## University Calculus: Early Transcendentals (3rd Edition)

$r \cos (\theta -\dfrac{\pi}{4})=3$
Conversion of polar coordinates and Cartesian coordinates are as follows: a) $r^2=x^2+y^2 \implies r=\sqrt {x^2+y^2}$ b) $\tan \theta =\dfrac{y}{x} \implies \theta =\tan^{-1} (\dfrac{y}{x})$ c) $x=r \cos \theta$ d) $y=r \sin \theta$ We are given that $\sqrt 2 x+\sqrt y=6$ This implies that $\cos (\dfrac{\pi}{4}) (r \cos \theta)+\sin (\dfrac{\pi}{4}) (r \sin \theta)=3$ or, $r[\cos (\dfrac{\pi}{4}) (\cos \theta)+\sin (\dfrac{\pi}{4}) (\sin \theta)]=3$ Thus, $r \cos (\theta -\dfrac{\pi}{4})=3$