Transform $x^3+y^3-3xy=0$












0














I was given the following task:




Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$




enter image description here



There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$



I've tried many things but did not get a closed form of $r(t)$:



$$x^3+y^3-3xy=0$$
$$x^3+y^3=3xy$$
$$x^2cdot x+y^2cdot y=3xy$$
$$x^2+y^2cdot t=3y$$
$$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
$$...$$



This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$






real-analysis







share|cite|improve this question



























    0














    I was given the following task:




    Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$




    enter image description here



    There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$



    I've tried many things but did not get a closed form of $r(t)$:



    $$x^3+y^3-3xy=0$$
    $$x^3+y^3=3xy$$
    $$x^2cdot x+y^2cdot y=3xy$$
    $$x^2+y^2cdot t=3y$$
    $$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
    $$...$$



    This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$






    real-analysis







    share|cite|improve this question

























      0












      0








      0







      I was given the following task:




      Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$




      enter image description here



      There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$



      I've tried many things but did not get a closed form of $r(t)$:



      $$x^3+y^3-3xy=0$$
      $$x^3+y^3=3xy$$
      $$x^2cdot x+y^2cdot y=3xy$$
      $$x^2+y^2cdot t=3y$$
      $$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
      $$...$$



      This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$






      real-analysis







      share|cite|improve this question













      I was given the following task:




      Calculate the area B, described by the curve: $$tag{x,y > 0}x^3+y^3-3xy=0$$




      enter image description here



      There are many ways to do this but my university told me to transform the curve into $$r(t); r=sqrt{x^2+y^2}; t = frac{y}{x}$$



      I've tried many things but did not get a closed form of $r(t)$:



      $$x^3+y^3-3xy=0$$
      $$x^3+y^3=3xy$$
      $$x^2cdot x+y^2cdot y=3xy$$
      $$x^2+y^2cdot t=3y$$
      $$dfrac{y^2cdot t}{y^2cdot t}x^2+y^2cdot t=3y$$
      $$...$$



      This keeps on going but I did not find a good result. I am very happy if someone could help me with transforming the curve into $r(t)$






      real-analysis




      real-analysis






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 20:55









      Finn Eggers

      364213




      364213






















          4 Answers
          4






          active

          oldest

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          3














          Tip:



          Replace $y$ with $tx$. You obtain
          $$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
          end{cases}$$
          Can you proceed?






          share|cite|improve this answer





















          • I didn't think of that. Thank you! I will try to proceed
            – Finn Eggers
            Nov 25 at 21:09



















          1














          I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.



          So I would go $$y=tx$$



          $$r^2=x^2+y^2=x^2(1+t^2)$$



          Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$



          Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.






          share|cite|improve this answer





















          • I am stuck at "square this to eliminate x". Could you clarify?
            – Finn Eggers
            Nov 25 at 21:18










          • @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
            – Mark Bennet
            Nov 25 at 21:22



















          1














          The transformation is close to polar coordinates, with $t=tantheta$.



          Hence plugging in the given equation,



          $$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$



          or



          $$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$






          share|cite|improve this answer































            0














            Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$



            Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
            You can solve for $r$ as a function of $t$






            share|cite|improve this answer





















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              4 Answers
              4






              active

              oldest

              votes








              4 Answers
              4






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              3














              Tip:



              Replace $y$ with $tx$. You obtain
              $$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
              end{cases}$$
              Can you proceed?






              share|cite|improve this answer





















              • I didn't think of that. Thank you! I will try to proceed
                – Finn Eggers
                Nov 25 at 21:09
















              3














              Tip:



              Replace $y$ with $tx$. You obtain
              $$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
              end{cases}$$
              Can you proceed?






              share|cite|improve this answer





















              • I didn't think of that. Thank you! I will try to proceed
                – Finn Eggers
                Nov 25 at 21:09














              3












              3








              3






              Tip:



              Replace $y$ with $tx$. You obtain
              $$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
              end{cases}$$
              Can you proceed?






              share|cite|improve this answer












              Tip:



              Replace $y$ with $tx$. You obtain
              $$x^3(1+t^3)-3x^2t=0iff begin{cases}x=0qquadtext{ or } \x(1+t^3)=3t
              end{cases}$$
              Can you proceed?







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 25 at 21:07









              Bernard

              118k638111




              118k638111












              • I didn't think of that. Thank you! I will try to proceed
                – Finn Eggers
                Nov 25 at 21:09


















              • I didn't think of that. Thank you! I will try to proceed
                – Finn Eggers
                Nov 25 at 21:09
















              I didn't think of that. Thank you! I will try to proceed
              – Finn Eggers
              Nov 25 at 21:09




              I didn't think of that. Thank you! I will try to proceed
              – Finn Eggers
              Nov 25 at 21:09











              1














              I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.



