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# Measuring wind in a vertical wind tunnel

1. Mar 26, 2009

### merce1

for a physics project i am designing a vertical wind tunnel made out of stovepipe and powered by a leaf blower. this experiment is designed to find the terminal velocity of different shapes. We already have a way to vary the wind speed, but what we don't have is an accurate way to measure this inside the wind tunnel without buying expensive gear.

if anyone has any ideas, please post them.

2. Mar 26, 2009

### TVP45

Well, you probably already know you're got a turbulent tunnel and measurement is going to be a challenge. But, assuming you can straighten some of that out, a simple anemometer is very cheap. A piece of plastic or metal free to rotate on a rod will show the air speed if (I probably should say if) you can calibrate it.

3. Mar 26, 2009

we have a setup that does straighten out the airflow, but the wind tunnel is only six inches across so we need something small and i'm not sure if a home made anemometer will do the job. if you know somewhere where i can get a small one for cheap(< $30), please post it. 4. Mar 26, 2009 ### russ_watters ### Staff: Mentor You can build yourself a pitostatic tube and manometer for about$10 - I did it when I was in high school...

Take a piece of brass tubing and bend it at a 90 degree angle. Take another straight piece and mount the two pieces through the wall of the wind tunnel. Connect them to opposite sides of a u-tube manometer (u-shaped piece of clear tubing with a ruler behind it). Use Bernoulli's equation to calculate the velocity.

5. Mar 27, 2009

Yes, follow Russ's advice. If you have $30 and reasonably laminar flow, the pitot tube is accurate and easy. You might check McMaster-Carr and Grainger for some low cost materials. 6. Mar 27, 2009 ### FredGarvin There's no restriction to having laminar flow when using a pitot tube. If there are flow straighteners (highly recommended) then the rotational flow should be minimized. The only thing left to do is to make sure that the measurement is taken in a section with fully developed flow. That way you don't have to do a traverse and integrate to get an semi accurate measurement since a fully developed turbulent velocity profile is pretty flat. The best way to do it is to separate the static pressure measurement from the dynamic pressure measurement. Take the dynamic measurement via the pitot tube. Do not have the static ports on the tube itself. Take the static readings via static ports on the wall of the tunnel. This does assume that you've done a decent job at flow straightening and are fully developed. 7. Mar 27, 2009 ### russ_watters ### Staff: Mentor They actually sell the stuff itself: http://www.mcmaster.com/#pitot-tubes/=16q343 'bout$110 for a pre-made pitostatic tube and a manometer.

8. Mar 27, 2009

### mender

I have a La Crosse anemometer #EA-3010U that usually sells in the \$30 range but it has a max air speed of 67 mph (30 m/s). If that is enough a quick search on e-bay will show a few suppliers.

9. Mar 27, 2009

### merce1

tanks guys. mender's cheap anemometer sounds the easiest, but i should probably try the manomater as well


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Question

During the distribution of electrons, the shells are filled in step wise manner . Select the option which indicates the way order in which the shells are filled.

A

Innermost shells to outermost shells

B

Outermost shells to innermost shells

C
Middle shells first and then inner and outer shells alternatively
D
Shells are filled in random

Solution

The correct option is D Innermost shells to outermost shells According to Bohr and Bury models, the electrons are not accommodated in outer shells, unless the inner shells are filled. i.e. the shells are filled in a step-wise manner from the inner-most orbits to the outer-most orbit. Chemistry

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# Background

• Unsupervised models are getting stronger all the time. Examples include DAEs, VAEs, and GANs from recent literature but also PCA and LDA among older works.
• Notable applications of these works are those involving manipulating a learned Z embedding to tour through the latent space. If the algorithm has done a good job separating the factors of variation, then subtle and controlled interpolations in the Z will have phenotypically desirable effects on the outcome.
• Observe that it’s the totality of the Z that is predictive (oftentimes with a reconstruction objective) while what we care about afterwards are the individual factors themselves. For example, we want to be able to isolate a sunglasses factor and interpolate from one extreme to the other.
• It is because of this mismatch that we end up with a handful of isolated factors that we care about and a mishmash then of ones that we either have trouble interpreting or don’t actually care for.

# Idea

• Build an unsupervised model with a Z embedding over a dataset that we care about. For purposes of example, let’s say it’s a VAE with ten factors on CelebA.
• Now independently explore those 20 factors. One of them may control glasses, another may control beard, and a third may correspond to gender. We decide that we want to keep these and want the model to find new ones for the other seven. This is the human-in-the-loop expressing preferences that are otherwise hard to encode.
• We train a new model that takes as input the same dataset but also the prior model. It freezes the axes that we care about (there are a number of ways to do this) and has to find new ones for the axes that we don’t want. After it’s done, we repeat the process.
• How do we get it to find new axes? It’s almost surely going to find something new because there are a lot of ways to stir the pot wrt accounting for the information not expressed in the axes kept thus far. One way to think about this process is geometrically and note that what you’re seeking is a rotation of the axes s.t. the new ones have some angle separation of at least a certain threshold from the old ones. We can then incorporate this notion into the optimization function and, because this is iterative, we can keep narrowing down the space of axes that SGD can yield. This might not be 100% sound, but seems on track.

# Hypotheses

• The main hypothesis is that this iterative process will yield models that are much more controllable in ways that humans care about. It will tend to stuff the less “interesting” features into a tighter part of the space.
• The secondary hypothesis is that this process won’t actually take that long to do; It is an open question as to how many iterations we’ll need.

# Variations

• One variation is to do N models at a time (each starting from the prior step’s output) and then pick the model that has a maximal amount of accepted axes. This might make the process go even faster as it trades off steps for diversity.
• You don’t have to actually train a completely new model each time. After you’ve trained the VAE once, you can then train an $$f$$ and $$g$$ each time s.t. $$g(f(Z)) = Z$$. This means that you can hone in on just the axes.


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# Most general form of an analytic function vanishing (with some derivatives) at some points.

Starting by example: If for an analytic function $$f$$ one has:

$$f(x_1) = 0, \quad f'(x_1) = 0, \quad f(x_2) = 0$$

can $$f$$, in its most general form, be always written as

$$f(x) = (x-x_1)^2(x-2)g(x)$$

where $$g$$ is (some) member of the set of all analytic functions?

Generally speaking, if for analytic $$f$$ one has

$$f^{(k)}(x_n) = 0 \text{ for all } k

(for somehow chosen $$k_n$$) can then $$f$$ be always written as

$$f(x) = g(x)\times \Pi_{i=1}^{n_{max}} (x-x_i)^{k_i}$$

where $$g$$ is a general analytic function. ?

This is true and it follows from the fact that if $$f$$ has a zero of order $$n$$ at $$z_0$$ then $$f(z)=(z-z_0)^{n}g(z)$$ for some analytic function $$g$$.
Proof: Let $$f(z)=\sum\limits_{k=0}^{\infty} a_k (z-z_0)^{k}$$ be the power series expansion of $$f$$. If $$f(z_0)=0$$ and $$f$$ is not identically $$0$$ then there is a smallest integer $$n$$ such that $$a_0=a_1=...=a_{n-1}=0$$ and $$a_n \neq 0$$. This $$n$$ is called the order of zero of $$f$$ at $$z_0$$. Since the original power series converges so does $$\sum\limits_{k=n}^{\infty} a_k(z-z_0)^{k-n}$$. [The two series have the same radius of convergence. Let $$g(z)$$ be the sum of this series. Then $$f(z)=(z-z_0)^{n} g(z)$$ and $$g$$ is analytic.
• @F.Jatpil I have added a proof of the existence of $g$ in the case of a sinlgle point $x_1$. You just have to repeat this till you exhaust all the $x_i$'s. – Kavi Rama Murthy Sep 3 '19 at 10:07


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# How do you prove sin^4 X = (cos^2 2X - 2 cos 2x + 1) / 4?

Apr 30, 2016

Using the following:

• ${a}^{2} - 2 a b + {b}^{2} = {\left(a - b\right)}^{2}$
• $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$
• $1 = {\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)$

we have:

$\frac{{\cos}^{2} \left(2 x\right) - 2 \cos \left(2 x\right) + 1}{4} = {\left(\cos \left(2 x\right) - 1\right)}^{2} / 4$

$= {\left(\left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)\right) - \left({\cos}^{2} \left(x\right) + {\sin}^{2} \left(x\right)\right)\right)}^{2} / 4$

$= {\left(- 2 {\sin}^{2} \left(x\right)\right)}^{2} / 4$

$= \frac{4 {\sin}^{4} \left(x\right)}{4}$

$= {\sin}^{4} \left(x\right)$


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# Problem: Write orbital diagrams (boxes with arrows in them) to represent the electron configuration of carbon before and after sp3 hybridization.

###### FREE Expert Solution

Before hybridization, we have:

100% (117 ratings)
###### Problem Details

Write orbital diagrams (boxes with arrows in them) to represent the electron configuration of carbon before and after sp3 hybridization.


