What happens if we keep $x$ and $y$ constant and change $\theta$ and $\rho$ in equation

$$x \cdot cos (\theta) + y \cdot sin (\theta) = \rho$$


I know the plot would look like a sine wave.

But, how would be the effect of $x$ and $y$ on the graph? What would be the relationship between (x, y) and the amplitude and phase of the wave?

I have plotted the equation using Wolfram Alpha. The plot doesn't make much sense.

  • $\begingroup$ You'll get a straight line $ax+by=c$ with $a=\cos\theta,b=\sin\theta, c=\rho$. $\endgroup$ – Mercy King May 27 '17 at 5:43
  • $\begingroup$ @anonymous The WA link you added shows a plot in $\theta, \rho$ with $x=y=5$ constant. $\endgroup$ – dxiv May 27 '17 at 5:45
  • $\begingroup$ @anonymous Sorry, demonstrate what? It's not clear what the question is that you mean to ask. $\endgroup$ – dxiv May 27 '17 at 5:48
  • $\begingroup$ So again, what is the question? The $\rho$ vs. $\theta$ plot for constant $x,y$ is a sine because of this. The $y$ vs. $x$ plot for constant $\theta,\rho$ is a straight line, as pointed out in the very first comment already. $\endgroup$ – dxiv May 27 '17 at 5:53
  • $\begingroup$ that was what I was looking for Guess that's good, though I don't know what that was. You may want to clarify both the title and the body of your question, in particular the keep θ and ρ constant and change x and y parts. $\endgroup$ – dxiv May 27 '17 at 6:01

$\rho=x\cos(\theta)+y\sin(\theta)=\sqrt{x^2+y^2}\bigg(\frac x{\sqrt{x^2+y^2}}\cos(\theta)+\frac{y}{\sqrt{x^2+y^2}}\sin(\theta)\bigg)$

Now the amplitude will be $A=\sqrt{x^2+y^2}$ and the phase $\phi$ verifies $\begin{cases} \sin(\phi)=\frac x{\sqrt{x^2+y^2}} \\ \cos(\phi)=\frac y{\sqrt{x^2+y^2}} \end{cases}$

By the sinus addition formula this becomes $\rho=A\sin(\theta+\phi)$.


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