Calculus (3rd Edition)

(a) $x^2+y^2=4$ matches (iv) $r=2$ (b) $x^2+(y-1)^2=1$ matches (iii) $r=2\sin \theta$ (c) $x^2-y^2=4$ matches (i) ${r^2}\left( {1 - 2{{\sin }^2}\theta } \right) = 4$ (d) $x+y=4$ matches (ii) $r\left( {\cos \theta + \sin \theta } \right) = 4$
(a) We have $x = r\cos \theta$ and $y = r\sin \theta$. So, $x^2+y^2=4$, $\begin{array}{l} {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta = 4\\ {r^2} = 4 \end{array}$ The answer is (iv) $r=2$. (b) We have $x = r\cos \theta$ and $y = r\sin \theta$. So, $x^2+(y-1)^2=1$, $\begin{array}{l} {r^2}{\cos ^2}\theta + {\left( {r\sin \theta - 1} \right)^2} = 1\\ {r^2}{\cos ^2}\theta + {r^2}{\sin ^2}\theta - 2r\sin \theta + 1 = 1\\ {r^2} - 2r\sin \theta = 0\\ r\left( {r - 2\sin \theta } \right) = 0 \end{array}$ The answer is (iii) $r=2\sin \theta$. (c) We have $x = r\cos \theta$ and $y = r\sin \theta$. So, $x^2-y^2=4$, $\begin{array}{l} {r^2}{\cos ^2}\theta - {r^2}{\sin ^2}\theta = 4\\ {r^2}\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) = 4 \end{array}$ Since ${\cos ^2}\theta - {\sin ^2}\theta = 1 - 2{\sin ^2}\theta$, the answer is (i) ${r^2}\left( {1 - 2{{\sin }^2}\theta } \right) = 4$. (d) We have $x = r\cos \theta$ and $y = r\sin \theta$. So, $x+y=4$, $r \cos \theta + r \sin \theta=4$ The answer is (ii) $r\left( {\cos \theta + \sin \theta } \right) = 4$.