www.mathcentre.ac.uk/.../Transposition%20or%20Re-arranging%20Formulae.mp4

Let's consider the formula
for the period of a simple

pendulum of length L is
formula is T period.

Is equal to 2π?

Times the square root of L
over G&L is the length of

the pendulum.

Now on Earth, we tend to
regard G's being fixed it.

There is a little with
altitude, but we tend to

think of it as being fixed.

Just supposing we had a pendulum
of a fixed length L, and we took

it somewhere else. Let's say the
moon or Mars.

Somewhere out of the solar

system. Gravity there would not
be the same and we might want to

measure gravity, so one of the
ways will be to take our

pendulum, set it swinging.

And measure the time.

Once we measured the time, we
could then use that to Calculate

G, but before we could do it, we
would need to know what G was in

terms of the rest of the symbols
in this formula. So we need to

rearrange this formula so that
it said G.

Equals.

Now it's what?

Goes instead of this question
mark that we're going to have a

look at in this video. We're
going to be looking at how we

transform formally, how we move
from an expression like this.

To another expression.

Involving exactly the same
variables, but one that helps us

answer the questions that were
after in terms of a different

variable. To do this?

Transformation of formula will
need all the techniques that we

had in terms of solving
equations, so the video solving

linear equations in one variable
might be very useful to have a

look at, but let's just recap by
having a look.

A simple linear equation.

So the one going to take
three X +5.

Equals 6
- 3 * 5

- 2 X.

Now we've got this equation we
want to solve it for X, so our

aim is to end up with X equals.

So one of the first things
we would do is multiply out

the bracket.

So we have minus three times by
5 is minus 15.

Minus three times by minus
2X is plus 6X.

Next, we can simplify this
bit. Here three X +5

equals 6X, and now six
takeaway 15 is minus 9.

Now, one of the things we would
want to do is to try and get all

the ex is together. So we were
three X here under 6X there, so

will take 3X away from each
side, so that's three X

takeaway. Three X +5 equals 6X
takeaway, 3X minus 9.

3X takeaway 3X the X is go.
We've known left and then we've

got this. Five is equal to six X
takeaway 3X. That's three X

again takeaway 9.

Now we need to get the constant
terms, the numbers together. So

to do that we would add 9 to
each side. So we 5 + 9 is equal

to three X minus 9 + 9.

So again, we're doing to the
same thing to both sides. Here

we took three X away from each
side. Here we're adding nine to

each side, so five and nine is
14 equals 3X, and we minus 9 +

9.0. Now we need to divide both
sides by three.

So we have 3X over three an 14
over 3, dividing both sides by

three, and so those three
canceled and we're left with X

equals 14 over 3. Now I've
written this out very fully.

I've written out every step that
I've said, but we wouldn't

normally expect to see all of
that written down.

Starting from here.

The next line we'd expect
to see is this one because

we say Take 3X away from
both sides and we'd expect

to see that as the result.

Then we'd say add 9 to both
sides, and So what we would

expect to see having done that
would be that.

And then we, say, divide both
sides by three.

And so we see that So what we
would see written down in our

exercise book or on a piece of
paper that carried the

solution to this equation will
be something like this.

OK, we've been through the
steps now of solving an

equation. These techniques,
particularly idea of a balance

of keeping the same on both
sides by doing the same thing

to both sides is what we're
going to do when we look at

the transformation of
formally.

So let's begin with our
first formula V equals U

plus AT. And the variable
that we're going to try and

find is TI.

T is the variable we're going to
find what is T expressed?

In terms of the rest of the
letters or symbols that there

are in this formula well, in
order to model this, what I'm

going to do is I'm going to
write down an equation which

looks like this, but it just got

numbers in. With a T there as
the unknown. So what we might

have is something like this.
Seven for V5 for you.

Let's say a tool
for A and then T.

