What is the polar form of #( 10,10 )#?

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Answer 1 · 2016-02-20T01:15:54

Use the following formulas:

#r=sqrt(x^2+y^2)#

#theta=tan^-1(y/x)#

The rectangular coordinate #(10,10)# lies in quadrant I. So, #theta# must fall between #0# and #90# degrees

First, solve for r:

#r=sqrt(x^2+y^2)=sqrt(10^2+10^2)=sqrt(2xx100)=10sqrt2#

Now, find #theta# ...

#theta=tan^-1(y/x)=tan^-1(10/10)=tan^-1(1)=45^o# or #(pi)/4#

Polar Form : #(r,theta)=(10sqrt2,(pi)/4)#

hope that helped