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On a particular day the velocity of the wind, in terms of \mathbf{i} and \mathbf{j}, is x\mathbf{i}-3\mathbf{j}, where x\in\mathbb{N}.

\mathbf{i} and \mathbf{j} are unit vectors in the directions East and North respectively.

To a man travelling due East the wind appears to come from a direction North \alpha^\circ West where \tan\alpha=2.

When he travels due North at the same speed as before, the wind appears to come from a direction North \beta^\circ West where \tan\beta=3/2.

Find the actual direction of the wind.


We start by writing


We have two equations.

Firstly, when the man travels due East, \overrightarrow{V_M}=a\mathbf{i} for some constant a. We have \overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j} and so


We know that this, with respect to the man, is coming from North \alpha^\circ West. This means we have:


And furthermore:

\displaystyle \tan\alpha=\frac{x-a}{3}\overset{!}{=}2

\Rightarrow x-a=6\Rightarrow a=x-6.

Now consider when the man travels due North, \overrightarrow{V_M}=a\mathbf{j}. We have \overrightarrow{V_W}=x\mathbf{i}-3\mathbf{j} still and so


We know that this, with respect to the man, is coming from North \beta^\circ West. This means we have:


And furthermore:

\displaystyle \tan\beta=\frac{x}{3+a}\overset{!}{=}\frac32

\underset{\times_{2(3+a)}}{\Rightarrow} 2x=9+3a,

but we have a=x-6 and so

2x=9+3(x-6)\Rightarrow x=9,

so that \overrightarrow{V_W}=9\mathbf{i}-3\mathbf{j}.

This means that the velocity of the wind looks like:


That is the wind comes from North \theta West. We have that \tan\theta=9/3\underset{\tan^{-1}}{\Rightarrow} \theta=\tan^{-1}(3)\approx 71.57^\circ, so the answer to the question asked is N 71.57^\circ W.


Here we present three solutions to the one problem. The vector solution is probably the slickest. The geometry solution here can be simplified by being less rigorous, and the coordinate geometry solution might be made easier by using the -b\pm\sqrt{\cdots} formula.

LCHL 2007, Q. 2(a)

Ship B is travelling west at 24 km/h. Ship A is travelling north at 32 km/h.
At a certain instant ship B is 8 km north-east of ship A.

(i) Find the velocity of ship A relative to ship B.

(ii) Calculate the length of time, to the nearest minute, for which the ships
are less than or equal to 8 km apart.

Solution to (i): We have that \overrightarrow{V_A}=32\mathbf{j} and \overrightarrow{V_B}=-24\mathbf{i} and so

\overrightarrow{V_{AB}}=\overrightarrow{V_A}-\overrightarrow{V_B}=32\mathbf{j} -(-24\mathbf{i})=24\mathbf{i}+32\mathbf{j}.

Vector Approach to (ii)

First of all we draw a picture. As we are talking relative to ship B we will put B at the origin. If ship B is 8 km north-east of ship A then ship A is 8 km south-west of ship B.


Where \overrightarrow{R_0} is the initial displacement of ship A relative to ship B, the displacement of ship A relative to ship B, as a function of time, is given by

\overrightarrow{R_{AB}}(t)=\overrightarrow{R_0}+t\cdot \overrightarrow{V_{AB}}.

Using some trigonometry — vertical and horizontal components of \overrightarrow{R_0} — we have


and so

\displaystyle \overrightarrow{R_{AB}(t)}=\left(24t-\frac{8}{\sqrt{2}}\right)\mathbf{i}+\left(32t-\frac{8}{\sqrt{2}}\right)\mathbf{j}.

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For those planning on focusing on questions one to five and ten:

Applied Maths some Notes

Occasionally, it might be useful to do as the title here suggests.

Two examples that spring to mind include:

  • solving a\cdot\cos\theta\pm b\cdot\sin\theta=c for \theta (relative velocity example with - below)
  • maximising a\cdot\cos\theta\pm b\cdot\sin\theta without the use of calculus

a\cdot \cos\theta- b\cdot\sin\theta

Note first of all the similarity between:

\displaystyle a\cdot \cos\theta-b\cdot \sin \theta\sim \sin\phi\cos\theta-\cos\phi\sin\theta.

This identity is in the Department of Education formula booklet.

The only problem is that a and b are not necessarily sines and cosines respectively. Consider them, however, as opposites and adjacents to an angle in a right-angled-triangle as shown:


Using Pythagoras Theorem, the hypotenuse is \sqrt{a^2+b^2} and so if we multiply our expression by \displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}} then we have something:

\displaystyle \frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}\cdot \left(a\cdot \cos\theta- b\cdot\sin\theta\right)

\displaystyle=\sqrt{a^2+b^2}\cdot \left(\frac{a}{\sqrt{a^2+b^2}}\cos\theta-\frac{b}{\sqrt{a^2+b^2}}\sin\theta\right)

=\sqrt{a^2+b^2}\cdot \left(\sin\phi\cos\theta-\cos\phi\sin\theta\right)=\sqrt{a^2+b^2}\sin(\phi-\theta).

