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**Unformatted text preview: **ement. −r ? = (−r )2 + (−r cos(θ + π ))
= (r )2 − r cos(θ + π ) ? Since r = −(r )2 − r cos(θ ) ? − −(r )2 − r cos(θ ) Since cos(θ + π ) = − cos(θ ) (r )2 + r cos(θ ) = (r )2 − r (− cos(θ ))
(r )2 + r cos(θ ) = (r )2 + r cos(θ ) Since both sides worked out to be equal, (−r , θ + π ) satisﬁes r = r2 + r cos(θ) which
2
means that any point (r, θ) which satisﬁes r2 = r2 + r cos(θ) has a representation
which satisﬁes r = r2 + r cos(θ), and we are done.
In practice, much of the pedantic veriﬁcation of the equivalence of equations in Example 11.4.3 is
left unsaid. Indeed, in most textbooks, squaring equations like r = −3 to arrive at r2 = 9 happens
without a second thought. Your instructor will ultimately decide how much, if any, justiﬁcation is
warranted. If you take anything away from Example 11.4.3, it should be that relatively nice things
in rectangular coordinates, such as y = x2 , can turn ugly in polar coordinates, and vice-versa. In the
next section, we devote our attention to graphing equ...

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