Find the point (r, $\theta$) in polar coordinates given the fact that when converted in Cartesian coordinates, the point is $(7,5)$. Use that to find the point $\left( 2r, \theta + \frac{\pi}{2} \right)$ in rectangular coordinates.

I found that $r = \sqrt{7^2+5^2} = \sqrt{74}$

However, I'm having trouble with $\theta$. I know that I would have to find $\tan \theta = \frac{5}{7}$, and then find $\theta$ from that, but I'm not sure how to find it so that it is in neat terms like $\frac{\pi}{6}$, so that it would be easier to add it to $\frac{\pi}{2}$.

Any hints?


Hint: The point $(7,5)$ is in the first quadrant of the cartesian plane. This way, you know that $0 < \theta < \pi/2$. You can also use the relations: $$\cos \theta = \frac{x}{r} = \frac{7}{\sqrt{74}}, \quad \sin \theta = \frac{y}{r} = \frac{5}{\sqrt{74}}.$$ The angle $\theta$ is uniquely determined by these relations. But I don't think we'll be able to get an exact value for it by hand.

Although adding $\pi/2$ to the angle is just a rotation, and doubling $r$ is just doubling each component... readily we have that the point $(2r, \theta + \pi/2)$ is $(-10, 14)$.

  • $\begingroup$ How do you think I should find $(2r, \theta + \frac{\pi}{2})$? $\endgroup$ – Mathy Person Jan 21 '15 at 17:43
  • 1
    $\begingroup$ I added a little spoiler in my answer. Didn't used these relations with sine and cosine.. just interpretating. $\endgroup$ – Ivo Terek Jan 21 '15 at 17:45

Given that:-

$P = (7, 5)$ in rectangular coordinates is equivalent to $P = [r, \theta]$ in polar coordinates.

Find $P’ = [r, \theta + \frac {\pi}{2}]$ in polar coordinates first.

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P’ is just a simple rotation of P through 90 degrees (in anti-clockwise direction). Its rectangular coordinates is then $P' = (-5, 7)$.

Therefore, $P’’ = [2r, \theta + \frac {\pi}{2}] = (-10, 14)$


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