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What I ask is if $1$ meter in $x$ direction is $2$ times bigger than $1$ meter in $y$ direction. What is the length of hypotenuse when for ex, $3$ in $x$ direction and $4$ in $y$ direction ?

I thought this when i was studying weighted least squares and there uses Mahalanobis distance. It is a very similar idea, but there uses the variance-covariance to compare scales of dimensions. I couldn't directly link variance to exact scale factor like $2$ in this example. I did something but i am not sure if it is right.

++ After thinking, i can rephrase better. Now i think of a moving object that moves with $V$ speed in $y$ direction and $2V$ speed in $x$ direction. If it goes along perpendicular axes, it would take $5.5$ time to move from one corner to another. What is time required if this object moves from one corner to another, diagonally?

Thanks in advance

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  • $\begingroup$ You may have something specific in mind, but to me at least, one metre being "bigger" in the $x$ direction than in the $y$ direction doesn't make sense. Could you give an example or two to say what you mean by this and how it relates to the question about a hypotenuse? $\endgroup$ – David Feb 1 '16 at 2:48
  • $\begingroup$ Thanks for your comment. After thinking, i can rephrase better. Now i think of a moving object that moves with V speed in y direction and 2V speed in x direction. If it goes along perpendicular axes, it would take 5.5 time to move from one corner to another. What is time required if this object moves from one corner to another, diagonally?.. And now, as solution i thought, 5* 9/25 length belongs to x axis, and 5* 16/25 length belongs to y axis. If so, 4.1 time required. Do you think is it true ? $\endgroup$ – xcvbnm Feb 1 '16 at 3:29
  • $\begingroup$ Wait, what are the coordinates of the points the object is travelling between? And how fast does it go when it moves diagonally? $\endgroup$ – David Feb 1 '16 at 3:34
  • $\begingroup$ My thinking was -it was just thinking without any real math- to think, how much of that diagonal path belongs to x and y axes. and i thought it would be related to their ratios of squares. it was just this. it may be false, i would like to see real solution $\endgroup$ – xcvbnm Feb 1 '16 at 3:39
  • $\begingroup$ points are in (3,0) and (0,4) and speed of object is somehow varying between in different axis. i dont know how it moves along diagonal but it moves. And when it goes through x axis, it goes with 2V speed and when it goes through y axis, it goes with V speed $\endgroup$ – xcvbnm Feb 1 '16 at 3:42
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One way to interpret what you are asking is to think of a change of coordinates. We have my "normal" coordinate system $(x,y)$ on $\Bbb R^2$ and you have another coordinate system $(z,y)$ with the transformation between your coordinates and mine as $z=2x$. The distance between two points $(z_1,y_1)$ and $(z_2,y_2)$ is $s=\sqrt{4(z_1-z_2)^2+(y_1-y_2)^2}$. It sounds like you are being perverse, but this may be useful. An example would be a crystal where the spacing in one axis is twice the spacing in the other and the coordinates now nicely count lattice positions. You have a space where the metric tensor is $\begin {bmatrix} 4&0\\0&1 \end {bmatrix}$

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  • $\begingroup$ So you say it is as if cutting x axis to half beforehand, then calculating the diagonal. I thought that diagonal is somehow is a mixture of both different scaled axes. You are right i think. Thanks. $\endgroup$ – xcvbnm Feb 1 '16 at 4:36
  • $\begingroup$ and future referance, distance is -in mahalanobis distance terms- [3 4]^T* inv([4 0; 0 1]) *[3 4]. Then exact scale factor refers to standard deviation in statistics $\endgroup$ – xcvbnm Feb 1 '16 at 4:39
  • $\begingroup$ Can you give comment in second interpretation of question? Maybe it is a different question, now i am not sure. also I am sorry if i am perverse $\endgroup$ – xcvbnm Feb 1 '16 at 4:47
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    $\begingroup$ A similar thought would apply to your question about speeds. I assume you now are back to the usual metric, just that the speed in $x$ is twice the speed in $y$. The two diagonals of a rectangle will take the same time to traverse. If the rectangle is $3$ in $x$ and $4$ in $y$ and the particle moves $1$ unit/sec in $y$ and $2$ unit/sec in $x$, you need $\sqrt {1.5^2+4^2}\approx 4.272$ to do the diagonal either way. $\endgroup$ – Ross Millikan Feb 1 '16 at 5:08

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