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# Coming on in leaps and BOUNDS

For this post, you need to know your basics in rounding numbers. Here is a recap

1. Pick out the digit in the last place that will survive the rounding process

33962

2. Look at the next digit… Is it a 0 1 2 3 or 4? If so take no action. Otherwise raise the underlined digit by 1. In this case a 9, so raise to 4
3. In both cases, remove all following digits after the underlined one

34

Now, for this subject of bounds we are going to look at this kinda back to front. We are told a number has been rounded to 2 decimal places. So what could that number be?

2.335 is the smallest number it could be. Any number smaller would be rounded to 2.33
2.34499999 is the largest number it could be. Any number larger would be rounded to 2.35.

2.335 and 2.3449999  are the upper and lower bounds to the number given the information we have.

Let’s look at something a bit more complicated

Now, if we know that I have ran 10.6m to 1 d.p  in  19.5 seconds to 1 d.p, then what is my highest possible speed and lowest possible speed, to 3 d.p ?

We are being asked for the ‘upper bound’ and ‘lower bound’ for the speed.

Remember  Speed = distance / time

Start by writing down the upper bound’ and ‘lower bound’ of my distance and time.

Upper Bound of distance is 10.6499m
Lower Bound of distance is 10.5500m
Upper Bound on Time is 19.5499 seconds
Lower Bound on Time is 19.4500 seconds
Now, this is where we have to think because to find the upper bound on the speed, we need the highest possible time BUT the lowest possible time. This is because as time goes up, speed comes down, for any given distance. I use the highest possible value for time in the lower bound calculation, for the same reason

So upper bound on Time = 10.6499/19.4500 = 0.547553

The lower bound on time = 10.5500/19.5499 = 0.539645

The bounds for the speed are [0.540, 0.548]  to 3 d p

Note that I used 4 d.p. in my calculations. I will always use one or more decimal places than required then round at the end. We risk losing accuracy otherwise.

# Applying the laws of motions

Recently my A-level mechanics student has had difficulty remembering  the SUVAT equations of motions. I wrote a blog entry on these a while ago.  I’m now trying to think of an easy way to remember them, which a struggle! The best advice is to use them regularly, or just write them out every day.

a = (v-u)/t – is just the definition of Acceleration
s = (u + v)/2 * t  is using the average speed in the old speed * time formula
The other three can be derived from these but in a hurry, better remembered

v2 = u 2 + 2as
s = ut + at2/2
s = vt – at2/2

All of these have different forms where the subject is different, but re-arranging formulas is easier than remembering 20 equations.

There are 5 variables here and five equations..  and each equation has one of the 5 variables missing.  This means we can find the other variables with just 3 pieces of information. Just use the equation with the 3 parts you have and the one you are looking for.

A car travels 20m from a standing start in 10 seconds at constant acceleration. Find this acceleration and the speed after those 10 seconds

We will work out these things separately with two different equations

The acceleration we can find with the 4th equation – the one without v

10 = 100a/2

This is made easier because the standing start means u =0 and we can disregard the first element

10 = 50a  so a = 0.2m/s

We now could use any of the equations to find s as we know everything else, but I do suggest we don’t use a in this calculation,, just in case we didn’t get that right first time – even though I can’;t see any mistake

So we are going to use s = (u + v)/2 * t

20 = v/2 * 10 = 5v

v = 4m/s

# Solving Sim equations by substitution

Today’s post is on a GCSE subject, but quite an advanced one…  Solving Simultaneous  equations by substitution

I say advanced; I think its not often covered, but I think it can be followed;  There might just be some exciting algebra to do

To solve these equations

4x + y =  9
3x – y =  5

we know we need to add the equations together, to get rid of the y.

Sometimes we have to be a bit cleverer, and multiply one of the equations by something. For example

4x + y =  9
3x –2y =  4.

We have to multiply the first equation by 2 so the number in front of the y is the same. Otherwise the y won’t disappear when we add.

This is called ‘solving by addition/subtraction’

It only works sometimes. This new method words more often. It’s called ‘solving by substitution’

Solving By Substitution

Did you solve the first equations? You should have got x = 2 and y = 1

Now let’s change the first equation so its y = something. We can -4x from both sides.  This gives us

y = 9 – 4x.

