnight then a linear model for the depth as a function of time may be written:

D = a t + b

where a and b are constants to be found from the other information in the problem or from assumptions. As there was no snow at midnight

0 = 0 + b or b = 0

It is stated that D = 19 when t = 10 , i.e.

19 = 10 a or a = 1.9

So the equation for estimating the depth of snow at different times is

D = 1.9 t

The model will predict a level of 30 cm at a time t given by

30 = 1.9 t or t = 30 1.9 = 15.789

i.e. 15.789 hours after midnight which is a little after 3.47 p.m.

Assuming that the snow build up is steady e.g. no drifting or change in precipitation, this suggests that ploughing should start about 3.45 p.m.

Exercises
  1. A cross-channel ferry usually takes 2 hours to make the 40 km crossing from England and France.
    1. What is the boat’s average speed?
    2. Derive a linear model connecting distance from England and time since leaving port. State any limitations of the model.
    3. According to this model, when will the boat be 15 km and 35 km from England?
  2. During one winter, the roads in the country district were completely free from snow when it started snowing at 2:30 a.m.and it snowed steadily all day. At 7:30 a.m.it was 14 cm deep. To save money, the local practice was to wait until the snow was 20 cm deep before ploughing the roads. Forecast when ploughing would start, stating any assumptions.
  3. In a drought, the population of a particular species of water beetle in a pond is observed to have halved when the volume of water in the pond has fallen by half. Make a simple assumption about the relationship between the beetle population and the volume of water in the pond and express this in symbols as an equation. What would your model predict for the population when the water volume is only one third of what it was originally.
  4. A firm produces a specialised instrument and, although it has the facilities to produce 100 instruments per week, it rarely produces more than 50. It is finding it difficult to assess the cost of producing the instruments and to set realistic prices. The firm’s accountant estimates that the firm pays out £5000 per week on fixed costs (overheads, salaries etc.) and that the additional cost of producing each instrument is £50.
    1. Derive and use a linear model for the variation in total costs with the quantity of instruments produced. State any limitation of this model.
    2. What is the model’s prediction for the cost of producing 80 instruments per week?
    1. A distance of 40 km is covered in 2 hours. So the average speed is 40 ∕ 2 = 20 km h − 1 .
    2. A linear model assumes that the boat is a point moving at constant speed and will only be valid for 2 hours (or 40 km) while the boat is travelling from England to France. It does not allow for variations in speed. If the distance from the English port at any time t hours is denoted by d km, then d = 20 t .
    3. When the boat is 15 km from England 15 = 20 t , so t = 15 ∕ 20 = 0.75 , so the boat is 0.75 hour (45 minutes) from port. When the boat is 35 km from England, 35 = 20 t , so t = 35 ∕ 20 = 1.75 , so the boat is 1.75 hour (1 hour 45 minutes) from port.
  2. Assume that there is no snow at 2:30 a.m.and that the rate of accumulation of snow is constant. Then, if the snow is 14 cm deep at 7:30 a.m., the rate of accumulation is 2.8 cm per hour. A linear model for the depth ( d cm) of snow t hours after 2:30 a.m. is d = 2.8 t . d will be 20 when 20 = 2.8 t , i.e. t = 20 ∕ 2.8 = 7.143 . This corresponds to about 9:39 a.m. So ploughing should start at about 9:40 a.m.
  3. Denote population by P and volume of pond by V . Then P is proportional to V so P = k V where k is a constant of proportionality. When V becomes V ∕ 3 , then P becomes P ∕ 3 .
    1. Denote the number of instruments made per week by N and the corresponding cost by £ C . Asume that C increases at a constant rate with N (i.e. C is proportional to N ). Then a linear model for total costs (£ T ) is T = 5000 + 50 N . This will be valid only for 0 ≤ N ≤ 100 .

      If N = 80 , then T = 5000 + 50 × 80 = 9000 .

    2. The predicted total cost is £9000.