#ph300

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Proyektor LG Probeam Usung Teknologi Laser

Proyektor LG Probeam Usung Teknologi Laser

Jakarta – LG Electronics Indonesia menghadirkan produk proyektor dalam jajaran Minibeam series, Probeam HF80JG. Produk ini diklaim sebagai proyektor pertama LG yang menggunakan teknologi laser.Untuk apa sih teknologi laser disematkan pada sebuah proyektor? IT Product Marketing LG Electronics Indonesia Ferdiansyah menyebutkan fungsi dari dari teknologi laser. “Teknologi laser membuat gambar yang…

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[thebkgruby]

LG PH300: A Minibeam LED Projector

LG PH300: A Minibeam LED Projector

I know I’m not alone. As I get older, my eyes have increasingly more difficulty seeing smaller screens. Reading the newspaper on my iPhone causes eye strain — unless I’m wearing glasses. Same thing watching movies and TV shows. Which is what initially intrigued me about LG’s PH300 Minibeam Projector, a $449 unit that weighs less than a pound and fits in the palm of an adult’s hand.

LG PH300…

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[teknoloji7com]

Dünyanın en küçük projektörü: LG HD MiniBeam!

Dünyanın en küçük projektörü: LG HD MiniBeam!

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[physicsandchemistryrevision]

Mathematics

Stokes' theorem

Cont'd from "Level surfaces", see "Curl of a vector" 

Simply put, Stokes' theorem states that the circulation of a vector field V around a surface is equal to the integral of the component of the curl of V which is in a plane perpendicular to the surface over the total area A of the surface. Therefore, a vector V will have circulation around a surface in the x-y plane equal to

circulation = ∬ (∇×V)z dA                      A                      

where, in this case, dA = dydx since the surface is in the x-y plane. (∇×V)z represents the z-component of curl(V).

—

For example, a vector field A = (x, x²y, xy²z) has a curl

∇×A = 2xyz i - y²z j + 2xy k

Now, to find the circulation C of A around a 2 by 1 rectangle in the x-y plane centred at the orgin, we take the z-component (i.e., the k part) of A and integrate as above, substituting dA = dxdy

C = ∬ 2xy dxdy       A                    

The limits are the area over which the integration happens. This will be the surface we're finding the circulation around. In this case, the rectangle is centred at the origin and has 2 units in the y-direction and 1 unit in the x-direction, implying that the limits are

0 ≤ x ≤ 1

0 ≤ y ≤ 2

 We now have the definite integral

      2  1             C = ∫  ∫ 2xy dxdy       0 0                   

which can be solved as follows!

      2        x = 1     C = ∫ [x²y]   ·   dy       0        x = 0   

      2       C = ∫ y dy       0          

                  y = 2 C = [1/2 · y²]                   y = 0

C = 2

—

Interestingly, Stokes' theorem was not initially presented as a formal publication, nor was it entirely his own work. Instead, the idea was suggested to him by Lord Kelvin. Stokes included it in his 1854 Smith's Prize exam as a question for his students to prove and the theorem quickly became attributed to Stokes after the potential of the question was realised. As a result, Stokes' theorem is less commonly known as the Kelvin-Stokes theorem.

[physicsandchemistryrevision]

Mathematics

Gradient of a scalar

See "Partial derivative of a multi-variable function"

The gradient of a scalar field returns a vector field describing the direction of the rate of change of a quantity. In the same way as the derivative of a graph gives the slope of the rate of change, the gradient quantity can determine whether this gradient is increasing or decreasing and in what way. The gradient of a general scalar field f(x,y,z,..,n) can be expressed as

∇f = ∂f/∂x i + ∂f/∂y j + ∂f/∂z k + ... + ∂f/∂n n

where n is an arbitrary unit vector.

The notation ∇f  is to imply the del operator ∇ acts as a vector differential operator on the scalar field. Commonly, grad(f) is used to notate the same idea.