Similarly, we have

a\cdot\cos\theta+b\cdot \sin\theta=\sqrt{a^2+b^2}\sin(\phi+\theta),

where \displaystyle\sin\phi=\frac{a}{\sqrt{a^2+b^2}}.

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Consider the following problem:

Two masses of 5 kg and 1 kg hang from a smooth pulley at the ends of a light inextensible string. The system is released from rest. After 2 seconds, the 5 kg mass hits a horizontal table:

i. How much further will the 1 kg mass rise?

ii. The 1 kg mass then falls and the 5 kg mass is jolted off the table. With what speed will the 5 kg mass begin to rise?

[6D Q. 4. Fundamental Applied Maths, 2nd Edition, Oliver Murphy]

It isn’t difficult to answer part i.: the answer is \displaystyle \frac89 g m.

However how to treat part ii.? First of all a picture to help us understand this problem:


The 1 kg mass has dropped under gravity through a distance of \displaystyle \frac89 g m. We can find the speed of the 1 kg mass using u=0,a=g,s=\frac89 g. Alternatively, we can use Conservation of (Mechanical) Energy.

Taking the final position as h=0, at its maximum height, the 1 kg mass has potential energy and no kinetic energy:

\text{PE}_0=mgh=1g\frac89 g=\frac{8}{9}g^2.

When it reaches the point where the string is once again taut, it has not potential energy but the potential energy it had has been transferred into kinetic energy:

\text{KE}_1=\frac12 mv^2=\frac12 v^2=\frac12 v^2,

and this must equal the potential energy \text{PE}_0:

\frac{8}{9}g^2=\frac12 v^2\Rightarrow v=\frac43g.

Now this is where things get trickier. My idea was to use conservation of momentum on the two particles separately. As this clever answer to this question shows, you can treat the 5 kg mass, string, and 1 kg mass as a single particle.

So the prior momentum is the mass of the 1 kg mass by \frac43 g:

p_0=m_0u=1\cdot \frac43 g=\frac43g.

The ‘after’ momentum is the mass of the 1 kg and 5 kg masses times the new velocity:

p_1=m_1v=6\cdot v.

By Conservation of Momentum, these are equal:

\frac43 g=6v\Rightarrow v=\frac{2}{9} g\text{ m s}^{-1}.

The following question gave a little grief:

Two smooth spheres of masses 2m and 3m respectively lie on a smooth horizontal table.

The spheres are projected towards each other with speeds 4u and u respectively.

i. Find the speed of each sphere after the collision in terms of e, the coefficient of restitution

ii. Show that the spheres will move in opposite directions after the collision if e>\frac13.

My contention is that the question erred in not specifying that the answers to part i. were to be in terms of e and u.


i. The following is the situation:


Let v_1 and v_2 be the velocities of the smaller respectively larger sphere after collision. Note that the initial velocity of the larger sphere is minus u.

Using conservation of momentum,


\Rightarrow 2m(4u)+3m(-u)=2mv_1+3mv_2

\Rightarrow 5u=2v_1+3v_2.


\displaystyle e(u_1-u_2)=-(v_1-v_2)\Rightarrow \frac{v_1-v_2}{u_1-u_2}=-e,


\displaystyle \frac{v_1-v_2}{4u-(-u)}=-e

\Rightarrow v_1-v_2=-5ue\Rightarrow v_1=v_2-5ue,

and so

5u=2(v_2-5ue)+3v_2\Rightarrow 5u=2v_2-10ue+3v_2,

\Rightarrow 5v_2=5u+10ue\Rightarrow v_2=u(1+2e).

\Rightarrow v_1=u(1+2e)-5ue=u+2ue-5ue=u(1-3e).


ii. v_2>0. If e>\frac13\Rightarrow 3e>1\Rightarrow 1-3e<0 and so


that is the particles move in opposite directions.



The First Problem

A car has to travel a distance s on a straight road. The car has maximum acceleration a and maximum deceleration b. It starts and ends at rest.

Show that if there is no speed limit, the time is given by

\displaystyle \sqrt{2s\cdot \frac{(a+b)}{ab}}.

Solution: We are going to use two pieces of information:

  • the area under the velocity-time graph is the distance travelled,
  • the slope of the velocity-time graph is the acceleration.

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To successfully analyse and solve the equations of Leaving Cert Applied Maths projectiles, one must be very comfortable with trigonometry.

