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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}.

This post follows on from this post where the following principle was presented:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

There are a number of subtleties here: basically sometimes you get extra ‘solutions’ (that are not solutions at all), and sometimes you can lose solutions.

Let us write the squaring function, e.g. 6\mapsto 36, x\mapsto x^2 by f(x)=x^2 and the square-rooting function by x\mapsto \sqrt{x}. It appears that (x^2,\sqrt{x}) are an inverse pair but not quite exactly. While

\displaystyle 81\overset{\sqrt{x}}{\mapsto} 9\overset{x^2}{\mapsto}81 and

\displaystyle 7\overset{x^2}{\mapsto}49\overset{\sqrt{x}}{\mapsto}7,

check out O.K. note that

-4\overset{x^2}{\mapsto}+16\overset{\sqrt{x}}{\mapsto}=+4\neq -4,

does not bring us back to where we started.

This problem can be fixed by restricting the allowable inputs to x^2 to positive numbers only but for the moment it is better to just treat this as a subtlety, namely while (\sqrt{x})^2=x, \sqrt{x^2}=\pm x… in fact I recommend that we remember that with an x^2 there will generally be two solutions.

The other thing we look out for as much as possible is that we cannot divide by zero.

There are other issues around such as the fact that \sqrt{x}>0, so that the equation \sqrt{x}=-2 has no solutions (no, x=4 is not a solution! Check.). This equation has no solutions.

Often, in context, these subtleties are not problematic. For example, equations with no solutions rarely arise and quantities might be positive so that if we have \pm\sqrt{a}, only +\sqrt{a} need be considered (for example, a might be a length).

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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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Quadratics are ubiquitous in mathematics. For the purposes of this piece a quadratic is a real-valued function q:\mathbb{R}\rightarrow \mathbb{R} of the form


where a,\,b,\,c\in \mathbb{R} such that a\neq 0. There is a little bit more to be said — particularly about the differences between a quadratic and a quadratic function but for those this piece is addressed to (third level: non-maths; all second level), the distinction is unimportant.


The basic object we study is the square function, s:\mathbb{R}\rightarrow \mathbb{R}, x\mapsto x^2:


All quadratics look similar to x^2. If a>0 then the quadratic has this \bigcup geometry. Otherwise it looks like y=-x^2 and has \bigcap geometry

The geometry dictates that quadratics can have either zero, one or two real roots. A root of a function is an input x such that f(x)=0. As the graph of a function is of the form y=f(x), roots are such that y=f(x)=0\Rightarrow y=0, that is where the graph cuts the x-axis. With the geometry of quadratics they can cut the x-axis no times, once (like s(x)=x^2), or twice.

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There are a number of ways of explaining why you cannot divide by zero. Here are my two favourites.

Any Set of Numbers Collapses to a Single Number

How old are you? Zero years old.

How tall are you? Zero metres old.

How many teeth do you have? Zero.

How many Superbowls has Tom Brady won? Zero

Yep, if you allow division by zero you only end up with one number to measure everything with.

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As I said in the previous post, there is a duality:

Points on a Curve (Geometry) \Leftrightarrow Solutions of an Equation (Algebra)

This means we can answer geometric questions using algebra and answer algebraic questions using geometry.


Consider the following two questions:

  1. Find the tangents to a circle \mathcal{C} of a given slope.
  2. Find the tangents to a circle \mathcal{C} through a given point.

Both can be answered using the duality principle.


Find the tangents to the circle

\mathcal{C}\equiv x^2+y^2-4x+6y-12=0

that are

(a) parallel to the line L\equiv 4x+3y+20=0

(b) through the point (-10,-5) [caution: the numbers here are disgusting]

Solution (a) i:

 First of all a sketch (and the remark that a tangent is a line):


Here we see the circle \mathcal{C} and the line L on the bottom left. The two tangents we are looking for are as shown. They have the same slope as L and have only one intersection with \mathcal{C}. These two pieces of information will allow us to find the equations of the tangents.

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Arguably, the three central concepts in the theory of differential calculus are that of a function, that of a tangent and that of a limit. Here we introduce functions and tangents.


When looking at differential calculus, two good ways to think about functions are via algebraic geometry and interdependent variables. Neither give the proper, abstract, definition of a function, but both give a nice way of thinking about them.

Algebraic Geometry Approach

Let us set up the plane, \Pi. We choose a distinguished point called the origin and a distinguished direction which we call ‘positive x‘. Draw a line through the origin in the direction of positive x. This is the x-axis. Choose a unit distance for the x-direction.

Now, perpendicular to the x-axis, draw a line through the origin. This is the y-axis. By convention positive y is anti-clockwise of positive x. Choose a unit distance for the y-direction.

This is the plane, \Pi:


Now points on the plane can be associated with a pair of numbers (a,b). For example, the point a distance one along the positive x and five along the negative y can be denoted by the coordinates (1,-5):


Similarly, I can take a pair of numbers, say (-1,3), and this corresponds to a point on the plane.

This gives a duality:

points on the plane \Leftrightarrow pairs of numbers

Now consider the completely algebraic objects


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This post follows on from this post where the logic for the below is discussed. I am not going to define here what easy means!

Here is the strategy/guiding principle:

Fundamental Principle of Solving ‘Easy’ Equations

Identify what is difficult or troublesome about the equation and get rid of it. As long as you do the same thing to both numbers (the “Lhs” and the “Rhs”), the equation will be replaced by a simpler equation with the same solution.

Read the rest of this entry »

There is a right way to think about equations and there is a wrong way to think about equations. Let us not speak of the wrong way…

The equations I have in mind are those equations written in the form


where the aim is to find all the real numbers x that ‘satisfy’ the equation.

We aren’t always taught the logic behind solving equations. The first thing to say is that many of us are trained to believe that this ‘=‘ means the ‘the answer is’. This is not what equals means. This may have happened to us because while young children our textbooks had stuff like


written in them… the ‘answer’ of course being eight and the = sign almost suggests that we have to ‘do something’ to 2+6. Of course, this is not what equals means, and while the pupil who writes


is correct, the pupil who writes e.g.


has written a statement just as true as 2+6=8.

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