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Question Number 10129 by Tawakalitu ayo mi last updated on 26/Jan/17
Write in polar form . (1) 3x − y + 5 = 0 (2) x^2 = (√(x^2 + y^2 ))
$$\mathrm{Write}\:\mathrm{in}\:\mathrm{polar}\:\mathrm{form}\:. \\ $$$$\left(\mathrm{1}\right)\:\:\mathrm{3x}\:−\:\mathrm{y}\:\:+\:\mathrm{5}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\mathrm{x}^{\mathrm{2}} \:=\:\sqrt{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} } \\ $$
Answered by icyfalcon999 last updated on 26/Jan/17
(1): 3(r cos θ)−(r sin θ)+5=0 3r cos θ−r sin θ=−5 r(3cos θ−sin θ)=−5 ((r(3 cos θ−sin θ))/(3 cos θ−sin θ))=((−5)/(3 cos θ−sin θ)) r=(5/(sin θ−3cos θ)) (2): x^2 =(√(x^2 +y^2 )) (rcos θ)^2 =r r^2 cos^2 θ=r ((r^2 cos^2 θ )/r)=(r/r) r cos^2 θ =1 ((r cos^2 θ )/(cos^2 θ ))=(1/(cos^2 θ)) r=(1/(cos^2 θ )) r=sec^2 θ
$$\left(\mathrm{1}\right): \\ $$$$\mathrm{3}\left({r}\:\mathrm{cos}\:\theta\right)−\left({r}\:\mathrm{sin}\:\theta\right)+\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{3}{r}\:\mathrm{cos}\:\theta−{r}\:\mathrm{sin}\:\theta=−\mathrm{5} \\ $$$${r}\left(\mathrm{3cos}\:\theta−\mathrm{sin}\:\theta\right)=−\mathrm{5} \\ $$$$\frac{{r}\left(\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right)}{\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta}=\frac{−\mathrm{5}}{\mathrm{3}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta} \\ $$$${r}=\frac{\mathrm{5}}{\mathrm{sin}\:\theta−\mathrm{3cos}\:\theta} \\ $$$$\left(\mathrm{2}\right): \\ $$$${x}^{\mathrm{2}} =\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} } \\ $$$$\left({r}\mathrm{cos}\:\theta\right)^{\mathrm{2}} ={r} \\ $$$${r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \:\theta={r} \\ $$$$\frac{{r}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta\:}{{r}}=\frac{{r}}{{r}} \\ $$$${r}\:\mathrm{cos}^{\mathrm{2}} \theta\:=\mathrm{1} \\ $$$$\frac{{r}\:\mathrm{cos}^{\mathrm{2}} \theta\:}{\mathrm{cos}^{\mathrm{2}} \theta\:}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta} \\ $$$${r}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta\:} \\ $$$${r}=\mathrm{sec}^{\mathrm{2}} \theta\: \\ $$
Commented by Tawakalitu ayo mi last updated on 26/Jan/17
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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