Projectile trigonometry all takes place in [0^\circ,90^\circ] so we should be able to work exclusively in right-angled-triangles (RATs), however I might revert to the unit circle for proofs (without using the unit circle, the definitions for zero and 90^\circ are found by using continuity).

Recalling that two triangles are similar if they have the same angles, the fundamental principle governing trigonometry might be put something like this:

Similar triangles differ only by a scale factor.

We show this below, but what this means is that the ratio of corresponding sides of similar triangles are the same, and if one of the angles is a right-angle, it means that if you have an angle, say 40^\circ, and calculate the ratio of, say, the length of the opposite to the length of the hypotenuse, that your answer doesn’t depend on how large your triangle is and so it makes sense to talk about this ratio for 40^\circ rather than just a specific triangle:


These are two similar triangles. The opposite/hypotenuse ratio is the same in both cases.

Suppose the dashed triangle is a k-scaled version of the smaller triangle. Then |A'B'|=k|AB| and |A'C'|=k|AC|. Thus the opposite to hypotenuse ratio for the larger triangle is

\displaystyle \frac{|A'B'|}{|A'C'|}=\frac{k|AB|}{k|AC|}=\frac{|AB|}{|AC|},

which is the same as the corresponding ratio for the smaller triangle.

This allows us to define some special ratios, the so-called trigonometric ratios. If you are studying Leaving Cert Applied Maths you know what these are. You should also be aware of the inverse trigonometric functions. Also you should be able to, given the hypotenuse and angle, find comfortably the other two sides. We should also know that sine is maximised at 90^\circ, where it is equal to one.

In projectiles we use another trigonometric ratio:

\displaystyle \sec(\theta)=\frac{\text{hypotenuse}}{\text{adjacent}}=\frac{1}{\cos\theta}.

Note \cos90^\circ=0, so that \sec90^\circ is not defined. Why? Answer here.

The Pythagoras Identity

For any angle \theta,


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In Ireland at least, we first encounter fractions at age 6-8. At this age, because of our maturity, while we might be capable of some conceptional understanding, by and large we are doing things by rote and, for example, multiplying fractions is just something that we do without ever questioning why fractions multiply together like that. This piece is aimed at second and third level students who want to understand why the ‘calculus’ of fractions is like it is.

Mathematicians can in a rigorous way, write down what a fraction is… this piece is pitched somewhere in between these constructions — perhaps seen in an undergraduate mathematics degree — and the presentation of fractions presented in primary school. It is closer in spirit to a rigorous approach but makes no claims at absolute rigour (indeed it will make no attempt at rigour in places). The facts are real number axioms.

Defining Fractions

We will define fractions in terms of integers and multiplication.

To get the integers we first define the natural numbers.

Definition 1: Natural Numbers

The set of natural numbers is the set of counting numbers


together with the operations of addition (+) and multiplication \times.

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Applied maths could be defined as the use of mathematics in studying natural phenomena. The branch of applied maths studied at Leaving Cert level is Newtonian mechanics. Mechanics is the study of systems under the action of forces. Newtonian mechanics is concerned with systems that can be adequately described by Newton’s Laws of Motion. Systems that aren’t adequately described by Newtonian mechanics include systems with speeds approaching the speed of light, systems of extremely small particles and systems with a large number of particles. Hence Leaving Cert Applied Maths is the study of simple macro-systems that have moderate speeds.

Applied maths is essentially a further study of the mathematics of chapters 6 through 13 in Real World Physics; following which more specific and involved questions than those of Leaving Cert Physics may be posed and answered. The emphasis in applied maths is more on problem-solving than anything and reflecting this, the need for rote-learning is almost non-existent. The course content itself could be presented on one A4 sheet. The skills required for the course include:
• Capacity for interpretation and visualisation
• Ability for strategic problem solving
• Competency in mathematics

Anyone who is strong in higher level maths (A or B standard) and is self-motivated can achieve great success in applied maths – especially in conjunction with LC Physics. There is then a three for the price of two and a half effect as applied maths will help your physics, physics will help your applied maths and maths will help your applied maths. For those looking at the bigger picture, if you are intending on pursuing any science or engineering course in college, having done LC Applied Maths will really stand to you in your daunting first year — most 1st year physics and applied maths modules broadly cover the material of LC Applied Maths.

The statistics over the last number of years are that typically over 90% of students taking applied maths do higher level, and of these about a quarter achieve an A, over half achieve an A or B and nearly two-thirds achieve an honour. 90% of students pass. So with strong maths and good, effective application to the task at hand, good grades are there for the taking.

Some LC Applied Math Notes and some Solutions to selected problems. Section 6.2 deals with the Fundamental Theorem of Calculus and Section 6.4 has a little integration. See Chapter 10 of these LC Maths Lecture Notes for some more on integration.

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