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What is the Cartesian form of #r^2+theta = -sin^2theta+4cot^3theta #?

Answer 1

Violet Anderson

Please see the explanation, because this breaks into 3 equations.

Multiply both sides by #r^5sin^3(theta)#:

#r^7sin^3(theta) + (theta)r^5sin^3(theta) = 4r^5cos^3(theta) - r^5sin^5(theta)#

Do some regrouping:

#r^4(rsin(theta))^3 + (theta)r^2(rsin(theta))^3 = 4r^2(rcos(theta))^3 - (rsin(theta))^5#

Substitute x for #(rcos(theta))# and y for #(rsin(theta))#:

#r^4y^3 + (theta)r^2y^3 = 4r^2x^3 - y^5#

Substitute #(x^2 + y^2)# for #r^2#

#(x^2 + y^2)^2y^3 + (theta)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5#

We cannot merely substitute #tan^-1(y/x)# for #theta#; we must specify that #x!=0 and y!=0# and then handle special cases for #x < 0#, #x > 0 and y>=0#, and #x > 0 and y < 0#

For #x > 0 and y>0#:

#(x^2 + y^2)^2y^3 + tan^-1(y/x)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5#

For #x < 0#:

#(x^2 + y^2)^2y^3 + (tan^-1(y/x) + pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5#

For #x > 0 and y < 0#

#(x^2 + y^2)^2y^3 + (tan^-1(y/x) + 2pi)(x^2 + y^2)y^3 = 4(x^2 + y^2)x^3 - y^5#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.