We are studying polar and rectangular equations in math. The last question on the homework was:
Write $r = \cos\theta + \sin\theta$ in rectangular form.
I have tried a few ways with no luck. I would love it if someone could tell me if this is possible, and explain the steps to convert it from polar to rectangular form. Thank you!
One example of what I have tried:
- $r = \cos\theta + \sin\theta$
- $r\cdot r = r\cos\theta + r\sin\theta$
Since $r^2 = x^2 + y^2$, $r\cos\theta = x$, and $r\sin\theta = y$,
- $x^2 + y^2 = x + y$
Step 3 is not equivalent to step 1 however, according to desmos here when zoomed in (a lot).
With further insight from the comments section, I have learned Desmos can become inaccurate when zoomed in very (very) closely, which is what caused me to believe my answer was incorrect.
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$\begingroup$What you tried is perfect, the two equations are almost equivalent, see e.g. the answer I gave here. In your case$$r^2=r\cos\theta+r\sin\theta\Longleftrightarrow\cases{r=\cos\theta+\sin\theta\\\text{or}\\r=0}$$
So the only caveat is what happens when $r=0$: the value of $\theta$ is undefined at the origin, so it does not seem clear whether or not the equation $r=\cos\theta+\sin\theta$ is satisfied.
However, you can look at it this way: the polar curve of equation $r={\color{grey}{(\cos\theta+\sin\theta)}}$ is the set of all points $({\color{grey}{(\cos\theta+\sin\theta)}}\cos\theta,{\color{grey}{(\cos\theta+\sin\theta)}}\sin\theta)$ when $\theta$ runs over $[0,2\pi]$, and this includes the point $\theta=\frac{3\pi}{4}$ where... $r=0$.
From this point of view, one can argue that the equation $r=\cos\theta+\sin\theta$ is satisfied when $r=0$ as well.
$\endgroup$ $\begingroup$You have $$ x^2+y^2 = x +y $$ which is $$ (x-1/2)^2+(y-1/2)^2=1/2 $$
The graph is a circle with center at $( 1/2, 1/2)$ and radius $ \sqrt 2 /2$.
The graph of $ r=\cos \theta +\sin \theta $ is the same circle and multiplying by $r$ does not add any new point to the graph.
Note that $r=0$ is a point of $r= \cos \theta +\sin \theta $ at $\theta = 3\pi/4$
$\endgroup$ $\begingroup$$$r= \cos \theta +\sin \theta $$
Just mutiply both sides by $r$
$$r^2=r \cos \theta + r \sin \theta $$
and recognize the terms
$$ x^2+y^2 = x +y $$
which is a cicle in its standard form going through the center:
$$ (x-\frac12)^2+(y-\frac12)^2=\frac12= \left(\frac{1}{\sqrt 2 } \right)^2 $$
which is same as the first equation with same graphs.
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