Mixing
Constant lengths (dimensional constants) and scaling
$begingroup$
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
asked Jan 3 at 14:14
user142523user142523
17611
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 14:14
user142523user142523
17611
$endgroup$
add a comment |
$begingroup$
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
asked Jan 3 at 14:14
user142523user142523
17611
$endgroup$
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
asked Jan 3 at 14:14
user142523user142523
17611
asked Jan 3 at 14:14
user142523user142523
17611
edited Jan 3 at 15:37
user142523
asked Jan 3 at 14:14
user142523user142523
17611
asked Jan 3 at 14:14
user142523user142523
17611
asked Jan 3 at 14:14
user142523user142523
17611
17611
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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 $A^prime B^prime$?
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
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 $A^prime B^prime$?
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
add a comment |
$begingroup$
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 $A^prime B^prime$?
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
add a comment |
$begingroup$
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 $A^prime B^prime$?
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
$endgroup$
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 $A^prime B^prime$?
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
answered Jan 3 at 14:23
mathcounterexamples.netmathcounterexamples.net
25.8k21954
25.8k21954
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
add a comment |
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
$L=sqrt{(x_a-x_b)^2+(y_a-y_b)^2}$; 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.
$endgroup$
– user142523
Jan 3 at 15:02
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
@user142523 $L$ is not a constant but the length is scaled by the constant of the scaling $C$.
$endgroup$
– mathcounterexamples.net
Jan 3 at 15:06
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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.
$endgroup$
– user142523
Jan 3 at 15:09
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
$begingroup$
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?
$endgroup$
– user142523
Jan 3 at 15:59
add a comment |
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