{
"query": {
"display": "cartesian to polar $$\\left(-2,\\:2\\right)$$",
"symbolab_question": "POLAR#polar (-2,2)"
},
"solution": {
"level": "PERFORMED",
"subject": "Pre Calculus",
"topic": "Polar Coordinates",
"subTopic": "Polar",
"default": "(2\\sqrt{2},-\\frac{π}{4}+π)"
},
"steps": {
"type": "interim",
"title": "Convert $$\\left(-2,\\:2\\right)\\:$$to polar coordinates:$${\\quad}\\left(2\\sqrt{2},\\:-\\frac{π}{4}+π\\right)$$",
"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": [
"$$x=-2$$",
"$$y=2$$"
]
},
{
"type": "step",
"primary": "$$r=\\sqrt{x^2+y^2}$$",
"result": "r=\\sqrt{\\left(-2\\right)^{2}+2^{2}}"
},
{
"type": "interim",
"title": "$$\\sqrt{\\left(-2\\right)^{2}+2^{2}}=2\\sqrt{2}$$",
"input": "\\sqrt{\\left(-2\\right)^{2}+2^{2}}",
"steps": [
{
"type": "step",
"primary": "Apply exponent rule: $$\\left(-a\\right)^{n}=a^{n},\\:$$if $$n$$ is even",
"secondary": [
"$$\\left(-2\\right)^{2}=2^{2}$$"
],
"result": "=\\sqrt{2^{2}+2^{2}}"
},
{
"type": "interim",
"title": "Simplify $$2^{2}+2^{2}:{\\quad}2^{3}$$",
"input": "2^{2}+2^{2}",
"result": "=\\sqrt{2^{3}}",
"steps": [
{
"type": "step",
"primary": "Add similar elements: $$2^{2}+2^{2}=2^{2}\\cdot\\:2$$",
"result": "=2^{2}\\cdot\\:2"
},
{
"type": "step",
"primary": "Apply exponent rule: $$a^b\\cdot\\:a^c=a^{b+c}$$",
"secondary": [
"$$2\\cdot\\:2^{2}=\\:2^{1+2}$$"
],
"result": "=2^{1+2}",
"meta": {
"practiceLink": "/practice/exponent-practice",
"practiceTopic": "Expand FOIL"
}
},
{
"type": "step",
"primary": "Add the numbers: $$1+2=3$$",
"result": "=2^{3}"
}
],
"meta": {
"interimType": "Algebraic Manipulation Simplify Title 1Eq"
}
},
{
"type": "step",
"primary": "Apply exponent rule: $$a^{b+c}=a^{b}a^{c}$$",
"result": "=\\sqrt{2^{2}\\cdot\\:2^{1}}",
"meta": {
"practiceLink": "/practice/exponent-practice",
"practiceTopic": "Expand FOIL"
}
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt{a\\cdot{b}}=\\sqrt{a}\\sqrt{b}$$",
"result": "=\\sqrt{2^{2}}\\sqrt{2^{1}}",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
},
{
"type": "step",
"primary": "Apply radical rule: $$\\sqrt[n]{a^n}=a,\\:\\quad$$ assuming $$a\\ge0$$",
"secondary": [
"$$\\sqrt{2^{2}}=2$$"
],
"result": "=2\\sqrt{2^{1}}",
"meta": {
"practiceLink": "/practice/radicals-practice",
"practiceTopic": "Radical Rules"
}
},
{
"type": "step",
"primary": "Apply rule $$a^{1}=a$$",
"secondary": [
"$$2^{1}=2$$"
],
"result": "=2\\sqrt{2}"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7z3fhNomdwT8bNJhnZYfuTo483d0gMOLJoaS8lTaeQ2l1g99dC9fj9sg0EHzBIRDRSYKD0JQy/1+rjuDyG8nfoWRLd2VwIqlBNByF6663syTgqXz7znVgX8AwDwo+S2NxlCXIjWblGR2IQ5UI4ATgirCI2sSeA74029n2yo277ZU="
}
},
{
"type": "step",
"result": "r=2\\sqrt{2}"
},
{
"type": "step",
"primary": "$$\\theta=\\arctan\\left(\\frac{y}{x}\\right)$$",
"result": "θ=\\arctan\\left(\\frac{2}{-2}\\right)"
},
{
"type": "interim",
"title": "Adjust $$\\theta$$ based on the quadrant of the point $$\\left(-2,\\:2\\right)$$",
"result": "θ=\\arctan\\left(\\frac{2}{-2}\\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": [
"$$\\left(-2,\\:2\\right)\\:$$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{2}{-2}\\right)+π"
}
],
"meta": {
"interimType": "Cartesian To Polar Adjust Theta 1Eq"
}
},
{
"type": "interim",
"title": "$$\\arctan\\left(\\frac{2}{-2}\\right)+π=-\\frac{π}{4}+π$$",
"input": "\\arctan\\left(\\frac{2}{-2}\\right)+π",
"steps": [
{
"type": "interim",
"title": "$$\\arctan\\left(\\frac{2}{-2}\\right)=-\\frac{π}{4}$$",
"input": "\\arctan\\left(\\frac{2}{-2}\\right)",
"steps": [
{
"type": "interim",
"title": "$$\\frac{2}{-2}=-1$$",
"input": "\\frac{2}{-2}",
"steps": [
{
"type": "step",
"primary": "Apply the fraction rule: $$\\frac{a}{-b}=-\\frac{a}{b}$$",
"result": "=-\\frac{2}{2}"
},
{
"type": "step",
"primary": "Apply rule $$\\frac{a}{a}=1$$",
"result": "=-1"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7LSSS6aBxb8E7wJa6sNoLfy061ljBSPJeENOw2efoSWuixCfF7o9L2hRx9Shm0jmvo3oe/oyhMy2+1TQhDBd2f4JcZRkmFFnFNVl6lkSFFQokt3WiGR7ZaCaXvz77bMjS"
}
},
{
"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/NPZQs3HZEx3diJX7HmpOA6MPBM0DnzlbPZjyKgy1eUCFsLd5C0Mhm8BpOdg5ThZH78RphVO1T0snFOPqKXL+S6MxVml7jiFdHr0SVom/kicRvwMrSC2Q+FeN7X71LW6sIifZ61+VlAtW4e5GbXcVxB3X2iQ="
}
},
{
"type": "step",
"result": "=-\\frac{π}{4}+π"
}
],
"meta": {
"solvingClass": "Solver",
"interimType": "Solver",
"gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7iGvFWxodO1NUsa8/NPZQs5XrF5rFYcU/YIOt39Nr1nIgJ/ZZA32ZInFBpDtxBfiKmWiTEpQjat3SO7/l2m58l3DewuNRqMoWEjVZFAQL7imgFJyBsDxkarmODbO7ahsZE+cBTgH9RlKiOcUvSyaOc+VILv5O4PTM/IFQWcFdOplpOMhL5TeSzbb0jLHXKnLC"
}
},
{
"type": "step",
"result": "θ=-\\frac{π}{4}+π"
},
{
"type": "step",
"primary": "The polar coordinates of $$\\left(-2,\\:2\\right)$$",
"result": "\\left(2\\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