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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)( - \sqrt 2 ,5\pi /4)\\(b)(1,5\pi /2)\\(c)(2, - 7\pi /6)\end{aligned}

(a)$$(x,y) = (1,1)$$

(b)$$(x,y) = (0,1)$$

(c)$$(x,y) = ( - \sqrt 3 ,1)$$

See the step by step solution

## Step 1: Convert polar coordinate to rectangular coordinate

\begin{aligned}{l}x &= r\cos \theta = - \sqrt 2 \cos \left[ {\frac{{5\pi }}{4}} \right] = \sqrt 2 \cos \left[ {\frac{{5\pi }}{4} - \pi } \right] = \sqrt 2 \cos \left[ {\frac{\pi }{4}} \right] = \sqrt 2 \times \frac{1}{{\sqrt 2 }} = 1\\y = r\sin \theta &= - \sqrt 2 \sin \left[ {\frac{{5\pi }}{4}} \right] = \sqrt 2 \sin \left[ {\frac{{5\pi }}{4} - \pi } \right] = \sqrt 2 \sin \left[ {\frac{\pi }{4}} \right] = \sqrt 2 \times \frac{1}{{\sqrt 2 }} = 1\end{aligned}

Remember that:

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

The point$$(1,1)$$ is plotted on the graph below:

## Step 2: Convert polar coordinate to rectangular coordinate

\begin{aligned}{l}x &= r\cos \theta = 2\cos \left[ { - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ { - \frac{\pi }{6}} \right] = - 2\cos \left[ {\frac{\pi }{6}} \right] = - 2 \times \frac{{\sqrt 3 }}{2} = - \sqrt 3 \\y &= r\sin \theta = 2\sin \left[ { - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ { - \frac{\pi }{6}} \right] = 2\sin \left[ {\frac{\pi }{6}} \right] = 2 \times \frac{1}{2} = 1\end{aligned}

Remember that:

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

And that:

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

The point$$( - \sqrt 3 ,1)$$ is plotted on the graph below:

## Step 3: Convert polar coordinate to rectangular coordinate

\begin{aligned}{l}x &= r\cos \theta = 2\cos \left[ { - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\cos \left[ { - \frac{\pi }{6}} \right] = - 2\cos \left[ {\frac{\pi }{6}} \right] = - 2 \times \frac{{\sqrt 3 }}{2} = - \sqrt 3 \\y &= r\sin \theta = 2\sin \left[ { - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ {\pi - \frac{{7\pi }}{6}} \right] = - 2\sin \left[ { - \frac{\pi }{6}} \right] = 2\sin \left[ {\frac{\pi }{6}} \right] = 2 \times \frac{1}{2} = 1\end{aligned}Remember that:

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

And that:

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

The point$$( - \sqrt 3 ,1)$$is plotted on the graph below: