If we select the ray l to it is in the hopeful x-axis, climate a allude P in the plane has both cartesian collaborates ( x,y) and also polar works with (r, q).

x = r cos(q) , y = r sin(q) |

r2 = x2+y2 and tan( q) = | y |

**(2)**

**EXAMPLE 3 convert the suggest ( 4, p/4) frompolar collaborates into cartesian coordinates, and also then display that (2) counter it earlier into polar .**

**Solution:**To execute so, we let r = 4 and let q = p/4 in (1) come obtain

x = 4cos | æè | p | |

r2 = x2 + y2 = 8 + 8 = 16, r = 4 |

If we substitute x = r cos(q) and also *y* = r sin(q) right into a curve *g*(*x,y*) = *k*, climate the result

g( rcos(q), rsin(q) ) = k |

*pullback*the the curve right into polar coordinates. The identification r2 = x2+y2 is regularly used in pulling a curve back into polar coordinates.Forexample, x2+y2 = R2 for a continuous

*R*> 0 has actually a pullback the

r2 = R2 Þ r = R |

r sin(q) = m r cos(q) Þ sin(q) = m cos(q) Þ tan(q) = m |

*polar name: coordinates transformation*

T( r, q ) = á r cos(q), r sin(q)ñ |

*r*to obtain a duty of the type

*r = f*(q) .EXAMPLE 4 convert the curve x2 + (y - 1)2 = 1 into polarcoordinates, and also then deal with for

*r*, if possible.

You are watching: X^2+y^2=r^2

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Solution: Expanding leads to x2 + y2 -2

*y*+ 1 = 1 , so the To execute so, we change y r2 and also let x = rcos( q) :

r2 - 2r sin(q) + 1 = 1 |

r = 2 sin(q) |