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Question with the prove of r(t) ande r'(t) is perpendicular.

  1. Aug 29, 2010 #1
    In the prove of [itex]\vec{r}(t) \;&\; \vec{r}'(t) \;[/itex] is perpendicular:

    [tex] \vec{r}(t) \;\cdot\; \vec{r}(t) \;=\; |\vec{r}(t)|^2 \;\Rightarrow\; \frac{d}{dt}[ \vec{r}(t) \;\cdot\; \vec{r}(t)] = \vec{r}(t) \;\cdot\; \vec{r}'(t) + \vec{r}(t) \;\cdot\; \vec{r}'(t) = \frac{d}{dt}[ |\vec{r}(t)|^2] [/tex]

    The book claimed since [itex]\; |\vec{r}(t)|^2 \;[/itex] is a constant, [itex]\frac{d}{dt}[ |\vec{r}(t)|^2] = 0[/itex].

    Question is why [itex]\; \frac{d}{dt}[ |\vec{r}(t)|^2] = 0\;[/itex]? Let [itex]\; \vec{r}(t) = x\hat{x} + y\hat{y}[/itex]

    [tex]\; \frac{d}{dt}[ |\vec{r}(t)|^2] = x^2 + y^2\;[/tex]

    x and y is not a constant!!! Why [itex]\; |\vec{r}(t)|^2 \;[/itex] is a constant?!!
     
  2. jcsd
  3. Aug 29, 2010 #2

    Mark44

    Staff: Mentor

    Does r(t) describe a circular path?
     
  4. Aug 29, 2010 #3
    this is what comes to my mind: If the path is circular then r(t) represents the radius of the circle at any point and the length of r(t) always equals a constant which is the radius. so its derivative must be zero everywhere. this is just another proof for this intuitive thing that the tangent line passing through a point on a circle is perpendicular to the radius of the circle.
     
  5. Aug 30, 2010 #4

    I found that the book did say if [itex]|\vec{r}(t)|[/itex] is constant.

    My bad.
     
    Last edited: Aug 30, 2010
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