# Proving Magnutude Of Vector Valued Function Is Constant

Lancelot59
Proving Magnitude Of Vector Valued Function Is Constant

I have a rather neat problem. I need to prove that the magnitude of this function:

$$F_{2}(x,y)=\frac{<-x,y>}{x^{2}+y^{2}}$$

is constant along circles centred about the origin. Now while proving that the magnitude is inversely proportional I had to get the magnitude, and it wound up being:

$$\sqrt{\frac{1}{(x^{2}+y^{2})}}$$

Which looks like the basic equation for a circle to me. That particular function will give elliptical level curves. I'm not sure how to go about this. Can I just say that because it's the inverse of a the form of an ellipse that if the level curve x^2+y^2 is equal to a constant c^2 then the magnitude will also be a constant because it has the same form?

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Gold Member
that's not what I get for the magnitude. Can you show me how you got that answer?

Lancelot59
This is how I did it:

First I multiplied in the scalar x^2+y^2.
$$=<\frac{-x}{x^{2}+y^{2}},\frac{y}{x^{2}+y^{2}}>$$

Then put it in the usual form:
$$=\sqrt{(\frac{-x}{x^{2}+y^{2}})^{2}+(\frac{y}{x^{2}+y^{2}})^{2}}$$

They have a common denominator, so I added them:
$$=\sqrt{\frac{x^{2}+y^{2}}{(x^{2}+y^{2})^{2}}$$

Which then simplifies down to:
$$=\sqrt{\frac{1}{x^{2}+y^{2}}$$

Gold Member
Well, your magnitude is equal to $$\frac{1}{r}$$, isn't it? Seems pretty obvious to me.
Well, your magnitude is equal to $$\frac{1}{r}$$, isn't it? Seems pretty obvious to me.