Easy Parametrics Q (1 Viewer)

kaz1

et tu
Joined
Mar 6, 2007
Messages
6,960
Location
Vespucci Beach
Gender
Undisclosed
HSC
2009
Uni Grad
2018
Timothy.Siu said:
7. 2x-y+4=0

x=t-2 y=2t

8. x2+y<sup>2</sup>=9

x=t y=sqareroot(9-t2)

i think...
u can just sub in random stuff

Isn't the parametric equation for a circle x=rcosθ and y=rsinθ
so it would be x=3cosθ and y=3sinθ but I'm not really sure.
 

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
there could be more than one answer when you convert cartesian equation to parametric.

the easiest way is to either make something the subject or use trig identities.
for 7, you could even have y=2t-4 and x=t
 

Fortian09

Member
Joined
May 19, 2008
Messages
120
Gender
Male
HSC
2009
u sure u just sub randoms? coz it doesnt seem right like i dont know the steps involved
 

jet

Banned
Joined
Jan 4, 2007
Messages
3,148
Gender
Male
HSC
2009
Yeah for a circle x^2 + y^2 = r^2
The parametrics are x=rcost, y=rsint

You can work this out using the triangle created by the x-axis, the radius and the height to the point. The parameter, t (or theta) will be the angle between the radius and the x-axis. (The length of the triangle on the x-axis is the projection of the radius on the x-axis)
 
Last edited:

Timothy.Siu

Prophet 9
Joined
Aug 6, 2008
Messages
3,449
Location
Sydney
Gender
Male
HSC
2009
kaz1 said:
Isn't the parametric equation for a circle x=rcosθ and y=rsinθ
so it would be x=3cosθ and y=3sinθ but I'm not really sure.
yeah i guess so, that wud make more sense, but u can really do anytihng as long as it works right?
 

Fortian09

Member
Joined
May 19, 2008
Messages
120
Gender
Male
HSC
2009
does anyone mind put all the parametric equations general forms plz?
that wud be a great help to everyone having trouble (probs only me)
 

azureus88

Member
Joined
Jul 9, 2007
Messages
278
Gender
Male
HSC
2009
What do u mean by general form? General form for parmetric equations is x=.... and y=.... where x and y are in terms of a parameter t right?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top