Suggested languages for you:

Americas

Europe

Q3E

Expert-verified
Found in: Page 522

### Essential Calculus: Early Transcendentals

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

# Plot the point whose polar coordinates are given. Then find the Cartesian coordinates of the point.\begin{aligned}{l}(a)(1,\pi )\\(b)(2, - 2\pi /3)\\(c)( - 2,3\pi /4)\end{aligned}

1. Rectangular coordinate:$$( - 1,0)$$
2. Rectangular coordinate:$$( - 1, - \sqrt 3 )$$
3. Rectangular coordinate:$$(x,y) = (\sqrt 2 , - \sqrt 2 )$$
See the step by step solution

## Step 1: Convert polar coordinate to rectangular coordinate

a. To convert polar coordinate to rectangular coordinate use following relation

\begin{aligned}{l}x = r\cos \theta \\y = r\sin \theta \\x = 1 \cdot \cos (\pi )\;\;\;{\rm{ Here given }}(r,\theta ) = (1,\pi )\\x = 1 \cdot - 1 = - 1\\y = r\sin \theta = 1 \cdot \sin \pi = 1 \cdot 0 = 0\end{aligned}

Rectangular coordinates corresponding to polar coordinate$$(1,\pi )$$is $$( - 1,0)$$

## Step 2: Convert polar coordinate to rectangular coordinate

b. To convert polar coordinate to rectangular coordinate use following relation

\begin{aligned}{l}x &= r\cos \theta \\y &= r\sin \theta \end{aligned}

Here given $$\left\langle {r,\theta } \right\rangle = (2, - \frac{{2\pi }}{3})$$

\begin{aligned}{l}x &= 2\cos \left( { - \frac{{2\pi }}{3}} \right)\\x &= 2 \cdot \cos \frac{{2\pi }}{3}\;\;\;\cos {\rm{ function is even function }}f( - x) = f(x)\\x &= 2 \cdot \frac{{ - 1}}{2} = - 1\\y &= 2\sin \left( { - \frac{{2\pi }}{3}} \right)\\y &= - 2 \cdot \sin \frac{{2\pi }}{3}\\{\rm{sin function is odd function }}f( - x) = - f(x)\\y = - 2 \cdot \frac{{\sqrt 3 }}{2} = - \sqrt 3 \end{aligned}

Rectangular coordinate corresponds to point$$\left( {2,\frac{{ - 2\pi }}{3}} \right){\rm{ is }}( - 1, - \sqrt 3 )$$

## Step 3: Convert polar coordinate to rectangular coordinate

(c)

\begin{aligned}{l}x &= r\cos \theta = - 2\cos \left( {\frac{{3\pi }}{4}} \right) = 2\cos \left( {\pi - \frac{{3\pi }}{4}} \right) = 2\cos \left( {\frac{\pi }{4}} \right) = 2 \times \frac{1}{{\sqrt 2 }} = \sqrt 2 \\y &= r\sin \theta = - 2\sin \left( {\frac{{3\pi }}{4}} \right) = - 2\sin \left( {\pi - \frac{{3\pi }}{4}} \right) = - 2\sin \left( {\frac{\pi }{4}} \right) = - 2 \times \frac{1}{{\sqrt 2 }} = - \sqrt 2 \end{aligned}

Remember that:

\begin{aligned}{l}\sin (\pi - \theta ) = \sin \theta \\\cos (\pi - \theta ) = - \cos \theta \end{aligned}

The point is plotted on the graph below:

Rectangular coordinate:$$(x,y) = (\sqrt 2 , - \sqrt 2 )$$