Eliminate the parametercartesian coordinates

In summary: If you already know the basics of calculus, you don't need to spend time learning pre-calculus. You should focus on the more important classes at your college.
  • #1
Null_
231
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.
 
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  • #2
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
 
  • #3
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.
 
  • #4
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..)
 
  • #5
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.
 
  • #6
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.
 

FAQ: Eliminate the parametercartesian coordinates

What is the purpose of eliminating the parameter in Cartesian coordinates?

The purpose of eliminating the parameter in Cartesian coordinates is to convert a parametric equation into a rectangular equation. This allows for a more simplified representation of the equation and makes it easier to graph and analyze.

How do you eliminate the parameter in Cartesian coordinates?

To eliminate the parameter in Cartesian coordinates, you can use algebraic manipulation and substitution. By solving for the parameter in one equation and substituting it into the other equation, you can eliminate the parameter and create a rectangular equation.

What are the benefits of using Cartesian coordinates over parametric equations?

Cartesian coordinates provide a simpler and more intuitive representation of equations, making it easier to understand and analyze. They also allow for easier graphing and visualization of the equation.

Are there any limitations to eliminating the parameter in Cartesian coordinates?

Eliminating the parameter in Cartesian coordinates may not always be possible for all parametric equations. Some equations may require advanced techniques or may not have a rectangular form.

Can eliminating the parameter in Cartesian coordinates change the shape of the graph?

Yes, eliminating the parameter in Cartesian coordinates can sometimes change the shape of the graph. This is because the rectangular equation may have different coefficients or constants than the original parametric equation.

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