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## Thomas' Calculus 13th Edition

$f'(1)=-1$
To find the slope of a function at any given point, all one has to do would be to look for the value of the derivative function at the same x value. Given that $f(x)=5x-3x^2$, then the derivative function by the power rule will be $f'(x)=5x^{1-1}-3(2)x^{2-1}=5-6x$ As such, the slope at the point $x=1$ would be $f'(1)=5-6(1)=-1$


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# ANAGRAMS

Student seminar

## Goal

There is a myriad of dimension functions in (homological) algebra, geometry and the theory of quadratic forms. There is no well-defined notion of dimension function, but let's say it is a numerical invariant of a ring, module or field. The goal would be to discuss some of these, and wherever applicable show how they are related.

The format of this seminar is different from the others:

• talks are somewhere between 15 minutes and 1 hour
• people can give several talks (or discuss several related dimensions at once)
• mostly definitions and facts, no difficult proofs
• provide pointers to the literature

## Schedule

February 7
Algebraic dimensions
in G.017, at 10h00
lectures by:
• Hoang Van Dinh: Krull dimension and transcendence degree (notes)
• Dennis Presotto: Gel'fand–Kirillov dimension (notes)
• Nikolaas Verhulst: Goldie (or uniform) dimension
• Julia Ramos González: Dimension of triangulated categories (notes)
February 14
Geometric dimensions and dimensions for quadratic forms
in G.017, at 10h00
lectures by:
• Erik Rijcken: Homological dimensions (notes)
• Pieter Belmans: Depth, and measuring singularities (notes)
• Andrew Dolphin: $u$-invariant, level, Pythagoras number

## References

In due time I might add these.

format: plain


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# Six trigonometric equations

Printable View

• July 21st 2010, 01:06 PM
FailingTrig
Six trigonometric equations
Given: cot theta = sqrt2/4 , pi < theta < 3pi/2
find the exact value of the other five trig functions of theta, leave answers in radical form.

could someone show me how to do this, when i did it i got:
sin theta= -2sqrt2/3
cos theta= -1/3
tan theta= 2/sqrt2
cot theta= sqrt2/4
sec theta= -3
csc theta= -3sqrt2/4

is this correct?

a friend in class said that it wasn't and that i had to start by doing cot theta= cos/sin cos = sqrt2 sin = 4....
• July 21st 2010, 02:47 PM
skeeter
image link doesn't work. some you got correct.

$\cot{t} = \frac{\sqrt{2}}{4}
$

noted that $t$ is in quad III ...

$\tan{t} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$

using the identity $\csc^2{t} = 1 + \cot^2{t}$ ...

$\csc{t} = -\frac{3}{2\sqrt{2}}$

$\sin{t} = -\frac{2\sqrt{2}}{3}$

$\cos{t} = -\frac{1}{3}$

$\sec{t} = -3$
• July 21st 2010, 02:52 PM
eumyang
$cot \theta = \frac{x}{y} = \frac{\sqrt{2}}{4}$, and we're given that θ is in quadrant III, so $x = -\sqrt{2}$ and $y = -4$

Use $x^2 + y^2 = r^2$ to find r:
$(-\sqrt{2})^2 + (-4)^2 = r^2$
$2 + 16 = r^2$
$18 = r^2$
$r = 3\sqrt{2}$

So...
$sin \theta = \frac{y}{r} = \frac{-4}{3\sqrt{2}} = ...$
$cos \theta = \frac{x}{r} = ...$
...
• July 21st 2010, 04:03 PM
Soroban
Hello, FailingTrig!

Here's another approach . . .

Quote:

Given: . $\cot\theta \,=\, \frac{\sqrt{2}}{4},\;\; \pi < \theta < \frac{3\pi}{2}$

Find the exact value of the other five trig functions of $\theta$.

We are given: . $\cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

That is, $\theta$ is in a right triangle with: . $adj = \sqrt{2},\;opp = 4$

Pythagorus says: . $hyp^2 \:=\:(\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$

Since $\theta$ is in Quadrant 3, only $\tan\theta$ and $\cot\theta$ are positive.

Therefore:

. . $\begin{array}{cccccc}
\sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm]
\cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm]
\tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm]
\cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm]
\sec\theta &=& -3 \\ \\[-3mm]
\csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

• July 21st 2010, 08:17 PM
FailingTrig
Quote:

Originally Posted by Soroban
Hello, FailingTrig!

Here's another approach . . .

We are given: . $\cot\theta \:=\:\dfrac{\sqrt{2}}{4} \:=\:\dfrac{adj}{opp}$

That is, $\theta$ is in a right triangle with: . $adj = \sqrt{2},\;opp = 4$

Pythagorus says: . $hyp^2 \:=\:(\sqrt{2})^2 + 4^2 \:=\:18 \quad\Rightarrow\quad hyp \:=\:3\sqrt{2}$

Since $\theta$ is in Quadrant 3, only $\tan\theta$ and $\cot\theta$ are positive.

Therefore:

. . $\begin{array}{cccccc}
\sin\theta &=& -\dfrac{4}{3\sqrt{2}} &=& -\dfrac{2\sqrt{2}}{3} \\ \\[-3mm]
\cos\theta &=& -\dfrac{\sqrt{2}}{3\sqrt{2}} &=& -\dfrac{1}{3} \\ \\[-3mm]
\tan\theta &=& \dfrac{4}{\sqrt{2}} &=& 2\sqrt{2} \\ \\[-3mm]
\cot\theta &=& \dfrac{\sqrt{2}}{4} \\ \\[-3mm]
\sec\theta &=& -3 \\ \\[-3mm]
\csc\theta &=& -\dfrac{3\sqrt{2}}{4}\end{array}$

Did it right then! Awesome, thanks for showing me everyone =)


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# Write equation

Type an equation and we'll make it into a nice image for you. Powered by MimeTeX, this feature results in nice, typeset equations! Here are some examples.