—

For example, if we have the scalar field s(x,y,z) = x²yz³, s has a single value at an arbitrary point (1,4,2) wherein s(1,4,2) = (1)²(4)(2)³ = 32. Therefore, we can apply the formula to find the gradient ∇s of the field at any point in the field, including at the point (1,4,2).

∇s = ∂s/∂x i + ∂s/∂y j + ∂s/∂z k

so, if we find the partial derivatives

∂s/∂x = 2xyz³

∂s/∂y = x²z³

∂s/∂z = 3x²yz²

and substitute these in to get

 ∇s = 2xyz³ i + x²z³ j + 3x²yz² k

and so 

∇s(1,4,2) = 2(1)(4)(2)³ i + (1)²(2)³ j + 3(1)²(4)(2)² k

∇s(1,4,2) = 64 i + 8 j + 48 k

[physicsandchemistryrevision]

Mathematics

Unit vectors

A unit vector is a vector of magnitude 1 that points in the direction of the original vector. Examples of this are the i, j and k unit vectors, which are directed along the x, y and z axes respectively.

If we consider a vector r = (x, y, z), the unit vector of r is given by

        r r = —       | r |

where the magnitude of r, | r |, can be represented by

           _________ | r | = √x² + y² + z²

The unit vector can also be represented using a vector matrix, wherein

[physicsandchemistryrevision]

Mathematics

Directional derivatives

Cont'd from "Euler's theorem on homogeneous functions"

A directional derivative (which we'll call Dd) at point (a,b) acting at angle αº from the x-axis can be described as

Dd = fx(a,b) cos α + fy(a,b) sin α

This is the value of the derivative when varying both x and y simultaneously. It is intended to find the derivative of vector quantities.

For example, the function

f(x,y) = x2y3

Has partial derivatives

fx(x,y) = 2xy3

fy(x,y) = 3x2y2

at the point (x,y) = (2,1)

fx(2,1) = 2(1)(2)3 = 2·8 = 16

fy(2,1) = 3(1)2(2)2 = 3·4 = 12

at an angle of 30º from the x-axis. The function therefore has the directional derivative

Dd = 16 cos 30 + 12 sin 30 = 16 · √3/2 + 12 · 1/2 = (16√3 + 12)/2

[physicsandchemistryrevision]

Mathematics

Euler's theorem on homogeneous functions

Cont'd from "Maxima, minima and saddle points of a multi-variable function"

If we have a function with multiple variables, its degree can be found by finding the sum of the powers of each term. For example, the function

f(x,y) = xnym + xnym

has degree n + m.

Consider, for example, the multi-variable function (1) as follows:

f(x,y,z) = x2yz + 2xy2z + 2x3y

We can see that the function is of 4th degree.

Euler deduced the relationship that a homogenous function f(x,y,z) of kth degree will satisfy the equation

x∂f/∂x + y∂f/∂y + z∂f/∂x = k · f(x,y,z)

We'll call this equation (2), which can be proved using function (1) by finding the relvant partial derivatives, substituting into euler's theorem and multiplying out.

∂f/∂x = 2xyz + 2y2z + 6x2y

∂f/∂y = x2z + 4xyz + 2x3

∂f/∂z = x2y + 2xy2

Substituting into equation (2), where we know k = 4 (discussed above),

4 f(x,y,z) = x (2xyz + 2y2z + 6x2y) + y (x2z + 4xyz + 2x3) + z (x2y + 2xy2)

= 2x2yz + 2xy2z + 6x3y + x2yz + 4xy2z + 2x3y + x2yz + 2xy2z

Collecting like terms

4 f(x,y,z) = 4x2yz + 8xy2z + 8x3y

If we factor out the 4, we see that

4x2yz + 8xy2z + 8x3y = 4 (x2yz + 2xy2z + 2x3y)

Since f(x,y,z) = x2yz + 2xy2z + 2x3y, it should be clear that

x∂f/∂x + y∂f/∂y + z∂f/∂x = 4f(x,y,z)

hence, confirming Euler's theorem on homogeneous functions.