{
"query": {
"display": "cartesian to polar $$\\left(-6,\\:6\\right)$$",
"symbolab_question": "POLAR#polar (-6,6)"
},
"solution": {
"level": "PERFORMED",
"subject": "Pre Calculus",
"topic": "Polar Coordinates",
"subTopic": "Polar",
"default": "(6\\sqrt{2},-\\frac{π}{4}+π)"
},
"steps": {
"type": "interim",
"title": "Convert $$\\left(-6,\\:6\\right)\\:$$to polar coordinates:$${\\quad}\\left(6\\sqrt{2},\\:-\\frac{π}{4}+π\\right)$$",
"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": [
"$$x=-6$$",
"$$y=6$$"
]
},
{
"type": "step",
"primary": "$$r=\\sqrt{x^2+y^2}$$",
"result": "r=\\sqrt{\\left(-6\\right)^{2}+6^{2}}"
},
{
"type": "interim",
"title": "$$\\sqrt{\\left(-6\\right)^{2}+6^{2}}=6\\sqrt{2}$$",
"input": "\\sqrt{\\left(-6\\right)^{2}+6^{2}}",
"steps": [
{
"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}+6^{2}}"
},
{
"type": "step",
"primary": "Add similar elements: $$6^{2}+6^{2}=6^{2}\\cdot\\:2$$",
"result": "=\\sqrt{6^{2}\\cdot\\:2}"
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt[n]{ab}=\\sqrt[n]{a}\\sqrt[n]{b},\\:\\quad$$ assuming $$a\\ge0,\\:b\\ge0$$",
"result": "=\\sqrt{2}\\sqrt{6^{2}}",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt[n]{a^n}=a,\\:\\quad$$ assuming $$a\\ge0$$",
"secondary": [
"$$\\sqrt{6^{2}}=6$$"
],
"result": "=6\\sqrt{2}",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7biV9ml2SRpXxmTVDdibQ9DK0pBe/w/bFZLAte0Z4NFl1g99dC9fj9sg0EHzBIRDRsBd6x7X0aSOq/iVcWwBgQWRLd2VwIqlBNByF6663syS0FACXQ8ffgFJOVJ80R2CYi63MftDXieSXGvY3RIujCbCI2sSeA74029n2yo277ZU="
}
},
{
"type": "step",
"result": "r=6\\sqrt{2}"
},
{
"type": "step",
"primary": "$$\\theta=\\arctan\\left(\\frac{y}{x}\\right)$$",
"result": "θ=\\arctan\\left(\\frac{6}{-6}\\right)"
},
{
"type": "interim",
"title": "Adjust $$\\theta$$ based on the quadrant of the point $$\\left(-6,\\:6\\right)$$",
"result": "θ=\\arctan\\left(\\frac{6}{-6}\\right)+π",
"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": [
"$$\\left(-6,\\:6\\right)\\:$$is in quadrant II"
]
},
{
"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{6}{-6}\\right)+π"
}
],
"meta": {
"interimType": "Cartesian To Polar Adjust Theta 1Eq"
}
},
{
"type": "interim",
"title": "$$\\arctan\\left(\\frac{6}{-6}\\right)+π=-\\frac{π}{4}+π$$",
"input": "\\arctan\\left(\\frac{6}{-6}\\right)+π",
"steps": [
{
"type": "interim",
"title": "$$\\arctan\\left(\\frac{6}{-6}\\right)=-\\frac{π}{4}$$",
"input": "\\arctan\\left(\\frac{6}{-6}\\right)",
"steps": [
{
"type": "interim",
"title": "$$\\frac{6}{-6}=-1$$",
"input": "\\frac{6}{-6}",
"steps": [
{
"type": "step",
"primary": "Apply the fraction rule: $$\\frac{a}{-b}=-\\frac{a}{b}$$",
"result": "=-\\frac{6}{6}"
},
{
"type": "step",
"primary": "Apply rule $$\\frac{a}{a}=1$$",
"result": "=-1"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s74AxHCcjix5J+1qnDESTGiS061ljBSPJeENOw2efoSWuixCfF7o9L2hRx9Shm0jmvo3oe/oyhMy2+1TQhDBd2f7W8tQVfH43jhwlaUu1xaaMkt3WiGR7ZaCaXvz77bMjS"
}
},
{
"type": "step",
"result": "=\\arctan\\left(-1\\right)"
},
{
"type": "step",
"primary": "Use the following property: $$\\arctan\\left(-x\\right)=-\\arctan\\left(x\\right)$$",
"secondary": [
"$$\\arctan\\left(-1\\right)=-\\arctan\\left(1\\right)$$"
],
"result": "=-\\arctan\\left(1\\right)"
},
{
"type": "interim",
"title": "Use the following trivial identity:$${\\quad}\\arctan\\left(1\\right)=\\frac{π}{4}$$",
"input": "\\arctan\\left(1\\right)",
"steps": [
{
"type": "step",
"primary": "$$\\begin{array}{|c|c|c|}\\hline x&\\arctan(x)&\\arctan(x)\\\\\\hline 0&0&0^{\\circ}\\\\\\hline \\frac{\\sqrt{3}}{3}&\\frac{\\pi}{6}&30^{\\circ}\\\\\\hline 1&\\frac{\\pi}{4}&45^{\\circ}\\\\\\hline \\sqrt{3}&\\frac{\\pi}{3}&60^{\\circ}\\\\\\hline \\end{array}$$"
},
{
"type": "step",
"result": "=\\frac{π}{4}"
}
],
"meta": {
"interimType": "Trig Trivial Angle Value Title 0Eq"
}
},
{
"type": "step",
"result": "=-\\frac{π}{4}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iGvFWxodO1NUsa8/NPZQs4gLgJHBBSi69c6EpZVxB+0DnzlbPZjyKgy1eUCFsLd5C0Mhm8BpOdg5ThZH78RphVO1T0snFOPqKXL+S6MxVml7jiFdHr0SVom/kicRvwMr+1fpi32V1YXfXgbjt9dcs1+VlAtW4e5GbXcVxB3X2iQ="
}
},
{
"type": "step",
"result": "=-\\frac{π}{4}+π"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iGvFWxodO1NUsa8/NPZQsyuB0ohw1W3AamtvmRCwJaIgJ/ZZA32ZInFBpDtxBfiKmWiTEpQjat3SO7/l2m58l3DewuNRqMoWEjVZFAQL7imgFJyBsDxkarmODbO7ahsZI4jic0qcjozLDLo6VpiNseVILv5O4PTM/IFQWcFdOplpOMhL5TeSzbb0jLHXKnLC"
}
},
{
"type": "step",
"result": "θ=-\\frac{π}{4}+π"
},
{
"type": "step",
"primary": "The polar coordinates of $$\\left(-6,\\:6\\right)$$",
"result": "\\left(6\\sqrt{2},\\:-\\frac{π}{4}+π\\right)"
}
]
}
}
Solution
cartesian to polar
Solution
Solution steps
Adjust based on the quadrant of the point
The polar coordinates of