 \large f^\prime(x)\ = \lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x} $\large+f^\prime(x)\+=+\lim_{\Delta+x\to0}\frac{f(x+\Delta+x)-f(x)}{\Delta+x}$


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# WebGL With Three.js: Basics

Difficulty:IntermediateLength:MediumLanguages:

3D graphics in the browser have been a hot topic ever since it was first introduced. But if you were to create your apps using plain WebGL, it would take ages. This is exactly why some really useful libraries have recently came about. Three.js is one of the most popular, and in this series I will show you how best to use it in order to create stunning 3D experiences for your users.

Before we begin, I do expect you to have a basic understanding of 3D space before you start reading this tutorial, as I won't be explaining stuff like coordinates, vectors, etc.

## Step 1: Preparation

First, create three files: index.html, main.js and style.css. Now, download Three.js (whole zip file with examples and source, or the JavaScript file alone, your choice). Now, open index.html and insert this code:

That's all you need in this file. Just a declaration of scripts and stylesheet. All the magic will happen in main.js, but before we get to that we need one more trick to make the app look good. Open style.css and insert this code:

This will position the canvas in the left-top corner, because by default the body will have 8px of margin. Now we can proceed with the JavaScript code.

## Step 2: The Scene and the Renderer

Three.js uses the concept of a display list. It means that all objects are stored in the list and then drawn to the screen.

Three.js uses the concept of a display list. This means that all objects are stored in the list and then drawn to the screen. Here, this is a THREE.Scene object. You need to add any object you want to be drawn on the screen to the scene. You can have as many scenes as you want, but one renderer can draw only one scene at once (of course you can switch the scene that is displayed).

The renderer simply draws everything from the scene to the WebGL canvas. Three.js also supports drawing on SVG or 2D Canvas, but we will focus on WebGL.

To get started, lets store the window's width and height in variables, we will use it later:

Now define the renderer and the scene:

The first line defines the WebGL renderer. You can pass the renderer's options in the first argument as a map. Here, we set the antialias to true, because we want the edges of objects to be smooth, not jagged.

The second line sets the renderer size to the size of the window, and in the third one we add the renderer's canvas element to the document (you can also do this using a library, like jQuery: \$('body').append(renderer.domElement)).

The last one defines the scene, no arguments needed.

## Step 3: The Cube

Now lets add something to be drawn. Let it be a cube, since it's the simplest 3D object. In Three.js the objects that are being drawn on the screen are called meshes. Each mesh has to have its own geometry and material. Geometry is a set of points that need to be connected in order to create the object. Material is simply the paint (or painting, but that is not the topic of this tutorial) that will cover the object. So, lets create our cube. Luckily for us there are some helper functions in Three.js for creating primitives (simple shapes):

As you can see, first we create the geometry. The arguments are defining a size of the cube: the width, height and depth.

Next, we define the cube's material. There are a few material types in Three.js, but this time we will use the THREE.MeshLambertMaterial, since we want to have some lighting later (this material uses Lambert's algorithm for light calculations). You can pass the options in the first argument as a map, the same as with the renderer - this is pretty much a rule for more complex objects in Three.js. Here, we only use color, which is passed as a hexadecimal number.

On the third line, we create a mesh using the geometry and material created earlier. Next, we rotate the cube by 45 degrees on the Y axis, to make it look better. We have to change degrees to radians, which is handled by the equation you probably remember from your high school physics class: Math.PI * 45 / 180. Finally, the cube is added to the scene.

Now you could open index.html in your browser to see the results, but you will see nothing because the scene is not rendered yet.

## Step 4: Camera!

To render something, first we need to add the camera to the scene, so the renderer knows from which point of view it should render stuff. There are a few types of cameras in Three.js, but you'll probably only use THREE.PerspectiveCamera. This type of camera is presenting the scene as we see our world. Lets create one:

"To render something, first we need to add the camera to the scene, so the renderer knows from which point of view it should render stuff."

Creating the camera is a bit more complicated than the rest of the things we've done so far. The first argument defines the FOV (field of view), the angle that can be seen from where the camera is. A FOV of 45 degrees looks natural. Next, we define the camera's ratio. This is always the width of the renderer divided by its height, unless you want to achieve some special effects. The last two numbers are defining how close and how far the object can be to the camera to be drawn.

Now we have to move the camera back and up a little, as all of the objects created in Three.js have their position set in the middle of the scene (x: 0, y: 0, z: 0) by default:

The z coordinate is positive in the direction of the viewer, so objects with a higher z position will appear closer to you (in this case, since we moved the camera, all of the objects will appear further away from you).

Now, lets add the camera to the scene and render it:

You add the camera just like you added the cube. The next line renders the scene using this camera. Now you can open the browser and you should see the following:

You should only be able to see the top of the cube. This is because we moved the camera up and it's still looking directly in front of it. This can be fixed by letting the camera know on what position it should look. Add this line after the lines setting the position of the camera:

The only argument passed in is a position on which the camera will look. Now, the scene looks better, but the cube is still black, no matter what color you've set when creating it:

## Step 5: Lights!

The cube is black, because there are no lights on the scene, so it's like a completely black room. You see a white background because there is nothing to draw apart from the cube. To avoid that, we will use a technique called skybox. Basically, we will add a big cube that will display the background of the scene (usually some far terrain if it's open space). So, lets create the box. This code should go before the renderer.render call:

This code is similar to the one that creates the cube. But this time the geometry is much bigger. We've also used THREE.MeshBasicMaterial since we don't need to light the skybox. Also, notice the additional argument passed to the material: side: THREE.BackSide. Since the cube will be displayed from the inside, we have to change the side that gets drawn (normally, Three.js draws only outside walls).

Now the rendered scene is completely black. To fix that we have to add light to the scene. We will use THREE.PointLight, which emits the light like a bulb. Add these lines after the skybox:

As you can see, we've created the point light with white color, then we are setting its position to be up and back a little, to light the front and the top of the cube. Finally the light is added to the scene like any other object. Open up the browser and you should see a colored, shaded cube:

But the cube is still pretty boring. Let's add some movement to it.

## Step 6: Action!

Now we will add some movement to the scene. Lets make the cube rotate around the Y axis. But first, we have to change the way that we render the scene. One renderer.render call, renders the current state of the scene once. So even if we animate the cube somehow, we will not see it move. To change that, we have to add the render loop to our app. This can be achieved using the renderAnimationFrame function, which was created specially for that purpose. It's supported in most of the major browsers, and for those which doesn't support it, Three.js comes with its own polyfill. So, lets change this:

to this:

Actually, there is no loop in there, because it would freeze the browser. The requestAnimationFrame function behaves a bit like setTimeout, but it's calling the function passed as quick as the browser is ready. So, nothing really changed in the displayed scene and the cube is still not moving. Lets fix that. Three.js comes with THREE.Clock which can be used to achieve smooth animation of objects. First, initialize it before the render function definition:

Now, each time you call clock.getDelta it will return the time since the last call, in milliseconds. This can be used to rotate the cube like this:

Add this line between the renderer.render and the requestAnimationFrame calls in the render function. It's simply subtracting the time passed from the cube's rotation on the Y axis (remember that it's in radians) to rotate the cube clockwise. Now open the browser and you should see your cube rotating clockwise smoothly.

## Conclusion

In this part of the series you learned how to prepare the scene, add objects and lights, and how to animate things. You can experiment with the app, add more or different objects, lights. It's up to you. Next time I will show you how to use textures and how to create some nice effects with particles. Don't forget to take a look at the documentation if you are having any problems.


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# Blync

There is funny little usb gadget called the Blync identified by VID: 0x1130, PID: 0x0001.

It's a glowing RGB cube, but it has two defects for me:

1. the driver is only available for Windows
2. only 7 colors are available

Let's fix it.

#### Software hack

That's the easy part.

1. Plug your Blync under Windows running in a virtual machine.
2. Start Wireshark under the Linux host
3. Dump USB packets
4. Think
5. Replay packets in Python

And Voilà: https://github.com/ticapix/blynux

#### Hardware hack

First, unscrew and remove the white top.

You will notice 5 RGB LEDs.

The LEDs look like this one PLCC6 SMD, LED SMP6-RGB. The Polarity ID or Cathode mark is in the bottom left corner, so lower pins are the ground. They even put labels to ease the identification 🙂 LR, LG, LB for, respectively, the Red, Green, Blue pins.

Take a Leonardo Pro Micro (! check that the boot loader is already installed !)

The Leonardo Pro Micro is a 5V board, but the LEDs need 2.1V for the Red and 3.3V for the Green and Blue so we need to add 3 resistors of value $R=\frac{(V_{in}-V_{forward})}{I_{forward}}$ and keep in mind that $I_{forward}$ is multiplied by 5.

In fact, the pins on the board can safely output only 40mA, which is less than the needed amount of 5*20mA=100mA. The proper solution would be to use a MOFSET like the IRF530.

I selected a resistor around $72 \Omega = \frac{(5-2.1)}{0.04}$ for red and $42 \Omega = \frac{(5-3.3)}{0.04}$ for green and blue.

Once this is done, solder the wires to the resistors to the PWM enable pins of the Leonardo. I choose pins 3, 5, 6.

Upload some code via the Arduino IDE and plug it in a USB port. The cube should glow in this order: Red $\rightarrow$ Yellow $\rightarrow$ Green $\rightarrow$ Cyan $\rightarrow$ Blue $\rightarrow$ Magenta $\rightarrow$ Red.

## 1 thought on “Blync”

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chemistry

# Derivation of Planks Black body spectrum equation

##### Planks Black body spectrum equation is given by,

Planks Spectral energy density, µ(ʋ,T)

$\dpi{120}&space;\fn_jvn&space;\large&space;\dpi{120}&space;\fn_jvn&space;U\left&space;(&space;\lambda,T&space;\&space;\right&space;)=&space;\frac{8\pi&space;h\nu&space;^{2}}{c^{3}}.&space;\frac{1}{e^{\frac{h\nu&space;}{KT}-1}}$

#### ♦DERIVATION

Consider a box of side ‘l’

Advertisement

The resonating modes of the electronegative radiation inside the box have the frequency, [docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/p2-2.odt” comments=”true”]

The energy of the proton is given by planks hypothesis as[docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/p3.odt” comments=”true”]

For a three dimensional box the energy expression will be[docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/p4-1.odt” comments=”true”]

The total energy, U, in a 1-D box is[docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/p5-2.odt” comments=”true”]

Therefor in a 3-D box the energy U will be[docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/p6-1.odt” comments=”true”]

Assuming the ni s to be continuous variables the summations can be replaced by integrals (fermi approximation)

Thus the total energy U is given by[docxpresso file=”https://learnwithstudy.com/wp-content/uploads/2018/04/plnk-last.odt” comments=”true”]

##### U(ʋ,T) is called the Black body spectrum and the equation is called Planks Black body spectrum equation

$\dpi{120}&space;\fn_jvn&space;\dpi{120}&space;\large&space;\frac{U}{l^{3}}=&space;\int_{0}^{\infty&space;}&space;U\left&space;(&space;\lambda&space;,T&space;\right&space;)d\lambda$

$\dpi{120}&space;\large&space;(d\nu&space;=&space;\frac{c}{\lambda&space;^{2}}d\lambda)$

where  $\dpi{120}&space;\fn_jvn&space;\large&space;\dpi{120}&space;\fn_jvn&space;U\left&space;(&space;\lambda,T&space;\&space;\right&space;)=&space;\frac{8\pi&space;h\nu&space;^{3}}{\lambda&space;^{3}}.&space;\frac{1}{e^{\frac{hc}{\lambda&space;KT}-1}}$

This spectral density function has units of energy per unit wavelength per unit volume.

Advertisement
##### Total energy

Evaluation of the integral to obtains the total energy inside the box yields


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## A curve which shows an overall negative slope may show

linear acceleration

any one of the above

negative acceleration

positive acceleration


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# 1. Rules of Differentiation

Throughout this module the following rules of differentiation will be assumed. (In the table of derivatives below, a is an arbitrary, non-zero constant.) [what is n? integer?]

y axn  sin (ax) cos (ax) eax  ln (ax)
dy/dx naxn − 1  a cos (ax) −a sin (ax) a eax  1/x

If a is any constant and u, v are two functions of x, then:

d/dx(u + v) d/dx(au) = du/dx + dv/dx = a du/dx

The stationary points of a function are those points where the gradient of the tangent (the derivative of the function) is zero.

Example 1: Find the stationary points of the functions:

(a)ƒ(x) = 3x2 + 2x − 9

Solution: The derivative is: dy/dx = 3 × 2x2−1 + 2 = 6x + 2

The stationary points are found by solving the equation:

dy/dx = 6x + 2 = 0

In this case there is only one solution, x = −3. Substituting this into the equation for ƒ(x), the corresponding y value is then:

y = 3(−3)2 + 2(−3) − 9 = 12

The stationary point is thus the point with coordinates (−3, 12).

(b)ƒ(x) = x3 − 6x2 + 9x − 2

Solution: The derivative is: dy/dx = 3x2 − 12x + 9

The stationary points are found by solving: 3x2 - 12x + 9 = 0. This is a quadratic equation (see the package on quadratic equations) and may be solved by factorising:

 0 = 3x2 − 12x + 9 = 3(x2 − 4x + 3) = 3(x − 3) (x − 1)

There are two solutions in this case, x = 3 and x = 1. If these are substituted into the function, the two stationary points will be found to be (3, −2) and (1, 2).

Exercise 1: Find the stationary points of the following functions:

(a)y = 16 − 6uu2

Solution: The derivative is: dy/du = −6 − 2u.

The stationary points are found by solving the linear equation:

−6 − 2u = 0

In this case there is only one solution u = −6 ⁄ 2 = −3. Substituting this into the expression for the function ƒ(u) = 16 − 6uu2, we find the corresponding y value:

y = 16 − 6 × (−3) − (−3)2 = 16 + 18 − 9 = 25

The stationary point is thus the point with coordinates (−3, 25).

(b)y = 3x2 − 4x + 7

Solution:The derivative is: dy/dx = 6x − 4.

The stationary points are found by solving the linear equation:

6x − 4 = 0

In this case there is only one solution x = 4 ⁄ 6 = 2 ⁄ 3. Substituting this this into the expression for the function ƒ(x) = 3x2 − 4x + 7, we find the corresponding y value:

y = 3 × (2 ⁄ 3)2 − 4 × 2 ⁄ 3 + 7 = 4 ⁄ 3 - 8 ⁄ 3 + 7 = 17 ⁄ 3

Therefore the function y = 3x2 − 4x + 7 has only one stationary point (2 ⁄ 3, 17 ⁄ 3).

(c)y = 1/3x3x2 − 3x + 2

Solution:The derivative is: dy/dx = x2 − 2x − 3.

The stationary points are found by solving the quadratic equation x2 − 2x − 3 = 0. Factorisation of this equation gives:

0 = x2 − 2x − 3 = (x + 1) (x − 3).

This yields two solutions, x = −1 and x = 3.

Substituting x = −1 into the function ƒ(x), the corresponding y value is:

y = 1 ⁄ 3 (−1)3 − (−1)2 − 3(−1) + 2 = −1 ⁄ 3 − 1 + 3 + 2 = 11 ⁄ 3

Similarly, substituting x = 3:

y = 1 ⁄ 3 (3)3 − (3)2 − 3(3) + 2 = 9 − 9 + 9 + 2 = −7

The two stationary points are thus (−1, 11 ⁄ 3) and (3, −7).

(d)y = x3 − 6x2 −15x + 16

Solution:The derivative is: dy/dx = 3 x2 − 12x − 15.

The stationary points are found by solving the quadratic equation 3x2 − 12x − 15 = 0. Factorising it:

0 = 3x2 − 12x − 15 = 3(x2 −4x − 5) = 3(x + 1) (x − 5)

We obtain two solutions, x = *minus;1 and x = 5.

Substituting x = −1 into the function ƒ(x), the corresponding y value is:

y = (−1)3 − 6 × (−1)2 −15 × (−1) + 16 = −1 − 6 + 15 + 1 = 24

Similarly, substituting x = 5:

y = (5)3 − 6 × (5)2 − 15 × (5) + 16 = −1 − 6 + 15 + 16 = −84

The two stationary points are thus (−1, 24) and (5, −84).

Click on questions to reveal solutions

# 2. Derivatives of Order 2 (and Higher)

In the package on Introductory Differentiation the height, h(t), in metres, of a ball thrown vertically at 10m s−1, was given by:

h(t) = 10t − 5t2

The velocity of the ball, vm s−1, after t seconds, was given by:

v(t) = dh/dt = 10 − 10t

The rate of change of velocity with time, which is the acceleration, is then given by a(t), where:

a(t) = dv/dt = −10m s−1

Since the acceleration was derived from h(t) by two successive differentiations, the resulting function, which in this case is a(t), is called the second derivative of h(t) with respect to t. In symbols:

a(t) = dv/dt = d/dt ( dh/dt ) = d2h/dt2     (Note the position of the superscripts)

Example 2: Find the first, second and third derivatives of the function: y = 3x6 − 2x + 1:

Solution:

The first derivative is given by:

y = dy/dx = 18x5 − 2

The second derivative is given by:

d2y/dx2 = d/dx ( dy/dx ) = 90x4

The third derivative is given by:

d3y/dx3 = d/dx ( d2y/dx2 ) = 360x3

Exercise 2: Find the first, second and third derivatives of the following functions:

(a)y = x3 + e2x

Solution:

The first derivative is given by:

dy/dx = 3 × x(3-1) + 2 × e2x = 3x2 + 2e2x

The second derivative is given by:

d2y/dx2 = d/dx ( dy/dx ) = 2 × 3x(2-1) + 2 × 2e2x = 6x + 4e2x

The third derivative is given by:

d3y/dx3 = d/dx ( d2y/dx2 ) = 6 + 8e2x
(b)y = sin (x)

Solution:

The first derivative is given by:

dy/dx = cos (x)

The second derivative is given by:

d2y/dx2 = d/dx ( dy/dx ) = −sin (x)

The third derivative is given by:

d3y/dx3 = d/dx ( d2y/dx2 ) = −cos (x)

Click on questions to reveal solutions

# 3. Maxima and Minima

 The figure on the right shows part of a function y = ƒ(x). The point A is a local maximum and the point B is a local minimum. At each of these points the tangent to the curve is parallel to the x-axis so the derivative of the function is zero. Both of these points are therefore stationary points of the function. The term local is used since these points are the maximum and minimum in this particular region, and there may be others outside this region.
 Further observations on the gradients of tangents to the curve are: to the left of A the gradients are positive (+) between A and B the gradients are negative (−) to the right of B the gradients are positive (+) About the local maximum point A the gradient changes from positive, to zero, to negative. The gradient is therefore decreasing. About the local minimum point B the gradient changes from negative, to zero, to positive. The gradient is therefore increasing.

The rate of change of a function is measured by its derivative. When the derivative is positive, the function is increasing, when the derivative is negative, the function is decreasing. Thus the rate of change of the gradient is measured by its derivative, which is the second derivative of the original function. In mathematical notation this is as follows:

At the point (a, b):

If dy/dx = 0 If dy/dx = 0
and d2y/dx2 < 0 and d2y/dx2 < 0
then the point (a, b) is a then the point (a, b) is a
local maximum local minimum

Example 3: Find the stationary point of the function y = x2 − 2x + 3 and hence determine the nature of this point.

Solution:

 If y = x2 − 2x + 3 then dy/dx = 2x − 2 and d2y/dx2 = 2 Now dy/dx = 2x − 2 = 0 when x = 1. Since d2y/dx2 = 2 > 0 for all values of x, this stationary point is a local minimum. Thus the function y = x2 − 2x + 3 has a local minimum at the point (1, 2). The figure on the right shows the function y = x2 − 2x + 3, with the local minimum point at (1,2) clearly visible.

Exercise 3: Determine the nature of the stationary points of the following functions:

(a)y = 16 − 6uu2 = 0

Solution: From the given function we have:

dy/du = −6 − 2u  and  d2y/du2 = −2

There is one stationary point when u = −3, which is obtained from the equation:

dy/du = −6 − 2u = 0

At this point y = 16 − 6 × (−3) − (−3)2 = 25. Since d2y/du2 = −2 < 0 for all values of u, this stationary point is a local maximum. Thus the function y = 16 − 6uu2 has a local maximum at the point (−3, 25).

(b)y = 3x2 − 4x + 7

Solution: From the given function we have:

dy/dx = 6x − 4  and  d2y/dx2 = 6

There is one stationary point when x = 2 ⁄ 3, which is the solution to the equation:

dy/dx = 6x − 4 = 0

At this point y = 3 × (2 ⁄ 3)2 − 4 × (2 ⁄ 3) + 7 = 17 ⁄ 3. Since d2y/dx2 = 6 > 0 for all values of x, this stationary point is a local minimum. Thus the function y = 3x2 4x + 7 has a local minimum at the point (2 ⁄ 3, 17 ⁄ 3).

Click on questions to reveal solutions

Quiz 1: Which of the following points is a local maximum of the function y = 2x3 − 15x2 − 36x + 6?

(a)(−3, 15) Wrong - please try again!
(b)(−6, 25) Wrong - please try again!
(c)(−2, 25) Wrong - please try again!
(d)(−1, 25)Correct - well done!

Explanation:

dy/dx = 6x2 − 30x − 36  and  d2y/dx2 = 12x − 30

There are two stationary points x = −1 and x = 6 which are obtained from solving the equation:

0 = dy/dx = 6x2 − 30x − 36 = 6 (x2 − 5x − 6 ) = 6 (x + 1) (x − 6)

At the point x = −1 the function value is y = 25 and its second derivative is:

d2y/dx2 = 12 × (−1) − 30 = −42 < 0

Therefore, the point (−1, 25) is a local maximum.

For the second stationary point, at x = 6, the second derivative is:

d2y/dx2 = 12 × (6) − 30 = 42 > 0

and y = −318. The point (6,−318) is therefore a local minimum.

Example 4: Find the stationary point of the function y = 2x3 − 9x2 + 12x − 3 and determine their nature.

Solution: From the given function we have:

dy/dx = 6x2 − 18x + 12   and  d2y/dx2 = 12x − 18

The stationary points are found by solving:

dy/dx = 6x2 − 18x + 12 = 0

Now 6x2 − 18x + 12 = 6 (x2 − 3x + 2), so the solution is found by solving x2 − 3x + 2 = 0, i.e. x = 1 and x = 2 (see the module on Quadratics). Note that when x = 1, y = 2 and when x = 2, y = 1.

