Asked by Nevaeh
Consider a point with rectangular coordinates (x,y).
If x<0 then the polar coordinates of the point are (r,θ) where r≥0 and −π/2≤θ<3π/2and:
r=
θ=
If x≥0 then the polar coordinates of the point are (r,θ) where r≥0 and −π/2≤θ<3π/2 and:
r=
θ=
For both r is r=sqrt(x^2+y^2) I'm sure what the angle would be.
If x<0 then the polar coordinates of the point are (r,θ) where r≥0 and −π/2≤θ<3π/2and:
r=
θ=
If x≥0 then the polar coordinates of the point are (r,θ) where r≥0 and −π/2≤θ<3π/2 and:
r=
θ=
For both r is r=sqrt(x^2+y^2) I'm sure what the angle would be.
Answers
Answered by
Nevaeh
I'm not sure about the angle
Answered by
Nevaeh
Your answers should be expressions in terms of x and y)
Answered by
oobleck
x<0 means QII,QIII so π/2≤θ<3π/2
x>0 means QI, QIV, so -π/2≤θ<π/2
In any case, tanθ = y/x unless x=0
x>0 means QI, QIV, so -π/2≤θ<π/2
In any case, tanθ = y/x unless x=0
Answered by
Nevaeh
The second ones theta is tan(theta) =y/x
But the first one is different
But the first one is different
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