How to translate $r=$ equations

Question

I have a polar graph function

$$r=2\cdot\sin(\theta)$$

How would I translate (shift) graph $r=2\cdot\sin(\theta)$ with a vector $(^x_y)$?

What I am looking for is a change to the polar function that moves the graph along $x$ by $x$ units and $y$ by $y$ units

I realised this is confusing for people. However, I do not understand how. I have always been told that an $r=$ graph is called a polar graph, and that the transformation of shifting is called translating

Are you sure that should be an $x$ inside the trig functions? Not $\theta$? — John Doe, Mar 16 '18 at 16:26
@JohnDoe Yes, I'm just used to pressing $x$ to represent $\theta$ — Xetrov, Mar 16 '18 at 16:28
Use $x=r\cos\theta,,,y=r\sin\theta$. I suppose it could be seen as off topic since you haven't included any attempts you made on this question. — John Doe, Mar 16 '18 at 16:30
@JohnDoe , that is an example. This is not for homework. I just want to know because my calculator will not translate graphs for me. I have a CASIO CFX 9850 PLUS — Xetrov, Mar 16 '18 at 16:32
I am not too sure what you mean by "translating graphs". If you wish to plot the graph, you could use an online tool like wolfram alpha. In general, I'd just manipulate the above co-ordinate changes if you want to get rid of polar co-ordinates. — John Doe, Mar 16 '18 at 16:38

score 1 · Answer 1 · answered Mar 16 '18 at 16:34

1

$$\binom xy=\binom{r\cos\theta}{r\sin\theta}=\binom{f(\theta)\cos\theta}{f(\theta)\sin\theta}$$

answered Mar 16 '18 at 16:34

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

I believe I just multiply it on? – Xetrov Mar 16 '18 at 16:37
For $x$, you multiply $r$ by $\cos$ and for $y$, you multiply by $\sin$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 16 '18 at 16:38
What am I doing to the function? – Xetrov Mar 16 '18 at 16:39
$r=f(\theta)$ is the function. Once you know $f$ and $\theta$ you can work out $x$ and $y$. – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Mar 16 '18 at 16:39
@TheSimpliFire OP has a graph $r=f(\theta)$, and wants to translate / shift this by a vector $(x,y)$. Unfortunately, the use of $x$ and $y$ as variables led people (myself included) to think that OP wanted to change to cartesian co-ordinates. I don't think this is the case – John Doe Mar 16 '18 at 18:19

Ng Chung Tak · Answer 2 · 2018-03-16T18:25:18.943

General equation of a circle in polar coordinates:

$$r^2-2rr_0\cos (\theta-\phi)+r_0^2=R^2$$

Now

$$r=2\sin \theta \implies r(r-2\sin \theta)=0 \implies r^2-2r\cos \left( \theta-\frac{\pi}{2} \right)+1=1$$

Hence, $$ \left \{ \begin{align} r_0\cos \phi &= x \\ r_0\sin \phi &= y+1 \end{align} \right.$$

The translated circle is $$r^2-2r[x\cos \theta+(y+1)\sin \theta]+x^2+(y+1)^2=1$$

The two roots of $r(\theta)$ represent the same circle but with different senses of rotation.

How to translate $r=$ equations

2 Answers2