CAVIN OJIJO

October 14, 2011

kinematics

Kinematics (from Greek κινεῖν, kinein, to move) is the branch of classical mechanics that describes the motion of bodies (objects) and systems (groups of objects) without consideration of the forces that cause the motion.^[1]^[2]^[3]^[4]

Kinematics is not to be confused with another branch of classical mechanics: analytical dynamics (the study of the relationship between the motion of objects and its causes), sometimes subdivided into kinetics (the study of the relation between external forces and motion) and statics (the study of the relations in a system at equilibrium). Kinematics also differs from dynamics as used in modern-day physics to describe time-evolution of a system.

The term kinematics is less common today than in the past, but still has a role in physics.^[5] (See analytical dynamics for more detail on usage). The term kinematics also finds use in biomechanics and animal locomotion.^[6] Further, mathematicians that include time as a parameter in geometry have developed the subject of kinematic geometry.

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Linear motion

Linear or translational kinematics^[7]^[8] is the description of the motion in space of a point along a line, also known as a trajectory or path.^{[note 1]} This path can be either straight (rectilinear) or curved (curvilinear).

[edit] Particle kinematics

Particle kinematics is the study of the kinematics of a single particle. The results obtained in particle kinematics are used to study the kinematics of collection of particles, dynamics and in many other branches of mechanics.

[edit] Position and reference frames

The position of a point in space is the most fundamental idea in particle kinematics. To specify the position of a point, one must specify three things: the reference point (often called the origin), distance from the reference point and the direction in space of the straight line from the reference point to the particle. Exclusion of any of these three parameters renders the description of position incomplete. Consider for example a tower 50 m south from your home. The reference point is home, the distance 50 m and the direction south. If one only says that the tower is 50 m south, the natural question that arises is "from where?" If one says that the tower is southward from your home, the question that arises is "how far?" If one says the tower is 50 m from your home, the question that arises is "in which direction?" Hence, all these three parameters are crucial to defining uniquely the position of a point in space.

Position is usually described by mathematical quantities that have all these three attributes: the most common are vectors and complex numbers. Usually, only vectors are used. For measurement of distances and directions, usually three dimensional coordinate systems are used with the origin coinciding with the reference point. A three-dimensional coordinate system (whose origin coincides with the reference point) with some provision for time measurement is called a reference frame or frame of reference or simply frame. All observations in physics are incomplete without the reference frame being specified.

[edit] Position vector

The position vector of a particle is a vector drawn from the origin of the reference frame to the particle. It expresses both the distance of the point from the origin and its sense from the origin. In three dimensions, the position of point A can be expressed as

$\mathbf{r}_A = (x_A,y_A,z_A),$

where x_A, y_A, and z_A are the Cartesian coordinates of the point. The magnitude of the position vector |r| gives the distance between the point A and the origin.

$|\mathbf{r}| = \sqrt{x_A^{\ 2} + y_A^{\ 2} + z_A^{\ 2}}.$

The direction cosines of the position vector provide a quantitative measure of direction. It is important to note that the position vector of a particle isn't unique. The position vector of a given particle is different relative to different frames of reference.

[edit] Rest and motion

Once the concept of position is firmly established, the ideas of rest and motion naturally follow. If the position vector of the particle (relative to a given reference frame) changes with time, then the particle is said to be in motion with respect to the chosen reference frame. However, if the position vector of the particle (relative to a given reference frame) remains the same with time, then the particle is said to be at rest with respect to the chosen frame. Note that rest and motion are relative to the reference frame chosen. It is quite possible that a particle at rest relative to a particular reference frame is in motion relative to the other. Hence, rest and motion aren't absolute terms, rather they are dependent on reference frame. For example, a passenger in a moving car may be at rest with respect to the car, but in motion with respect to the road.

[edit] Path

A particle's path is the locus between its beginning and end points which is reference-frame dependent. The path of a particle may be rectilinear (straight line) in one frame, and curved in another.

[edit] Displacement

Displacement is a vector describing the difference in position between two points, i.e. it is the change in position the particle undergoes during the time interval. If point A has position r_A = (x_A,y_A,z_A) and point B has position r_B = (x_B,y_B,z_B), the displacement r_AB of B from A is given by

$\mathbf{r}_{AB} = \mathbf{r}_B - \mathbf{r}_A = (x_B-x_A,y_B-y_A,z_B-z_A).$

Geometrically, displacement is the shortest distance between the points A and B. Displacement, distinct from position vector, is independent of the reference frame. This can be understood as follows: the positions of points is frame dependent, however, the shortest distance between any pair of points is invariant on translation from one frame to another (barring relativistic cases).

The distance traveled is always greater than or equal to the displacement.

[edit] Distance

Distance is a scalar quantity, describing the length of the path between two points along which a particle has travelled.

When considering the motion of a particle over time, distance is the length of the particle's path and may be different from displacement, which is the change from its initial position to its final position. For example, a race car traversing a 10 km closed loop from start to finish travels a distance of 10 km; its displacement, however, is zero because it arrives back at its initial position.

If the position of the particle is known as a function of time (r = r(t)), the distance s it travels from time t₁ to time t₂ can be found by

$s = \int_{t_1}^{t_2} |d\mathbf{r}| = \int_{t_1}^{t_2} ds =\int_{t_1}^{t_2} \sqrt{dx^2 + dy^2 + dz^2} = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}\; dt.$

The formula utilizes the fact that over an infinitesimal time interval, the magnitude of the displacement equals the distance covered in that interval. This is analogous to the geometric fact that infinitesimal arcs on a curved line coincide with the chord drawn between the ends of the arc itself.

[edit] Velocity and speed

Average velocity is defined as

$\overline{\mathbf{v}} = \frac {\Delta \mathbf{r}}{\Delta t} \ ,$

where Δr is the change in position and Δt is the interval of time over which the position changes. The direction of v is same as the direction of the change in position Δr as Δt>0.

Velocity is the measure of the rate of change in position with respect to time, that is, how the distance of a point changes with each instant of time. Velocity also is a vector. Instantaneous velocity (the velocity at an instant of time) can be defined as the limiting value of average velocity as the time interval Δt becomes smaller and smaller. Both Δr and Δt approach zero but the ratio v approaches a non-zero limit v. That is,

$\mathbf{v} = \lim_{\Delta t\rightarrow0}\frac{\Delta\mathbf{r}}{\Delta t} = \frac {d \mathbf{r}}{d t} \, ,$

where dr is an infinitesimally small displacement and dt is an infinitesimally small length of time.^{[note 2]} As per its definition in the derivative form, velocity can be said to be the time rate of change of position. Further, as dr is tangential to the actual path, so is the velocity.

As a position vector itself is frame dependent, velocity is also dependent on the reference frame.

The speed of an object is the magnitude |v| of its velocity. It is a scalar quantity:

$|\mathbf{v}| = \left|\frac {d \mathbf{r}}{d t} \right| = \frac {d s}{d t}$

The distance traveled by a particle over time is a non-decreasing quantity. Hence, ds/dt is non-negative, which implies that speed is also non-negative.

[edit] Acceleration

Average acceleration (acceleration over a length of time) is defined as:

$\overline{\mathbf{a}} = \frac {\Delta \mathbf{v}}{\Delta t} \ ,$

where Δv is the change in velocity and Δt is the interval of time over which velocity changes.

Acceleration is the vector quantity describing the rate of change with time of velocity. Instantaneous acceleration (the acceleration at an instant of time) is defined as the limiting value of average acceleration as Δt becomes smaller and smaller. Under such a limit, a → a.

$\mathbf{a} = \lim_{\Delta t \rightarrow 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac {d \mathbf{v}}{d t} \, ,$

where dv is an infinitesimally small change in velocity and dt is an infinitesimally small length of time.

[edit] Types of motion based on velocity and acceleration

If the acceleration of a particle is zero, then the velocity of the particle is constant over time and the motion is said to be uniform. Otherwise, the motion is non-uniform.

If the acceleration is non-zero but constant, the motion is said to be motion with constant acceleration. On the other hand, if the acceleration is variable, the motion is called motion with variable acceleration. In motion with variable acceleration, the rate of change of acceleration is called the jerk.

