Constant lengths (dimensional constants) and scaling












0














Suppose in the $xy$-plane we have defined the constant length $L$. This can be a fixed radius of a circle; or a boundary condition or any condition such, that $L$ has dimension of "meters" and is constant.



Now, scale the plane
$$x'=Cx$$
$$y'=Cy$$



Does this mean that in the new, primed coordinates, $L'=CL$? Or does $L$ stay exactly the same, $L'=L$?



I want to understand how fixed-length objects behave when the plane is scaled down (that is, when every graph of a function y=y(x) is scaled down).



It is confusing, because all variable-related objects (functions, curves, etc) in the plane must be scaled down; but $L$, being a constant, is not related to any variables, yet is defined in the plane. Does $L$ scale down also?






calculus geometry transformation constants







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    0














    Suppose in the $xy$-plane we have defined the constant length $L$. This can be a fixed radius of a circle; or a boundary condition or any condition such, that $L$ has dimension of "meters" and is constant.



    Now, scale the plane
    $$x'=Cx$$
    $$y'=Cy$$



    Does this mean that in the new, primed coordinates, $L'=CL$? Or does $L$ stay exactly the same, $L'=L$?



    I want to understand how fixed-length objects behave when the plane is scaled down (that is, when every graph of a function y=y(x) is scaled down).



    It is confusing, because all variable-related objects (functions, curves, etc) in the plane must be scaled down; but $L$, being a constant, is not related to any variables, yet is defined in the plane. Does $L$ scale down also?






    calculus geometry transformation constants







    share|cite|improve this question



























      0












      0








      0







      Suppose in the $xy$-plane we have defined the constant length $L$. This can be a fixed radius of a circle; or a boundary condition or any condition such, that $L$ has dimension of "meters" and is constant.



      Now, scale the plane
      $$x'=Cx$$
      $$y'=Cy$$



      Does this mean that in the new, primed coordinates, $L'=CL$? Or does $L$ stay exactly the same, $L'=L$?



      I want to understand how fixed-length objects behave when the plane is scaled down (that is, when every graph of a function y=y(x) is scaled down).



      It is confusing, because all variable-related objects (functions, curves, etc) in the plane must be scaled down; but $L$, being a constant, is not related to any variables, yet is defined in the plane. Does $L$ scale down also?






      calculus geometry transformation constants







      share|cite|improve this question















      Suppose in the $xy$-plane we have defined the constant length $L$. This can be a fixed radius of a circle; or a boundary condition or any condition such, that $L$ has dimension of "meters" and is constant.



      Now, scale the plane
      $$x'=Cx$$
      $$y'=Cy$$



      Does this mean that in the new, primed coordinates, $L'=CL$? Or does $L$ stay exactly the same, $L'=L$?



      I want to understand how fixed-length objects behave when the plane is scaled down (that is, when every graph of a function y=y(x) is scaled down).



      It is confusing, because all variable-related objects (functions, curves, etc) in the plane must be scaled down; but $L$, being a constant, is not related to any variables, yet is defined in the plane. Does $L$ scale down also?






      calculus geometry transformation constants




      calculus geometry transformation constants






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday

























      asked yesterday









      user142523

      17611




      17611






















          1 Answer
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          oldest

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          1














          Hint



          Take a segment $AB$ of length $L$ based on coordinates of $A$ and $B$. Compute the image of $A to A^prime$ and $B to B^prime$ by the scaling. What is the length of the segment $A^prime B^prime$?






          share|cite|improve this answer





















          • $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
            – user142523
            yesterday












          • @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
            – mathcounterexamples.net
            yesterday










          • I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
            – user142523
            yesterday










          • are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
            – user142523
            yesterday











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Hint



          Take a segment $AB$ of length $L$ based on coordinates of $A$ and $B$. Compute the image of $A to A^prime$ and $B to B^prime$ by the scaling. What is the length of the segment $A^prime B^prime$?






          share|cite|improve this answer





















          • $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
            – user142523
            yesterday












          • @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
            – mathcounterexamples.net
            yesterday










          • I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
            – user142523
            yesterday










          • are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
            – user142523
            yesterday
















          1














          Hint



          Take a segment $AB$ of length $L$ based on coordinates of $A$ and $B$. Compute the image of $A to A^prime$ and $B to B^prime$ by the scaling. What is the length of the segment $A^prime B^prime$?






          share|cite|improve this answer





















          • $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
            – user142523
            yesterday












          • @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
            – mathcounterexamples.net
            yesterday










          • I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
            – user142523
            yesterday










          • are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
            – user142523
            yesterday














          1












          1








          1






          Hint



          Take a segment $AB$ of length $L$ based on coordinates of $A$ and $B$. Compute the image of $A to A^prime$ and $B to B^prime$ by the scaling. What is the length of the segment $A^prime B^prime$?






          share|cite|improve this answer












          Hint



          Take a segment $AB$ of length $L$ based on coordinates of $A$ and $B$. Compute the image of $A to A^prime$ and $B to B^prime$ by the scaling. What is the length of the segment $A^prime B^prime$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered yesterday









          mathcounterexamples.net

          24.9k21853




          24.9k21853












          • $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
            – user142523
            yesterday












          • @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
            – mathcounterexamples.net
            yesterday










          • I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
            – user142523
            yesterday










          • are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
            – user142523
            yesterday


















          • $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
            – user142523
            yesterday












          • @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
            – mathcounterexamples.net
            yesterday










          • I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
            – user142523
            yesterday










          • are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
            – user142523
            yesterday
















          $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
          – user142523
          yesterday






          $L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; now transform $x_i ^{prime} = Cx_i$ and $y_i ^{prime} = Cy_i$, where $i in {a,b}$ and I get $L'=CL$. The confusion comes from "how you actually scale". Because in this example, $L$ seems not like a constant, but more like $L=L(x,y)$... A "constant" length should not be tied to the variables $x,y$, right? Where is my thinking wrong.
          – user142523
          yesterday














          @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
          – mathcounterexamples.net
          yesterday




          @user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
          – mathcounterexamples.net
          yesterday












          I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
          – user142523
          yesterday




          I was thinking of $L$ as a constant term that has a value that is constant or cannot change, because it does not contain any modifiable variables.
          – user142523
          yesterday












          are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
          – user142523
          yesterday




          are you referring to the units of measurement, "meter". Are you saying, because the units are scaled down, this is why the dimensional constant $L$ is scaled down too?
          – user142523
          yesterday


















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