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The following vector field $$ \vec G(x,y) = \dfrac{-y\hat \imath + x\hat \jmath}{\sqrt{x^2+y^2}} $$ shows vectors that are tangent to circle centered at the origin. However, the text I am using also mentions that the magnitudes of these vectors are equal to their distances from the origin.

My question is, wouldn't you have to mutliply both components by $\sqrt{x^2+y^2} $ rather than divide?


I now see that $|\vec G| = 1$, something I should have noticed before. My new approach to get the magnitudes to equal to the distances of the points from the origin:

$$ |\vec G_1| = \sqrt{k^2\cdot(y^2 + x^2)} = \sqrt{y^2+x^2} \implies k=1 \implies \vec G_1(x,y) = -y\hat \imath + x\hat \jmath$$

So is the text (as I have written it) simply incorrect or is there something else behind it?

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  • $\begingroup$ I may disagree with your text. The length of $G$ is constantly 1. $\endgroup$ – Randall Sep 18 '17 at 1:27
  • $\begingroup$ Do you see any other reason for this? I have edited the question. $\endgroup$ – Rithwik Sudharsan Sep 18 '17 at 2:34
  • $\begingroup$ The text would be correct without the denominator: the length of the vector $-yi + xj$ is in fact equal to its distance from the origin. $\endgroup$ – Randall Sep 18 '17 at 19:48
  • $\begingroup$ Thanks! I speculated so, I must have read the text wrong if it's not incorrect. $\endgroup$ – Rithwik Sudharsan Sep 21 '17 at 4:59

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