[edit] Integral relations

The above definitions can be inverted by mathematical integration to find:

$\mathbf{v}(t) =\mathbf{v}_0 + \int_{t_0}^t \mathbf{a}(t) \; dt$

$\begin{align} \mathbf{r}(t) &=\mathbf{r}_0 + \int_{t_0}^t \mathbf{v}(t) \; dt \\ &= \mathbf{r}_0 + \mathbf{v}_0 t + \int_{t_0}^t \left[\int_{t_0}^{t} \mathbf{a}(t) dt \right]\; dt \\ \end{align}$

[edit] Kinematics of constant acceleration

Many physical situations can be modeled as constant-acceleration processes, such as projectile motion.

Integrating acceleration a with respect to time t gives the change in velocity. When acceleration is constant both in direction and in magnitude, the point is said to be undergoing uniformly accelerated motion. In this case, the integral relations can be simplified:

$\mathbf{v}(t) = \int_0^{t} \mathbf{a} \; dt' = \mathbf{v}_0 + \mathbf{a}t.$

$\begin{align} \mathbf{r}(t) &= \mathbf{r}_0 + \int_0^t \mathbf{v} \; dt' = \mathbf{r}_0 + \int_0^t (\mathbf{v}_0 + \mathbf{a} t) \; dt' \\ &= \mathbf{r}_0 + \mathbf{v}_0 t + \tfrac{1}{2} \mathbf{a} t^2. \end{align}$

Additional relations between displacement, velocity, acceleration, and time can be derived. Since a = (v − v₀)/t,

$\mathbf{r}(t) = \mathbf{r}_0 + \left(\frac{\mathbf{v}+ \mathbf{v_0}}{2}\right) t .$

By using the definition of an average, this equation states that when the acceleration is constant average velocity times time equals displacement.

A relationship without explicit time dependence may also be derived for one-dimensional motion. Noting that at = v − v₀,

$(\mathbf{r} - \mathbf{r_0}) \cdot \mathbf{a} t = \left( \mathbf{v} - \mathbf{v}_0 \right) \cdot \frac{\mathbf{v} + \mathbf{v}_0}{2} t \ ,$

where · denotes the dot product. Dividing the t on both sides and carrying out the dot-products:

$2(\mathbf{r} - \mathbf{r_0}) \cdot \mathbf{a} = v^2 - v_0^{\ 2}.$

In the case of straight-line motion, (r - r₀) is parallel to a. Then

$v^2= v_0^2 + 2 a(r-r_0).$

This relation is useful when time is not known explicitly.

[edit] Relative velocity

Main article: Relative velocity

To describe the motion of object A with respect to object B, when we know how each is moving with respect to a reference object O, we can use vector algebra. Choose an origin for reference, and let the positions of objects A, B, and O be denoted by r_A, r_B, and r_O. Then the position of A relative to the reference object O is

$\mathbf{r}_{A/O} = \mathbf{r}_{A} - \mathbf{r}_{O} \,\!$

Consequently, the position of A relative to B is

$\mathbf{r}_{A/B} = \mathbf{r}_A - \mathbf{r}_B = \mathbf{r}_A - \mathbf{r}_O - \left(\mathbf{r}_B-\mathbf{r}_O\right) = \mathbf{r}_{A/O}-\mathbf{r}_{B/O} \ .$

The above relative equation states that the motion of A relative to B is equal to the motion of A relative to O minus the motion of B relative to O. It may be easier to visualize this result if the terms are re-arranged:

$\mathbf{r}_{A/O} = \mathbf{r}_{A/B} + \mathbf{r}_{B/O} \ ,$

or, in words, the motion of A relative to the reference is that of B plus the relative motion of A with respect to B. These relations between displacements become relations between velocities by simple time-differentiation, and a second differentiation makes them apply to accelerations.

For example, let Ann move with velocity $\mathbf{V}_{A}$ relative to the reference (we drop the O subscript for convenience) and let Bob move with velocity $\mathbf{V}_{B}$ , each velocity given with respect to the ground (point O). To find how fast Ann is moving relative to Bob (we call this velocity $\mathbf{V}_{A/B}$ ), the equation above gives:

$\mathbf{V}_{A} = \mathbf{V}_{B} + \mathbf{V}_{A/B} \,\! .$

To find $\mathbf{V}_{A/B}$ we simply rearrange this equation to obtain:

$\mathbf{V}_{A/B} = \mathbf{V}_{A} -\mathbf{V}_{B} \,\! .$

At velocities comparable to the speed of light, these equations are not valid. They are replaced by equations derived from Einstein's theory of special relativity.

[show]Example: Rectilinear (1D) motion

[show]Example: Projectile (2D) motion

Kinematics is the study of how things move. Here, we are interested in the motion of normal objects in our world. A normal object is visible, has edges, and has a location that can be expressed with (x, y, z) coordinates. We will not be discussing the motion of atomic particles or black holes or light.

We will create a vocabulary and a group of mathematical methods that will describe this ordinary motion. Understand that we will be developing a language for describing motion only. We won't be concerned with what is causing or changing the motion, or more correctly, the momentums of the objects. In other words, we are not concerned with the action of forces within this topic.

[edit] Rotational motion

Main article: Circular motion

Figure 1: The angular velocity vector Ω points up for counterclockwise rotation and down for clockwise rotation, as specified by the right-hand rule. Angular position θ(t) changes with time at a rate ω(t) = dθ/dt.

Rotational or angular kinematics is the description of the rotation of an object.^[9] The description of rotation requires some method for describing orientation, for example, the Euler angles. In what follows, attention is restricted to simple rotation about an axis of fixed orientation. The z-axis has been chosen for convenience.

Description of rotation then involves these three quantities:

Angular position: The oriented distance from a selected origin on the rotational axis to a point of an object is a vector r ( t ) locating the point. The vector r(t) has some projection (or, equivalently, some component) r_⊥(t) on a plane perpendicular to the axis of rotation. Then the angular position of that point is the angle θ from a reference axis (typically the positive x-axis) to the vector r_⊥(t) in a known rotation sense (typically given by the right-hand rule).
Angular velocity: The angular velocity ω is the rate at which the angular position θ changes with respect to time t:

$\mathbf{\omega} = \frac {\mathrm{d}\theta}{\mathrm{d}t}$

The angular velocity is represented in Figure 1 by a vector Ω pointing along the axis of rotation with magnitude ω and sense determined by the direction of rotation as given by the right-hand rule.

Angular acceleration: The magnitude of the angular acceleration α is the rate at which the angular velocity ω changes with respect to time t:

$\mathbf{\alpha} = \frac {\mathrm{d}\mathbf{\omega}}{\mathrm{d}t}$

The equations of translational kinematics can easily be extended to planar rotational kinematics with simple variable exchanges:

$\omega_{\mathrm{f}} = \omega_{\mathrm{i}} + \alpha t\!$

$\theta_{\mathrm{f}} - \theta_{\mathrm{i}} = \omega_{\mathrm{i}} t + \tfrac{1}{2} \alpha t^2$

$\theta_{\mathrm{f}} - \theta_{\mathrm{i}} = \tfrac{1}{2} (\omega_{\mathrm{f}} + \omega_{\mathrm{i}})t$

$\omega_{\mathrm{f}}^2 = \omega_{\mathrm{i}}^2 + 2 \alpha (\theta_{\mathrm{f}} - \theta_{\mathrm{i}}).$

Here θ_i and θ_f are, respectively, the initial and final angular positions, ω_i and ω_f are, respectively, the initial and final angular velocities, and α is the constant angular acceleration. Although position in space and velocity in space are both true vectors (in terms of their properties under rotation), as is angular velocity, angle itself is not a true vector.

[edit] Point object in circular motion

[edit] Coordinate systems

In any given situation, the most useful coordinates may be determined by constraints on the motion, or by the geometrical nature of the force causing or affecting the motion. Thus, to describe the motion of a bead constrained to move along a circular hoop, the most useful coordinate may be its angle on the hoop. Similarly, to describe the motion of a particle acted upon by a central force, the most useful coordinates may be polar coordinates. Polar coordinates are extended into three dimensions with either the spherical polar or cylindrical polar coordinate systems. These are most useful in systems exhibiting spherical or cylindrical symmetry respectively.

