Vector Function

A vector function is a function of the form
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with
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Continuity of Vector Function

The vector function $f$ is continuous at $t_0$ if $f_1,f_2,f_3$ are continuous at $t_0.$

Theorem $f$ is continuous at $t_0$ if and only if for each $\epsilon >0$ there exists $\delta >0$ such that MATH for MATH

Proof.
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If
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MATH

MATH
then MATH If MATH then
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MATH

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Differentiability of a Vector Function

A vector function $f$ is differentiable at $t_0$ if $f_1,f_2,f_3$ are differentiable at $t_0.$ In this case
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This is so since
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Differentiation Formula

Proof. Note that MATH Since the square root function MATH is differentiable for $x>0,$ it follows that the composite
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is differentiable. Differentiating the identity MATH we obtain
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so we have
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Integral

If $f$ is continuous on the interval $[a,b],$ we define
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Properties of Integral

Curve in $\QTR{bf}{R^3}$

Let the position of the curve be given by
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Its velocity vector is MATH and its acceleration vector is MATH Assume that MATH The unit tangent vector is
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Suppose that MATH The principal normal vector is
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Since $T(t)\cdot T(t)=1,$ it follows that
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This shows that the principal normal vector is perpendicular to the unit tangent vector.

Arc Length

The arc length of $r(u)$ for $u\in [t_0,t]$ is given by
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It follows from the Fundamental Theorem of Calculus that $s$ is differentiable and
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for all $t.$ The scalar function $s^{\prime }$ is called the speed of $r(t).$ Thus
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It follows that MATH for all $t,$ i.e., $s$ is an increasing function. Therefore the inverse function $s^{-1}$ of $s$ exists. (Thus $s^{-1}(t)$ is the unique $u$ such that $s(u)=t.$) The position vector $r$ may be regarded as a function of the arc length $s.$ Consider the composition
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Taking the derivative of $r$ with respect to $s:$
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Curvature

The curvature of a curve is the magnitude of the change in the tangent $T$ per unit length:
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Since
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it follows that
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and so
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Thus the curvature may be computed as function of $t.$

We have
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so
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MATH

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Consequently
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Normal

The tangent vector $T$ satisfies $T\cdot T=1.$ When this identity is differentiated with respect to $s,$ we obtain
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Therefore MATH At a point where $\kappa $MATH is nonzero, the principal unit normal is given by
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Thus
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Acceleration

Since the velocity $v$ satisfies
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the acceleration $a$ satisfies
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In this way, the quantity $\frac{d^2s}{dt^2}$ is expressed as the tangential component of acceleration while the quantity MATH is expressed as the normal component of acceleration. Since
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the tangential component of acceleration is
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The computation
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and the fact MATH show that
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Therefore, the normal component of acceleration may be expressed as
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Curvature for a Plane Curve

Suppose that the plane curve is given by $(x(t),y(t)).$ Regard it as a space curve MATH its curvature equals
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MATH

Suppose that the plane curve is the graph $y=f(x)$ of the function $f(x).$ It can be regarded as the parametric curve
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Hence its curvature equals
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Geometric Interpretation


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$\phi =$the angle that the tangent makes with the x-axis

Since MATH it follows that MATH so
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It follows from the chain rule that
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Hence
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Radius,Center,Circle of Curvature

The reciprocal
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of the curvature is called the radius of curvature. A point at a distance $\rho $ from the curve in the direction of principal normal is called the center of curvature. The circle with center at center of curvature and radius $\rho $ is called the osculating circle (also called the circle of curvature.) Thus the position vector of the center of curvature is MATH

Center of Curvature of a Plane Curve

Show that the center of curvature $(x_{c},y_{c})$ of a plane curve $(x,y)$ is given by
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Binormal

Let $T$ be that tangent and $B$ the normal of a curve. The vector
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is called the binormal. These vectors satisfy
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Differentiating both sides of the relation $B\cdot T=0$ with respect to $s$ we have:
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Therefore
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Differentiating both sides of the relation $B\cdot B=1$ with respect to $s$ we have:
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Therefore
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It follows that the vector $\frac{dB}{ds}$ is perpendicular to both $T$ and $B;$ we may therefore write
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for some scalar function $\tau $ which is known as the torsion of the curve. We now show how this function may be computed as function of $t:$Differentiating both sides of
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with respect to $t$ we obtain
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Since
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and since
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MATH

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it follows that
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Also we have
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Therefore
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Consequently
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Formula of Frenet

Differentiating $N\cdot N=1$ with respect to $s$ we have
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Therefore $\frac{dN}{ds}$ can be expressed as
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for some scalars MATH To determine these scalars we have
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and
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Therefore
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Summery:
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