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Proof of r(t) and r'(t) orthogonal on a sphere

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    if a curve has the property that the position vector r(t) is always perpendicular to the tangent vector r'(t). show that the curve lies on a sphere with center the origin.

    2. Relevant equations



    3. The attempt at a solution

    I'm not quite sure how to prove this.

    I can write r'(t)[itex]\bullet[/itex]r(t)=0 but expansion doesn't yield anything seemingly useful.

    What direction should I look in?
     
  2. jcsd
  3. Sep 29, 2011 #2
    I suggest you look at the definition of magnitude. It is closely related to the Cartesian inner product you are working with. Can you use the fact of orthogonality that they gave you to show something about r(t)'s magnitude for all t? What will the magnitude be if r(t) is a sphere?
     
  4. Sep 29, 2011 #3

    vela

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    Hint: Consider the quantity [tex]\frac{d}{dt}(\vec{r}\cdot\vec{r})[/tex]
     
  5. Sep 29, 2011 #4
    Hmm.
    Lineintegral1:
    Ok, I can rewrite what I have written.

    [tex]
    \vec{r}(t) \cdot \vec{r}'(t)=0
    [/tex]
    [tex]
    \left\|\vec{r}(t) \right\|\left\|\vec{r}'(t)\right\|cos\theta=0
    [/tex]

    ||r(t)|| has to equal the radius of the sphere. However, isen't that what I'm suppose to be proving given that r(t) and r'(t) are orthogonal.

    vela:
    I'm not sure how [tex]\frac{d}{dt}(\vec{r}\cdot\vec{r})[/tex]

    helps me rearrange what I already have. i know that [tex]\vec{u} \cdot \vec{u}=\left\|\vec{u}\right\|^2[/tex].

    However, in my rewritten version of the dot product I have the product of the magnitude of two different vectors.
     
  6. Sep 29, 2011 #5
    This is a great start at jotting down the important information. You are now ready to solve the problem. You already stated that,

    [tex]||\vec{r}||^2=\vec{r}\cdot\vec{r}[/tex]

    So, if you can show that the derivative of the magnitude squared is zero everywhere, then the magnitude is constant. If it is constant (as you recognized), it is the radius of a sphere.
     
  7. Sep 29, 2011 #6
    Ok, I stared at the problem for a good amount of time and I think I have it. I managed to manipulate a few things...

    [tex]2( \vec{r}(t) \cdot \vec{r}'(t))=\frac{d}{dt}(\vec{r}(t)\cdot\vec{r}(t))=\frac{d}{dt}(\left\|\vec{r}(t)\right\|^2)=0
    [/tex]
    so it must be true then that also

    [tex]\frac{d}{dt}(\left\|\vec{r}(t)\right\|)=0[/tex]

    Because [tex]\left\|\vec{r}(t)\right\|[/tex] does not change for any value of t, r(t) is a constant length for any curve therefore the possible curves can only lie on a sphere centered at the origin.

    Did I miss anything?
     
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