What is the Cartesian form of #-2r^3 = -theta+cot(theta)- cos(theta) #?

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Answer 1 · 2017-03-19T22:20:22

#-2(x^2+y^2)^(3/2)=-tan^-1(y/x)+x/y-x/sqrt(x^2+y^2)#

The polar notation arises from the following triangle:

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This tells us that #r=sqrt(x^2+y^2)#. We can use the triangle with legs #x# and #y# and angle #theta# to rewrite any trigonometric functions.

Here, #cot(theta)=1/tan(theta)=x/y# and #costheta=x/r=x/sqrt(x^2+y^2)#.

#theta# can be written many ways but the simplest is to note that #tan(theta)=y/x#, so #theta=tan^-1(y/x)#.

We then see the equation's transformation from polar to Cartesian form:

#-2r^3=-theta+cot(theta)-cos(theta)#

#-2(x^2+y^2)^(3/2)=-tan^-1(y/x)+x/y-x/sqrt(x^2+y^2)#

Which is in Cartesian form.