{
"query": {
"display": "cartesian to polar $$\\left(-6,\\:0\\right)$$",
"symbolab_question": "POLAR#polar (-6,0)"
},
"solution": {
"level": "PERFORMED",
"subject": "Pre Calculus",
"topic": "Polar Coordinates",
"subTopic": "Polar",
"default": "(6,0)"
},
"steps": {
"type": "interim",
"title": "Convert $$\\left(-6,\\:0\\right)\\:$$to polar coordinates:$${\\quad}\\left(6,\\:0\\right)$$",
"steps": [
{
"type": "definition",
"title": "Definition",
"text": "To convert Cartesian coordinates $$\\left(x,\\:y\\right)\\:$$to Polar coordinates $$\\left(r,\\:\\theta\\right)\\:$$apply:<br/>$$r=\\sqrt{x^2+y^2}\\quad\\theta=\\arctan\\left(\\frac{y}{x}\\right)$$",
"secondary": [
"$$x=-6$$",
"$$y=0$$"
]
},
{
"type": "step",
"primary": "$$r=\\sqrt{x^2+y^2}$$",
"result": "r=\\sqrt{\\left(-6\\right)^{2}+0^{2}}"
},
{
"type": "interim",
"title": "$$\\sqrt{\\left(-6\\right)^{2}+0^{2}}=6$$",
"input": "\\sqrt{\\left(-6\\right)^{2}+0^{2}}",
"steps": [
{
"type": "step",
"primary": "Apply rule $$0^{a}=0$$",
"secondary": [
"$$0^{2}=0$$"
],
"result": "=\\sqrt{\\left(-6\\right)^{2}+0}"
},
{
"type": "step",
"primary": "Apply exponent rule: $$\\left(-a\\right)^{n}=a^{n},\\:$$if $$n$$ is even",
"secondary": [
"$$\\left(-6\\right)^{2}=6^{2}$$"
],
"result": "=\\sqrt{6^{2}+0}"
},
{
"type": "step",
"primary": "$$6^{2}+0=6^{2}$$",
"result": "=\\sqrt{6^{2}}"
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt[n]{a^n}=a,\\:\\quad$$ assuming $$a\\ge0$$",
"result": "=6",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7biV9ml2SRpXxmTVDdibQ9AGB+TmibiLKcE/g/B3ZUFJ1g99dC9fj9sg0EHzBIRDRjBGJGUdgcV1QE8NtKLFEFUxjDTDYT3JYIM83Fw7N7a8dGquUFU5QxZRrJogEQx6OJLd1ohke2Wgml78++2zI0g=="
}
},
{
"type": "step",
"result": "r=6"
},
{
"type": "step",
"primary": "$$\\theta=\\arctan\\left(\\frac{y}{x}\\right)$$",
"result": "θ=\\arctan\\left(\\frac{0}{-6}\\right)"
},
{
"type": "interim",
"title": "Adjust $$\\theta$$ based on the quadrant of the point $$\\left(-6,\\:0\\right)$$",
"result": "θ=\\arctan\\left(\\frac{0}{-6}\\right)",
"steps": [
{
"type": "definition",
"title": "Point location",
"text": "If x>0, y>0, then the point is in quadrant I<br/>If x<0, y>0, then the point is in quadrant II<br/>If x<0, and y<0, then the point is in quadrant III<br/>If x>0, and y<0, then the point is in quadrant IV",
"secondary": [
"$$\\left(-6,\\:0\\right)\\:$$is in quadrant "
]
},
{
"type": "step",
"primary": "If in quadrant II or III, add $$\\pi$$ to $$\\theta$$<br/>If in quadrant IV, add $$2\\pi$$ to $$\\theta$$",
"result": "θ=\\arctan\\left(\\frac{0}{-6}\\right)"
}
],
"meta": {
"interimType": "Cartesian To Polar Adjust Theta 1Eq"
}
},
{
"type": "interim",
"title": "$$\\arctan\\left(\\frac{0}{-6}\\right)=0$$",
"input": "\\arctan\\left(\\frac{0}{-6}\\right)",
"steps": [
{
"type": "step",
"primary": "Apply rule $$\\frac{0}{a}=0,\\:a\\ne\\:0$$",
"result": "=\\arctan\\left(0\\right)"
},
{
"type": "step",
"primary": "Use the following trivial identity:$${\\quad}\\arctan\\left(0\\right)=0$$",
"secondary": [
"$$\\begin{array}{|c|c|c|}\\hline x&\\arctan(x)&\\arctan(x)\\\\\\hline 0&0&0^{\\circ}\\\\\\hline \\frac{\\sqrt{3}}{3}&\\frac{\\pi}{6}&30^{\\circ}\\\\\\hline 1&\\frac{\\pi}{4}&45^{\\circ}\\\\\\hline \\sqrt{3}&\\frac{\\pi}{3}&60^{\\circ}\\\\\\hline \\end{array}$$"
],
"result": "=0"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iGvFWxodO1NUsa8/NPZQsyiqRYqjsiql3XsGcKuSD/sDnzlbPZjyKgy1eUCFsLd5uA0vbJw0XRhSBCaIjQqkFh5cNKRCR7MHGsj3p+K2bBCtZVbroCr9IztZxhM3IEuvsIjaxJ4DvjTb2fbKjbvtlQ=="
}
},
{
"type": "step",
"result": "θ=0"
},
{
"type": "step",
"primary": "The polar coordinates of $$\\left(-6,\\:0\\right)$$",
"result": "\\left(6,\\:0\\right)"
}
]
}
}
Solution
cartesian to polar
Solution
Solution steps
Adjust based on the quadrant of the point
The polar coordinates of