What is the polar form of #( 0,0 )#?
3 Answers
Explanation:
Well, this coordinate point is the origin of both the Cartesian and polar coordinate planes, so the polar form is also
Nevertheless, we can solve this if we'd like, using the equations
-
#r^2 = x^2 + y^2# -
#theta = arctan(y/x)#
We know:
-
#x=0# -
#y=0#
So we have
#r^2 = 0^2 + 0^2#
#ul(r = 0#
The coordinate is thus
#color(blue)(ulbar(|stackrel(" ")(" "(0,0)" ")|)#
#(r,theta) -= (0,0)# .
Well, consider the conversion:
#x = rcostheta#
#y = rsintheta#
#(x,y) = (0,0) = (rcostheta, rsintheta)#
We necessarily have that
#0 = rcostheta = rsintheta#
If
- if
#sintheta = 0# ,#costheta = pm1# , and... - if
#costheta = 0# ,#sintheta = pm1# .
Thus, we must have that
However, even when
#overbrace((0","0))^((x","y)) harr overbrace((0","theta))^((r","theta))#
Despite that, since the distance from the origin is zero, we can choose
Explanation:
By convention we choose
Ordinarily the ambiguity does not represent an issue, however one must take extreme caution when converting from rectangular to polar coordinates in integration, as the Jacobian:
# J = | ( (partial r)/(partial x), (partial r)/(partial y) ), ( (partial theta)/(partial x), (partial theta)/(partial y) ) | #
is not defined at the origin.
Similarly if we are examining limits at the origin we need to be very careful about the value of