Change of coordinates vs change of shape












0












$begingroup$


It is an elementary fact that we are able to change coordinates system to new one. for example in Cartesian coordinates $x^2+y^2=1$ illustrates a circle. Changing to polar coordinates, this equation becomes $r^2=1$ with the same graph.



But the graph of $z=x^3-3xy^2$ in Cartesian coordinates and Cylindrical coordinates seems to be different?!



$(x,y,z): z=x^3-3xy^2$: see its graph



$(t,r,w): w=r^3sin(t)(-3+4sin^2(t))$: see its graph



What's wrong? Am I misunderstanding the concept of coordinates?






calculus graphing-functions coordinate-systems cylindrical-coordinates







share|cite|improve this question











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  • 2




    $begingroup$
    I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 4:26










  • $begingroup$
    Ohh. I have a typo in my question. I'll fix it.
    $endgroup$
    – C.F.G
    Jan 6 at 4:34










  • $begingroup$
    Its not important that $x=rsin(t)$ or $x=rcos(t)$.
    $endgroup$
    – C.F.G
    Jan 6 at 4:36






  • 1




    $begingroup$
    @C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
    $endgroup$
    – Jackozee Hakkiuz
    Jan 6 at 5:17












  • $begingroup$
    So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
    $endgroup$
    – C.F.G
    Jan 6 at 5:28


















0












$begingroup$


It is an elementary fact that we are able to change coordinates system to new one. for example in Cartesian coordinates $x^2+y^2=1$ illustrates a circle. Changing to polar coordinates, this equation becomes $r^2=1$ with the same graph.



But the graph of $z=x^3-3xy^2$ in Cartesian coordinates and Cylindrical coordinates seems to be different?!



$(x,y,z): z=x^3-3xy^2$: see its graph



$(t,r,w): w=r^3sin(t)(-3+4sin^2(t))$: see its graph



What's wrong? Am I misunderstanding the concept of coordinates?






calculus graphing-functions coordinate-systems cylindrical-coordinates







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 4:26










  • $begingroup$
    Ohh. I have a typo in my question. I'll fix it.
    $endgroup$
    – C.F.G
    Jan 6 at 4:34










  • $begingroup$
    Its not important that $x=rsin(t)$ or $x=rcos(t)$.
    $endgroup$
    – C.F.G
    Jan 6 at 4:36






  • 1




    $begingroup$
    @C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
    $endgroup$
    – Jackozee Hakkiuz
    Jan 6 at 5:17












  • $begingroup$
    So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
    $endgroup$
    – C.F.G
    Jan 6 at 5:28
















0












0








0





$begingroup$


It is an elementary fact that we are able to change coordinates system to new one. for example in Cartesian coordinates $x^2+y^2=1$ illustrates a circle. Changing to polar coordinates, this equation becomes $r^2=1$ with the same graph.



But the graph of $z=x^3-3xy^2$ in Cartesian coordinates and Cylindrical coordinates seems to be different?!



$(x,y,z): z=x^3-3xy^2$: see its graph



$(t,r,w): w=r^3sin(t)(-3+4sin^2(t))$: see its graph



What's wrong? Am I misunderstanding the concept of coordinates?






calculus graphing-functions coordinate-systems cylindrical-coordinates







share|cite|improve this question











$endgroup$




It is an elementary fact that we are able to change coordinates system to new one. for example in Cartesian coordinates $x^2+y^2=1$ illustrates a circle. Changing to polar coordinates, this equation becomes $r^2=1$ with the same graph.



But the graph of $z=x^3-3xy^2$ in Cartesian coordinates and Cylindrical coordinates seems to be different?!



$(x,y,z): z=x^3-3xy^2$: see its graph



$(t,r,w): w=r^3sin(t)(-3+4sin^2(t))$: see its graph



What's wrong? Am I misunderstanding the concept of coordinates?






calculus graphing-functions coordinate-systems cylindrical-coordinates




calculus graphing-functions coordinate-systems cylindrical-coordinates






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 4:34







C.F.G

















asked Jan 6 at 4:23









C.F.GC.F.G

1,4301821




1,4301821








  • 2




    $begingroup$
    I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 4:26










  • $begingroup$
    Ohh. I have a typo in my question. I'll fix it.
    $endgroup$
    – C.F.G
    Jan 6 at 4:34










  • $begingroup$
    Its not important that $x=rsin(t)$ or $x=rcos(t)$.
    $endgroup$
    – C.F.G
    Jan 6 at 4:36






  • 1




    $begingroup$
    @C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
    $endgroup$
    – Jackozee Hakkiuz
    Jan 6 at 5:17












  • $begingroup$
    So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
    $endgroup$
    – C.F.G
    Jan 6 at 5:28
















  • 2




    $begingroup$
    I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
    $endgroup$
    – Lord Shark the Unknown
    Jan 6 at 4:26










  • $begingroup$
    Ohh. I have a typo in my question. I'll fix it.
    $endgroup$
    – C.F.G
    Jan 6 at 4:34










  • $begingroup$
    Its not important that $x=rsin(t)$ or $x=rcos(t)$.
    $endgroup$
    – C.F.G
    Jan 6 at 4:36






  • 1




    $begingroup$
    @C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
    $endgroup$
    – Jackozee Hakkiuz
    Jan 6 at 5:17












  • $begingroup$
    So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
    $endgroup$
    – C.F.G
    Jan 6 at 5:28










2




2




$begingroup$
I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:26




$begingroup$
I get $x^3-3xy^3=r^3cos t(cos^2t-3rsin^3t)$.
$endgroup$
– Lord Shark the Unknown
Jan 6 at 4:26












