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Parametric - Cartesian

x = cot0
y = 2sin0
Find the cartesian equation
I have got as far as x = 2 root(1-y^2/4)/y not sure whether it's right though?
Cheers for any help :smile:
Reply 1
Avatar for nmudz_009
nmudz_009
22
Original post by matthew769
x = cot0
y = 2sin0
Find the cartesian equation
I have got as far as x = 2 root(1-y^2/4)/y not sure whether it's right though?
Cheers for any help :smile:


Use identity cosec=cot + 1
Reply 2
Avatar for matthew769
matthew769
OP
18
Original post by nmudz_009
Use identity cosec=cot + 1


Poo, it was a class test and I haven't memorised the formulas from C3 trig yet :frown:
Thanks for the swift reply :smile:
Reply 3
Avatar for nmudz_009
nmudz_009
22
Original post by matthew769
Poo, it was a class test and I haven't memorised the formulas from C3 trig yet :frown:
Thanks for the swift reply :smile:


Lol no problem:smile: This is actually a simple question, so you should learn them for c4- there's plenty of problems involving trig
(edited 12 years ago)
Reply 4
Avatar for matthew769
matthew769
OP
18
Original post by nmudz_009
Lol no problem:smile: This is actually a simple question, so you should learn them for c4- there's plenty of problems involving trig


Yeah I will have to :redface:. Just out of interest, would I get any marks for what I've written since it still gives cot0 essentially?
Reply 5
Avatar for nmudz_009
nmudz_009
22
Original post by matthew769
Yeah I will have to :redface:. Just out of interest, would I get any marks for what I've written since it still gives cot0 essentially?


X=cot(theta) is already given in the question isn't it? You do not get marks for stating what they already told you. I think you would get your first mark for indicating the connection between sine and cot. I.e stating the identity
Reply 6
Avatar for matthew769
matthew769
OP
18
Original post by nmudz_009
X=cot(theta) is already given in the question isn't it? You do not get marks for stating what they already told you. I think you would get your first mark for indicating the connection between sine and cot. I.e stating the identity


Ah ok, but surely there's more than one way of proving it/manipulating it.
By showing that cot0 = cos0/sin0, but using x and y expressions, would that not be acceptable?
Reply 7
Avatar for nmudz_009
nmudz_009
22
Original post by matthew769
Ah ok, but surely there's more than one way of proving it/manipulating it.
By showing that cot0 = cos0/sin0, but using x and y expressions, would that not be acceptable?


Oh I'm sorry I didnt read the expression you wrote in the op properly :tongue:. yh either identity wud be fine. I wud always state the identity on the side as well tho, so that it's clear.

So what you've written in the OP is perfectly fine except it could be written more simply. Sin(theta)+cos(theta) also equals one :smile:
(edited 12 years ago)
Reply 8
Avatar for matthew769
matthew769
OP
18
Original post by nmudz_009
Oh I'm sorry I didnt read the expression you wrote in the op properly :tongue:. yh either identity wud be fine. I wud always state the identity on the side as well tho, so that it's clear.

So what you've written in the OP is perfectly fine except it could be written more simply. Sin(theta)+cos(theta) also equals one :smile:


Ok thanks for the replies :smile: +repped.
Reply 9
Avatar for TenOfThem
TenOfThem
18
There seem to be a lot of squares missing on this thread
Original post by matthew769
Ok thanks for the replies :smile: +repped.


Sure, I got x=(1-y/2) / y/2 btw using (cos)/sin=cot
Reply 11
Avatar for Slumpy
Slumpy
16
Original post by TenOfThem
There seem to be a lot of squares missing on this thread


My thinking also.
Reply 12
Avatar for matthew769
matthew769
OP
18
Original post by TenOfThem
There seem to be a lot of squares missing on this thread


Original post by Slumpy
My thinking also.


I noticed, but ignored as I understand what is meant :smile:
Original post by matthew769
I noticed, but ignored as I understand what is meant :smile:


phew :smile:

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