 When x = 1, d2y/dx2 = 12(1) − 18 = −6 < 0, so (1, 2) is a local maximum. When x = 2, d2y/dx2 = 12 (2) − 18 = 6 > 0, so (2, 1) is a local minimum. The local maximum point at (1, 2) and the local minimum point at (2, 1) are clearly visible in the graph of the given function.

Exercise 4: Determine the nature of the stationary points of the following functions:

(a)y = 16 − 6uu2 = 0

Solution: From the given function we have:

dy/du = −6 − 2u   and   d2y/ du2 = −2

There is one stationary point when u = −3 which is obtained from the equation:

dy/du = −6 − 2u = 0

At this point y = 16 − 6 × (−3) − (−3)2 = 25. Since d2y/du2 = −2 < 0 for all values of u, this stationary point is a local maximum. Thus the function y = 16 − 6uu2 has a local maximum at the point (−3, 25).

(b)y = 1/3s2 −12s + 32

Solution: We have:

dy/ds = 2/3s − 12   and   d2y/ ds2 = 2/3

There is only one stationary point, when s = 18, which is obtained from the equation:

dy/ds = 2/3s − 12 = 2/3(s − 18) = 0

When s = 18 the function ƒ(s) takes the value y = −76. Since its second derivative d2y/ds2 = 2/3 is positive for all values of s, the stationary point (18, −76) is a local minimum.

(c)y = x3 − 6x2 − 15x + 16

Solution: We have:

dy/dx = 3x2 − 12x − 15   and   d2y/ dx2 = 6x − 12

There are two stationary points x = −1 and x = 5, which are obtained from solving the quadratic equation:

dy/dx = 3x2 − 12x − 15 = 3 (x2 − 4x − 5) = 3 (x + 1) (x − 5) = 0

When x = −1 the function value is y = 24 and its second derivative d2y/dx2 = 6 × (−1) − 12 = −18 < 0. The point (−1, 24) is a local maximum.

At the second point, x = 5, the function value is y = −84 and the second derivative d2y/dx2 = 6 × (5) − 12 = 18 > 0. The point (5, −84) is a local minimum.

(d)y = x3 − 12x + 12

Solution: We have:

dy/dx = 3x2 − 12   and   d2y/ dx2 = 6x

There are two stationary points x = −2 and x = 2, which are obtained from solving the equation:

dy/dx = 3x2 − 12 = 3 (x2 − 4) = 3 (x + 2) (x − 2)

When x = −2 the function value is y = 28 and its second derivative d2y/dx2 = 6 × (−2) = −12 < 0. The point (−2, 28) is a local maximum.

At the second point, x = 2, the function value is y = −4 and the second derivative d2y/dx2 = 6 × (2) = 12 > 0. The point (2, −4) is a local minimum.

Quiz 2: Given the function y = x + 1 + 1 ⁄ x, which of the following statements is true?

(a)(1, 3) is a local maximum Wrong - please try again!
(b)(1, −3) is a local minimum Wrong - please try again!
(c)(−1, 1) is a local minimum Wrong - please try again!
(d)(−1, −1) is a local maximum Correct - well done!

Explanation:

We have:   dy/ dx = 1 − x−2 = 1 − 1/x2   and   d2y / dx2 = −(−2)x−3 = 2/x3

The stationary points are found from the equation:

dy/dx = 1 − 1/x2 = 1 − x2/x2 = 0

and this is zero when 1 − x2 = 0. There are two solutions in this case, x = −1 and x = 1.

When x = 1 the function value is y = 1 and its second derivative is d2y/dx2 = 2/(−1)3 = −2 < 0. The point (−1, −1) is a local maximum.

When x = 1 the function value is y = 3 and the second derivative is d2y/dx2 = 2/(1)3 = 2 > 0. The point (1, 3) is a local minimum.

Quiz 3: Given the function y = x2/2 − cos (x), which of the following statements is true?

(a)(π2, π28) is a local maximum Wrong - please try again!
(b)(π2, π28) is a local maximum Wrong - please try again!
(c)(0, −1) is a local minimum Correct - well done!
(d)(0, −1) is a local maximum Wrong - please try again!

Explanation:

We have:  dy/dx = x + sin (x)  and  d2y/dx2 = 1 + cos (x)

The stationary points are found from the equation:

dy/dx = x + sin (x) = 0

Since sin (0) = 0, a solution to this equation is x = 0. At this point the function value is:

y = 0 −cos (0) = 0 − 1 = −1

Its second derivative when x = 0 is:

d2y/dx2 = 1 + cos (0) = 1 + 1 = 2 > 0.

The point (0, −1) is therefore a local minimum of the given function.

# 4. Quiz on Maxima and Minima

In each of the following cases, choose the correct option:

1. If y = 3x2 + 6x + 2  then:
(a)(1, 11) is a local minimum
(b) (−1, −1) is a local minimum
(c)(1, 11) is a local maximum
(d)(−1, −1) is a local maximum.
2. The function x3/3 + x2/2 − 2 x + 4 has a:
(a)local minimum when x = 2
(b)local maximum when x = −2
(c)local maximum when x = 1
(d)local minimum when x = −1
3. The function y = sin (x) + cos (x) has a:
(a)local maximum when x = π4
(b)local minimum when x = π4
(c)local maximum when x = π2
(d)local minimum when x = π2


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# Simplify and express with positive indices

Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)

But it needs to be simplifyed, and expressed with positive indices. Now as far as my knowledge takes me, this can't be simplifyed, but then I am probably wrong. If anybody could help to solve, that would be great.

