# Put the calculator down!

Posted on 27-05-15 in Physics

That instinctive reach for the calculator is something I've found to be near universal for GCSE and A level students. What makes this dangerous is how automatic it becomes. Even simple calculations cause students to doubt their own ability to count. Add in any amount of complexity and you have a recipe for silly mistakes and order of magnitude errors.

In this (hopefully short) post, I'll list a few tips. Hopefully a few you'll find useful!

## 1. Rearrangement before substitution

Even if the question is asking for something simple, rearrange the formula you are using before substituting numbers. In more complex cases you can find that common terms cancel out, saving you time and effort. Importantly this will also reduce the chance of 'losing' numbers or powers.

Finally as I stress to most of my students, I'm rarely interested in the final answer. I am however very interested in how you got to it! By writing out the rearranged equation it makes it easier to see what you've done and spot mistakes easily.

For example

Q) Light is incident at $$30^\circ$$ to the boundary between glass $$n=1.5$$ and water $$n=1.33$$. What is the outgoing angle?

A) $$n_1\sin\theta_1 = n_2\sin\theta_2$$

$$\theta_2 = \arcsin(\frac{n_1}{n_2}\sin\theta_1)$$

$$\theta_2 = \arcsin(\frac{1.5}{1.33}0.5)$$

Now grab the calculator to find the final answer

$$\theta_2 = 86^\circ$$

How did I know $$\sin(30^\circ)=0.5$$? See further down this post!

## 2. Cast everything into standard form

So you've found your final formula. Now we need to put the numbers in. We can make our life immeasurably easier if we substitute in numbers in using standard form.

Standard form is where we remove the power of tens from the number and ensure everything is in SI units. So 150 $$\mu$$m becomes $$1.5 \times 10^{-4}$$ m. There are two advantages. Firstly we get more practice recalling $$\mu = 10^{-6}$$ (which many students do get wrong!). Secondly it's trivial to collect powers of ten using our fabulous knowledge of indices. Consider the following simple example - no calculator needed!

Q) Estimate the mass of a swimming pool (density of water = 1000 kgm$${}^{-3}$$)

A) mass = density * volume ($$m=\rho v$$)

volume = length * width * height ($$v=xyz$$)

$$\therefore m = \rho xyz$$

$$m = (1\times10^3)(2.5\times10)(1\times10)(2\times10^{0})$$

m = $$5\times10^5$$ kg

It is also worth noting that the order of magnitude, the number above the ten, gives us an idea of whether our final answer is correct. For the above question does 100 metric tonnes sound reasonable for a swimming pool? I'd say so.

## 3. Common numbers & patterns

More often that you might expect questions will try to make your life easier. Common patterns and easy simplifications are rife on physics examination questions. This is because physicists wish to (unsurprisingly) test your physics, not your mathematics. Whilst there are too many possibilities for me to list them all, I will try to list a couple.

First off most multiplication and division problems should be easy for you to complete mentally. Some of this is a matter of practice and sometimes there are tricks we can do. For instance $$49\times3$$ might at first glance appear difficult. We can recognise this is $$50\times3 - 3$$ and suddenly the problem isn't difficult at all. Further cases include $$\sqrt{4^2}$$, which cancels back to just being 4. Consider another example.

Q) Simplify $$\sqrt{\frac{1024}{64}}$$

A) $$\sqrt{\frac{2^{10}}{2^6}}$$

$$\sqrt{2^4} = (2^4)^{0.5} = 2^2 = 4$$

Obviously this kind of 'pattern spotting' doesn't come over night, but it's an incredibly useful skill that will pay dividends over time. Furthermore this is a good way to speed up your mathematics reliably. I'll leave the fanatics to googling tricks to improve this further. The bottom line is this :

if you could do it in primary school, you don't need a calculator for it.

