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forward as the reverse process. We could solve r2 = x2 + y 2 for r to get r = ± x2 + y 2
and solving tan(θ) = x requires the arctangent function to get θ = arctan x + πk for
integers k . Neither of these expressions for r and θ are especially user-friendly, so we opt for
a second strategy – rearrange the given polar equation so that the expressions r2 = x2 + y 2 ,
r cos(θ) = x, r sin(θ) = y and/or tan(θ) = x present themselves.
(a) Starting with r = −3, we can square both sides to get r2 = (−3)2 or r2 = 9. We may now
substitute r2 = x2 + y 2 to get the equation x2 + y 2 = 9. As we have seen,10 squaring an
equation does not, in general, produce an equivalent equation. The concern here is that
the equation r2 = 9 might be satisﬁed by more points than r = −3. On the surface, this
appears to be the case since r2 = 9 is equivalent to r = ±3, not just r = −3. However,
any point with polar coordinates (3, θ) can be represented as (−3, θ + π ), which means
any point (r, θ) whose polar coordinates satisfy the relation r =...
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