Curvature vs acceleration? (calc III)

In summary, curvature is a measure of how quickly and abruptly a curve changes with respect to its arc length. The normal vector is the change in the tangent vector with respect to a parameter, but there is a difference between the parameter and the arc length. The arc length is not always proportional to the parameter, and measures the acceleration of the curve. The difference between dT/dt and dT/ds is that one is a vector giving the angular velocity and the other is the same vector with a new magnitude, measuring the curvature. This confirms the statement that curvature is an intrinsic property of the curve itself and is independent of any other parameter.
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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.
 
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  • #2
Yes, ##S## is the arc length: ##S(t)=\int_0^t|T(\tau)|\,d\tau## so ##S## is not always proportional to ##t##
 
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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?
 
  • #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.
 
  • #5
iScience said:
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)?

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.
 
  • #6
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 realize 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.
 
  • #7
gopher_p said:
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.

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.
 

1. What is the difference between curvature and acceleration?

Curvature refers to the rate at which a curve changes direction, while acceleration refers to the rate at which an object's velocity changes. In other words, curvature is a measure of how "curvy" a curve is, while acceleration is a measure of how quickly an object is changing its speed or direction of motion.

2. How are curvature and acceleration related?

Curvature and acceleration are related because an object's acceleration is directly linked to the curvature of its path. This is because acceleration is caused by a change in direction, and the greater the change in direction, the greater the curvature of the path.

3. Can you have curvature without acceleration?

No, you cannot have curvature without acceleration. As mentioned before, acceleration is caused by a change in direction, and therefore, any curvature in a path will result in acceleration. However, it is possible to have acceleration without noticeable curvature if the change in direction is very small or gradual.

4. How is curvature calculated?

Curvature is calculated by taking the derivative of the tangent vector to a curve. In other words, it is the rate of change of the direction of the tangent vector. This value can also be expressed as the reciprocal of the radius of the circle that best fits the curve at a specific point.

5. What is the significance of curvature and acceleration in calculus?

Curvature and acceleration are important concepts in calculus because they help us understand the behavior of curves and objects in motion. They are also used in many real-world applications, such as in engineering, physics, and geometry. In addition, they are essential in understanding and solving problems in higher-level math courses, such as differential geometry and differential equations.

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