[edit] Fixed rectangular coordinates

In this coordinate system, vectors are expressed as an addition of vectors in the x, y, and z direction from a non-rotating origin. Usually i, j, k are unit vectors in the x-, y-, and z-directions.

The position vector, r, the velocity vector, v, and the acceleration vector, a are expressed using rectangular coordinates in the following way:

$\mathbf{r} = x\, \hat {\mathbf{i}} + y \, \hat {\mathbf{j}} + z \, \hat {\mathbf{ k}} \, \!$

$\mathbf{v} = \dot {\boldsymbol{ r}} = \dot {x} \, \hat {\mathbf{ i}} + \dot {y} \, \hat {\mathbf{ j}} + \dot {z} \, \hat {\mathbf{ k}} \, \!$

$\mathbf{a} = \ddot {\boldsymbol{ r}} = \ddot {x} \, \hat {\mathbf{ i}} + \ddot {y} \, \hat {\mathbf{ j}} + \ddot {z} \, \hat {\mathbf{ k}} \, \!$

Note: $\dot {x} = \frac{\mathrm{d}x}{\mathrm{d}t}$ , $\ddot {x} = \frac{\mathrm{d}^2x}{\mathrm{d}t^2}$

[edit] Two-dimensional rotating reference frame

[edit] Derivatives of unit vectors

The position, velocity, and acceleration vectors of a given point can be expressed using these coordinate systems, but we have to be a bit more careful than we do with fixed frames of reference. Since the frame of reference is rotating, the unit vectors also rotate, and this rotation must be taken into account when taking the derivative of any of these vectors. If the coordinate frame is rotating at angular rate ω in the counterclockwise direction (that is, Ω = ω k using the right hand rule) then the derivatives of the unit vectors are as follows:

$\dot{\hat {\mathbf{ i}}} = \omega \hat {\mathbf{ k}} \times \hat {\mathbf{ i}} = \omega\hat {\mathbf{ j}}$

$\dot{\hat {\mathbf{ j}}} = \omega \hat {\mathbf{ k}}\times \hat {\mathbf{ j}} = - \omega \hat {\mathbf{ i}}$

[edit] Position, velocity, and acceleration

Given these identities, we can now figure out how to represent the position, velocity, and acceleration vectors of a particle using this reference frame.

[edit] Position

Position is straightforward:

$\boldsymbol{ r} = x \ \hat {\mathbf{ i}} + y \ \hat {\mathbf{ j}}$

It is just the distance from the origin in the direction of each of the unit vectors.

[edit] Velocity

Velocity is the time derivative of position:

$\mathbf{v} = \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} = \frac{\mathrm{d} (x \ \hat {\mathbf{ i}})}{\mathrm{d}t} + \frac{\mathrm{d} (y \ \hat {\mathbf{ j}})}{\mathrm{d}t}$

By the product rule, this is:

$\mathbf{v} = \dot x \ \hat {\mathbf{ i}} + x \dot{\ \hat {\mathbf{ i}}} + \dot y \ \hat {\mathbf{ j}} + y \dot{ \ \hat {\mathbf{ j}}}$

Which from the identities above we know to be:

$\mathbf{v} = \dot x \ \hat {\mathbf{ i}} + x \omega \ \hat {\mathbf{ j}} + \dot y \ \hat {\mathbf{ j}} - y \omega \ \hat {\mathbf{ i}} = (\dot x - y \omega) \ \hat {\mathbf{ i}} + (\dot y + x \omega) \ \hat {\mathbf{ j}}$

or equivalently

$\mathbf{v} = (\dot x \ \hat {\mathbf{ i}} + \dot y \ \hat {\mathbf{ j}}) + (y \dot{ \hat {\mathbf{ j}}} + x \dot{\hat {\mathbf{ i}}}) = \mathbf{v}_{rel} + \boldsymbol{\Omega} \times \mathbf{r}$

where v_rel is the velocity of the particle relative to the rotating coordinate system.

[edit] Acceleration

Acceleration is the time derivative of velocity.

We know that:

$\boldsymbol{ a} = \frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{ v} = \frac{\mathrm{d} \boldsymbol{ v}_{rel}}{\mathrm{d}t} + \frac{\mathrm{d}}{\mathrm{d}t} \boldsymbol{\Omega} \times \boldsymbol{r}$

Consider the $\stackrel{\frac{ \mathrm{d} } { \mathrm{d} t }}{}$ $\boldsymbol{ v}_{rel}$ part. $\boldsymbol{ v}_{rel}$ has two parts we want to find the derivative of: the relative change in velocity ( $\boldsymbol{ a}_{rel}$ ), and the change in the coordinate frame

( $\boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$ ).

$\frac{\mathrm{d} \boldsymbol{ v}_{rel}}{\mathrm{d}t} = \boldsymbol{ a}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

Next, consider $\stackrel{\frac{\mathrm{d}}{\mathrm{d}t}}{}$ $(\boldsymbol{\Omega} \times\boldsymbol{ r})$ . Using the chain rule:

$\frac{\mathrm{d} (\boldsymbol{\Omega} \times \boldsymbol{ r})}{\mathrm{d}t} = \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times \dot{\boldsymbol{ r}}$

$\dot{\boldsymbol{ r}}=\boldsymbol{ v}=\boldsymbol{ v}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ r}$ from above:

$\frac{\mathrm{d} (\boldsymbol{\Omega} \times \boldsymbol{ r})}{\mathrm{d}t} = \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times\boldsymbol{ r}) + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

So all together:

$\boldsymbol{ a} = \boldsymbol{ a}_{rel} + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel} + \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{ r}) + \boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}$

And collecting terms:^[10]

$\boldsymbol{ a} = \boldsymbol{ a}_{rel} + 2(\boldsymbol{\Omega} \times \boldsymbol{ v}_{rel}) + \dot{\boldsymbol{\Omega}} \times \boldsymbol{ r} + \boldsymbol{\Omega} \times (\boldsymbol{\Omega} \times \boldsymbol{ r})\ .$

[edit] Kinematic constraints

This section requires expansion.

A kinematic constraint is any condition relating properties of a dynamic system that must hold true at all times. Below are some common examples:

[edit] Rolling without slipping

An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass,

$\boldsymbol{ v}_G(t) = \boldsymbol{\Omega} \times \boldsymbol{ r}_{G/O}$ .

For the case of an object that does not tip or turn, this reduces to v = R ω.

[edit] Inextensible cord

This is the case where bodies are connected by an idealized cord that remains in tension and cannot change length. The constraint is that the sum of lengths of all segments of the cord is the total length, and accordingly the time derivative of this sum is zero. See Kelvin and Tait^[11]^[12] and Fogiel.^[13] A dynamic problem of this type is the pendulum. Another example is a drum turned by the pull of gravity upon a falling weight attached to the rim by the inextensible cord.^[14] An equilibrium problem (not kinematic) of this type is the catenary.^[15]

[edit] See also

	Science portal
	Physics portal

[edit] Notes

^ In mathematics, a line refers to a straight trajectory, and a curve to a trajectory which may have curvature. In mechanics and kinematics, "line' and "curve" both refer to any trajectory, in particular a line may be a complex curve in space. Any position along a specified trajectory can be described by a single coordinate, the distance traversed along the path, or arc length. The motion of a particle along a trajectory can be described by specifying the time dependence of its position, for example by specification of the arc length locating the particle at each time t. The following words refer to curves and lines:
- "linear" (= along a straight or curved line;
- "rectilinear" (= along a straight line, from Latin rectus = straight, and linere = spread),
- "curvilinear" (=along a curved line, from Latin curvus = curved, and linere = spread).
^ Because magnitude of dr is necessarily the distance between two infinitesimally spaced points along the trajectory of the point, it is the same as an increment in arc length along the path of the point, customarily denoted ds.

[edit] References

Moon, Francis C. (2007). The Machines of Leonardo Da Vinci and Franz Reuleaux, Kinematics of Machines from the Renaissance to the 20th Century. Springer. ISBN 9781402055980.

