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# FP2 Polar Coordinates Question watch

1. The question is:

The straight line lambda passes through the point A with polar co-ordinations (p,alpha) and it perpendicular to OA.
Prove that the polar equation of lambda is p = r cos(theta - alpha)
Use the expansion of cos(theta - alpha) to find the cartesian equation of lambda

I have absolutely no idea, any advice/ help?
2. (Original post by ComputerMaths97)
The question is:

The straight line lambda passes through the point A with polar co-ordinations (p,alpha) and it perpendicular to OA.
Prove that the polar equation of lambda is p = r cos(theta - alpha)
Use the expansion of cos(theta - alpha) to find the cartesian equation of lambda

I have absolutely no idea, any advice/ help?
I waited a while before replying in case someone else helped you (because their help would be better than mines) but if you don't mind doing the question back to front you could convert polar to Cartesian and then convert back at the end.

I have what I think is the solution if you want to see it

Posted from TSR Mobile
3. (Original post by DylanJ42)
I waited a while before replying in case someone else helped you (because their help would be better than mines) but if you don't mind doing the question back to front you could convert polar to Cartesian and then convert back at the end.

I have what I think is the solution if you want to see it

Posted from TSR Mobile
You've waited an hour???
4. (Original post by Major Wilson)
You've waited an hour???
yea obviously? ive been refreshing the page continuously for an hour waiting for my chance to help someone with their maths homework, it's what i live for
Spoiler:
Show
jk I done homework and had dinner, came back and still no replies soo
5. (Original post by DylanJ42)
I waited a while before replying in case someone else helped you (because their help would be better than mines) but if you don't mind doing the question back to front you could convert polar to Cartesian and then convert back at the end.

I have what I think is the solution if you want to see it

Posted from TSR Mobile
I managed to show it, but I really don't understand why it works.

I drew a new point (r,theta) (polar co-ordinates), with theta > alpha and r such that the perpendicular line through A also goes through this new point. So I had a right angled triangle, hypotenuse r, adjacent p with angle = theta-alpha. This meant that r = p/cos(theta-alpha) therefore required answer. However why does this work? Doesn't make much sense to me as to why finding this gives us a polar equation?
6. (Original post by ComputerMaths97)
I managed to show it, but I really don't understand why it works.

I drew a new point (r,theta) (polar co-ordinates), with theta > alpha and r such that the perpendicular line through A also goes through this new point. So I had a right angled triangle, hypotenuse r, adjacent p with angle = theta-alpha. This meant that r = p/cos(theta-alpha) therefore required answer. However why does this work? Doesn't make much sense to me as to why finding this gives us a polar equation?

This is kind of just repeating what you are saying, but who knows maybe it will help.

This topic is quite hard to explain actually (probably shows lack of understanding on my side)

but basically it's saying that the distance r (the red line in my diagram) will get longer the bigger the angle between theta and alpha gets.

Actually, if you think if of (theta - alpha) being 90, iethe black and red line being perpendicular, then you would have;

r x 0 = p

which is not satisfied for any value of r, no matter how big, this makes sense when you think of the graph, it is saying that no matter how long you draw the red line (ie how big you make r) it will still not touch the line perpendicular to OA.

I doubt this helps but just keep thinking about it and it will just click
7. The lack of replies is because you posted it in the wrong forum, it's been moved to the correct one now.
8. (Original post by DylanJ42)

This is kind of just repeating what you are saying, but who knows maybe it will help.

This topic is quite hard to explain actually (probably shows lack of understanding on my side)

but basically it's saying that the distance r (the red line in my diagram) will get longer the bigger the angle between theta and alpha gets.

Actually, if you think if of (theta - alpha) being 90, iethe black and red line being perpendicular, then you would have;

r x 0 = p

which is not satisfied for any value of r, no matter how big, this makes sense when you think of the graph, it is saying that no matter how long you draw the red line (ie how big you make r) it will still not touch the line perpendicular to OA.

I doubt this helps but just keep thinking about it and it will just click
Okay and this is what polar equations represent on a cartesian graph, how r changes as theta changes? I just kept thinking of it on a polar graph and I got all mixed up and confused. This definitely helped though thanks, I just seem to keep losing marks on polar co-ordinates questions whenever it steers away from the standard draw this simple polar curve and find the area under it. Anything else I didn't have a clue
9. (Original post by Zacken)
The lack of replies is because you posted it in the wrong forum, it's been moved to the correct one now.
I thought I put it in A level lol oops, thanks!
10. (Original post by ComputerMaths97)
I thought I put it in A level lol oops, thanks!
You did. It should have been in maths.
11. (Original post by ComputerMaths97)
Okay and this is what polar equations represent on a cartesian graph, how r changes as theta changes? I just kept thinking of it on a polar graph and I got all mixed up and confused. This definitely helped though thanks, I just seem to keep losing marks on polar co-ordinates questions whenever it steers away from the standard draw this simple polar curve and find the area under it. Anything else I didn't have a clue
Basically r represents a straight line and theta represents the angle from the horizontal

So if you wanted to plot r = cosx you could do this if you really wanted to;

I obviously just picked easy angles, however you could do every 5 degrees and draw straight lines for length r for each angle, it would take ages but it would work

You maybe could pick something like r = 2 - cos(x) and attempt to draw it using a protractor and ruler, it would probably help understanding (i will actually be taking my own advice and doing one )

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