Now what would we do to solve
this equation? The first thing

we try and do is get the TS on
their own and so to do that,

we take 5 away from each side.
So this would be 2. Over here

equals 2 T, so having done
that there, let's do it here.

V minus U taking you away from
each side.

Now we would divide both sides
by two over here.

In order to end up with just T
on its own, so let's do the

same. Here. A is what multiplies
by T, so we need to divide

everything on this side V minus
U over a equals T. That's it.

Notice everything here is over
a, not just one part of it, but

both the whole expression.

The minus U over a.

Let's take another
one. V squared equals

U squared plus 2A S.

And this time, let's say the
symbol that I'm going to try and

find all the variables that I'm
going to try and find in terms

of all the others, is you.

So that's the case. Again,
I'm going to write down an

equation over here which
looks like this one.

So let's have.

25

equals U squared.

+9.

Weather 9 is in place of this
lump here of algebra.

So what would we do here? Our
first step would be to take

nine away from both sides. So
let's do that. 25 - 9 equals

U squared and that gets us EU
squared on its own. So let's

do that over here. Let's take
this lump away from both

sides. 20 squared minus 2A S
is equal to U squared.

Now what I would want to do now
is to take the square root of

both sides 'cause I want just
you and here I've got you

squared. So I need the square
root of 25 - 9 and I want the

square root of all of that not
square root of 25 minus the

square root of 9. I want the
square root of 25 - 9, so again

I've got to do the same over
here I want the square root of.

All. Love it.

Let's just check why I need the
square root of all of this.

25 - 9 is 16, so that 16 equals
U squared. Take the square root

of both sides. Four is equal to
you and we're happy. That's the

answer. But what if I do the
square root of 25 minus the

square root of nine? Well,
that's 5 - 3 is 2, which is not

the answer very definitely not
the answer, so we need to take

the square root of everything,
not just each little piece. And

so this over here has to be the
square root of all of that.

I've been working so far with
the formula that are for uniform

acceleration, so let's continue
with that and take.

Another one of them S equals
UT plus 1/2 AT squared, and

this time it's a.

Then I'm going to be looking for
can I rearrange this formula?

Can I transform it so it says a

equals? Again, let me model it
with an equation. Let's say that

S is 21, but U times by
T is 15 + 1/2 of a

Times by 9.

And it's this a the time after
first of all. Let's isolate the

term with the unknown in it. So
that's the ater. So let's take

15 away from both sides. So I'm
21 - 15 is equal to 1/2 of

a times by 9. So again, let's do
that here. Let's take this lump

away. S minus Utah equals 1/2.

AT squared, so again, my first
steps are to try and isolate the

variable. The term that I'm

looking for. Now this one
looks a bit complicated. We

got a half. There would be
nice to get rid of the half,

so let's multiply both sides
by two. If we do it at this

side, that's just a times by
9. We do it this side 2 * 21

- 15 and it multiplies all
of it. So let's do the same

here. Multiply both sides by
2, two times S minus UT.

Equals AT squared.

Going back to the equation we
got equals a Times by 9. I just

want a on its own and so I must
divide both sides by 9.

Here the thing that's doing
the multiplying it's T

squared. So let's divide both
sides by T squared and in the

same way as I divided
everything by 9, I've got to

divide everything here by T
squared.

And that gives me a.

Notice I haven't made any
effort to cancel because T

is not a common factor. It
is not a common factor.

No, we've been working with the
equations that are to do with

uniform acceleration. But there
are some other kinds of

equations that we need to gain
practice at and in order to

develop these skills, I'm going
to start with some made up

equations that don't have
physical applications in the

real world, but we will come
back once we've developed those

skills to looking at some real
equations that do contain

physical applications that we
will be able to transform, but

we need to develop some skills
to begin with. So first of all.

Let's take this expression,
let's call it rather than a

formula.

So then we have Y times 2X plus
one equals X Plus One and the

thing that we're going to be
after is X. Can we rearrange it

so it says X equals?

