Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 12 - Parametric Equations, Polar Coordinates, and Conic Sections - 12.3 Polar Coordinates - Exercises - Page 619: 41


$(x^2+y^2)^2= x^3 - 3xy^2$

Work Step by Step

We have \begin{align*} \cos 3\theta &=\cos(\theta+2\theta)\\ &=\cos\theta \cos(2\theta)-\sin\theta \sin(2\theta)\\ &=\cos\theta (\cos^2\theta-\sin^2\theta)-2\sin^2\theta \cos\theta\\ &= \cos^3\theta-3\sin^2\theta \cos\theta. \end{align*} Now, we get the following \begin{align*} r &=\cos 3\theta\\ &= \cos^3\theta-3\sin^2\theta \cos\theta\\ &=\frac{x^3}{r^3}-\frac{3xy^2}{r^3}. \end{align*} That is $$r^4= x^3 - 3xy^2 \Longrightarrow (x^2+y^2)^2= x^3 - 3xy^2.$$
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