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Found in: Page 528

### Essential Calculus: Early Transcendentals

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

# 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 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$${\rm{(3/2,5\pi /6)}}$$.