# Help with odd partial derivatives in velocity $\bar v^2 = \dot x ^2+\dot y^2$

I am doing a physics -course Tfy-0.2061. My teacher claims that this is velocity squared, $\bar v^2 = \dot x ^2+\dot y^2$. I cannot understand why it is not $\bar v^2 = (\dot x +\dot y)^2$.

If distance is $\bar d = \bar x + \bar y$. Then velocity is $\partial_t \bar d = \dot x + \dot y$. Now just square it to get $$\bar v^2 = \dot x^2 +2\dot x\dot y +\dot y^2 \not = \dot x^2 +\dot y^2.$$

What does my teacher mean by velocity $\bar v^2 = \dot x ^2+\dot y^2$?

P.s. the goal was to do something called "nopeuden radiaalinen komponentti" that probably means radial component of velocity. I don't just understand what it means, some angular velocity? I am doing the exercise 3b here, sorry not in English.

Trial 1

The only way that my teacher can be correct is if $y_0=0$ and $x_0=0$ because

-
Distance is not x+y –  DarenW Sep 21 '12 at 7:09
@hhh you say it's a vector sum. But if those are vectors, how on earth are you squaring them? –  Robert Mastragostino Sep 21 '12 at 7:30
@hhh exactly. $v=\dot{x}+\dot{y}$, but $|v|\neq |\dot{x}|+|\dot{y}|$. The first statement is about vectors, not their norms, so squaring it doesn't make sense. –  Robert Mastragostino Sep 21 '12 at 16:01
@RobertMastragostino yes but I meant this $|v|^2=|\dot x|^2+|\dot y|^2$, norm-2 -- not norm-1, although it is also a length measure. –  hhh Sep 21 '12 at 18:44

This is shorthand for $$|v|^2=v\cdot v=\dot{x}^2+\dot{y}^2,$$ i.e. a statement about the length of the velocity vector whose components are $(\dot{x},\dot{y})$.

-

Distance is a vector quantity. Write it as $d=xi+yj$ where $i=(1,0)$, $j=(0,1)$. So $v=d'=x'i+y'j$. Now $v^2$ makes no sense - you can't square a vector. What makes sense is $\|v\|^2$ and that, the way vectors work, is $(x')^2+(y')^2$.

-

Displacement and velocity both are vector quantities.

So, displacement $d=x\vec i+y\vec j$ where $\vec i$ and $\vec j$ are two unit-vectors perpendicular to each other.

velocity $v=\dot x\vec i+\dot y\vec j$

Now, square of velocity is actually dot-product $|\vec v|^2=\vec v\cdot\vec v=\dot x^2+\dot y^2$

In fact, square of distance is actually dot-product $|\vec d|^2=\vec d\cdot\vec d=x^2+y^2$ not $x^2+y^2+2xy$

-