Well gain, let me
try and model this.

With an equation. So all
I need to do to model

this as an equation is
replaced the Y by three.

And if I was to solve this
inequation, my first step would

be to multiply out the bracket.
So let's do that. 6X plus three

is equal to X plus one.

So let's do that over here.
Multiply out this bracket. So

that's two XY Plus Y.

Is equal to X plus one.

Our next step with the equation
will be to get all the excess

together. So I would take X away
from both sides, so that would

be 6X takeaway X +3, and taking
the X away from this side just

leaves me with one. So let's do
that here. Let's take this X

away from both sides, so have
two XY minus X Plus Y.

Is equal to 1.

Now I would naturally want to
combine these two in some way.

6X minus X is just 5X.

I can't really do that at this
side, but what I can do is

gather together the terms in X
by taking out X as a common

factor. So we take out X from
this as a common factor. The

other factor is 2 Y minus.

And taking X out
of there is one.

Plus Y equals
1.

What now? Well now I've got my
ex is together over here. Let's

write this as 5X plus 3 equals
1. My next step will be to leave

the X term on its own by taking
three away from each side.

So let's do that here. Lead
the Exterm on its own by

taking the other term. That's
why away from both sides.

Now I'm just going to simplify
this. Five X equals minus two,

and in order to solve this to
get a value of XI need to divide

both sides by this number 5. So
that's X equals minus two over

5. So if I come back to this
this term in the bracket, two Y

minus one is the term that's
multiplying the X, and so I need

to divide. Both sides by so X
is equal to 1 minus Y over

2Y plus. Sorry made a mistake
there. Two Y minus one.

I just look again at what we've
done here. We've mimic the

solving of an equation. First,
we multiplied out the brackets.

Then we got the terms together

in X. The variable
that we were after.

Having got those terms together.

We then isolated them so that
they were on their own.

So we need to bear that in mind
and follow it through.

Let's take another expression Y
over Y plus X.

+5 is equal to X.

This time. We're going to be
finding why we're going to be

getting Y equals, and we want
some lump of numbers and X is

over this sigh.

So let's model this with an
equation Y over Y plus 3

+ 5 is equal to 3.

Look at this side. Here we've
got an algebraic fraction.

Y over Y plus 3Y plus three is
the denominator. It's in the

bottom of the fraction.

So in order to get rid of that,
we need to multiply everything

in this equation by this
denominator. So I'm going to

write that out in full Y over Y,
plus three. Put that in a

bracket times Y plus 3.

Plus five times Y.

Three is equal to three times Y
plus three. Let me just

emphasize we have to do the same
thing to both sides, so here

I've had to multiply everything
on both sides of the equation by

this term Y plus 3.

Here those will cancel out,
just leaving me with why, so

let's do that at this site.
Let's multiply everything by

this term Y plus X so we know
this first one. When we've

multiplied it by wiper sex is
just going to give us Why.

Plus five times by Y
plus X is equal to

X times by Y plus
X.

Coming back to the equation
here, faced with a lot of

brackets, we know the first
thing we would do is multiply

out the brackets. So let's do
that, why?

Plus five times by Y is 5
Y five times by three is 15

is equal 2 three times by Y
is 3 Y and three times by

three is 9.

Do the same here, why?

+5 Y.
Five times by X Plus 5X is

equal to X times by Y is
XY&X times by X is X.

Square.

On this side now we would like
to get all of our wise together.

So we've Y plus

5Y6Y. Take away 3 Y
so we've six. Why

already and we're going
to take away 3Y. Plus

15 is equal to 9, so
let's gather all the

wise together Y +5. Y
is 6 Y.

I need to bring this term over,
so I'll take XY away from both

sides minus XY.

And then let's just write down
the other terms.

Now here I simplify this. I'd
have six Y takeaway 3 Y and that

would just leave me with three Y
plus 15 equals 9. I can't do

that here, but the thing that I
can do is bring them much closer

together by taking out why.