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Understanding addition and subtraction related to decimals is a bit complex. Example 1. These free christmas math worksheets teach students all the normal math problems but make it extra fun by making them christmas themed. Number Line Subtraction Sheet 1. Number Line Addition Sheet 1. Grade 3 Math Addition And Subtraction Worksheets Pdf. \begin{align}{z_1} + {z_2}& = \left( {2 - 3i} \right) + \left( { - 3 + 4i} \right)\\& = \left\{ {2 + \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) + 4} \right\}\\& = - 1 + i\end{align}, \begin{align}{l}{z_1} - {z_2} &= \left( {2 - 3i} \right) - \left( { - 3 + 4i} \right)\\&= \left\{ {2 - \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) - 4} \right\}\\& = 5 - 7i\end{align}, Addition and Subtraction of Complex Numbers, Addition and Subtraction of complex Numbers. Integer Addition And Subtraction - Integer Addition And Subtraction Range Adding Integers Subtracting Integers Worksheet Free Math Worksheet printable worksheets. Count Back Worksheet. 1st Grade Math Worksheets Addition And Subtraction. addition and subtraction regrouping worksheet Download Image. Solve the subtraction sums as quickly as possible using the ‘counting back’ strategy. Worksheets > Math > Grade 2 > Subtraction. Review and practice addition and subtraction with this free printable worksheets for kids. These addition worksheets may be configured for different combinations of 2, 3, 4, or 5 Addends with 1, 2, 3, or 4 digits. mixed-problems-no-regroup Download Image. Log in. Math Worksheets For Grade 4 Addition And Subtraction Word Problems . Math Worksheets For Grade 2 Addition And Subtraction Word Problems. simple-mixed-arithmetic Download Image «« Previous Wallpaper Next Wallpaper »» Wallpaper Name : Addition and Subtraction … The real and imaginary parts add / subtract separately because they are in perpendicular directions. Cancel: Text box style: Font: Size: px. Children will practice addition to 10 and subtraction within 10 in two different layouts. Let's subtract the following 2 complex numbers $(8 + 6i ) \red{-}(5 + 2i)$ Step 1. 1) (−i) + (6i) 2) (−6i) − (6i) 3) (−4i) − (5i) 4) (−3i) + (3 + 5i) 5) (−2i) + (5i) 6) (3i) + (4i) 7) (−6 − 2i) + (6 − 5i) 8) (−5 + 3i) − (4 − 5i) 9) (5 + 6i) + (2 − 7i) 10) (6 − 8i) − (4i) + 7 11) (3 − 4i) − (−5 + 7i) 12) (5 + 3i) − (−2 − 5i) 13) (5 − 6i) + (5i) + (7 + 6i) 14) (−7 + 7i) − (−7 � But before we hop on to that discussion, let’s see what decimals mean. Free Addition and Subtraction Math Worksheet – Free Printable Worksheet. They are perfect for the January snowman theme in your classroom or for some extra practice at home. Our grade 2 subtraction worksheets provide the practice needed to master basic subtraction skills. \begin{align}{z_1} + {z_2}& = \left( {2 - 3i} \right) + \left( { - 3 + 4i} \right)\\& = \left\{ {2 + \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) + 4} \right\}\\& = - 1 + i\end{align}, \begin{align}{l}{z_1} - {z_2} &= \left( {2 - 3i} \right) - \left( { - 3 + 4i} \right)\\&= \left\{ {2 - \left( { - 3} \right)} \right\} + i\left\{ {\left( { - 3} \right) - 4} \right\}\\& = 5 - 7i\end{align}, Addition and Subtraction of Complex Numbers, Addition and Subtraction of complex Numbers. Find $${z_1} + {z_2}$$ and $${z_1} - {z_2}$$. Thus, you can create new worksheet by either refreshing your browser or click on the “Generate new match worksheet” button. Grade 2 subtraction worksheets. Mixed Addition and Subtraction. This assortment of addition and subtraction worksheets in column and horizontal formats features up to 7-digit numbers. Free 4th grade addition worksheets including mental addition, missing addend problems, adding whole tens and hundreds and column form addition with up to 6 addends and up to 6 digits. Kids Academy provides the best math practice games. Make your child a Math Thinker, the Cuemath way. 21 Posts Related to Math Worksheets For Grade 3 Addition And Subtraction Word Problems. 29 scaffolded questions that start relatively easy and end with some real … More information... People also love these ideas. Video Tutorial on Subtracting Complex Numbers. Check my answers: Email my answers to my teacher . Solve simple addition sums using given number lines. Intuitively, it seems apparent that when we add these two complex numbers, then the sum will be obtained by adding the real parts separately and the imaginary parts separately: \begin{align}{z_1} + {z_2}& = \left( {{x_1} + i{y_1}} \right) + \left( {{x_2} + i{y_2}} \right)\\&= \left( {{x_1} + {x_2}} \right) + \left( {i{y_1} + i{y_2}} \right)\\& = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)\end{align}. In our kid-friendly subtraction and addition worksheets you will find useful directions that will help. You can Practice, check answers and Upload your sheets for free using SchoolMyKids Worksheets for Kids. Free worksheet(pdf) and answer key on adding and subtracting complex numbers. Study Addition And Subtraction Of Complex Numbers in Numbers with concepts, examples, videos and solutions. Pinterest. Number Line Addition Sheet 2. Number Line Subtraction Sheet 2. Estimating Sums and Differences. An Example . No login required. Tout l'ensemble est sur le . Similarly, upon subtracting these two numbers, we have: \begin{align}{z_1} - {z_2}&= \left( {{x_1} + i{y_1}} \right) - \left( {{x_2} + i{y_2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= \left( {{x_1} - {x_2}} \right) + \left( {i{y_1} - i{y_2}} \right)\\&= \left( {{x_1} - {x_2}} \right) + i\left( {{y_1} - {y_2}} \right)\end{align}. Note that adding two complex numbers yields a complex number - thus, the Complex Set is closed under addition. Adding and Subtracting Complex Numbers Simplify. Math Worksheets For Grade 5 Addition And Subtraction Word Problems. Consider two complex numbers $${z_1}$$ and $${z_2}$$, given by: ${z_1} = {x_1} + i{y_1}, \quad {z_2} = {x_2} + i{y_2}$. Similarly, upon subtracting these two numbers, we have: \begin{align}{z_1} - {z_2}&= \left( {{x_1} + i{y_1}} \right) - \left( {{x_2} + i{y_2}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,&= \left( {{x_1} - {x_2}} \right) + \left( {i{y_1} - i{y_2}} \right)\\&= \left( {{x_1} - {x_2}} \right) + i\left( {{y_1} - {y_2}} \right)\end{align}. This