Related Calculus and Beyond Homework Help News on Phys.org
lanedance
Homework Helper
maybe you could start by trying breaking things down into prime components eg.
$$6^{x+2} = 3^{x+2}2^{x+2}$$

I'm sorry but that example made no sense to me at all? I need more explanation.

lanedance
Homework Helper
well say you have, separate powers of 2 multiplied toegther, you can simplify by
$$2^{a}2^{b} = 2^{a+b}$$

so break down each number into its prime components, then multiply similar numbers together
6 = 3.2
4 = 2.2 etc.

HallsofIvy
Homework Helper
The rules of exponents,

$a^xa^y= a^{x+y}$ (which lanedance gave) and $(a^x)^y= a^{xy}$, are usually learned long before taking Calculus!

You will also need to know that $4= 2^2$, $6= 3(2)$, $9= 3^2$, and $12= 2^2 (3)$. Break everything down into powers of 2 and 3 and use the laws of exponents.

Could somebody please do this as an example, this is just the first one in the list I have to do.

Thankyou

lanedance
Homework Helper
you've got all the info, why not try it out?

All this made no sence to me either. Can someone show an example?

lanedance
Homework Helper
do for the first terms
$$6^{x+2}.4^{2x-4} = (3.2)^{x+2}.(2.2)^{2x-4} = (3^{x+2}.2^{x+2})(2^{2x-4}. 2^{2x-4}) = (3^{x+2})(2^{x+2}.2^{2x-4}. 2^{2x-4}) = (3^{x+2})(2^{(x+2) + (2x-4) + (2x-4)}) = (3^{x+2})(2^{5x+6})$$

then for terms that are divided, the power will be negative

sorry to be a bother but, that really just confused me even more.
If u can could you please show the whole example, and explain everything?
That would help heeps.
THANKS...:)

lanedance
Homework Helper
i think you've got way more than enough to attempt something... even though its not your post, have a try & i'll guide you the through

basically all you need to do is the same for the last 4 terms
$3^{5-x}$ - this is in lowest form
$2^{x-6}$ - this is in lowest form
$\frac{1}{12^{4x+3}} = 12^{-4x-3} = (2.3.3)^{-4x-3}$

and then i've leave the last for you... then multiply all the terms together & collect exponents

i think you've got way more than enough to attempt something...
There may be heeps on info, but it makes no sence to me, thats y i wanted an example so i could attempt questions just like that.

lanedance
Homework Helper
ok so what doesn't make sense?

post #9 is in effect a simple example

the whole thing makes sence, can u tell me what answer u got, cus i got 1.5x^3.
Is this correct???

wait thats not it, never mind.
can u please show me how u did ur working out cus im completly stuck?

lanedance
Homework Helper
I have in post 9..

part of the confusion may be how its written:
Ok, so I have this Expression, (6x+2 x 42x-4 x 35-x x 2x-6)/(124x+3 x 92x-3)
QUOTE]

I take it to actually mean teh lower crosses are "multlipy", whilst the x's in the exponents are the variable "x", giving below
$$(6^{+2}4^{2x-4} x 3{5-x}x 2{x-6})/(12{4x+3}x 9{2x-3})$$

that's open to interpretation

sorry but thats even more confusing. can u show me how u worked it out?

$$\frac{6^{(x + 2)} \ \cdot \ 4^{(2x - 4)} \ \cdot \ 3^{(5 - x)} \ \cdot \ 2^{(x - 6)}}{12^{(4x + 3)} \ \cdot \ 9^{(2x - 3)}}$$

Is that the equation??? (the dot means multiplication).

Start off by recognizing that you can do things like;

$$6^{(x + 2)} = 6^x \cdot 6^2$$

Play with this one first, compare the values of the Left Hand Side & The Right Hand side by giving different x values, you'll see something surprising, then generalise this idea.

We can continue to try to make sense of what doesn't make sense to you. Which concept is confusing: prime factorization or the laws of exponents alluded to by HallsofIvy and lanedance?

This will sound dumb but i dont even really know wat that means, thats y i wanted the answer, or the working so i can use it withother similar questions.

I think you should watch this video

http://khanexercises.appspot.com/video?v=zM_p7tfWvLU [Broken]

It's number 53 in the list underneath it, if you understand it then watch numbers 54 & 55 after it.

Come back after you watch these and read people's comments again, I bet everything will be clearer.

Then you should watch the videos on logarithms, ( I think I saw you in another thread with logarithm questions)

Last edited by a moderator:
the video helped a little, but not with this question. it just makes me more fustrated. i cant do it no matter how many time u interpret it.

If you watched 53,54 & 55 and then look at my first post here that should explain everything.

You've also got about 3 other ways to do this problem just given to you from people on this thread.

I can only guess that you haven't studied this stuff before or haven't looked at a good book explaining it.

There are these notes here = http://tutorial.math.lamar.edu/Classes/Alg/Alg.aspx that explicitly taught me how to do these problems ages ago, just read the 'Preliminaries' chapter, but really all you need to do is study this page here http://tutorial.math.lamar.edu/Classes/Alg/IntegerExponents.aspx that is from that link, I've just checked & the method to do this problem is taught explicitly.

If you give studying this page a shot & still can't do it then I'll show you step by step but you have to give this a shot first to learn for yourself.

I thought the video would just remind you of something you'd learned but you must not have ever seen this before, (or just understood it).

The above chapter I gave you should be 100% all you need to master this problem.

Last edited by a moderator:


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# PSNR

PSNR stands for Peak Signal-to-Noise Ratio. It is used as a measure of video quality. It is expressed in decibels. It's defined as

where I is the reference image, K is the image under test, and MSE is the Mean Squared Error between the two:

Where M is the number of pixels in a frame (width · height)

The double-Σ term states that (j,k) runs over all the pixels, summing the square of the difference between reference image I and test image K.


