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Curvature vs acceleration? (calc III)

  1. Sep 18, 2013 #1
    i asked this question before, but i didn't ask it quite right so i didn't get a satisfactory answer..

    curvature is define as how quickly/ abruptly a curve changes with respect to its arc length.

    okay so the normal vecor (N = T ') is the change in the tangent vector of a curve with respect to some parameter t.

    but conceptually i don't get the difference. if the parameter t = Const. nothing moves (everything is still) and so the space curve doesn't get to traverse freely through space. but when it does get to traverse, it traverses the path of the function it is set by and so, wouldn't [itex]\frac{dT}{dS}[/itex] [itex]\alpha[/itex] [itex]\frac{dT}{dt}[/itex] (variable 'S' = arc length)? i mean, when there is a sharp turn, the T(t) value with respect to the length of the curve will be high and also.. the change in the tangent vector with respect to the parameter will be high. i just don't see the difference between the two. they're both quantities involving the change in the tangent vector. and S is always proportional to t. except.. i understand dT/dt is. but i don't understand the difference between dT/dt and dT/dS. i already looked up curvature in my book, notes, and google, i still don't get the difference though.
  2. jcsd
  3. Sep 18, 2013 #2
    Yes, ##S## is the arc length: ##S(t)=\int_0^t|T(\tau)|\,d\tau## so ##S## is not always proportional to ##t##
  4. Sep 18, 2013 #3
    i don't get it, with the parameter (τ) that allows the space curve to traverse, if S is not proportional to τ, this implies that the arc length gets shorter with increasing τ. how is this possible? this would almost be like.. erasing an already existing curve. i can't make any other intuitive sense out of this based upon your response. is what i proposed correct though? the idea of erasing?
  5. Sep 19, 2013 #4
    When you're driving and you turn your car, you can feel yourself being pulled in a direction opposite the direction in which you are turning. If you go through the same turn at a higher speed, you feel that pull more. If you go more slowly, you feel less of a pull. The pull you feel is acceleration. It changes depending on your speed.

    The curve itself does not change. If you want to have a measure of how much the road curves, you need to have a parameter that is independent of speed.
  6. Sep 19, 2013 #5


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    No. If the parameterized curve does not have constant speed then its acceleration vector will have a non -zero component in the tangent direction. It will not be parallel to the normal.

    Try differentiating the speed function.
  7. Sep 19, 2013 #6


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    If ##T' = {dT \over dt}## is the vector rate of change of the unit vector of velocity, then ##|T'|## is the angular velocity of the unit vector of velocity. This is not obvious: ##ΔT## is a chord whose related arc has length ##θ##, the angle of rotation (##T## is a unit vector). ##|ΔT| ≈ θ## with equality in the limit.

    ##{ds \over dt} = |v|## is the speed of the particle, so ##{|T'| \over |v|} = |{dT \over ds}| = \kappa## is the ratio of these, the angular velocity divided by the speed. And if you realise that the angular velocity is proportional to the speed, ##\kappa## is just the constant of proportionality.

    So now, what is the difference between ##dT \over dt## and ##dT \over ds##? One is a vector giving the angular velocity of rotation and the plane and orientation of that rotation, and the other is the same vector with a new magnitude, angular velocity divided by speed, which is just the constant of their proportionality. It measures the curvature of course. The direction is in fact irrelevant, we only care about the magnitude.
  8. Sep 19, 2013 #7
    okay, i think this answers my question. probably just needs some time to sink in..
    but this agrees with what my professor said. he stated that the curvature is an intrinsic property of the curve itself and that it was independent of any other parameter the curve is put in terms of.
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