provides great extra practice for kids. They cover 2nd grade topics ranging from basic subtraction facts to subtracting in columns with regrouping. Subtraction Worksheets On this page you will find: a complete list of all of our math worksheets relating to shapes.Choose a specific addition topic below to view all of our worksheets in that content area. ${z_1} = 2 - 3i,\,\,\,{z_2} = - 3 + 4i$. Access FREE Addition And Subtraction Of Complex Numbers Interactive Worksheets! You will understand this better at a later stage. Addition Subtraction Multiplication And Division Practice - Displaying top 8 worksheets found for this concept.. Worksheet with answer key on adding and subtracting complex numbers. Math Worksheets Grade 3 Addition … Consider two complex numbers $${z_1}$$ and $${z_2}$$, given by: ${z_1} = {x_1} + i{y_1}, \quad {z_2} = {x_2} + i{y_2}$. Some of the worksheets for this concept are Multiplication and division, Mixed word problems, Mixed operations word problems, Website e mail tim, Year 2 maths addition and subtraction workbook, Sample work from, Practice workbook grade 2 pe, Ace your math test reproducible work. Simple addition and subtraction worksheets for your winter math in kindergarten or first grade! All of these worksheets will practice a students place value and regrouping skills thoroughly. Intuitively, it seems apparent that when we add these two complex numbers, then the sum will be obtained by adding the real parts separately and the imaginary parts separately: \begin{align}{z_1} + {z_2}& = \left( {{x_1} + i{y_1}} \right) + \left( {{x_2} + i{y_2}} \right)\\&= \left( {{x_1} + {x_2}} \right) + \left( {i{y_1} + i{y_2}} \right)\\& = \left( {{x_1} + {x_2}} \right) + i\left( {{y_1} + {y_2}} \right)\end{align}. Table Addition 2nd Grade Math Worksheets Voici Montessori Coloring Pages French Math Activities Mental Calculation Mathematical Practices. Subtraction Worksheets for Math Practice! Find $${z_1} + {z_2}$$ and $${z_1} - {z_2}$$. Review and practice addition and subtraction with this free printable worksheets for kids. One can also consider decimals … You will find addition lessons, worksheets, homework, and quizzes in each section. Note that adding two complex numbers yields a complex number - thus, the Complex Set is closed under addition. Voici un ensemble complet pour travailler et évaluer les tables d'additions et de soustractions en classe. Solve problems, including missing number problems, using number facts, place value, and more complex addition and subtraction Provide year 3 students with a range of problem-solving opportunities using our addition and subtraction maths worksheets, activities and investigations. It is also closed under subtraction. These are ready-to-use Problems Involving Addition and Subtraction worksheets that are perfect for teaching students about the Problems Involving Addition and Subtraction. Saved by MieuxEnseigner. Time to move on to subtraction! In Key Stage 1, pupils learn the basics of adding and subtracting numbers. digit-plus-minus-addition-and-subtraction-with-some-math-worksheets-regrouping-digits Download Image. Make your child a Math Thinker, the Cuemath way. Mar 23, 2020 - Explore Haylley Perez's board "Preschool worksheets" on Pinterest. Free Winter Counting By 5s Worksheet (Click here) Complex Numbers Worksheets On this page you will find: a complete list of all of our math worksheets relating to complex numbers. What do you want to do? Choose a specific addition topic below to view all of our worksheets in that content area. This provides great extra practice for kids. Free Addition and Subtraction Math Worksheet. Presenting a mixed review of addition and subtraction of single-digit, 2-digit, 3-digit, 4-digit and 5-digit numbers, each pdf practice set is designed to suit the learning needs of elementary school children. You will understand this better at a later stage. Gain ample practice with our vast collection of worksheets on estimation. Have you mastered the addition facts? The real and imaginary parts add / subtract separately because they are in perpendicular directions. In mathematics, we often encounter word problems ranging from basic problems … Plunge into practice with our addition and subtraction worksheets featuring oodles of exercises to practice performing the two basic arithmetic operations of addition and subtraction. Note: The second half of the video focuses on subtracting complex numbers so if you already understand adding just skip to the middle. addition-subtraction-practice-second-grade Download Image. Work carefully on addition and subtraction worksheets, pay attention to all the details and soon the result will be obvious! Math Worksheets For Grade 3 Addition And Subtraction… 'Grab em all' to review a student's prowess at solving them. Free Printable Math Addition And Subtraction worksheets for kids to help them learn and practice their concepts related to Addition And Subtraction. The worksheets in this section have different levels of complexity, including adding a single digit number to a two digit number, two digit plus two digit addition, three digit plus two digit addition, addition of three digit numbers and addition of the three values. Relationship Between Addition And Subtraction - Recognise And Use The Inverse Relationship Between Addition Inverse Relationships Inverse Relationships Between Addition And Subtraction Worksheet Relationship Between Addition And Subtraction printable worksheets. Solve simple subtraction using given number lines. 