$begingroup$
Ohh. I have a typo in my question. I'll fix it.
$endgroup$
– C.F.G
Jan 6 at 4:34




$begingroup$
Ohh. I have a typo in my question. I'll fix it.
$endgroup$
– C.F.G
Jan 6 at 4:34












$begingroup$
Its not important that $x=rsin(t)$ or $x=rcos(t)$.
$endgroup$
– C.F.G
Jan 6 at 4:36




$begingroup$
Its not important that $x=rsin(t)$ or $x=rcos(t)$.
$endgroup$
– C.F.G
Jan 6 at 4:36




1




1




$begingroup$
@C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
$endgroup$
– Jackozee Hakkiuz
Jan 6 at 5:17






$begingroup$
@C.F.G It is important. Besides that, Wolfram is not graphing you input in cylindrical coordinates.
$endgroup$
– Jackozee Hakkiuz
Jan 6 at 5:17














$begingroup$
So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
$endgroup$
– C.F.G
Jan 6 at 5:28






$begingroup$
So, How to plot this in Cylindrical coordinates? I think It is not important. The resulting graph must be equal ( $(t,r,w): w=r^3cos(t)(-3+4cos^2(t))$).
$endgroup$
– C.F.G
Jan 6 at 5:28












1 Answer
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oldest

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$begingroup$

Indeed the surface should be the same regardless of the choice of coordinates. In particular, setting $x = r cos t$ as more commonly done, or alternatively $x = sin t$ necessarily yield the same result.



Programming-wise, there's so built-in cylindrical plot in Wolfram, as they do it in the more general PareametricPlot3D.



What matters is to have the correct expression for the height of the surface accordingly (be it called $z$ or $w$).



For ${x, y } = {r cos t, r sin t}$, we have the height $z = r^3 cos t left( cos^2 t - 3 sin^2 t right)$



For ${x, y } = {r sin t, r cos t}$, we have the height $w = r^3 sin t left( sin^2 t - 3 cos^2 t right)$



The expression in the question post is a different surface and cannot be obtained via trigonometric substitution of any kind. (no matter how you swap the coordinates: this, or that)






share|cite|improve this answer









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    $begingroup$

    Indeed the surface should be the same regardless of the choice of coordinates. In particular, setting $x = r cos t$ as more commonly done, or alternatively $x = sin t$ necessarily yield the same result.



    Programming-wise, there's so built-in cylindrical plot in Wolfram, as they do it in the more general PareametricPlot3D.



    What matters is to have the correct expression for the height of the surface accordingly (be it called $z$ or $w$).



    For ${x, y } = {r cos t, r sin t}$, we have the height $z = r^3 cos t left( cos^2 t - 3 sin^2 t right)$



    For ${x, y } = {r sin t, r cos t}$, we have the height $w = r^3 sin t left( sin^2 t - 3 cos^2 t right)$



    The expression in the question post is a different surface and cannot be obtained via trigonometric substitution of any kind. (no matter how you swap the coordinates: this, or that)






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Indeed the surface should be the same regardless of the choice of coordinates. In particular, setting $x = r cos t$ as more commonly done, or alternatively $x = sin t$ necessarily yield the same result.



      Programming-wise, there's so built-in cylindrical plot in Wolfram, as they do it in the more general PareametricPlot3D.



      What matters is to have the correct expression for the height of the surface accordingly (be it called $z$ or $w$).



      For ${x, y } = {r cos t, r sin t}$, we have the height $z = r^3 cos t left( cos^2 t - 3 sin^2 t right)$



      For ${x, y } = {r sin t, r cos t}$, we have the height $w = r^3 sin t left( sin^2 t - 3 cos^2 t right)$



      The expression in the question post is a different surface and cannot be obtained via trigonometric substitution of any kind. (no matter how you swap the coordinates: this, or that)






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Indeed the surface should be the same regardless of the choice of coordinates. In particular, setting $x = r cos t$ as more commonly done, or alternatively $x = sin t$ necessarily yield the same result.



        Programming-wise, there's so built-in cylindrical plot in Wolfram, as they do it in the more general PareametricPlot3D.



        What matters is to have the correct expression for the height of the surface accordingly (be it called $z$ or $w$).



        For ${x, y } = {r cos t, r sin t}$, we have the height $z = r^3 cos t left( cos^2 t - 3 sin^2 t right)$



        For ${x, y } = {r sin t, r cos t}$, we have the height $w = r^3 sin t left( sin^2 t - 3 cos^2 t right)$



        The expression in the question post is a different surface and cannot be obtained via trigonometric substitution of any kind. (no matter how you swap the coordinates: this, or that)






        share|cite|improve this answer









        $endgroup$



        Indeed the surface should be the same regardless of the choice of coordinates. In particular, setting $x = r cos t$ as more commonly done, or alternatively $x = sin t$ necessarily yield the same result.



        Programming-wise, there's so built-in cylindrical plot in Wolfram, as they do it in the more general PareametricPlot3D.



        What matters is to have the correct expression for the height of the surface accordingly (be it called $z$ or $w$).



        For ${x, y } = {r cos t, r sin t}$, we have the height $z = r^3 cos t left( cos^2 t - 3 sin^2 t right)$



        For ${x, y } = {r sin t, r cos t}$, we have the height $w = r^3 sin t left( sin^2 t - 3 cos^2 t right)$



        The expression in the question post is a different surface and cannot be obtained via trigonometric substitution of any kind. (no matter how you swap the coordinates: this, or that)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 6 at 6:31









        Lee David Chung LinLee David Chung Lin

        4,39341242




        4,39341242






























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