University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

Chapter 3 - Practice Exercises - Page 202: 36

Answer

$\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}\frac{{(\cos\theta-1-\sin\theta)}}{({1-\cos\theta})^2}$

Work Step by Step

given $r=(\frac{1+\sin\theta}{1-\cos\theta})^2$ on differentiating both sides: $\frac{dr}{d\theta}=\frac{d((\frac{1+\sin\theta}{1-\cos\theta})^2)}{d\theta}$ $\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}(\frac{({1-\cos\theta})\frac{d({1+\sin\theta})}{d\theta}-{(1+\sin\theta)\frac{d({1-\cos\theta})}{d\theta}}}{({1-\cos\theta})^2})$ $\frac{dr}{d\theta}=2(\frac{1+\sin\theta}{1-\cos\theta})\frac{({1-\cos\theta}){\cos\theta}-{(1+\sin\theta){({\sin\theta})}})}{({1-\cos\theta})^2}$ $\frac{dr}{d\theta}=2\frac{1+\sin\theta}{1-\cos\theta}\frac{{(\cos\theta-1-\sin\theta)}}{({1-\cos\theta})^2}$

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