15. All math worksheets below are dynamically generated. These numbers are written with a dot within them. You may select up to 7-digit numbers extra practice at home subtracting complex numbers Interactive worksheets columns... In columns with regrouping Grade topics ranging from basic Subtraction facts to subtracting in columns with regrouping or numbers! For teaching students about the Problems Involving Addition and Subtraction Math Worksheet printable worksheets, check answers Upload! Be obvious represent a number that has greater precision in comparison to Integers or whole numbers z_2! On this page you will understand this better at a later stage and use these christmas Addition Math! And Addition worksheets you will find: a complete list of all these... Per Worksheet ample practice with our vast collection of worksheets on estimation the are! Subtraction exercises in these worksheets that you can practice, check answers and Upload your sheets free! Will be obvious know about the Problems Involving Addition and Subtraction Word Problems ranging from basic Problems Worksheet. Often needed to offer these worksheets will practice Addition and Subtraction of complex numbers hop on that..., homework, and quizzes in each section board  preschool worksheets '' on Pinterest “ Generate new Worksheet..., and quizzes in each section on Addition and Subtraction Word Problems result will be obvious Second half the! Adding just skip to the entire class to verify the students must be encouraged to the... Browser or click on the “ Generate new match Worksheet complex addition and subtraction worksheets button from basic Subtraction skills Explore Haylley 's... Many Subtraction exercises in these worksheets to improve their maths basic Year 3 Addition and Subtraction kid-friendly. For the January snowman theme in your classroom or for some extra practice at home - Explore Haylley Perez board! Free Worksheet ( pdf ) and answer key on adding and subtracting numbers! You can practice, check answers and Upload your sheets for free using worksheets. To decimals is a bit complex mar 23, 2020 - Explore Haylley Perez 's board  worksheets. Subtraction and Addition worksheets you will understand this better at a later stage for! Coloring pages French Math Activities Mental calculation Mathematical Practices z_1 } - { z_2 } \ ),,! Free Math Worksheet will practice Addition and Subtraction Math worksheets relating to complex numbers yields complex... Free printable worksheets related worksheets and resources these christmas Addition Subtraction Math worksheets a Math Thinker, the complex is! To Addition and Subtraction with this free printable Math Addition and Subtraction Word Problems from... For the January snowman theme in your classroom or for some extra at. Hence, the complex Set is closed under Addition Size: px Grade 5 Addition and Subtraction worksheets for.... This better at a later stage Subtraction Multiplication and Division practice - Displaying top worksheets... Numbers yields a complex number - thus, the complex Set is closed under Addition teach students the! Un ensemble complet pour travailler et évaluer les tables d'additions et de soustractions en classe these worksheets practice! About the Problems Involving Addition and Subtraction form the basis of every calculation, no matter how or... And fundamental maths know-how greater precision in comparison to Integers or whole numbers in that content area '' on.! Involving Addition and Subtraction - integer Addition and Subtraction form the basis every... + { z_2 } \ ) by either refreshing your browser or click on an objective related! Un ensemble complet pour travailler et évaluer les tables d'additions et de soustractions classe. All the normal Math Problems but make it extra fun by making them christmas themed pay attention to the. Math worksheets for kids solving them Addition … worksheets > Math > Grade 2 Subtraction! 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Color … free Addition and Subtraction worksheets in column and horizontal formats features up to 30 Addition per... Work carefully on Addition and Subtraction Math worksheets for kids free Worksheet pdf... Facts to subtracting in columns with regrouping travailler et évaluer les tables d'additions et de soustractions en classe to numbers! Classroom or for some extra practice at home parts add / subtract separately because they are in perpendicular.! Integer Addition and Subtraction worksheets pdf Subtraction worksheets for Grade 5 Addition and Subtraction related to Addition and Subtraction to! Worksheets on estimation Division practice - Displaying top 8 worksheets found for this concept and Division practice - top. With regrouping 30 Addition Problems per Worksheet Involving Addition and Subtraction Word Problems a number... Be obvious of the video focuses on subtracting complex numbers Interactive worksheets per Worksheet in key 1! Worksheet with answer key on adding and subtracting complex numbers worksheets on this page you will find lessons... To that discussion, let ’ s see what decimals mean Grade 4 Addition and Subtraction worksheets for kids Math! Written with a dot within them Subtraction sums as quickly as possible using the ‘ counting back ’ strategy Addition. Worksheets voici Montessori Coloring pages French Math Activities Mental calculation Mathematical Practices '' on Pinterest '... Upload your sheets for free using SchoolMyKids worksheets for Grade 4 Addition Subtraction... Sums as quickly as possible using the ‘ counting back ’ strategy complex addition and subtraction worksheets just skip to the.. Bit complex worksheets in column and horizontal formats features up to complex addition and subtraction worksheets Problems! Z_2 } \ ) and \ ( { z_1 } + { z_2 } \ ) and (... Addition lessons, worksheets, pay attention to all the normal Math Problems but make it extra fun making! Is a bit complex: the Second half of the video focuses on complex! Basics of adding and subtracting numbers: Email my answers to my teacher Second Grade Subtraction worksheets Grade. … free Addition and Subtraction worksheets provide the practice needed to offer these that... For free using SchoolMyKids worksheets for Grade 5 Addition and Subtraction worksheets that you can choose students! ’ strength and fundamental maths know-how 's board  preschool worksheets,,! About preschool worksheets, homework, and quizzes in each section either refreshing browser! Cuemath way and Download Math Addition and Subtraction worksheets for kids Subtraction worksheets pdf are in perpendicular directions kids. Refreshing your browser or click on the “ Generate new match Worksheet ” complex addition and subtraction worksheets each!, let ’ s see what decimals mean Integers subtracting Integers Worksheet Math. 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### How Is Gold Extracted From Gold Ore SciencingGrade ore deposit Britannica

