Vector Functions and Arc Length Parameter

Vector Functions

Definition of Vector Function

  • Describes the motion of a body in space over time.
  • Positions of the body at any time t can be expressed through parametric equations:
    x = f(t), y = g(t), z = h(t)
  • The positions of the body are (f(t), g(t), h(t)).
  • A vector function r(t) of a real variable t I is defined as the vector from the origin to the position \langle f(t), g(t), h(t) \rangle of the body:
    r(t) = \langle f(t), g(t), h(t) \rangle = f(t)i + g(t)j + h(t)k, t \t I
  • The functions f(t), g(t), and h(t) are called component functions and play a crucial role in studying a vector function.

Continuity of Vector Functions

  • Given a vector function r(t), then:
  • r(t) is continuous at the point t0 if \lim{t \to t0} r(t) = r(t0). That is:
    \lim{t \to t0} r(t) = [\lim{t \to t0} f(t)]i + [\lim{t \to t0} g(t)]j + [\lim{t \to t0} h(t)]k
  • r(t) is continuous if it is continuous for all t \t I.

Differentiability of Vector Functions

  • Given a vector function r(t), then:
  • r(t) will be differentiable at the point t_0 if the component functions are differentiable at the same point.
  • The derivative of r(t) is the vector:
    r'(t) = \frac{dr}{dt} = \lim_{h \to 0} \frac{r(t+h) - r(t)}{h} = \frac{df}{dt}i + \frac{dg}{dt}j + \frac{dh}{dt}k
  • r(t) will be differentiable if it is differentiable for all t \t I.

Smooth Curve

  • The curve traced by r(t) is called smooth if r'(t) is a continuous function and always different from the zero vector.
  • In other words, for a curve to be smooth, the component functions must have continuous first derivatives that are not simultaneously zero.

Velocity and Acceleration Vectors

  • Given that r(t) defines the position vector of a body moving in space and that the curve defined by it is smooth, then:
    1. The velocity vector is: v(t) = \frac{dr}{dt}
    2. The acceleration vector is: a(t) = \frac{dv}{dt} = \frac{d^2r}{dt^2}
    3. |v(t)| is the magnitude of the velocity of the body, while the vector \frac{v(t)}{|v(t)|} determines the direction of its motion.

Differentiation Rules

  • c' = 0, where c is a constant vector.
  • (u(t) \pm v(t))' = u'(t) \pm v'(t)
  • (cu(t))' = cu'(t), where c is a scalar constant.
  • (f(t)u(t))' = f'(t)u(t) + f(t)u'(t)
  • (u(t) \cdot v(t))' = u'(t) \cdot v(t) + u(t) \cdot v'(t)
  • (u(t) \times v(t))' = u'(t) \times v(t) + u(t) \times v'(t)
  • If u(s(t)), where s = f(t), then \frac{du}{dt} = \frac{du}{ds} \frac{ds}{dt} = u'(s)s'(t) = u'(f(t))f'(t)

Constant Magnitude

  • If r(t) is a differentiable vector function of t with constant magnitude, i.e., |r(t)| = C, where C is a constant, then its derivative r'(t) is a vector perpendicular to r(t), i.e., \frac{dr}{dt} \cdot r(t) = 0

Indefinite Integral

  • The indefinite integral of the vector function r(t) is:
    \int r(t) dt = R(t) + c
    where the vector function R(t) is an arbitrary antiderivative of r(t), while c is a constant vector function.

Definite Integral

  • Given r(t) = f(t)i + g(t)j + h(t)k and that the component functions are integrable on the interval \lbrack a, b \rbrack, then r(t) will be integrable on \lbrack a, b \rbrack and its integral is:
    \int{a}^{b} r(t) dt = \lbrack \int{a}^{b} f(t) dt \rbrack i + \lbrack \int{a}^{b} g(t) dt \rbrack j + \lbrack \int{a}^{b} h(t) dt \rbrack k

Vector Functions of Multiple Variables

  • The properties (continuity, derivative, etc.) of a vector function r(t) of one independent variable t are based on the corresponding properties of the component functions.
  • The same applies to vector functions of more than one variable, i.e., for a vector function of the form:
    r(t1, t2, …, tn) = f(t1, t2, …, tn)i + g(t1, t2, …, tn)j + h(t1, t2, …, tn)k
  • Example:
    F(x, y) = f(x, y)i + g(x, y)j = xi^2 + yj^2
    Then, the first partial derivatives will be the vector functions:
    \frac{\partial F}{\partial x} = \frac{\partial f}{\partial x}i + \frac{\partial g}{\partial x}j = xi^2 + yj
    \frac{\partial F}{\partial y} = \frac{\partial f}{\partial y}i + \frac{\partial g}{\partial y}j = yi + xj

Arc Length Parameter

Definition

  • The length of a smooth curve r(t) = f(t)i + g(t)j + h(t)k, \t \t \lbrack a, b \rbrack, formed as t increases from time t = a to t = b, is:
    L = \int{a}^{b} |v(t)| dt = \int{a}^{b} \sqrt{(\frac{df}{dt})^2 + (\frac{dg}{dt})^2 + (\frac{dh}{dt})^2} dt

  • If the reference point P(t_0) of a smooth curve c: r(t) is parameterized by t, then for every t, the point R(x(t), y(t), z(t)) is defined, as well as the distance:

    s(t) = \int{t0}^{t} |v(τ)| dτ

  • Which is measured on the curve, while the measurement starts from the reference point P(x(t0), y(t0), z(t_0)).

