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Essential Calculus: Early Transcendentals
Found in: Page 522
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

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 by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

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:\((x,y) = (\sqrt 2 , - \sqrt 2 )\)

Most popular questions for Math Textbooks

Sketch the curve with the given polar equation by first sketching the graph\({\rm{r}}\) as a function of\({\rm{\theta }}\) Cartesian coordinates.

\({\rm{r = cos5\theta }}\)

Sketch the curve and find the area that it encloses.

\({\rm{r = 4 + 3sin\theta }}\)

Write a polar equation of a conic with the focus at the origin and the given data.

Parabola, vertex \({\rm{(4,3}}\pi {\rm{/2)}}\)

The Cartesian coordinates of a point are given.

  1. Find polar coordinates \({\rm{(r,\theta )}}\)of the point, where\({\rm{r > 0}}\) and\({\rm{0}} \le {\rm{\theta < 2\pi }}\).
  2. Find polar coordinates \({\rm{(r,\theta )}}\)of the point, where \(r{\rm{ < }}0\)and\({\rm{0}} \le {\rm{\theta < 2\pi }}\).

(a) \(\left( {{\rm{2, - 2}}} \right)\) (b) \(\left( {{\rm{ - 1,}}\sqrt {\rm{3}} } \right)\)

Find the exact length of the polar curve.

\({\rm{r}} = {{\rm{\theta }}^{\rm{2}}}{\rm{,0}} \le {\rm{\theta }} \le {\rm{2\pi }}\)
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