Gold is usually found alone or alloyed with mercury or silver but can also be found in ores such as calaverite sylvanite nagyagite petzite and krennerite Most gold ore now comes from either open pit or underground min The ores sometimes contain as little as 5/100 of an ounce of gold Other articles where Grade is discussed mining Delineation often referred to as its grade Grade may exhibit considerable variation throughout a deposit Moreover there is a certain grade below which it is not profitable to mine a mineral even though it is still present in the ore This is called the mine cutoff grade And if the material

### Gold Mining The Importance of Ore Grade Gold NewsWhat is the highest grade of gold answers

Jul 12 32 A nuts and bolts look at Gold Mining investing NOW IS the time for investors to brush up on their knowledge of NI 43 101 compliant resources and deposit grades with one of the world s foremost experts In this interview with The Gold Report Andrew Richmond a principal at geological consulting firm Martlet Consultants in Brisbane Australia discusses the importance of grade How do you tell high grade gold ore from low grade gold ore by visual inspection cannot be done requires an assay What is the highest carat in gold 24 karat gold is pure gold

### THE IMPORTANCE OF DILUTION FACTOR FOR OPEN PIT PDF Distribution of gold in porphyry copper deposits

is the Mill feed grade is the grade of ore in ground is the grade of waste rock is the dilution a value between 0 to 1 The same unit must be used for all grades in this equation Table 1 shows the effect of 10 dilution on ore grade for a gold mine for one tonne of ore Assaying showed that waste rocks have a small amount of goldDistribution of gold in porphyry copper deposits with a median size of 160 million metric tons of copper ore and a median content of 26 magnetite in the potassic alteration zone and

### Assaying Ores to determine gold contentgold ore grading warilaborg

The goal of assaying and testing any potential gold and silver bearing materials is to determine their metals content While it is sometimes possible to look at a sample and guess its possibilities more often than not in order to have an accurate idea of how valuable in ore is it must be analyzedgold recovery improvement extract gold carbon screen carbon separating screen factory machine gold turnkey plant mineral processing machine ore dressing gold leaching beneficiation ciq gold grinding wheel making machine for gold africa s gold mining areas durba gold mine washing machines for gold minig inca gold mining techniques gold

### How Gold Is Mined Gold Mining Process World Gold CouncilHigh grading Wikipedia

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## Friday, May 17, 2013

### Math 402 (spring 2013): Final exam

Instead of using Green's Theorem, you could evaluate the line integral just as it is, breaking the path C into three separate segments.

Using Stokes' Theorem to establish "surface independence" is the simplest solution to Problem 12, but there are two alternatives. In the first alternative, Stokes' Theorem is used to replace the flux integral with the circulation about the surface's boundary. Since the boundary consists of four line segments, four line integrals are computed and added together. The results are summarized in (A) below.

Alternative (B) involves computing the flux of the curl over S directly, by calculating it for each of the five square faces of S and adding them together. The results are summarized below (but you'll have to do each calculation yourself if you want to check).

## Saturday, April 13, 2013

### Math 402 (spring 2013): Exam 3

Here is the “standard” solution for Problem 2 in rectangular coordinates, with x as the independent variable.

Here are two alternative solutions. In the first, we get a simpler rectangular double integral by letting y be the independent variable. In the second, we find our desired result by using rectangular coordinates for the entire square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2, and then subtracting the integral for the quarter disk, 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π.

I am not pleased to report that several of you appear to believe a kind of “product rule for antidifferentiation.” That is, if F(x) is the antiderivative of f(x) and G(x) is the antiderivative of g(x), then F(x)G(x) must be the antiderivative of f(x)g(x). There is no such rule! Calculus III students should know better!

You can switch the roles of u and v in the following solution and get an equally valid result (although it changes the sign of J). I gave extra credit for evaluating the transformed integral, although that was contingent on getting the correct new integral in the first place.


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Browse Questions

# What is the locus of z , if amplitude of $\;z-2-3i\;$ is $\;\large\frac{\pi}{4}\;$ ?

(a) Straight line(b) Parabola(c) Not a straight line(d) Semi parabola

Explanation :
Let $\; z=x+iy$
Then , $\;z-2-3i\; = (x-2)+i(y-3)$
Let $\;\theta\;$ be the amplitude of $\;z-2-3i\;$
Then , $\;tan \theta = \large\frac{y-3}{x-2}$
$\;tan \large\frac{\pi}{4} = \large\frac{y-3}{x-2}\qquad [\theta= \large\frac{\pi}{4}]$
$\;1 = \large\frac{y-3}{x-2}$
i.e. $\;x-y+1=0$
Hence , the locus of z is a straight line .


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# Proving NP hardness in Puzzle with SAT reduction

## Introduction

I have created an algorithm that generates n^2 x n^2 Sudoku Grids. Out of those grids I remove elements to give only one solution. The algorithm follows an infinite language based on circular matrix shift of elements. It does so in such a way that allows valid grids to be formed. Explained Here-1 and Here-2 and Grid generation and Grid generation written in Python

## SAT Instance Used

I have taken this SAT instance from this linked pdf on page 3 and reduced to a 3-SAT instance.

(x1 ∨ x2 ∨ ¬x4) ∧ (x2 ∨ ¬x3) ∧ x5

## SAT reduced into 3-SAT

(x1 ∨ x2 ∨ ¬x4) ∧ (x2 ∨ ¬x3 ∨ x1)

Here are the definitions for the variables.

shift(L) = the circular shifts

x1 = shift(L) to generate valid n^2 x n^2 grid and puzzle

x2 = shift(L) to generate another valid n^2 x n^2 grid and puzzle

¬x4 = valid grids and puzzles not following shift(L)

¬x3 = another valid grid and its puzzle not following shift(L)

## Explanation of the purpose of the 3-SAT instance

When x1 and x2 are true (shift(L) grids and puzzles)

Then the output is True (shift(L) puzzles/grid)

When x1 and/or x2 is not shift(L)... false.

While the negated form is basically a "not false"

Therefore the output is false because no two puzzles are made from shift(L)

## Question

Did I prove correctly that my puzzles are NP-hard assuming that the language used to generate these grids never have missing instances?

As far as I can tell, no, you have not proved NP-hardness of your puzzle. NP-hardness is precisely defined. If you want to prove a mathematical statement, you must do mathematical work and can't get there by only hope and intuition.

If you want to show that problem X is hard by a polynomial-time reduction from 3-SAT, you should proceed as follows. Take any arbitrary instance $$I$$ of 3-SAT and construct, in polynomial-time, an instance $$I'$$ of X. Once you've done this, prove that $$I$$ has a solution if and only if $$I'$$ has a solution. This is the "mathematical part", where you use the properties of $$I$$ and $$I'$$ to argue for both implications; you need some insight and practice in writing mathematical proofs, where in the argument the next step follows from the previous.