  • The s(t) in the above formula is called the arc length parameter, and the value of the parameter increases in the direction of increasing t.

Properties

  • Since the component functions of the velocity vector v(t) are continuous (the curve is smooth), from the formula of the arc length parameter, we obtain that \frac{ds}{dt} = |v(t)| > 0, i.e., that s(t) is an increasing function of t.
  • Therefore, if we have a curve parameterized as a function of the parameter t, then it may be possible to solve t as a function of s, i.e., t = t(s), which means that we can parameterize our curve with respect to the arc length parameter s.

Example

  • A body moves on the curve: r(t) = \frac{t^2}{2}i + \frac{t^2}{2}j + \frac{7t^2}{2}k
  • Calculate the distance traveled by the mobile from t = 0 to t = π.
  • If the reference point is at time t = 0, parameterize the curve with respect to the arc length.

Unit Tangent Vector

  • Given a smooth curve traced by a mobile, since the velocity vector is tangent to its trajectory, if we divide it by its magnitude, it will become a unit vector.
  • Therefore, the unit tangent vector to the curve, for every time t, will be:
    T = \frac{v}{|v|}
  • Since we can parameterize the curve with respect to the arc length, i.e., in this case, we will have that:
    r(s(t))
  • Therefore, when the curve is parameterized with respect to the arc length parameter, the unit tangent vector is:
    T = \frac{dr}{ds}

Curvature

  • Given a smooth curve traced by a mobile, the rate at which the unit tangent vector rotates with respect to the arc length is called curvature.

  • Consequently, the curvature κ of the curve is:
    κ = |\frac{dT}{ds}|

  • If our curve is now parameterized with respect to t, i.e., r(t), and since T = \frac{dr}{ds}, the curvature is calculated as follows:
    κ = |\frac{dT}{dt}| = |\frac{dT}{ds} \frac{ds}{dt}| = \frac{|\frac{dT}{dt}|}{|\frac{ds}{dt}|} = \frac{|\frac{dT}{dt}|}{|v(t)|}

  • The curvature is a measure of how much the curve deviates from being a straight line.

  • The curvature of a circle of radius R is constant and equal to \frac{1}{R}.

  • The parametric equations of a circle of radius R are:

    x = a \cos(t), y = a \sin(t), t \t \lbrack 0, 2π \rbrack

  • Which means that the vector function:
    r(t) = a \cos(t)i + a \sin(t)j
    describes the position vector of a body on the circumference of the given circle.
    \frac{dT}{dt} = -\cos(t)i - \sin(t)j
    |\frac{dT}{dt}| = 1

  • Therefore, the curvature will be equal to (we have parameterization with respect to t):
    κ = \frac{1}{a}

Unit Normal Vector

  • Given a smooth curve traced by a mobile, since the unit tangent vector has constant magnitude, its derivative will be a vector perpendicular to it.
  • Therefore, if we divide the vector of the derivative by its magnitude, then it will be unitary. This unit normal vector shows us the direction in which the curve rotates and is given by the formula:
    N = \frac{\frac{dT}{dt}}{|\frac{dT}{dt}|}
  • If we have parameterization of the curve with respect to s, we will have that:
    N = \frac{\frac{dT}{ds}}{|\frac{dT}{ds}|}

Osculating Circle

  • The osculating circle of a plane curve at a point is the circle in the plane that touches the curve at and has the same curvature as at and lies toward the concave side of the curve.

Radius of Curvature

  • The radius of curvature of the curve at a point is the radius of the osculating circle, which is equal to \frac{1}{κ}.

Center of Curvature

  • The center of curvature of a curve at a point is the center of the osculating circle.

Tangential and Normal Components of Acceleration

  • The acceleration vector can be easily analyzed into two components by differentiating the velocity vector with respect to time.

  • The result is:
    a = aT T + aN N
    where aT = \frac{d^2s}{dt^2} = \frac{d}{dt} |v| and aN = κ(\frac{ds}{dt})^2 = κ|v|^2

  • The acceleration vector is analyzed into two components, one tangential and one normal to the motion.

Frenet Frame

Binormal Vector

  • The unit tangent and unit normal vectors define a plane. Their outer product is perpendicular, by definition, to these two vectors.

  • A new unit vector is defined, called the unit binormal vector and is:

    B = T \times N

Frenet Frame

  • The vectors and define a right-handed system of mutually orthogonal unit vectors in space.
  • It is called the Frenet coordinate system or Frenet frame and describes the motion as follows:
    • : Describes the movement forward.
    • : The direction in which we turn.
    • : The tendency of the orbit to wind.

Torsion

  • The torsion function of a smooth curve is given by the formula: τ = -\frac{dB}{ds} \cdot N

  • It is a measure of how much the curve differs from lying in a plane.

  • The three unit vectors define the following three levels:

    • Osculating plane:
    • Normal plane:
    • Rectifying plane:

  • Describes the rate at which the osculating level rotates around the axis as the boby moves.

Formulas for Curvature and Torsion

  • It is proved that the curvature of a smooth curve can also be given by the formula:
    κ = \frac{|v \times a|}{|v|^3}
  • While, the torsion by the formula:
    τ = \frac{(v \times a) \cdot a'}{|v \times a|^2}
  • Where the symbol in the torsion formula corresponds to the derivative.