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Tangent vector

  1. Jun 21, 2015 #1
    I was reading about the tangent vector at a point on a curve.
    It is formulated as r' = Lim Δt→0 [r(t+Δt) - r(t)] / Δt (sorry for the misrepresentation of the 'Lim Δt→0 ')
    where r(t) is a position vector to the curve and t is a parameter and r' is the derivative of r(t).
    All I can infer from the formulation is that the tangent is the rate of change of a position vector at a point in question. But then a tangent vector cannot have its components to be rates of change. Because then the defined tangent vector is the rate of change of tangent vector and not the tangent vector itself.

    Therefore I think that the tangent vector should be formulated as:
    Lim Δt→0 [r(t+Δt) - r(t)].
    This would have both the direction and magnitude at a point (or in a very very small interval in this case as Δt→0 ). This could also give us the equation of the tangent at the point.

    My question is simple. Why would we represent a vector (tangent vector) at a point by the rate of change of position vector at the point? Shouldn't it just be the difference between two position vectors with Δt→0?

    Please tell me if what I think of the formulation is correct.

    Thanks
     
    Last edited: Jun 21, 2015
  2. jcsd
  3. Jun 21, 2015 #2

    FactChecker

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    No. Without dividing by Δt, the limit will always be 0 if the function r is continuous. So that would not tell you anything. You want to know the change or r per change in t. So dividing is necessary.
     
  4. Jun 21, 2015 #3
    Then why is the rate of change called the tangent vector itself?
     
  5. Jun 21, 2015 #4

    FactChecker

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    Any rate has to be change of r per unit time (or per unit of something else). So you must divide by the amount of time that gave that change. You must divide by Δt. You will have to think about it in those terms. It can't be said any other way.
     
  6. Jun 21, 2015 #5
    But the rate of change at a point can never be a tangent at that point. It has to be integrated to obtain the equation of the tangent.
     
  7. Jun 21, 2015 #6

    Svein

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    Sorry, you are wrong.
    Example: Let [itex]\vec{r}(t)=(a\cdot\cos(t), a\cdot\sin(t)) [/itex]. Then [itex]\vec{\dot{r}}(t)=(-a\cdot\sin(t), a\cdot\cos(t)) [/itex]. Now the first equation describes a circle with radius a. The second is the rate of change of that equation. Now, a tangent to a circle will always be normal to the radius at that point. If you calculate the scalar product of both expressions, you will see that it is 0, which again means that they are normal to each other.
    Rates of change of what?
    Now you have completely lost me. If you integrate the rate of change of the curve expression, surely you will get the curve expression back (plus or minus an arbitrary constant)?
     
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