MHB How to Convert Polar Coordinates to Rectangular Coordinates?

  • Thread starter Thread starter karush
  • Start date Start date
  • Tags Tags
    Polar Rectangular
AI Thread Summary
To convert the expression $2\sin(\theta) - 3\cos(\theta)$ into rectangular coordinates, the relationships $x = r\cos(\theta)$ and $y = r\sin(\theta)$ are essential. Substituting $\sin(\theta)$ and $\cos(\theta)$ in terms of $x$ and $y$ leads to the expressions $2\sin(\theta) = \frac{2y}{r}$ and $-3\cos(\theta) = -3\frac{x}{r}$. The conversion process involves recognizing that $r = \sqrt{x^2 + y^2}$, which allows for the elimination of $r$ from the equation. Ultimately, the goal is to express the original polar equation entirely in terms of $x$ and $y$. The discussion emphasizes the importance of careful substitution and manipulation to achieve the correct rectangular form.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

Convert to rectangular coordinates
I'm clueless?
 
Mathematics news on Phys.org
karush said:
$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

Convert to rectangular coordinates
I'm clueless?
[math]x = r~cos(\theta) \text{ and } y = r~sin(\theta)[/math]

and
[math]r = \sqrt{x^2 + y^2}[/math]

Can you finish?

-Dan
 
$r=\sqrt{9\cos^2\left({\theta}\right)+4\sin^2\left({\theta}\right)}$
That's it?
 
karush said:
$r=\sqrt{9\cos^2\left({\theta}\right)+4\sin^2\left({\theta}\right)}$
That's it?

Not even close. Look at Dan's post in more detail. If $\displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*}$ then what is $\displaystyle \begin{align*} \cos{(\theta)} \end{align*}$?
 
$\cos\left({\theta}\right)=\frac{x}{-3}$
 
karush said:
$\cos\left({\theta}\right)=\frac{x}{-3}$

No! You should NOT be substituting into the original expression yet!

If $\displaystyle \begin{align*} x = r\cos{(\theta)} \end{align*}$ then $\displaystyle \begin{align*} \cos{(\theta)} = \frac{x}{r} \end{align*}$. This is closer to being in a rectangular form, but we still have an "r" there. What are we going to do with that?
 
Sorry but I'm lost?
 
karush said:
Sorry but I'm lost?

Can you think of a way to relate $r$ to $x$ and $y$?
 
$r^2=-3x+2y=x^2+y^2$
 
Last edited:
  • #10
I was thinking of the relationship Dan posted above:

$$r=\sqrt{x^2+y^2}$$ :D
 
  • #11
Completed the square
$\left({x}^{2}+\frac{3}{2}\right)^2+\left(y-1\right)^2=\frac{13}{4}$

If $r=\sqrt{x^2+y^{2}}$
then the radical is in the way.
 
  • #12
karush said:
$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

Convert to rectangular coordinates
I'm clueless?
Try this. We know that [math]y = r~sin(\theta)[/math]. This means that [math]sin(\theta) = \frac{y}{r}[/math]. So the first term in your expression is
[math]2~sin(\theta) = 2 \cdot \frac{y}{r}[/math]

Since [math]r = \sqrt{x^2 + y^2}[/math],
[math]2~sin(\theta) = 2 \cdot \frac{y}{r} = 2 \cdot \frac{y}{\sqrt{x^2 + y^2}}[/math]

What does the second term come out to be?

-Dan
 
  • #13
I don't see why that would be easier?
 
  • #14
Easier than what? You posted this earlier:
karush said:
$r^2=-3x+2y=x^2+y^2$

But your original problem was not an equation. Your problem was to express [math]2~sin(\theta) - 3~cos(\theta)[/math] in rectangular coordinates. How did you get [math]r^2 = -3x + 2y[/math]?

-Dan
 

Similar threads

Back
Top