Question: How change cartesian to polar step by step?

I did some example to change them but i didn't get the same result as book did it and i try to figure out by hand i get another result i know all of them are true but polar is for making the simple shap for ploting so How i can get the simple shape and if possible How do step by step like using totur we have for polar or not? 


 

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(SolveTools)

with(plots); with(plottools)

with(VectorCalculus)

with(Student:-LinearAlgebra)

with(Student:-MultivariateCalculus)

S := x^2+y^2-1

x^2+y^2-1

(1)

convert(S, polar)

polar(abs(x^2+y^2-1), argument(x^2+y^2-1))

(2)

ChangeOfVariables(S, [cartesian[x, y], polar[r, theta]])

r^2*cos(theta)^2+r^2*sin(theta)^2-1

(3)

solve(%, r)

1, -1

(4)

 

 

T := x^4+2*x^2*y^2+y^4+2*x^3+2*x*y^2-y^2

x^4+2*x^2*y^2+y^4+2*x^3+2*x*y^2-y^2

(5)

ChangeOfVariables(T, [cartesian[x, y], polar[r, theta]])

r^4*cos(theta)^4+2*r^4*cos(theta)^2*sin(theta)^2+r^4*sin(theta)^4+2*r^3*cos(theta)^3+2*r^3*cos(theta)*sin(theta)^2-r^2*sin(theta)^2

(6)

simplify(%)

r^2*(-sin(theta)^2+r*(r+2*cos(theta)))

(7)

solve(%, r)

0, 0, -cos(theta)+(cos(theta)^2+sin(theta)^2)^(1/2), -cos(theta)-(cos(theta)^2+sin(theta)^2)^(1/2)

(8)

R := -3*x^2+y^2-4*x-1

-3*x^2+y^2-4*x-1

(9)

ChangeOfVariables(R, [cartesian[x, y], polar[r, theta]])

-3*r^2*cos(theta)^2+r^2*sin(theta)^2-4*r*cos(theta)-1

(10)

simplify(%)

-4*r^2*cos(theta)^2-4*r*cos(theta)+r^2-1

(11)

solve(%, r)

-1/(2*cos(theta)+1), -1/(2*cos(theta)-1)

(12)

NULL


 

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