How to Convert Polar Coordinates to Rectangular Coordinates?

[karush]

$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

Convert to rectangular coordinates
I'm clueless?

 

[topsquark]

$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

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

 

[karush]

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

 

[Prove It]

$r=\sqrt{9\cos^2\left({\theta}\right)+4\sin^2\left({\theta}\right)}$
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 $\displaystyle \begin{align*} \cos{(\theta)} \end{align*}$?

 

[karush]

$\cos\left({\theta}\right)=\frac{x}{-3}$

 

[Prove It]

$\cos\left({\theta}\right)=\frac{x}{-3}$

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?

 

[karush]

Sorry but I'm lost?

 

[MarkFL]

Sorry but I'm lost?

Can you think of a way to relate $r$ to $x$ and $y$?

 

[karush]

$r^2=-3x+2y=x^2+y^2$

 

[MarkFL]

I was thinking of the relationship Dan posted above:

$$r=\sqrt{x^2+y^2}$$ :D

 

[karush]

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.

 

[topsquark]

$2\sin\left({\theta}\right)-3\cos\left({\theta}\right)$

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

 

[karush]

I don't see why that would be easier?

 

[topsquark]

Easier than what? You posted this earlier:

$r^2=-3x+2y=x^2+y^2$

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