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# The norm of an operator

Let $\rho(x)$ be a weight function in a unit sphere, such that $$\begin{array}{l} \displaystyle 1. \rho(x)\ge 0,\int_{\mathbb{R}^n}\rho(x)=1\\ \displaystyle 2. \rho(x)\in \mathrm{C}_{0}^{\infty}(\mathbb{R}^n) \\ \displaystyle 3. \rho(x)=0 \mbox{ if |x|\ge1} \end{array}$$ We define $\rho_{\varepsilon}(x)=\frac{1}{\varepsilon}\rho(\frac{x}{\epsilon})$,$u_{\varepsilon}(x)=(\rho_{\varepsilon}*u)(x)$.

$\Omega$ is the domain of $u$,when $x\not\in \Omega$, $u(x)=0$.

Is it true that the operator $A_{\epsilon}:L^{p}(\Omega)\to L^{p}(\Omega)$ $$A_{\epsilon}(u)= u_\varepsilon$$ satisfies: $$||A_{\epsilon}-I||\ge 1$$ Moreover, $$||A_{\epsilon}-I||=1$$

I know that $$\int_{B_{\epsilon}(0)}u(x-y)\rho_{\varepsilon}(y)dy=\int_{\Omega}u(z)\rho_{\varepsilon}(x-z)dz$$ But I am confused by the way to $\sup\frac{||Ax||}{||x|}$ , I am not familiar with convolution. Thanks for your help.

-


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In this situation, we will let $$r$$ be the magnitude of $$z$$ (that is, the distance from $$z$$ to the origin) and $$\theta$$ the angle $$z$$ makes with the positive real axis as shown in Figure $$\PageIndex{1}$$. for f/g one needs g(z 0) 6= 0. In other words, it is the original complex number with the sign on the imaginary part changed. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). If a = 0 a = 0 and b b is not equal to 0, the complex number is called a pure imaginary number. If b is 0, it simply returns a. For example, you could rewrite i as a real part-- 0 is a real number-- 0 plus i. If we add to this set the number 0, we get the whole numbers. What are complex numbers? How to perform operations with and graph complex numbers. But the complex number 1 = 1+0i has this property. 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Focus: Circuits for Light

Physics 6, 91
Devices that process light signals the way traditional circuits process electronic signals can now work in a technologically important wavelength range.

Nanoscale structures called metatronic circuits process light waves the way conventional circuit elements process electric currents, but metatronics could be made much smaller and faster than conventional electronics. Now a team reporting in Physical Review Letters has built the first metatronic circuit elements that work for near-infrared light, the wavelength range of fiber optics and other silicon-based technology. The team’s findings could lead to ultrafast, light-based information processing fully integrated with a wide range of high-tech devices.

To continue the trend toward smaller and faster information processing circuits, researchers may soon need to harness light as the information carrier, replacing electric currents. To create circuit elements for light waves, Nader Engheta of the University of Pennsylvania and his colleagues are turning to a class of structures known as metamaterials, which have light-manipulating properties not found in naturally occurring materials.

Last year Engheta’s research group used silicon nitride to create the first metamaterial circuit elements [1]. Their circuitry processed light in the mid-infrared range, with wavelengths of $8$ to $14$ micrometers. Their structure consisted of an array of parallel nanorods, which are long blocks a few hundred nanometers wide, separated by air gaps and arranged horizontally. Illuminated from above, the rods and air gaps respond to the light’s fast-oscillating electric field in ways similar to inductors and capacitors responding to a more slowly oscillating voltage in a conventional circuit.

Inductors and capacitors change the timing and amplitude of an oscillating voltage in a way that varies with frequency, and the strength of their effects depends in part on their shapes. A conventional array of these circuit elements can act as a “band-stop” filter that blocks transmission of voltage signals within a limited frequency range or as a “bandpass” filter that allows only a certain frequency range through. Similarly, the rods and gaps in the “metatronic” circuit can partially block or transmit a range of wavelengths of light, depending on the orientation of the light’s electric field and other factors. The central wavelength and width of the affected range depend on the width and height of the rods. Similar filters are essential to information processing techniques such as multiplexing, which allows an optical fiber to carry many streams of data, each with a slightly different wavelength. But a more long-term goal is to design new light-processing circuits using optical capacitors, inductors, and perhaps other elements.

While the silicon nitride structures worked for mid-infrared light, they were unable to filter shorter-wavelength (and thus higher-frequency) light in the near-infrared range. This range includes $1.55$ micrometers, the wavelength at which most telecommunications optical fibers operate.

So Engheta and his colleagues have now made band-stop and bandpass filters from indium tin oxide, which has the right properties in the near-infrared range. The researchers also discovered new ways to vary the wavelength range and central wavelength of the filters, beyond simply changing the shape and spacing of the nanorods. They added nickel chromium on top of the rods or lead sulfide in the gaps to alter the structure’s inductance or capacitance and thus shift the band toward shorter or longer wavelengths. Engheta believes techniques like these may allow future engineers to design metatronic circuits to order. “I am optimistic that in the near future, we’ll be able to customize this,” he says.

“This paper represents a major step forward along the technological road map,” says electrical engineer George Eleftheriades of the University of Toronto. “We see here the evolution of ‘metatronics’ from an abstract theoretical concept to an actual technology.” He says that integrating optical circuitry with telecommunications fibers and other photonic devices is a major goal for designers of photonic information processing technology. But before that happens, we still need an optical transistor, which Eleftheriades says is currently “the key missing element.”

–Gabriel Popkin

Gabriel Popkin is a freelance science writer in Mount Rainier, Maryland.

References

1. Y. Sun, B. Edwards, A. Alù, and N. Engheta, “Experimental Realization of Optical Lumped Nanocircuits at Infrared Wavelengths,” Nature Mater. 11, 208 (2012)

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