Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

enter image description here

share|cite|improve this question
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
up vote 2 down vote accepted

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 $(\dot{x},\dot{y})$.

share|cite|improve this answer

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 $(x')^2+(y')^2$.

share|cite|improve this answer

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 $x^2+y^2+2xy$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.