^ Edmund Taylor Whittaker & William McCrea (1988). A Treatise on the Analytical Dynamics of Particles and Rigid Bodies. Cambridge University Press. Chapter 1. ISBN 0521358833.
^ Joseph Stiles Beggs (1983). Kinematics. Taylor & Francis. p. 1. ISBN 0891163557.
^ O. Bottema & B. Roth (1990). Theoretical Kinematics. Dover Publications. reface. ISBN 0486663469.
^ Thomas Wallace Wright (1896). Elements of Mechanics Including Kinematics, Kinetics and Statics. E and FN Spon. Chapter 1.
^ See, for example: Russell C. Hibbeler (2009). "Kinematics and kinetics of a particle". Engineering Mechanics: Dynamics (12th ed.). Prentice Hall. p. 298. ISBN 0136077919., Ahmed A. Shabana (2003). "Reference kinematics". Dynamics of Multibody Systems (2nd ed.). Cambridge University Press. ISBN 0521544114., P. P. Teodorescu (2007). "Kinematics". Mechanical Systems, Classical Models: Particle Mechanics. Springer. p. 287. ISBN 1402054416.
^ A. Biewener (2003). Animal Locomotion. Oxford University Press. ISBN 19850022X.
^ James R. Ogden & Max Fogiel (1980). The Mechanics Problem Solver. Research and Education Association. p. 184. ISBN 0878915192.
^ R. Douglas Gregory (2006). Classical Mechanics: An Undergraduate Text. Cambridge UK: Cambridge University Press. Chapter 2. ISBN 0521826780.
^ R. Douglas Gregory (2006). Chapter 16. Cambridge: Cambridge University. ISBN 0521826780.
^ R. Douglas Gregory (2006). pp. 475-476. Cambridge: Cambridge University. ISBN 0521826780.
^ William Thomson Kelvin & Peter Guthrie Tait (1894). Elements of Natural Philosophy. Cambridge University Press. p. 4. ISBN 1573929840.
^ William Thomson Kelvin & Peter Guthrie Tait (1894). op. cit.. p. 296.
^ M. Fogiel (1980). "Problem 17-11". The Mechanics Problem Solver. Research & Education Assoc.. p. 613. ISBN 0878915192.
^ Irving Porter Church (1908). Mechanics of Engineering. Wiley. p. 111. ISBN 1110365276.
^ Morris Kline (1990). Mathematical Thought from Ancient to Modern Times. Oxford University Press. p. 472. ISBN 0195061365.

[edit] External links

Java applet of 1D kinematics
Flash animated tutorial for 1D kinematics
Physclips: Mechanics with animations and video clips from the University of New South Wales
physicsfunda.googlepages.com, Kinematics for High School ant IIT JEE level
Kinematic Models for Design Digital Library (KMODDL)
Movies and photos of hundreds of working mechanical-systems models at Cornell University. Also includes an e-book library of classic texts on mechanical design and engineering.

Look up kinematics in Wiktionary, the free dictionary.

[show]v · d · eKinematics

kinemetics in also a branch of physics

October 13, 2011

he Mie solution to Maxwell's equations (also known as the Lorenz–Mie solution, the Lorenz–Mie–Debye solution or Mie scattering) describes the scattering of electromagnetic radiation by a sphere. The solution takes the form of an analytical infinite series.

The term "Mie theory" is used on occasion but it is misleading, since it does not refer to an independent physical theory or law. The phrase "the Mie solution (to Maxwell's equations)" is therefore preferable. Currently, the term "Mie solution" is also used in broader contexts, for example when discussing solutions of Maxwell's equations for scattering by stratified spheres or by infinite cylinders, or generally when dealing with scattering problems solved using the exact Maxwell equations in cases where one can write separate equations for the radial and angular dependence of solutions.

A modern formulation of the Mie solution to the scattering problem on a sphere can be found in many books, e.g., in J. A. Stratton's Electromagnetic Theory.^[1] In this formulation, the incident plane wave as well as the scattering field is expanded into radiating spherical vector wave functions. The internal field is expanded into regular spherical vector wave functions. By enforcing the boundary condition on the spherical surface, the expansion coefficients of the scattered field can be computed.

For particles much larger or much smaller than the wavelength of the scattered light there are simple and excellent approximations that suffice to describe the behaviour of the system. But for objects whose size is similar to the wavelength (e.g., water droplets in the atmosphere, latex particles in paint, droplets in emulsions including milk, and biological cells and cellular components) this more exact approach is necessary.^[2]

The Mie solution^[3] is named after its developer, German physicist Gustav Mie. Danish physicist Ludvig Lorenz and others independently developed the theory of electromagnetic plane wave scattering by a dielectric sphere.

The formalism allows the calculation of the electric and magnetic fields inside and outside a spherical object and is generally used to calculate either how much light is scattered, the total optical cross section, or where it goes, the form factor. The notable features of these results are the Mie resonances, sizes that scatter particularly strongly or weakly.^[4] This is in contrast to Rayleigh scattering for small particles and Rayleigh–Gans–Debye scattering (after Lord Rayleigh, R. Gans and Peter Debye) for large particles. The existence of resonances and other features of Mie scattering, make it a particularly useful formalism when using scattered light to measure particle size.

[edit] Mie scattering codes

Mie solutions are implemented in number of codes written in different computer langauges such as Fortran, Matlab, Mathematica. These solutions are in terms of infinite series and include calculation of scattering phase function, extinction, scattering, and absorption efficiencies, and other parameters such as asymmetry of parameter or radiation torque. Current useage of "Mie solution" indicate series approximation to solution of Maxwell's equations. There are several known objects which allow such a solutions: spheres, concentric spheres, infinite cylinders, cluster of spheres and cluster of cylinders, there are also known series solutions for scattering on ellipsoidal particles. For list of these specialized codes examine these articles

Codes for electromagnetic scattering by spheres - solutions for single sphere, coated spheres, multilayer sphere, cluster of spheres
Codes for electromagnetic scattering by cylinders - solutions for single cylinder, multilayer cylinders, cluster of cylinders

[edit] Approximations

[edit] Rayleigh approximation (scattering)

The change of sky colour at sunset (red nearest the sun, blue furthest away) is caused by Rayleigh scattering by atmospheric gas particles which are much smaller than the wavelengths of visible light. The grey/white colour of the clouds is caused by Mie scattering by water droplets which are of a comparable size to the wavelengths of visible light.

Rayleigh scattering describes the elastic scattering of light by spheres which are much smaller than the wavelength of light. The intensity of the scattered radiation is given by

$I = I_0 \left( \frac{ 1+\cos^2 \theta }{2 R^2} \right) \left( \frac{ 2 \pi }{ \lambda } \right)^4 \left( \frac{ n^2-1}{ n^2+2 } \right)^2 \left( \frac{d}{2} \right)^6,$

where R is the distance between the particle and the observer, θ is the scattering angle, n is the refractive index of the particle, and d is the diameter of the particle.

It can be seen from the above equation that Rayleigh scattering is strongly dependent upon the size of the particle and the wavelengths. The intensity of the Rayleigh scattered radiation increases rapidly as the ratio of particle size to wavelength increases. Furthermore, the intensity of Rayleigh scattered radiation is identical in the forward and reverse directions.

The Rayleigh scattering model breaks down when the particle size becomes larger than around 10% of the wavelength of the incident radiation. In the case of particles with dimensions greater than this, Mie's scattering model can be used to find the intensity of the scattered radiation. The intensity of Mie scattered radiation is given by the summation of an infinite series of terms rather than by a simple mathematical expression. It can be shown, however, that Mie scattering differs from Rayleigh scattering in several respects; it is roughly independent of wavelength and it is larger in the forward direction than in the reverse direction. The greater the particle size, the more of the light is scattered in the forward direction.

