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Vector calculus velocity question

  1. Mar 8, 2012 #1
    Hi, I am not fully understanding how to express velocity as a vector equation. In an example my professor gave in class, a particle moves (in a straight line) from point A to point B (starting from t=0) and traveling at 6 m/s. The solution included a unit vector. Does that mean all vector equations describing velocity must include a unit vector?

    The solution was (initial point)+[<vector>/unit vector](speed)
  2. jcsd
  3. Mar 8, 2012 #2
    No, they don't. From what I've perceived from your example:

    [itex]\vec{r}[/itex]=[itex]\vec{r}[/itex]0+[itex]\vec{v}[/itex]t → [itex]\vec{r}[/itex]=[itex]\vec{A}[/itex]+6t[itex]\hat{r}[/itex]

    Because when t=0, [itex]\vec{r}[/itex]0=[itex]\vec{A}[/itex].

    And [itex]\vec{r}[/itex]B=[itex]\vec{A}[/itex]+6tB[itex]\hat{r}[/itex], with [itex]\vec{r}[/itex]B=[itex]\vec{B}[/itex]. Taking one derivative:


    Being [itex]\hat{v}[/itex] a unit vector (normed vector), like [itex]\hat{r}[/itex].
    It's not obligatory to represent, in a vector equation describing velocity, the unit vector [itex]\hat{v}[/itex]. You can always replace it by [itex]\frac{\vec{v}}{|\vec{v}|}[/itex], which is the same thing.
    However, it's advised to use them because they facilitate geometrical description of equations.
    For example, the Newton's law of gravitational force can be written like this:

    [itex]\vec{F}[/itex]g=-G(m1m2)/(r)3 [itex]\vec{r}[/itex]

    Instead of the usual:

    [itex]\vec{F}[/itex]g=-G(m1m2)/(r)2 [itex]\hat{r}[/itex]=|[itex]\vec{F}[/itex]g|[itex]\hat{r}[/itex]

    The first one may lead you to think that [itex]\vec{F}[/itex]g [itex]\propto[/itex] r-3 but it's not, as you certainly know.
    I hope I've cleared your doubt about unit vectors, if not other users may, or else explain yourself better.
  4. Mar 8, 2012 #3
    Thanks for your explanation, Mathaholic. If A=(1, 2, 3), and the vector parallel to motion is <4, 5, 6>, and the speed is 6m/s, is it incorrect to express the velocity as

    r'(t)=A+<4, 5, 6>6t

    If there is an initial point, a speed and a vector, what am I missing if this is incorrect?
  5. Mar 8, 2012 #4
    If |[itex]\vec{v}[/itex]| is constant through time and [itex]\vec{v}[/itex]=[itex]\vec{const}[/itex] (straight line, as you said before), then the vector velocity is expressed as:


    Now, if there's a vector parallel to the movement (thus, to the vector velocity) and the speed is 6 m/s. Firstly, you note that the magnitude of (4,5,6) is not equal to 6, otherwise, that'd be your vector velocity. But if (4,5,6) is parallel to the velocity, then:


    You need to determine the x,y,z coordinates of [itex]\vec{v}[/itex] using that expression and knowing its magnitude (6 m/s) and most importantly, calculating λ.
    And then you'll be able to express the vector velocity as:


    And yes, it's incorrect to express the vector velocity as:


    The RHS has units of m/s and LHS has at least units of m (but never m/s!). Secondly, you cannot sum points (A) with vectors of velocity, you can only do that with vectors of position ([itex]\vec{r}[/itex]).
  6. Mar 9, 2012 #5
    Hi Mathoholic if you're still around! Would the correct vector equations be

    r'(t)=[<4,5,6>/square root77](6t)? That would make the equation v(t)=speed(velocity unit vector)?

    r(t)= (1,2,3)+[<4,5,6>/square root77](6t)

    Hope I've got it?
  7. Mar 9, 2012 #6
    The vector position is well written. As for the vector velocity, not so much. If the speed is constant through time, then the vector velocity doesn't have the parameter time (t), as it is also the 1st derivative of the position. As for the rest, it's all good.
  8. Mar 9, 2012 #7
    Thanks, yes, definitely, the t would not be there if I just took the derivative... appreciate your patience! See you around!
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