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Essential Calculus: Early Transcendentals
Found in: Page 528
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

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Short Answer

Find all points of intersection of the given curves.

\({\rm{r = 1 + sin\theta ,}}\;\;\;{\rm{r = 3sin\theta }}\)

The result of the point of intersection of the curve is pole\({\rm{(3/2,\pi /6)}}\)and\({\rm{(3/2,5\pi /6)}}\).

See the step by step solution

Step by Step Solution

Step 1: Curves overlap.

Solve\({\rm{\theta }}\) for by making the two equations equal.

\({\rm{1 + sin\theta = 3sin\theta }}\;\;\;\)

Since \({\rm{sin\theta }}\)act as positive and \({\rm{\theta }}\)act as QII or QI

\(\begin{aligned}{c}{\rm{2sin\theta = 1}}\\{\rm{sin\theta = 1/2}}\\{\rm{\theta = \pi /6,5\pi /6}}\end{aligned}\)Now plug each value into one of the two equations to get the\({\rm{r}}\)value, and hence the polar coordinates, where the two curves overlap.

\(\begin{aligned}{c}{\rm{\theta = \pi /6}} \Rightarrow {\rm{r = 1 + sin(\pi /6)}}\\{\rm{ = 1 + 1/2}}\\{\rm{ = 3/2}}\\{\rm{r = 3sin(\pi /6)}}\\{\rm{ = 3(1/2)}}\\{\rm{ = 3/2}}\;\;\;\end{aligned}\)

\(\begin{aligned}{c}{\rm{\theta = 5\pi /6}} \Rightarrow {\rm{r = 1 + sin(5\pi /6)}}\\{\rm{ = 1 + 1/2}}\\{\rm{ = 3/2}}\\{\rm{r = 3sin(5\pi /6)}}\\{\rm{ = 3(1/2)}}\\{\rm{ = 3/2}}\end{aligned}\)

The intersections points are \({\rm{(3/2,\pi /6)}}\) and\({\rm{(3/2,5\pi /6)}}\).

Step 2: Point of the junction.

When the pole\({\rm{r = 0}}\),

\(\begin{aligned}{c}{\rm{r = 1 + sin\theta = 0}}\\ \Rightarrow {\rm{sin\theta = - 1}}\\ \Rightarrow {\rm{\theta = 3\pi /2}}\\{\rm{r = 3sin\theta = 0}}\\ \Rightarrow {\rm{sin\theta = 0}}\\ \Rightarrow {\rm{\theta = 0}}\end{aligned}\)

Both curves pass through the pole although at different \({\rm{\theta }}\)values, indicating that the pole is a point of junction.

Therefore, the point of intersection of the curve is pole\({\rm{(3/2,\pi /6)}}\)and\({\rm{(3/2,5\pi /6)}}\).

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