Now, this is where the ‘substitution’ comes in.  We change the y in the other equation –

3x – (9-4x) = 5

We re-arrange this, remembering that – -4x = + 4x!
3x – 9 +4x = 5
7x = 14; x = 2..  and y = 1

Now that looks similar but more complicated. In fact I would recommend solving this equation the way we knew before.

So why am I showing you ‘substitution’?

To show the result is the same, and to get ready for when the other method doesn’t work.

For example
2x2  + y2  =11
y – x = 2

Here we have ‘squared’ in the mix, and this makes it hard to see how we can get rid of terms by adding or subtracting to two equations. This is where substitution comes into its own

Change the 2nd equation to say y = x + 2 (I’ve just added the x to both sides)
Substitution y = x + 2 into the first

2x2 + (x + 2)2 = 11

We have to a little work here, multiplying the brackets out and rearranging the terms
2x2 + x2 + 4x + 4 = 11
3x2 + 4x – 7 = 0  (I’ve combined the x2 terms and taken 11 from both sides)
Factorising this gives
(3x  + 7 ) (x  – 1)

That could be quite a hard factorisation to see straight off, though it helps that not many pairs of numbers multiply together to make -7. A little ‘trial and error’ may help.
So x = 1 or x = -7/3

Now remember y = x + 2 so

When  x = 1, y = 3..  That’s quite easy to see

When x = -7/3, y = -1/3. Just add on 2;

# Differentiation of the Ln function.

OK  a spot of A-level maths comes to the blog

A few things to remember:

1. Ln x – The Natural log of x – is the inverse function Footnote 1 of ex
2. d/dx ex = ex  – That’s what so special about ex

3. Remember the result from last time that Now how can we use these three facts to find d/dx ln x?

Let f(x) = y = Ln x . We want f’(x) = dy/dx Footnote 2
Raise e to both sides

ey = eln x  = x  (Using 1 above)

so  x = ey  so dx/dy = ey  (Using 2 above)

so dy/dx = 1/ ey

ey = x so substitute this in the give f’(x) in terms of x and we get 1/x.

d/dx  ln x = 1/x.

Footnotes

1 I use the word function here but caution is required. You must be careful with the domain of Ln x if it’s going to meet the criteria of a Function – It must be defined as x≥0

2. Ok so I’m mixing my calculus formats here! Both have their uses, maybe the subject for another blog entry.

# Mixed fractions: Multiplying and adding How do we multiply mixed fractions.

We could treat the parts separately.  I mean, its the multiplication of two mixed numbers and we’ve been asked to give the answer as a mixed number..  so we can use multiply each part of the first number with each part of the second.

1 x 3     +   1 x 1/3   + 3/4 x 3  +  3/4 x 1/3

= 3  + 1/3 + 9/4 + 1/4

Then we need to get the three fractions over the same number, 12

= 3 + (4 + 27 + 3)/12 = 3 + 34/12 = 3 + 17/6 = 3 + 2 5/6 = 5 5/6

But actually that’s actually quite a lot of work,  and changing these mixed numbers into ‘top-heavy’* fractions  and back again can save time…. In this pic you’ll see I’ve converted the two numbers into 7/4 and 10/3 – done just a little cancelling, and then multiplied

The answer is 35/6   which we can convert back into 6 5/6. This  the same answer as before,  which is always a relief!

We can also add these two mixed numbers. I think the first method is easier then.  Don’t convert into top heavy fractions. We don’t need to multiply all parts together.  Just add the whole numbers. Then add the fractions separately, which does mean looking for a common denominator

1 + 3 = 4
3/4 + 1/3 = 9/12 + 4/14 = 13/12 = 1 1/12

So the answer is 5 1/12

If you would like some more practice in working with mixed numbers, look at this document

*The proper word for top heavy fractions is ‘vulgar’ fractions and I’m sure you’ll here a teacher use this word..  Its just not a words I like for numbers..  all numbers are fun, none of them vulgar at all  IMHO!!