The blue colour of the sky results from Rayleigh scattering, as the size of the gas particles in the atmosphere is much smaller than the wavelength of visible light. Rayleigh scattering is much greater for blue light than for other colours due to its shorter wavelength. As sunlight passes through the atmosphere, its blue component is Rayleigh scattered strongly by atmospheric gases but the longer wavelength (e.g. red/yellow) components are not. The sunlight arriving directly from the sun therefore appears to be slightly yellow while the light scattered through rest of the sky appears blue. During sunrises and sunsets, the Rayleigh scattering effect is much more noticeable due to the larger volume of air through which sunlight passes.

In contrast, the water droplets which make up clouds are of a comparable size to the wavelengths in visible light, and the scattering is described by Mie's model rather than that of Rayleigh. Here, all wavelengths of visible light are scattered approximately identically and the clouds therefore appear to be white or grey.

[edit] Rayleigh Gans Approximation

The Rayleigh Gans Approximation is an approximate solution to light scattering when the relative refractive index of the particle is close to unity, and its size is much smaller in comparison to the wavelength of light divided by |m − 1|, where m is the refractive index.

[edit] Anomalous diffraction approximation of van de Hulst

Anomalous diffraction approximation is valid for large and optically soft spheres. Extinction efficiency in this approximation is given by

$Q = 2 - \frac{4}{p} \sin{p} + \frac{4}{p^2} (1-\cos{p})$ ,

where Q is the efficiency factor of scattering, which is defined as the ratio of the scattering cross section and geometrical cross section πa²;
p = 4πa(n – 1)/λ has a physical meaning of the phase delay of the wave passed through the centre of the sphere;
a is the sphere radius, n is the ratio of refractive indices inside and outside of the sphere, and λ the wavelength of the light.
This set of equations, first described by nan de Hulst (1957). ^[4].

[edit] Applications

Mie theory is very important in meteorological optics, where diameter-to-wavelength ratios of the order of unity and larger are characteristic of many problems regarding haze and cloud scattering. A further application is in the characterization of particles via optical scattering measurements. The Mie solution is also important for understanding the appearance of common materials like milk, biological tissue and latex paint.

[edit] Atmospheric science

Mie scattering occurs when the particles in the atmosphere are the same size as the wavelengths being scattered. Dust, pollen, smoke and water vapour are common causes of Mie scattering which tends to affect longer wavelengths. Mie scattering occurs mostly in the lower portions of the atmosphere where larger particles are more abundant, and dominates when cloud conditions are overcast.

[edit] Cancer detection and screening

Mie theory has been used to determine if scattered light from tissue corresponds to healthy or cancerous cell nuclei using angle-resolved low-coherence interferometry.

[edit] Particle sizing

Mie theory has been used in the detection of oil concentration in polluted water.^[5]^[6]

[edit] Parasitology

It has also been used to study the structure of Plasmodium falciparum, the most important cause of malaria.^[7]

[edit] See Also

[edit] References

^ J. A. Stratton: Electromagnetic Theory, New York: McGraw-Hill, 1941.
^ C. F. Bohren,D.R. Huffmann: Absorption and scattering of light by small particles.New York, Wiley-Interscience,2010 ISBN 3527406646
^ G. Mie, "Beiträge zur Optik trüber Medien, speziell kolloidaler Metallösungen," Leipzig, Ann. Phys. 330,377–445(1908)English translation, American translation
^ ^a ^b H. C. van de Hulst: Light scattering by small particles, New York, John Wiley and Sons, 1957
^ He, L (2003). "Rapid in situ determination of total oil concentration in water using ultraviolet fluorescence and light scattering coupled with artificial neural networks". Analytica Chimica Acta 478: 245. doi:10.1016/S0003-2670(02)01471-X.
^ Lindner, H (2001). "Measurements on Concentrated Oil in Water Emulsions Using Static Light Scattering". Journal of Colloid and Interface Science 242: 239. doi:10.1006/jcis.2001.7754.
^ Serebrennikova, Yulia M.; Patel, Janus; Garcia-Rubio, Luis H. (2010). "Interpretation of the ultraviolet-visible spectra of malaria parasite Plasmodium falciparum". Applied Optics 49 (2): 180. Bibcode 2010ApOpt..49..180S. doi:10.1364/AO.49.000180. PMID 20062504.

[edit] Further reading

Kerker, M. (1969). The scattering of light and other electromagnetic radiation. New York: Academic.
Barber, P. W.; Hill, S. S. (1990). Light scattering by particles: Computational methods. Singapore: World Scientific. ISBN 9971508133.
Mishchenko, M.; Travis, L.; Lacis, A. (2002). Scattering, Absorption, and Emission of Light by Small Particles. New York: Cambridge University Press. ISBN 052178252X.
Frisvad, J.; Christensen, N.; Jensen, H. (2007). "Computing the Scattering Properties of Participating Media using Lorenz-Mie Theory". ACM Transactions on Graphics 26 (3): 60. doi:10.1145/1276377.1276452.
Wriedt, Thomas (2008). "Mie theory 1908, on the mobile phone 2008". Journal of Quantitative Spectroscopy & Radiative Transfer 109 (8): 1543–1548. Bibcode 2008JQSRT.109.1543W. doi:10.1016/j.jqsrt.2008.01.009.

[edit] External links

JMIE (2D C++ code to calculate the analytical fields around an infinite cylinder, developed by Jeffrey M. McMahon)
Collection of light scattering codes
www.T-Matrix.de. Implementations of Mie solutions in FORTRAN, C++, IDL, Pascal, Mathematica and Mathcad
ScatLab. Mie scattering software for Windows.
Online Mie solution calculator is available, with documentation in German and English.
Online Mie scattering calculator produces beautiful graphs over a range of parameters.
phpMie Online Mie scattering calculator written on PHP.
Mie resonance mediated light diffusion and random lasing

The square thread form is a common thread form for leadscrews. It gets its name from the square cross-section of the thread.^[1]

The greatest advantage of square threads is that they have a much higher intrinsic efficiency than acme threads. Due to the lack of a thread angle there is no radial pressure, or bursting pressure, on the nut. This also increases the nut life.^[1]

The greatest disadvantage is the difficulty in machining such a thread. The single-point cutting tools or taps and dies used to cut the thread cannot have efficient rake and relief angles (because of the square form), which makes the cutting slow and difficult. Square threads also cannot carry as much load as a trapezoidal thread, because the root of the square thread is smaller. Also, there is no way to compensate for wear on the nut, so it must be replaced when worn out.^[1]

[edit] Classification

Square threads are defined as follows by ISO standards:

$Sq\,60 \times 9$

where Sq designates a square thread, 60 is the nominal diameter in millimeters, and 9 is the pitch in millimeters. When there is no suffix it is a single start thread. If there is a suffix then the value after the multiplication sign is the lead and the value in the parentheses is the pitch. For example:

$Sq\,60 \times 18 (P9) LH$

would denoted two starts, as the lead divided by the pitch is two. The "LH" denotes a left hand thread.^[2]

Standard pitches for metric diameters
Nominal diameter [mm]	Pitch [mm]
^[2]
22, 24, 26, 28	5
30, 32, 36	6
40, 44	7
48, 50, 52	8
55, 60	9
65, 70, 75, 80	10
85, 90, 95, 100	12

[edit] Mechanics

Main article: Leadscrew mechanics

[edit] References

[edit] Notes

^ ^a ^b ^c Bhandari, p. 203.
^ ^a ^b Bhandari, p. 205.

[edit] Bibliography

Bhandari, V B (2007), Design of Machine Elements, Tata McGraw-Hill, ISBN 9780070611412, http://books.google.com/?id=f5Eit2FZe_cC .

OJIJO CAVIN: he general technique for solving bigger-than-quadr...

OJIJO CAVIN: he general technique for solving bigger-than-quadr...: he general technique for solving bigger-than-quadratic polynomials is pretty straightforward, but the process can be time-consuming. The fir...