The thing that I'm after as a
common factor. So let's take why

out of these two terms. Why
brackets? Why times by 6 Y means

I must have a 6 in their minus
XY means I must have a minus X

there plus 5X equals X squared.

Now over here I'd isolate this
term in why by taking 15 away

from both sides 3 Y is equal
to. Now this is 9 - 15,

so I've got to do the same
here. Take 5X away from each

side, isolating this term in YY
times 6 minus X is equal to

X squared minus.

5X.

This side now I've got 3 Yi,
just need Y so I would divide

everything on this side by
three. So I've got to do the

same with this. I've got to
divide everything on this side

by this term 6 minus X so we
have X squared minus 5X.

Divided by 6 minus X.

Put it all in a bracket to keep
it together. Notice I'm

attempting no canceling 'cause
there is no common factor in

this numerator and this
denominator. They do not share a

common factor, so that's my
answer and I finished. I've got

Y equals a lump of algebra
involving X is and numbers.

We started with a real formula.
We started with the formula for

a simple pendulum.

Of length Lt equals
2π square root.

L over G and the problem that we
posed was what was G in terms of

T? So G.

After G equals.

Let's have a look at a simple
equation. Let's say it's 10

equals 2π and I'll keep the two
Pike's. Part is just a number,

so 2π is just a number.

Square root
of 3 over G.

Now as an equation, this one is
a bit tricky because the G is

trapped inside this square root
sign. We need to bring it out

and the way to get a square root
sign to disappear, so to speak,

is to square both sides, reverse
the operation if you like, so

will square. Both sides of this
equation so 10 multiplied by

itself. I'm going to write it as
10 squared equals 2π multiplied

by itself, 'cause we're squaring
the whole of both sides times by

a nice square this.

Three over G.

Now we need to do the same here,
square the whole of both sides.

So that will be T squared.

2π all squared
L over G.

Coming back to the equation,
jeez, in the denominator I don't

want it there. I want it to say
G equals I want G upstairs, so

to speak, so I have to get rid
of it out of the denominator and

so to do that I must multiply
both sides by G.

So I've got 10 squared times.
G is equal to 2π squared

times three, so let's do that
here. Multiply both sides by G.

So I have T squared times
by G is equal to 2π

all squared times by L.

Gee, I want on its own. I want
to say G equal, so I must divide

by this thing 10 squared. So G
is equal to 2π all squared times

3, all over 10 squared.

And I can work that out. Let's
do the same with this G equals

2π all squared times by L, and
I'm getting rid of this T

squared by dividing everything
by T squared.

And we can leave it like that.

We can make it look a little
bit nicer if we want to buy

spotting that we've got a
square here under Square

there, and it might be nice
to combine those two.

Squaring process is by
writing 2π over T all

squared times by L.

Basically this is the answer
that we finished with.

Let's take another couple of.

Examples of real formula that we
might want to be able to

manipulate. So let's have a look
at the lens Formula One over F.

Is equal to one over U.

This one over V.

Let's say it's you.

That we're after we want you

equals. Well, let me write
down an equation.

Similar.

One over 3 equals 1 over U plus
one over 5.

I want to isolate this term
first in one over you.

To do that, I'm going to take
this away from both sides. 1/3

takeaway 150 is equal to one

over you. So let's do that.
Isolate this term in you.

And so to do that, we take away
one over V from both sides, so

we want over F.

Take away one over V is equal to
one over you.

Now, faced with problems
like this, it's very, very

tempting to simply turn
everything upside down.

Well, I just have a think about
that. This says 1/3 - 1/5. Now a

third is bigger than 1/5.

So 1/3 - 1/5 is a positive
number, so you at the very least

has got to be positive.

Watch what happens if I just
turn everything upside down.

3 - 5 equals U, which tells me
that minus two is equal to you.

