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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

Identify the curve by finding a Cartesian equation for the curve.

\({\rm{\theta = }}{{\rm{\pi }} \mathord{\left/

{\vphantom {{\rm{\pi }} 3}} \right.

\kern-\nulldelimiterspace} 3}\)

The curve is\({\rm{y = }}\sqrt {\rm{3}} {\rm{x}}\).

See the step by step solution

Step by Step Solution

Step 1: Find a Cartesian equation for the curve and use it to identify it.

Given: \({\rm{\theta = }}\frac{{\rm{\pi }}}{{\rm{3}}}\)

Take the tan from both sides and combine them.

\({\rm{tan\theta = tan}}\frac{{\rm{\pi }}}{{\rm{3}}}\)

Known value, \({\rm{tan\theta = }}\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}\)

\(\frac{{{\rm{sin\theta }}}}{{{\rm{cos\theta }}}}{\rm{ = }}\sqrt {\rm{3}} \)

Both the numerator and denominator should be multiplied by\({\rm{r}}\).

\(\frac{{{\rm{rsin\theta }}}}{{{\rm{rcos\theta }}}}{\rm{ = }}\sqrt {\rm{3}} \)

Keep in mind that: \({\rm{rcos\theta = x\& rsin\theta = y}}\)

\(\begin{aligned}{l}\frac{{\rm{y}}}{{\rm{x}}}{\rm{ = }}\sqrt {\rm{3}} \\{\rm{y = }}\sqrt {\rm{3}} {\rm{x}}\end{aligned}\)

Step 2: Result.

Therefore, the curve is\({\rm{y = }}\sqrt {\rm{3}} {\rm{x}}\).

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