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How does taking the derivative of unit vector squared prove the following...? watch

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    Hi,

    please see attachment.
    Does anyone understand what the slide is trying to get across? I understand the maths, but I don't understand how taking the derivative of unit vector squared proves the derivative of the unit vector of velocity is perpendicular to the unit vector of velocity.

    cheers
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    (Original post by SWIMSHALLOW)
    Hi,

    please see attachment.
    Does anyone understand what the slide is trying to get across? I understand the maths, but I don't understand how taking the derivative of unit vector squared proves the derivative of the unit vector of velocity is perpendicular to the unit vector of velocity.

    cheers
    Because their dot product is zero.
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    The unit vector isn't squared. It's their dot product that's being considered: u.u .
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    (Original post by SWIMSHALLOW)
    Hi,

    please see attachment.
    Does anyone understand what the slide is trying to get across? I understand the maths, but I don't understand how taking the derivative of unit vector squared proves the derivative of the unit vector of velocity is perpendicular to the unit vector of velocity.
    As others have pointed out, for any two vectors:

    \bold{a} \cdot \bold{b} = 0 \Rightarrow \bold{a} \perp \bold{b}

    since we have

    ab\cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2}

    But you should try to understand this intuitively (or graphically maybe).

    If \bold{u}(t) is a unit vector, then its length can't change, by definition. So if it varies with time, only its angle can change. So let it turn through an angle \Delta \theta in time \Delta t.

    Now draw \bold{u}(t), \bold{u}(t+\Delta t) as arrows of unit length originating from some point P. Then \Delta \bold{u} = \bold{u}(t+\Delta t) - \bold{u}(t) is the arrow from \bold{u}(t) to \bold{u}(t+\Delta t).

    As \Delta t \to 0, \Delta \bold{u} tends to make a right angle with the other two vectors more and more closely (draw diagram and label angles to see this - the angles in question are \frac{\pi}{2} -\frac{\Delta \theta}{2}), so the change of the unit vector is perpendicular to itself in the limit, and thus so is its rate of change, as that is simply \frac{\Delta \bold{u}}{\Delta t}, a vector divided by a scalar, which doesn't change the direction of the vector.
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    (Original post by atsruser)
    As others have pointed out, for any two vectors:

    \bold{a} \cdot \bold{b} = 0 \Rightarrow \bold{a} \perp \bold{b}

    since we have

    ab\cos\theta = 0 \Rightarrow \theta = \frac{\pi}{2}

    But you should try to understand this intuitively (or graphically maybe).

    If \bold{u}(t) is a unit vector, then its length can't change, by definition. So if it varies with time, only its angle can change. So let it turn through an angle \Delta \theta in time \Delta t.

    Now draw \bold{u}(t), \bold{u}(t+\Delta t) as arrows of unit length originating from some point P. Then \Delta \bold{u} = \bold{u}(t+\Delta t) - \bold{u}(t) is the arrow from \bold{u}(t) to \bold{u}(t+\Delta t).

    As \Delta t \to 0, \Delta \bold{u} tends to make a right angle with the other two vectors more and more closely (draw diagram and label angles to see this - the angles in question are \frac{\pi}{2} -\frac{\Delta \theta}{2}), so the change of the unit vector is perpendicular to itself in the limit, and thus so is its rate of change, as that is simply \frac{\Delta \bold{u}}{\Delta t}, a vector divided by a scalar, which doesn't change the direction of the vector.
    Thanks for the reply bud, appreciate it

    so basically what you're saying is as delta t tends to 0, the angle between the two vectors delta u(t) and delta u (t+delta t) gets smaller and smaller, and ideally when the angle between them is 0, delta u is idealy perpendicular to the two vectors...?
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    (Original post by SWIMSHALLOW)
    Thanks for the reply bud, appreciate it

    so basically what you're saying is as delta t tends to 0, the angle between the two vectors delta u(t) and delta u (t+delta t) gets smaller and smaller, and ideally when the angle between them is 0, delta u is idealy perpendicular to the two vectors...?
    More or less, yes, though your terminology is a bit off (replace "ideally" with "ultimately" or "in the limit").

    Draw a picture of what I described and all should be clear.
 
 
 
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