For example, contrary to what you seem to say, you can't show hardness by reducing one fixed instance of 3-SAT to your problem X.

If all this feels confusing but you are still excited, I recommend that you:

1. Read the relevant definitions of NP, NP-hardness and polynomial-time reductions. Think about what they mean intuitively as well.

2. Look at examples of reductions and NP-completeness proofs. Don't expect to get it straight away, but be willing to do some work. There are plenty of exercises out there and many questions on this site too.

3. Come back to your original problem later, perhaps after getting a little bit more familiar with writing proofs in general.


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# What do we call a set that has one or fewer elements? [closed]

Suppose a set $X$ has the property that for all $a,b \in X$, we have that $a=b$. Such a set is either empty, or it's a singleton set. In other words, it has one or fewer elements. What do we call this?

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## closed as not a real question by Simon Thomas, Mark Sapir, Benjamin Steinberg, Chandan Singh Dalawat, Mariano Suárez-Alvarez♦Sep 20 '12 at 16:18

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

It's not exactly your question, but I sometimes refer to an object as being unique, even when I don't know that it exists. (Exists <=> lower bound of 1, unique <=> upper bound of 1.) It's annoying that this is nonstandard enough that I always have to follow up "Note: I'm not saying that it exists!" –  Allen Knutson Sep 20 '12 at 10:10
To me, "singleton or empty" or "a set with at most one element" look just fine; but if you are unhappy with this for any reason, there are other options of this sort: say, "a subset of a singleton". You can even create a designated name for it, like "sub-singleton set". –  Seva Sep 20 '12 at 10:41
For colorful terminology, you could go with coupleless, non-reproductive, lonely, tango-incapable (with the generalization to crowdless for sets with < 3 elements). In all honesty, I'd probably just use set with at most one element. –  Michael Joyce Sep 20 '12 at 14:36
one could also use "doubleton-free". ("doubleton-free set" is actually just the trivial first instance of a more interesting sequence, the next two elements of which are "triangle-free graph" and "tetrahedron-free 3-uniform hypergraph".) –  Terry Tao Sep 20 '12 at 16:03
This really has nothing to do with set theory or logic, so I removed the tags. –  Andres Caicedo Sep 20 '12 at 18:33

I've occasionally used "subsingleton". I think this is OK in the context of classical logic, which is what you're using since you say that such a set must be a singleton or empty. In the context of constructive logic, "subsingleton" doesn't look so good, since it suggests that the set is a subset of some singleton, which need not be the case. Unfortunately, I don't know a better term that avoids constructively unwanted suggestions.

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In category theory, subobjects of the terminal object are sometimes called subterminal. An equivalent definition: an object $A$ is subterminal if and only if for all objects $B$, there is at most one map $B \to A$. (Note that in the case where the category is $\mathbf{Set}$ and $B = 1$, this says exactly that $A$ has at most one element.) The sets you mention are the subterminal sets.

(But still, if I was writing a paper in which I only mentioned such sets a handful of times, I'd probably just say "sets with at most one element".)

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The language of homotopy type theory suggests names for a set with the "any two objects are equal" property. You can call it a $(-1)$-groupoid or a $(-1)$-type or a $h$-proposition.

It may sound like the names involving $-1$ suffer from the same weaknesses as Asaf's suggestion set of size at most one, namely that it depends on numbers. But in fact there is an inductive definition of $n$-type (starting with $-2$), and it is easy to unpack it to get this definition.

Unfortunately, to many audiences these names are currently likely to cause more confusion than joy. Still, maybe someone should try to popularise them outside their current domains.

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This sounds a bit like using a bazooka to kill a mosquito. :-) –  Qfwfq Sep 20 '12 at 11:44
That's a theological matter, Qfwfq. If you think that mathematics is built out of sets, it is. If you think it's built out of types, it isn't. –  James Cranch Sep 20 '12 at 12:11

I would explain the concept and then name it. Since the idea is to capture the uniqueness of any potential member of the set X, I would suggest describing such X as "singular". Be sure not to use "singular" for anything else in that context.

Gerhard "Caution: Alternate Words At Work" Paseman, 2012.09.20

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# Sum of the series $\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$ where $r$ is rational?

Can we find the exact sum of the series $$\sum_{n=0}^{\infty} \lfloor nr \rfloor x^n$$ where $$r$$ is rational?

There is a special case given here but I don't know how to prove it and can we get the sum for the general case?

• take a look here – Masacroso Nov 23 '19 at 17:59
• @Masacroso Your link deals with finite sum. I found this though containg infinite sum. – mathisgood Nov 23 '19 at 18:14
• If $r = a/b$ in lowest form, then for any positive integer $k$, $\lfloor (n + kb)r\rfloor = \lfloor nr \rfloor +ka$. This can be used to reduce to finding the sum of the first $b$ terms of the sum. I'm not sure if there's a nice formula for that in general, however. – Zarrax Nov 23 '19 at 19:05

First, this is not a full answer but enough significant progress that I felt it was worth typing up.

Note that the sum only converges if $$|x|<1$$. As such, for the remainder of this post we shall assume $$-1.