But we just agreed that you had
to be positive and not negative.

You can't do that.

These are fractions, and so
because their fractions we have

to combine them as fractions,
which means we have to find a

common denominator and a common
denominator for three and five

is a number that both three and
five will divide into.

Easiest number is 3 times by 5.

So three Zing to three times by
5, which is 15 goes five times.

So I found multiplied 3 by 5. I
must multiply by this one by 5,

so that's 5 minus.

I've multiplied 5 by three, so I
multiply this one by three

equals one over. You know if I
did that here, I've got to do it

over here. So I want
to common denominator.

Here the common denominator I
talk was three times by 5, so

let's take F times by vis.

I've multiplied F.

By the so I must multiply the

one. By faith.

Minus.

I've multiplied the V by F.

So I must multiply the one by F.

Now if I come back over here,
let me just simplify this. This

is 2 over 15 is one over you,
and now I've got a complete

fraction on both sides. I can
turn it upside down and say

that's what you is.

Now there was a lot of
calculation went on here. I

can't do that calculation here,
but I have got a complete

fraction here so I can turn it.

Upside down.

To give me you.

So let's take as our final
example the time dilation

formula from relativity.

The formula is T
equals T, not.

Divided by.

1. Minus. V
squared over C squared to the

power. 1/2 or square root.

So let's take some numbers and
in fact what we're going to be

after is we're going to be
looking for this expression

here, V over C. So I just
write that down there. That's

what we're going to be looking
for. Can we get V over C in

terms of T&T Nord?

So let's have 6
equals 5 over 1

minus X squared.

To the half. Now solving this.

What would we do? Well, we've
got a square root. Let's Square

both sides in order to get rid
of that square root. So that

would be 6 squared equals 5
squared all over 1 minus X

squared. So let's do that here.
Square both sides.

T squared. Equals TN
squared all over 1

minus V squared over
C squared.

Now, well here the term I want
in X is in the denominator. I

need it upstairs, so let's
multiply both sides of this

equation by one minus X
squared. That would be 6

squared times 1 minus X
squared is equal to 5 squared.

So let's do that here.

T squared times, 1
minus V squared over

C squared is equal
to T Nord squared.

Now I want the X squared bit.

And it would be nice perhaps to
multiply out the bracket, but

there's a slightly quicker way.
I can divide both sides by 6

squared, and that's nice. 'cause
it keeps the square bits

together, so to speak. So let me
do that one minus X squared is

equal to 5 squared over 6
squared. So I'm going to do that

here. Divide both sides by T
squared, 1 minus V squared over

C squared. Is equal to
TN squared over T squared?

Now.

Running out of paper here.
So what I'm going to do is

turn over and write this one
down at the top of the next

page, and this one as well.

So we've got 1 minus V
squared over. C squared is

equal to T not squared over
T squared.

And here, with one minus X
squared is equal to 5 squared

over 6 squared.

It's The X squared that we're
after so I can add X squared

both sides, so this is one
equals 5 squared over 6 squared

plus X squared. So let me add
V squared over C squared to each

side. One is equal to T not
squared over T squared plus B

squared over C squared.

Now I want the X
squared on it so.

So take this away from both
sides. 1 - 5 squared over

6 squared is equal to X
squared, so 1 minus T, not

squared over T squared is equal
to V squared over C squared.

Taking this away from both

sides. Almost there now it's
X that I want not X squared

selects. Take the square root of

both sides. So do the
same here. The oversee that I

want, so let's take the square
root of both sides.

And that's it. We have found
the ratio of the velocity to

the speed of light in terms
of the two times.

So. That concludes this video on
transformation of formula thing

to do is to treat it always as
though it was an equation that

you were trying to solve.

In terms of the variable that
you want, X equals G equals and

you just perform the steps as
though you were solving an

equation. If you keep that in
mind, then most of your problems

should be taken care of.