Eliminate the parametercartesian coordinates

  • Thread starter Thread starter Null_
  • Start date Start date
  • Tags Tags
    Coordinates
Click For Summary
The discussion focuses on eliminating the parameter from the equations x = 2 cos θ and y = 3 sin θ to find a Cartesian equation of the curve. Participants emphasize the importance of using the trigonometric identity cos²(θ) + sin²(θ) = 1 to correctly derive the relationship between x and y. One user expresses concerns about their lack of precalculus knowledge affecting their calculus performance, while another suggests considering a precalc course for a stronger foundation. Ultimately, the user successfully derives the correct equation, which is an ellipse: (x²/4) + (y²/9) = 1. The conversation highlights the challenges of transitioning into calculus without a solid precalc background.
Null_
Messages
227
Reaction score
0

Homework Statement


You are given the following.
x = 2 cos θ, y = 3 sin θ, -π/2<= θ<= π/2
Eliminate the parameter to find a Cartesian equation of the curve.


Homework Equations



n/a

The Attempt at a Solution


x=2cosθ ; cosθ=y
y=3sinθ ; sinθ=x

so..

x=2y and 2y-x=0
y=3x and 3x-y=0
2y-x=3x-y
3y=4x
1=4x/3y


I'm pretty sure I went wrong somewhere...I have a feeling the equation is supposed to be a curve, not just a line. I'm in calculus I in college right now, but I basically had no pre-calc in high school, so I need to learn it while I go. Thanks.
 
Physics news on Phys.org
Null_ said:



The Attempt at a Solution


x=2cosθ ; cosθ=y
y=3sinθ ; sinθ=x




x and y are given in therms of θ. x=2cosθ, y=3sinθ .

Why do you think that x=sinθ and y=cosθ?

Eliminate θ from the original equations, using some relation between sinθ and cosθ.

ehild
 
Null_ said:

Homework Statement


You are given the following.
x = 2 cos θ, y = 3 sin θ, -π/2<= θ<= π/2
Eliminate the parameter to find a Cartesian equation of the curve.


Homework Equations



n/a

The Attempt at a Solution


x=2cosθ ; cosθ=y
y=3sinθ ; sinθ=x
You're given x = 2cosθ and y = 3sinθ. Where did cosθ = y and sinθ = x come from?
Null_ said:
so..

x=2y and 2y-x=0
y=3x and 3x-y=0
2y-x=3x-y
3y=4x
1=4x/3y


I'm pretty sure I went wrong somewhere...I have a feeling the equation is supposed to be a curve, not just a line. I'm in calculus I in college right now, but I basically had no pre-calc in high school, so I need to learn it while I go. Thanks.

Right, these parametric equations do not define a line, but do define a curve. The trick for this problem is to work with your two given parametric equations and use the trig identity cos2(A) + sin2(A) = 1.

Regarding your comment about no precalc in high school and learning it as you go, you might want to reconsider and think about taking a precalc course at your college instead of the calculus class you're currently in. Otherwise, you will be needing to work twice as hard as most of the other students in the class you're in. Although you might understand the general ideas that are being presented, you are probably going to have a lot of difficulties with the details that involve the algebra and trig that make up precalculus. These form the foundation on which calculus is built. If they aren't solid, you'll have many difficulties with calculus.
 
Mark44 said:
You're given x = 2cosθ and y = 3sinθ. Where did cosθ = y and sinθ = x come from?


Right, these parametric equations do not define a line, but do define a curve. The trick for this problem is to work with your two given parametric equations and use the trig identity cos2(A) + sin2(A) = 1.

Regarding your comment about no precalc in high school and learning it as you go, you might want to reconsider and think about taking a precalc course at your college instead of the calculus class you're currently in. Otherwise, you will be needing to work twice as hard as most of the other students in the class you're in. Although you might understand the general ideas that are being presented, you are probably going to have a lot of difficulties with the details that involve the algebra and trig that make up precalculus. These form the foundation on which calculus is built. If they aren't solid, you'll have many difficulties with calculus.

Thank you very much! I had been learning the unit circle and definitions of sine cosine and tangent and had a momentarily lapse of judgement and applied the unit definition to this problem. I see where I went wrong now, and I got the answer correct ( ((x^2)/4) + ((y^2)/9)=1 )

See, I had a horrible high school experience. I took AP Calc AB [the highest math available to us] and had a horrible teacher. However, I took it upon myself to learn the material (something that I wished I had done in pre-calc) and ended up getting a 4 on the exam which could have placed me into calc II. I decided to stay back in Calc I. So, I know the calculus already, and I figured that while everyone else has a solid pre-calc base, I already know the calculus. I hope to spend more time on pre-calc in the class than calc.

Knowing this, would you still recommend that I drop back to pre-calc or should I continue learning it on my own? (I learn better on my own anyway..)
 
Well, that makes things different. If you are pretty strongly motivated to fill in your precalc (including trig?) gaps, then maybe you can be successful in the class your are in.
 
I've decided that I'm staying in the class. I'm pretty confident that I can learn trig within a few week's study.
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
4
Views
2K
Replies
4
Views
2K
Replies
7
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K