My Math Forum Polar to Cartesian Coordinate Conversion Help
 User Name Remember Me? Password

 Algebra Pre-Algebra and Basic Algebra Math Forum

 April 13th, 2012, 09:51 AM #1 Member   Joined: Feb 2012 Posts: 78 Thanks: 0 Polar to Cartesian Coordinate Conversion Help Convert the following point from a polar to cartesian coordinate. $(-1,-\frac{3\pi }{4})$ $x= -1\cos(-\frac{3\pi }{4})$= $-1(\frac{1}{\sqrt{2}})$ $x=-\frac{1}{\sqrt{2}}$ $y= -1\sin(-\frac{3\pi }{4})$= $-1(-\frac{1}{\sqrt{2}})$= $y=\frac{1}{\sqrt{2}}$ My answer: $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ However this is wrong. Where did I screw up?
April 13th, 2012, 10:32 AM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Polar to Cartesian Coordinate Conversion Help

Hello, swm06!

Quote:
 $\text{Convert to cartesian coordinates: }\:\left(-1,\:-\frac{3\pi }{4}\right)$

$\text{W\!e have: }\:r \,=\,-1,\;\theta \,=\,-\frac{3\pi}{4}$

$x \;=\;r\cos\theta \;=\;(-1)\,\!\cos\left(-\frac{3\pi}{4}\right) \;=\;(-1)\left(-\frac{1}{\sqrt{2}}\right) \;=\;\frac{1}{\sqrt{2}}$

$y \;=\;r\sin\theta \;=\;(-1)\,\!\sin\left(-\frac{3\pi}{4}\right) \;=\;(-1)\left(-\frac{1}{\sqrt{2}}\right) \;=\;\frac{1}{\sqrt{2}}$

$\text{Cartesian coordinates: }\:\left(\frac{1}{\sqrt{2}},\;\frac{1}{\sqrt{2}}\r ight)$

April 13th, 2012, 10:34 AM   #3
Member

Joined: Nov 2009

Posts: 67
Thanks: 0

Re: Polar to Cartesian Coordinate Conversion Help

Quote:
 Originally Posted by swm06 Convert the following point from a polar to cartesian coordinate. $(-1,-\frac{3\pi }{4})$ $x= -1\cos(-\frac{3\pi }{4})$= $-1(\frac{1}{\sqrt{2}})$ $x=-\frac{1}{\sqrt{2}}$ $y= -1\sin(-\frac{3\pi }{4})$= $-1(-\frac{1}{\sqrt{2}})$= $y=\frac{1}{\sqrt{2}}$ My answer: $(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$ However this is wrong. Where did I screw up?
Hi swm06,

You missed a sign when you found your x-coordinate. $\cos -\frac{3 \pi}{4}$ is negative. When you multiply by -1, it becomes positive.

$\cos -\frac{3 \pi}{4}=-\frac{1}{\sqrt{2}}$

That will make $x=\frac{1}{\sqrt{2}}$

You should rationalize your denominators in the end to arrive at $$$\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}$$$

[color=#FF0000]A couple of minutes too slow for Soroban![/color]

 Tags cartesian, conversion, coordinate, polar

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post fotoni Algebra 4 July 30th, 2013 05:42 PM tancm Algebra 0 November 1st, 2012 05:07 AM emath Calculus 2 July 24th, 2012 08:46 AM DoodleGal Linear Algebra 2 May 18th, 2010 10:27 AM timissar Algebra 4 April 29th, 2009 07:51 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top