October 11, 2011

TEC 101: BASIC ENGINEERING MATHEMATICS I
Course Content
1.Algebra:unions,intersections,complements.Algebraic structures such as rational indices,multiplication,addition and partial fractions.
2.Polynomial functions:constant,linear,quadratic,division,remainder factor;functions and mappings;inverse,constant,step,even,odd,composite.
3.Trigonometric and hyperbolic functions,logarithms and exponential functions.
4.Vectors:scalars and vectors,components,addition,multiplication,vector spaces.Ration theorem,scalar and vector products, unit vectors,geometric interpretations,applications to mechanics.
5.Matrices:matrix algebra,determinants,rank of a matrix,transpose,inverse of an n x n matrix,eigenvalues,eigenvectors.
6.Solution of linear equations,crammer's rule, elementary row operations;Gauss' elimintion method;lower-upper decomposition.Solutions of homogenous equations.
References
1. Engineerin Mathematics, 5th edition by K.A Stroud
2. Mathematics for scientists n Engineers, vol 1 by Phiri PA Vuma D and Luboobi L.S.
3.Further Pure maths 1 vol 1 bz Sam Boardman,Tony Clouph,David Evans

wedges and screws

Wedges and screws

Wedge: A wedge is in general a triangular object which is placed between two objects to either hold them in place or is used to move one relative to the other. For example, the following shows a wedge under a block that is supported by the wall.

If the force P is large enough to push the wedge forward, then the block will rise and the following is an appropriate free-body diagram. Note that for the wedge to move one needs to have slip on all three surfaces. The direction of the friction force on each surface will oppose the slipping.

Since before the wedge can move each surface must overcome the resistance to slipping, one can assume that

These equations and the equations of equilibrium are combined to solve the problem. If the force P is not large enough to hold the top block from coming down, then the wedge will be pushed to the left and the appropriate free-body diagram is the following. Note that the only change is the direction of the frictional forces. A similar analysis to the above yield the solution to the problem.

Screws: One can consider a screw and a bolt as a combination of two wedges. One wedge is obtained from opening the helical treads of the screw and the other will come from opening the helical threads of the bolt. For example, if one opens one revolution of the thread of a screw having a lead of l and mean thread diameter 2r, one gets the following wedge where is the lead angle.

Now consider the situation where a screw is in a bolt or threaded hole. In the picture the screw is being pushed into the hole as the screwing moment is trying to unscrew it. Depending on the lead angle, the magnitude of the axial load W, and the magnitude of the applied moment M, either the axial load will dominate and the screw will move down or the moment will dominate and the screw will move up.

If the axial moment is sufficient to unscrew the screw, then the frictional force will oppose moving up of the screw threads and one will use the following free-body diagram for the wedge.

If the axial load is large enough to screw the screw into the bolt, then the frictional force will oppose the downward motion of the screw threads and one must use the following free-body diagram.

One can reverse the axial load W to be pulling the screw out or one can reverse the direction of the screwing moment. These cases can be studied in a similar way. If you need to know how much axial force W or screwing moment M is needed to make the screw turn in a given way, then you assume that the threads are slipping and set . Take the static coefficient of friction if you need the initial value to get the screw started turning, and take the kinetic coefficient of friction if you would like to know what is needed to keep the screw turning.

Self-locking screw: A screw is considered self-locking if the lead angle is selected such that in the absence of a screwing moment (i.e., M=0) the frictional force will remain less than so that the threads will not slip relative to each other. This can be studied using the following free-body diagram. The maximum thread angle for the screw to be self-locking is given by setting .

he general technique for solving bigger-than-quadratic polynomials is
pretty straightforward, but the process can be time-consuming.

The first step is to apply the Rational Roots Test to the polynomial
to get a list of values that might possibly be solutions to the
polynomial equation. You can follow this up with an application of
Descartes' Rule of Signs, if you like, to narrow down which possible
zeroes might be best to check. Of course, if you've got a graphing
calculator, it's a good idea to do a quick graph, since x-intercepts
of the graph are the same as zeroes of the equation. Seeing where the
graph looks like it crosses the axis can quickly narrow down your list
of possible zeroes.

Once you've found a value you want to test, you use synthetic division
to see if you can get a zero remainder. If you get a zero remainder,
you've not only found a zero, but you've also reduced your polynomial
by one degree. Remember that synthetic division is, among other
things, division, so checking if x = a is a solution is the same as
dividing out the linear factor x – a. This means that you should not
return to the original polynomial for your next computation (for
finding the other zeroes); you should instead work with the output of
the synthetic division. It's smaller, so it's easier to work with.

You should not be surprised to see some complicated solutions to your
polynomials (that is, solutions containing square roots or complex
numbers, or both); these zeroes will come from applying the Quadratic
Formula to the last (quadratic) factor of your polynomial. Here's how
the process plays out:

* Find all the zeroes of the following polynomial: 2x5 + 3x4 –
30x3 – 57x2 – 2x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is check to see if x = 1
or x = –1 is a root, because these are the simplest roots to test for.
This isn't an "official" first step, but it can often be a timesaver,
because you can just look at the powers and the numbers. When x = 1,
the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1
isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so
x = –1 is a root, and I can take care of it right away:

synthetic division by x = -1

This leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x
+ 24. (Since I've divided out the factor x + 1, I've reduced the
degree of the polynomial by 1. That's how I know this is a degree-four
polynomial.) Now I'll apply the Rational Roots Test to get a list of
potential zeroes to try:

±1/2, 1, 3/2, 2, 3, 4, 6, 8, 12, 24

From experience, I've learned that most of these exercises have
their zeroes near the middle of the list, rather than at the extremes.
This isn't always true, of course, but it's usually better to stay
away from the larger numbers. In this case, I won't start off by
trying stuff like x = –24 or x = 12. Instead, I'll start out with
smaller values like x = 2. And I can narrow down my options further by
"cheating" and looking at the graph:

graph of y = 2x^4 + x^3 - 31x^2 - 26x + 24

This is a fourth-degree polynomial, so it has, at most, four
x-intercepts, and I can see all four of them on the graph. It looks
like one of the zeroes is around –3.5, but –7/2 isn't on the list that
the Rational Roots Test gave me, so this must be an irrational root.
I'll leave it until last.

It also looks like there may be zeroes near –1.5 and 0.5. But
the clearest solution looks to be at x = 4 and since whole numbers are
easier to work with than fractions, x = 4 would probably be a good
value to try:

synthetic division with x = 4

The zero remainder says that x = 4 is a root. The bottom row of
the synthetic division tells me that I'm now left with factoring 2x3 +
9x2 + 5x – 6. Looking at the constant term "6", I can see that x =
±24, ±12, ±8, and –4 won't work as rational roots (even if I didn't
already know from the graph), so I can cross them off of my list.
(Always check the numbers as you go. The Rational Roots Test can give
a very long list of possibilities, and it can be helpful to notice
that some of those values can be ignored, especially if you don't have
a graphing calculator to "cheat" with.) Comparing the remaining values
on the list with the intercepts on the graph, I'll try x = 1/2:

synthetic division with x = 1/2

The remainder isn't zero, so that test root didn't work. This
means that the zero close to x = 1/2 on the graph must be irrational;
I'll find it when I apply the Quadratic Formula later. For now, I'll
try x = –3/2: Copyright © Elizabeth Stapel 2005-2011 All Rights
Reserved

synthetic division with x = -3/2

The division came out evenly, leaving me with the polynomial 2x2
+ 6x – 4. Since I'm looking for the zeroes of the polynomial, what I
really have here is 2x2 + 6x – 4 = 0. Dividing through by 2 to get
smaller numbers gives me x2 + 3x – 2 = 0, to which I can apply the
Quadratic Formula:

x = [-3 ± sqrt(17)]/2, or about -3.5 and 0.5

Then the complete solution is:

x = -1, [-3 +/- sqrt(17)]/2, -3/2, 4

Asking you to find the zeroes of a polynomial means the same thing as
asking you to find the solutions to a polynomial equation. The zeroes
of a polynomial are the values of x that make the polynomial equal to
zero. So the above problem could have been stated along the lines of
"Find the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 = 0" or "Find
the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x = –24", and the answers
would have been the exact same list of x-values.

You can use these same techniques to factor bigger-than-quadratic
polynomials....