## 4. Angles and Triangles

There are many rules relating angles with triangles. For physics (and indeed A level mathematics) we need to look at a couple of specific triangles. First however let's look at a pattern which comes along again and again - pythagorean triples. The first four are given below :

• 3, 4, 5
• 5, 12, 13
• 8, 15, 17
• 7, 24, 25

So why are these useful? In the first instance if we're given a triangle with two sides labelled we can quickly identify the third. Perhaps much more sneakily we can also do trigonometry. For instance if $$\sin{x} = \frac{3}{5}$$, we can see that $$\cos{x}=\frac{4}{5}$$. For another example have a look at

$$\sin{\theta} = \frac{15}{17}, \cos{\theta}= \frac{8}{17}, \tan{\theta} = \frac{8}{15}$$

This can be very useful in many cases. Firstly, we often use sines and cosines to decompose vectors in mechanics. Using this method we can decompose without any loss of precision. Consider the following question

Q) A footpath up a hill rises by 6 m every 8 m. A walker travels at 1 m/s up the slope.

i) How fast does the climber rise?

ii) The hill is 200 m high, how long does it take the climber to reach the summit?

To answer this we will need to draw a triangle of the footpath. Next we need to recognise that a right-angled triangle is a (hidden) Pythagorean triple. Can you see how?

The answer to part i) is 0.6 m/s. See if you can figure this out! Part ii) should be simple for those who remember the relationship between speed, distance and time. I'll explain both parts at the end of this post.

Next let's first look at two simple triangles. These are shown below.

In the equilateral triangle (on the left) we set each side to have a length of 1. In the isosceles triangle (on the right) we set the two identical sides to have a length of 1. Recalling that there is $$180^\circ$$ in a circle and some symmetry afforded by the triangles we can identify the angles inside. For the left triangle, each angle is $$60^\circ$$ and on the right, two of the angles must be $$45^\circ$$. Make sure you understand this step! With a little bit of Pythagoras we can label these triangles to give us

Earlier in this post I said that $$\sin(30^\circ) = 0.5$$ and now we can see why. Sine is defined as the ratio of the opposite to the hypotenuse. Which, for $$\sin(30^\circ)$$ is $$\frac{0.5}{1}$$.

Any time we need to find 30, 45 or 60 degrees we should recall these triangles.

Q) The final height in metres is given by $$x$$. Simplify $$x = 20\sin(45^\circ)$$

A) $$\therefore x = 20 \times \frac{1}{\sqrt{2}}$$

$$x = 10 \times \frac{2}{\sqrt{2}}$$

$$x = 10\sqrt{2}$$ m

## 4. Learning to Approximation

Finally if we're only interested in roughly how much an answer is, we can approximate. This is a valuable method for determining quickly if a more detailed calculation has come out on the right order of magnitude. In this case we substitute into our equations only the order of magnitudes, not the actual numbers. This is known as a fermi estimation.

Whilst not recommended during an exam, this can be a quick way to validate an answer or find a good estimate very quickly. A great example of one can be found on the what if xkcd blog.

Finally there are two more numbers that we can memorise to save time.

• The sun is 500 light seconds away from the earth.
• The number of seconds in a year is $$\pi\times10^7$$.

The error in the number of seconds in a year calculated this way is less than 0.5%! Very handy! For instance

Q) How many metres to Alpha Centauri? (Alpha Centauri is 4.4 Ly away). The speed of light is $$3\times10^8$$ m/s

A) Distance = $$(\pi\times10^7)(4.4\times10^0)(3\times10^8)$$

Distance = $$(3)(4.4)(\pi)\times10^{15}$$

Therefore Distance = $$4.1\times10^{16}$$ m

## Summary

There are two reasons to avoid using a calculator. Firstly, mental arithmetic is going to be quicker. Secondly, I've found it remarkably common for students to make simple mistakes inputting simple calculations into a calculator. This is especially annoying when there's no need to do so!

The main thing to take away from this post is to think...

'do I need to use a calculator really?'

If you've exhausted the above steps, then go for it!