Converting a Cartesian equation to polar form

In summary, to convert the Cartesian equation x^2/9 + y^2/4 = 1 to polar form, multiply both sides by 36 to get rid of fractions, then break up the sin^2 term and solve for r to get r^2(4 + 5sin^2(t))=36.
  • #1
rdioface
11
0

Homework Statement


Convert the following Cartesian equation to polar form.
x^2/9 + y^2/4 = 1

Homework Equations


r*cos(t)=x
r*sin(t)=y
r=Sqrt(x^2 + y^2)
y/x = Arctan(t)

The Attempt at a Solution


I get ugly looking things like r^2(cos^2(t)/9 + sin^2(t)/4) = 1 but being a simple ellipse (edit: duh) I expect a cleaner answer.
 
Last edited:
Physics news on Phys.org
  • #2
rdioface said:

Homework Statement


Convert the following Cartesian equation to polar form.
x^2/9 + y^2/4 = 1

Homework Equations


r*cos(t)=x
r*sin(t)=y
r=Sqrt(x^2 + y^2)
y/x = Arctan(t)

The Attempt at a Solution


I get ugly looking things like r^2(cos^2(t)/9 + sin^2(t)/4) = 1 but being a simple hyperbola I expect a cleaner answer.
It's actually an ellipse, not a hyperbola.

Multiply both sides by 36 to get rid of the fractions. After that you get
4r2cos2(t) + 9r2sin2(t) = 36

You can break up the sin2 term into 4r2sin2(t) + 5r2sin2(t). Does that give you any ideas?
 
  • #3
That gets me to r^2(4 + 5sin^2(t))=36. Are there any further steps to be done?
 
  • #5
Alright thanks, it seemed like there might have been some crazy trig identity I was missing or something.
 

Related to Converting a Cartesian equation to polar form

1. What is the process for converting a Cartesian equation to polar form?

The first step is to substitute x = rcosθ and y = rsinθ into the Cartesian equation. Then, use trigonometric identities to simplify the equation and solve for r in terms of θ. Finally, rewrite the equation in polar form using r = f(θ).

2. Can all Cartesian equations be converted to polar form?

No, not all Cartesian equations have a corresponding polar form. Some equations may have restrictions or limitations that prevent them from being expressed in polar coordinates.

3. What are the advantages of using polar coordinates over Cartesian coordinates?

Polar coordinates can be more useful when dealing with circular or symmetric shapes, as the equations are simpler and more intuitive. They can also be more convenient for certain calculations, such as finding the distance or angle between two points.

4. How do you determine the domain and range of a polar equation?

The domain of a polar equation is typically the range of values for θ, which can be determined by any restrictions or limitations on the equation. The range is the set of possible values for r, which may also be limited by the equation or the shape it represents.

5. Is there a specific method for converting polar equations to Cartesian form?

Yes, the process is similar to converting from Cartesian to polar form. Start by substituting r = √(x² + y²) into the polar equation. Then, use trigonometric identities to simplify and solve for either x or y in terms of r and θ. Finally, rewrite the equation in Cartesian form as y = f(x).

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
349
  • Calculus and Beyond Homework Help
Replies
1
Views
894
  • Calculus and Beyond Homework Help
Replies
1
Views
979
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
802
  • Calculus and Beyond Homework Help
Replies
1
Views
7K
  • Calculus and Beyond Homework Help
Replies
1
Views
913
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
833
Back
Top