By now, you should be familiar with variables and exponents, and you
may have dealt with expressions like 3x4 or 6x. Polynomials are sums
of these "variables and exponents" expressions. Each piece of the
polynomial, each part that is being added, is called a "term".
Polynomial terms have variables which are raised to whole-number
exponents (or else the terms are just plain numbers); there are no
square roots of variables, no fractional powers, and no variables in
the denominator of any fractions. Here are some examples:
6x –2 This is NOT
a polynomial term... ...because the variable has a negative exponent.
1/x2 This is NOT
a polynomial term... ...because the variable is in the denominator.
sqrt(x) This is NOT
a polynomial term... ...because the variable is inside a radical.
4x2 This IS a polynomial term... ...because it obeys all the rules.

Here is a typical polynomial:

terms

Notice the exponents on the terms. The first term has an exponent of
2; the second term has an "understood" exponent of 1; and the last
term doesn't have any variable at all. Polynomials are usually written
this way, with the terms written in "decreasing" order; that is, with
the largest exponent first, the next highest next, and so forth, until
you get down to the plain old number.

Any term that doesn't have a variable in it is called a "constant"
term because, no matter what value you may put in for the variable x,
that constant term will never change. In the picture above, no matter
what x might be, 7 will always be just 7.

The first term in the polynomial, when it is written in decreasing
order, is also the term with the biggest exponent, and is called the
"leading term".

The exponent on a term tells you the "degree" of the term. For
instance, the leading term in the above polynomial is a "second-degree
term" or "a term of degree two". The second term is a "first degree"
term. The degree of the leading term tells you the degree of the whole
polynomial; the polynomial above is a "second-degree polynomial". Here
are a couple more examples:

* Give the degree of the following polynomial: 2x5 – 5x3 – 10x + 9

This polynomial has four terms, including a fifth-degree term, a
third-degree term, a first-degree term, and a constant term.

This is a fifth-degree polynomial.

* Give the degree of the following polynomial: 7x4 + 6x2 + x

This polynomial has three terms, including a fourth-degree term,
a second-degree term, and a first-degree term. There is no constant
term.

This is a fourth-degree polynomial.

When a term contains both a number and a variable part, the number
part is called the "coefficient". The coefficient on the leading term
is called the "leading" coefficient.

terms

In the above example, the coefficient of the leading term is 4; the
coefficient of the second term is 3; the constant term doesn't have a
coefficient. Copyright © Elizabeth Stapel 2000-2011 All Rights
Reserved

The "poly" in "polynomial" means "many". I suppose, technically, the
term "polynomial" should only refer to sums of many terms, but the
term is used to refer to anything from one term to the sum of a
zillion terms. However, the shorter polynomials do have their own
names:

* a one-term polynomial, such as 2x or 4x2, may also be called a
"monomial" ("mono" meaning "one")
* a two-term polynomial, such as 2x + y or x2 – 4, may also be
called a "binomial" ("bi" meaning "two")
* a three-term polynomial, such as 2x + y + z or x4 + 4x2 – 4, may
also be called a "trinomial" ("tri" meaning "three")

I don't know if there are names for polynomials with a greater numbers
of terms; I've never heard of any names other than what I've listed.

Polynomials are also sometimes named for their degree:

* a second-degree polynomial, such as 4x2, x2 – 9, or ax2 + bx +
c, is also called a "quadratic"
* a third-degree polynomial, such as –6x3 or x3 – 27, is also
called a "cubic"
* a fourth-degree polynomial, such as x4 or 2x4 – 3x2 + 9, is
sometimes called a "quartic"
* a fifth-degree polynomial, such as 2x5 or x5 – 4x3 – x + 7, is
sometimes called a "quintic"

There are names for some of the higher degrees, but I've never heard
of any names being used other than the ones I've listed.

By the way, yes, "quad" generally refers to "four", as when an ATV is
referred to as a "quad bike". For polynomials, however, the "quad"
from "quadratic" is derived from the Latin for "making square". As in,
if you multiply length by width (of, say, a room) to find the area in
"square" units, the units will be raised to the second power. The area
of a room that is 6 meters by 8 meters is 48 m2. So the "quad" refers
to the four corners of a square, from the geometrical origins of
parabolas and early polynomials.

Evaluation

"Evaluating" a polynomial is the same as evaluating anything else: you
plug in the given value of x, and figure out what y is supposed to be.
For instance:

* Evaluate 2x3 – x2 – 4x + 2 at x = –3

I need to plug in "–3" for the "x", remembering to be careful
with my parentheses and the negatives:

2(–3)3 – (–3)2 – 4(–3) + 2
= 2(–27) – (9) + 12 + 2
= –54 – 9 + 14
= –63 + 14
= –49

Always remember to be careful with the minus signs!

The Factor Theorem

The Factor Theorem is a result of the Remainder Theorem, and is based
on the same reasoning. If you haven't read the lesson on the Remainder
Theorem, review that topic first, and then return here.

As the Remainder Theorem points out, if you divide a polynomial p(x)
by a factor x – a of that polynomial, then you will get a zero
remainder. Let's look again at that Division Algorithm expression of
the polynomial:

p(x) = (x – a)q(x) + r(x)

If x – a is indeed a factor of p(x), then the remainder after division
by x – a will be zero. That is:

p(x) = (x – a)q(x)

In terms of the Remainder Theorem, this means that, if x – a is a
factor of p(x), then the remainder, when we do synthetic division by x
= a, will be zero.

The point of the Factor Theorem is the reverse of the Remainder
Theorem: If you synthetic-divide a polynomial by x = a and get a zero
remainder, then, not only is x = a a zero of the polynomial (courtesy
of the Remainder Theorem), but x – a is also a factor of the
polynomial (courtesy of the Factor Theorem).

Just as with the Remainder Theorem, the point here is not to do the
long division of a given polynomial by a given factor. This Theorem
isn't repeating what you already know, but is instead trying to make
your life simpler. When faced with a Factor Theorem exercise, you will
apply synthetic division and then check for a zero remainder.

* Use the Factor Theorem to determine whether x – 1 is a factor of
f (x) = 2x4 + 3x2 – 5x + 7.

For x – 1 to be a factor of f (x) = 2x4 + 3x2 – 5x + 7, the
Factor Theorem says that x = 1 must be a zero of f (x). To test
whether x – 1 is a factor, I will first set x – 1 equal to zero and
solve to find the proposed zero, x = 1. Then I will use synthetic
division to divide f (x) by x = 1. Since there is no cubed term, I
will be careful to remember to insert a "0" into the first line of the
synthetic division to represent the omitted power of x in 2x4 + 3x2 –
5x + 7:

completed division: 2 2 5 0 7

Since the remainder is not zero, then the Factor Theorem says that:

x – 1 is not a factor of f (x).

* Using the Factor Theorem, verify that x + 4 is a factor of
f (x) = 5x4 + 16x3 – 15x2 + 8x + 16.

If x + 4 is a factor, then (setting this factor equal to zero
and solving) x = –4 is a root. To do the required verification, I need
to check that, when I use synthetic division on f (x), with x = –4, I
get a zero remainder:

completed division: 5 –4 1 4 0

The remainder is zero, so the Factor Theorem says that:

x + 4 is a factor of 5x4 + 16x3 – 15x2 + 8x + 16.

In practice, the Factor Theorem is used when factoring polynomials
"completely". Rather than trying various factors by using long
division, you will use synthetic division and the Factor Theorem. Any
time you divide by a number (being a potential root of the polynomial)
and get a zero remainder in the synthetic division, this means that
the number is indeed a root, and thus "x minus the number" is a
factor. Then you will continue the division with the resulting smaller
polynomial, continuing until you arrive at a linear factor (so you've
found all the factors) or a quadratic (to which you can apply the
Quadratic Formula).

If x = –2 is a zero, then x + 2 = 0, so x + 2 is a factor.
Similarly, if x = 1/3 is a zero, then x – 1/3 = 0, so x – 1/3 is a
factor. By giving me two of the zeroes, they have also given me two
factors: x + 2 and x – 1/3.

Since I started with a fourth-degree polynomial, then I'll be
left with a quadratic once I divide out these two given factors. I can
solve that quadratic by using the Quadratic Formula or some other
method.

