Stitz-Zeager_College_Algebra_e-book

Hence argz 2k k is an integer of these values

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ight forward as the reverse process. We could solve r2 = x2 + y 2 for r to get r = ± x2 + y 2 y y 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 , y 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 =...
View Full Document

Ask a homework question - tutors are online