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Easy polar coordinates question

Hello, this is a really easy question but I'm not sure if I'm working things out correctly here.

C:r=sin⁡θ+3cos⁡θ, 0≤θ≤π2\displaystyle C: r =\sin\theta + \sqrt3\cos\theta, \ 0\leq \theta \leq \frac{\pi}{2}
At the point P on C, the tangent to C is perpendicular to the initial line. Find the polar coordinates of P.

Now, what I've done is that

If the tangent is perpendicular to θ=0⇔dxdθ=0\displaystyle \text{If the tangent is perpendicular to} \ \theta = 0 \Leftrightarrow \frac{\mathrm{d} x}{\mathrm{d} \theta} = 0

x=rcos⁡θ=(sin⁡θ+3cos⁡θ)cos⁡θ\displaystyle x = r\cos\theta = \left ( \sin\theta + \sqrt3\cos\theta \right )\cos\theta

dxdθ=cos⁡2θ−sin⁡2θ−23sin⁡θcos⁡θ=cos⁡2θ−3sin⁡2θ\displaystyle \frac{\mathrm{d} x}{\mathrm{d} \theta} = \cos^{2}\theta - \sin^{2}\theta -2\sqrt3\sin\theta\cos\theta = \cos2\theta - \sqrt3\sin2\theta

Now, after some simple trig after letting dxdθ=0\frac{\mathrm{d} x}{\mathrm{d} \theta} = 0, I've found that θ=π12\theta = \frac{\pi}{12} and that the required polar coordinates is P(6+22,π12) \mathrm{P}\left ( \frac{\sqrt{6}+\sqrt{2}}{2}, \frac{\pi}{12} \right ). Can anyone confirm since I don't have the answers to this? :colondollar:
(edited 8 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by aymanzayedmannan
Can anyone confirm since I don't have the answers to this? :colondollar:


Yep, that's fine. :-)
Reply 2
Avatar for Ayman!
Ayman!
OP
15
Original post by Zacken
Yep, that's fine. :-)


merci beaucoup, mon ami :biggrin:
Reply 3
Avatar for TeeEm
TeeEm
19
am I too late?

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