The Factor Theorem says that I don't have to do the long
division with the known factors of x + 2 and x – 1/3. Instead, I can
use synthetic division with the associated zeroes –2 and 1/3. Here is
what I get when I do the first division with x = –2:

completed divison: bottom row: 3 –1 3 –1 0

The remainder is zero, which is expected because they'd told me
at the start that –2 was a known zero of the polynomial. Rather than
starting over again with the original polynomial, I'll now work on the
remaining polynomial factor of 3x3 – x2 + 3x – 1 (from the bottom line
of the synthetic division). I will divide this by the other given
zero, x = 1/3:

completed division: bottom row: 3 0 3 0

This leaves me with the quadratic 3x2 + 3, which I can solve:

3x2 + 3 = 0
3(x2 + 1) = 0
x2 + 1 = 0
x2 = –1
x = ± i

If the zeroes are x = –i and x = i, then the factors are x –
(–i) and x – (i), or x + i and x – i. I need to remember that I
divided off a "3" when I solved the quadratic; it is still part of the
polynomial, and needs to be included as a factor. Then the
fully-factored form is:

3x4 + 5x3 + x2 + 5x – 2 = 3(x + 2)(x – 1/3)(x + i)(x – i)

REMAINDER THEOREM

The Remainder Theorem
The Remainder Theorem is useful for evaluating polynomials at a given
value of x, though it might not seem so, at least at first blush. This
is because the tool is presented as a theorem with a proof, and you
probably don't feel ready for proofs at this stage in your studies.
Fortunately, you don't "have" to understand the proof of the Theorem;
you just need to understand how to use the Theorem.

The Remainder Theorem starts with an unnamed polynomial p(x), where
"p(x)" just means "some polynomial p whose variable is x". Then the
Theorem talks about dividing that polynomial by some linear factor x –
a, where a is just some number. Then, as a result of the long
polynomial division, you end up with some polynomial answer q(x) (the
"q" standing for "the quotient polynomial") and some polynomial
remainder r(x).

As a concrete example of p, a, q, and r, let's look at the polynomial
p(x) = x3 – 7x – 6, and let's divide by the linear factor x – 4 (so a
= 4):

completed division: quotient x^2 + 4x + 9, remainder 30

So we get a quotient of q(x) = x2 + 4x + 9 on top, with a remainder of
r(x) = 30.

You know, from long division of regular numbers, that your remainder
(if there is one) has to be smaller than whatever you divided by. In
polynomial terms, since we're dividing by a linear factor (that is, a
factor in which the degree on x is just an understood "1"), then the
remainder must be a constant value. That is, when you divide by "x –
a", your remainder will just be some number.

The Remainder Theorem then points out the connection between division
and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3 = 12.
If you get a remainder, you do the multiplication and then add the
remainder back in. For instance, since 13 ÷ 5 = 2 R 3, then 13 = 5 × 2
+ 3. This process works the same way with polynomials. That is:

If p(x) / (x – a) = q(x) with remainder r(x),

then p(x) = (x – a) q(x) + r(x).

(Technically, this "if - then" statement is the "Division Algorithm
for Polynomials". But the Algorithm is the basis for the Remainder
Theorem.)

Since (x^3 – 7x – 6) / (x – 4) = x2 + 4x + 9 with remainder 30,

then x3 – 7x – 6 = (x – 4) (x2 + 4x + 9) + 30.

The Remainder Theorem says that we can restate the polynomial in terms
of the divisor, and then evaluate the polynomial at x = a. But when x
= a, the factor "x – a" is just zero! Then evaluating the polynomial
at x = a gives us:

p(a) = (a – a)q(a) + r(a)
= (0)q(a) + r(a)
= 0 + r(a)
= r(a)

But remember that the remainder term r(a) is just a number! So the
value of the polynomial p(x) at
x = a is the same as the remainder you get when you divide that
polynomial p(x) by x – a. In terms of our concrete example:

p(4) = (4 – 4)((4)2 + 4(4) + 9) + 30
= (0)(16 + 16 + 9) + 30
= 0 + 30
= 30

But you gotta think: Okay, fine; the value of the polynomial p(x) at x
= a is the remainder r(a) when you divide by x – a, but who wants to
do the long division each time you have to evaluate a polynomial at a
given value of x?!? You're right; this would be overkill. Fortunately,
that's not what they really want you to do.

When you are dividing by a linear factor, you don't "have" to use long
polynomial division; instead, you can use synthetic division, which is
much quicker. In our example, we would get:

completed division: bottom row: 1 4 9 30

Note that the last entry in the bottom row is 30, the remainder from
the long division (as expected) and also the value of p(x) = x3 – 7x
– 6 at x = 4. And that is the point of the Remainder Theorem: There is
a simpler, quicker way to evaluate a polynomial p(x) at a given value
of x, and this simpler way is not to evaluate p(x) at all, but to
instead do the synthetic division at that same value of x. Here are
some examples:

* Use the Remainder Theorem to evaluate f (x) = 6x3 – 5x2 + 4x –
17 at x = 3.

First off, even though the Remainder Theorem refers to the
polynomial and to long division and to restating the polynomial in
terms of a quotient, a divisor, and a remainder, that's not actually
what I'm meant to be doing. Instead, I'm supposed to be doing
synthetic division, using "3" as the divisor:

completed division: bottom row: 6 13 43 112

Since the remainder (the last entry in the bottom row) is 112,
then the Remainder Theorem says that:

f (3) = 112.

* Using the Remainder Theorem, find the value of f (–5), for f
(x) = 3x4 + 2x3 + 4x.

I need to do the synthetic division, remembering to put zeroes
in for the powers of x that are not included in the polynomial:

completed division: bottom row: 3 -13 65 -321 1605

Since the remainder is 1605, then, thanks to the Remainder
Theorem, I know that:

f (–5) = 1605.

* Use the Remainder Theorem to determine whether x = 2 is a zero of
f (x) = 3x7 – x4 + 2x3 – 5x2 – 4

For x = 2 to be a zero of f (x), then f (2) must evaluate to
zero. In the context of the Remainder Theorem, this means that my
remainder, when dividing by x = 2, must be zero:

completed division: bottom row: 3 6 12 23 48 91 182 360

The remainder is not zero. Then x = 2 is not a zero of f (x).

* Use the Remainder Theorem to determine whether x = –4 is a solution of
x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8 = 0

For x = –4 to be a solution of f (x) = x6 + 5x5 + 5x4 + 5x3 +
2x2 – 10x – 8 = 0, it must be that f (–4) = 0. In the context of the
Remainder Theorem, this means that the remainder, when dividing by x =
–4, must be zero:

completed division: bottom row: 1 1 1 1 -2 -2 0

The remainder is zero. Then x = –4 is a solution of the given equation.

October 10, 2011

Scholarship

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Pages

October 14, 2011

Linear motion

[edit] Particle kinematics

[edit] Position and reference frames

[edit] Position vector

[edit] Rest and motion

[edit] Path

[edit] Displacement

[edit] Distance

[edit] Velocity and speed

[edit] Acceleration

[edit] Types of motion based on velocity and acceleration

[edit] Integral relations

[edit] Kinematics of constant acceleration

[edit] Relative velocity

[edit] Rotational motion

[edit] Point object in circular motion

[edit] Coordinate systems

[edit] Fixed rectangular coordinates

[edit] Two-dimensional rotating reference frame

[edit] Derivatives of unit vectors

[edit] Position, velocity, and acceleration

[edit] Position

[edit] Velocity

[edit] Acceleration

[edit] Kinematic constraints

[edit] Rolling without slipping

[edit] Inextensible cord

[edit] See also

[edit] Notes

[edit] References

[edit] External links

October 13, 2011

[edit] Mie scattering codes

[edit] Approximations

[edit] Rayleigh approximation (scattering)

[edit] Rayleigh Gans Approximation

[edit] Anomalous diffraction approximation of van de Hulst

[edit] Applications

[edit] Atmospheric science

[edit] Cancer detection and screening

[edit] Particle sizing

[edit] Parasitology

[edit] See Also

[edit] References

[edit] Further reading

[edit] External links

[edit] Classification

[edit] Mechanics

[edit] References

[edit] Notes

[edit] Bibliography

October 11, 2011

October 10, 2011

October 09, 2011

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