              So I would go $$y=tx$$



              $$r^2=x^2+y^2=x^2(1+t^2)$$



              Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$



              Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.






              share|cite|improve this answer





















              • I am stuck at "square this to eliminate x". Could you clarify?
                – Finn Eggers
                Nov 25 at 21:18










              • @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
                – Mark Bennet
                Nov 25 at 21:22
















              1














              I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.



              So I would go $$y=tx$$



              $$r^2=x^2+y^2=x^2(1+t^2)$$



              Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$



              Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.






              share|cite|improve this answer





















              • I am stuck at "square this to eliminate x". Could you clarify?
                – Finn Eggers
                Nov 25 at 21:18










              • @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
                – Mark Bennet
                Nov 25 at 21:22














              1












              1








              1






              I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.



              So I would go $$y=tx$$



              $$r^2=x^2+y^2=x^2(1+t^2)$$



              Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$



              Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.






              share|cite|improve this answer












              I think you need to square the equation at some stage, because $r$ is expressed in terms of even powers of $x$ and $y$ and $t$ is homogeneous, while the original equation is not, and includes odd powers.



              So I would go $$y=tx$$



              $$r^2=x^2+y^2=x^2(1+t^2)$$



              Substitute the first of these in the original equation $$(1+t^3)x^3=3tx^2$$



              Now, either $x=0$ or $(1+t^3)x=3t$. Now square this to eliminate $x$ and get an expression in terms of $r^2$ and then you can take the square root confident that $r$ is positive.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 25 at 21:07









              Mark Bennet

              80.1k981179




              80.1k981179












              • I am stuck at "square this to eliminate x". Could you clarify?
                – Finn Eggers
                Nov 25 at 21:18










              • @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
                – Mark Bennet
                Nov 25 at 21:22


















              • I am stuck at "square this to eliminate x". Could you clarify?
                – Finn Eggers
                Nov 25 at 21:18










              • @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
                – Mark Bennet
                Nov 25 at 21:22
















              I am stuck at "square this to eliminate x". Could you clarify?
              – Finn Eggers
              Nov 25 at 21:18




              I am stuck at "square this to eliminate x". Could you clarify?
              – Finn Eggers
              Nov 25 at 21:18












              @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
              – Mark Bennet
              Nov 25 at 21:22




              @FinnEggers well you get $x^2$ in terms of $t$ and $r$ from the second equation - or you could just take the square root of the second equation to get $x$ directly.
              – Mark Bennet
              Nov 25 at 21:22











              1














              The transformation is close to polar coordinates, with $t=tantheta$.



              Hence plugging in the given equation,



              $$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$



              or



              $$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$






              share|cite|improve this answer




























                1














                The transformation is close to polar coordinates, with $t=tantheta$.



                Hence plugging in the given equation,



                $$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$



                or



                $$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$






                share|cite|improve this answer


























                  1












                  1








                  1






                  The transformation is close to polar coordinates, with $t=tantheta$.



                  Hence plugging in the given equation,



                  $$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$



                  or



                  $$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$






                  share|cite|improve this answer














                  The transformation is close to polar coordinates, with $t=tantheta$.



                  Hence plugging in the given equation,



                  $$r^3(cos^3theta+sin^3theta)-3r^2costhetasintheta=0$$



                  or



                  $$r=frac{3costhetasintheta}{cos^3theta+sin^3theta}=frac{3dfrac1{sqrt{t^2+1}}dfrac t{sqrt{t^2+1}}}{dfrac1{(sqrt{t^2+1})^3}+dfrac{t^3}{(sqrt{t^2+1})^3}}=3tfrac{sqrt{t^2+1}}{t^3+1}.$$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 25 at 22:35

























                  answered Nov 25 at 21:16









                  Yves Daoust

                  124k671221




                  124k671221























                      0














                      Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$



                      Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
                      You can solve for $r$ as a function of $t$






                      share|cite|improve this answer


























                        0














                        Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$



                        Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
                        You can solve for $r$ as a function of $t$






                        share|cite|improve this answer
























                          0












                          0








                          0






                          Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$



                          Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
                          You can solve for $r$ as a function of $t$






                          share|cite|improve this answer












                          Polar coordinates with $$x=rcos theta, y=rsin theta , t=tan theta$$



                          Results in $$ r(1+t^3)-3tsqrt {1+t^2}=0$$
                          You can solve for $r$ as a function of $t$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 25 at 21:09









                          Mohammad Riazi-Kermani

                          40.6k42058




                          40.6k42058






























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