Transposing Formulae

Revisiting this topic now and finding it difficult to get the order correct. Have two questions:

Make "f" the subject of:
R/r = √(f+P/f-P)

And

Make "a" the subject of:
R/ω= √(4L^2−3a^2/L^2−a^2)

Reply 1

nerak99

Original post by mercedes36

Clues:
All of these are a matter of collecting terms and it is a game really.
In your first one we want f and we can see that two f terms are trapped in the root.
First, square both sides to get the root away and leave (R.r)^2 on the left.

Now we have f in a denominator as well as a numerator on RHS f+P/f-P so multiply through by f

(R/r)^2 f = f^2 +P -Pf

Now the fact we have f^2 means that this a quadratic

so rearrange to get 0 = f^2-Pf - (R/r)^2f+P which is the same as f^2-Pf - (R/r)^2f+P = 0

Take f out of terms 2 and 3 to get f^2 - (P + (R/r)^2) f+P = 0 (Watch out for the sign change we had to do in the ()

Next you have to either complete the square or use the formula to get something like f = some stuff including plus or minus a square root.

You can check answer to rearrange formula by putting simple numbers in the original expression and checking that these numbers still work together in your answer.

I hope that you have some idea about how to proceed.

(edited 4 years ago)

Reply 2

mercedes36

Original post by nerak99

I have attached what you said to do (I didn't continue through to the end) and when you said (I underlined it in your statement above) to multiply through by f, how would it move the "-p" from the denominator to the numerator (see in red on the attachment what I thought it would leave).

Sorry - been a while since I've done these.

Transposing F.JPG29.9KB

Reply 3

nerak99

The way you write the formula in your OP it looked like

(\frac{R}{r})=\sqrt{f +\frac{ P}{f} - P}

Thats what I worked with. Does that help?

Working with

(\frac{R}{r})=\sqrt{\frac{f +P}{f - P}}

Proceeds like

(\frac{R}{r})^2={\frac{f +P}{f - P}}

Then multiply through by (f-P)

(\frac{R}{r})^2(f-P)={f +P}

Expand the bracket on the LHS and then collect f term to the left .

Take a common factor of f out of the LHS and then finally divide through by whatever was multiplying the f on the LHS

(edited 4 years ago)

Reply 4

mercedes36

Original post by nerak99

The way you write the formula in your OP it looked like

(\frac{R}{r})=\sqrt{f +\frac{ P}{f} - P}

Thats what I worked with. Does that help?

Sorry I didn't realise I see why you thought that. (I don't know how to make the formula look like what you wrote)

Unfortunately it does not help... I'm not sure why I am not understanding this.

Thank you for taking the time to try help me either way !-

Reply 5

nerak99

Original post by mercedes36

I have expanded significantly on my post to try to help. The way you presented it just needed a bracket to make it understandable in line. As things get complex you have to use Latex for which TSR have a help file somewhere on how to do it.

(edited 4 years ago)

Reply 6

mercedes36

Original post by nerak99

Ah yes I already did that and I even expanded it but it got real messy so I resorted to TSR.
Thanks for your help.

Reply 7

mercedes36

Original post by nerak99

Would you be able to check where I keep going wrong? I really do not know what I'm not getting right.
How to deal with the f-f?

Transposing F.JPG36.8KB

Reply 8

nerak99

Just your last line is wrong really. we should get ...

f((\frac{R}{r}^2)-1)= P+P(\frac{R}{r})^2=P(1+(\frac{R}{r}^2) )

Unparseable latex formula:

f=P\frac{1+(\frac{R}{r}^2)}{ ( \frac{R}{r}^2)-1}}

Unparseable latex formula:

f=P\frac{(\frac{R}{r}^2)+1}{ ( \frac{R}{r}^2)-1}}

I reckon that is right

Reply 9

mercedes36

Original post by nerak99

Just your last line is wrong really. we should get ...

f((\frac{R}{r}^2)-1)= P+P(\frac{R}{r})^2=P(1+(\frac{R}{r}^2) )

Unparseable latex formula:

f=P\frac{1+(\frac{R}{r}^2)}{ ( \frac{R}{r}^2)-1}}

Unparseable latex formula:

f=P\frac{(\frac{R}{r}^2)+1}{ ( \frac{R}{r}^2)-1}}

I reckon that is right

But isn't f-f = 0? That bit baffles me.

Reply 10

nerak99

Original post by mercedes36

But isn't f-f = 0? That bit baffles me.

Mmmm. There is no f-f.
If I have something like f(a-1) and expand the bracket, I have af -f. There is no f - f

I am am not entirely sure where you are getting the f - f from but I think it is an error with the common factor in a bracket manipulation.