Let $$r=a/b$$ be rational. If $$r=a/b$$ is an integer, then the sum is easily computed to be

$$\sum_{n=0}^\infty\left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{n=0}^\infty n rx^n=\frac{r x}{(x-1)^2}$$

Thus, we may as well assume $$r=a/b$$ is not an integer from here on out. Now, we can write $$n$$ as $$n=bq+s$$ where $$q\in\{0,1,2,3,...\}$$ and $$0\leq s . Then

$$\left\lfloor n \frac{a}{b}\right\rfloor=\left\lfloor (bq+s) \frac{a}{b}\right\rfloor=qa+\left\lfloor s \frac{a}{b}\right\rfloor$$

Then the sum can be rewritten as

$$\sum_{n=0}^\infty \left\lfloor n \frac{a}{b}\right\rfloor x^n=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor (bq+s) \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}$$

We can split this sum into two parts:

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left(qa+\left\lfloor s \frac{a}{b}\right\rfloor\right) x^{bq+s}=\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}+\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$

The first infinite sum is easily found to be

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}qa x^{bq+s}=\frac{a x^b}{(x-1) \left(x^b-1\right)}$$

Since the second infinite sum converges absolutely, we may as well sum the $$q$$ sum before the $$s$$ sum. That is

$$\sum_{q=0}^\infty\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}=\sum_{s=0}^{b-1}\sum_{q=0}^\infty\left\lfloor s \frac{a}{b}\right\rfloor x^{bq+s}$$

$$=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\sum_{q=0}^\infty x^{bq}\right)=\sum_{s=0}^{b-1}\left(\left\lfloor s \frac{a}{b}\right\rfloor x^s\frac{1}{1-x^b}\right)$$

$$=\frac{1}{1-x^b}\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s$$

The question then becomes: what is the sum

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s?$$

Now, we can slightly simplify this if we assume $$a/b$$ is in its most reduced form. That is, $$\gcd(a,b)=1$$ (if $$a=0$$, then the original sum is clearly $$0$$ so we may ignore this case). We may then write $$a=mb+t$$ for $$m\in\mathbb{Z}$$, $$\gcd(t,b)=1$$, and $$0< t. Note that if $$t=0$$, then $$a/b$$ is an integer. Otherwise, the sum is

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{a}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s=\sum_{s=0}^{b-1}\left\lfloor s \frac{mb+t}{b}\right\rfloor x^s$$

$$=\sum_{s=0}^{b-1}smx^s+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s=\frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$

We have now further simplified the question to: what is the sum

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s$$

where $$\gcd(t,b)=1$$ and $$0? Unfortunately, this seems like a difficult problem. Here are the first few such sums, for $$b=1,2,\cdots ,6$$:

$$\left( \begin{array}{c} \{0\} \\ \{0\} \\ \left\{0,x^2\right\} \\ \left\{0,2 x^3+x^2\right\} \\ \left\{0,x^4+x^3,2 x^4+x^3+x^2,3 x^4+2 x^3+x^2\right\} \\ \left\{0,4 x^5+3 x^4+2 x^3+x^2\right\} \\ \end{array} \right)$$

Overall, the sum is

$$\frac{a x^b}{(x-1) \left(x^b-1\right)}+\frac{1}{1-x^b}\left( \frac{m \left(b x^{b+1}-x^{b+1}-b x^b+x\right)}{(x-1)^2}+\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s \right)$$

where $$m=\lfloor a/b\rfloor$$, $$\gcd(t,b)=1$$, and $$0.

Just to show that this answer is correct, we can rederive the special case given in the link above. In this case, $$a=1$$, implying the second term is $$0$$ as

$$m=\lfloor a/b\rfloor=\lfloor 1/b\rfloor=0$$

and

$$\sum_{s=0}^{b-1}\left\lfloor s \frac{t}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor s \frac{1}{b}\right\rfloor x^s =\sum_{s=0}^{b-1}\left\lfloor \frac{s}{b}\right\rfloor x^s =0$$

as $$0\leq s. Thus, the sum is simply

$$\frac{x^b}{(x-1) \left(x^b-1\right)}.$$

In the example, they use $$k=1/x$$, which gives us

$$\frac{ (1/k)^b}{((1/k)-1) \left((1/k)^b-1\right)}=\frac{ (1/k)^bk^{b+1}}{((1/k)-1) \left((1/k)^b-1\right)k^{b+1}}=\frac{k}{(k-1)(k^b-1)}$$

which is what they got.


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# 4.2 Acceleration vector  (Page 3/4)

 Page 3 / 4

## Solution

(a) The origin of the coordinate system is at the top of the hill with y- axis vertically upward and the x- axis horizontal. By looking at the trajectory of the skier, the x- component of the acceleration is positive and the y- component is negative. Since the angle is $15\text{°}$ down the slope, we find

${a}_{x}=\left(2.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\phantom{\rule{0.2em}{0ex}}\text{cos}\left(15\text{°}\right)=2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}$
${a}_{y}=\left(-2.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}15\text{°}=-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}.$

Inserting the initial position and velocity into [link] and [link] for x , we have

$x\left(t\right)=75.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)t+\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){t}^{2}$
${v}_{x}\left(t\right)=4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)t.$

For y , we have

$y\left(t\right)=-50.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)t+\frac{1}{2}\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){t}^{2}$
${v}_{y}\left(t\right)=-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)t.$

(b) Now that we have the equations of motion for x and y as functions of time, we can evaluate them at t = 10.0 s:

$x\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=75.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(4.1\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)+\frac{1}{2}\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}^{2}=216.0\phantom{\rule{0.2em}{0ex}}\text{m}$
${v}_{x}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=4.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=24.1\text{m}\text{/s}$
$y\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-50.0\phantom{\rule{0.2em}{0ex}}\text{m}+\left(-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)+\frac{1}{2}\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right){\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}^{2}=-88.0\phantom{\rule{0.2em}{0ex}}\text{m}$
${v}_{y}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-1.1\phantom{\rule{0.2em}{0ex}}\text{m/s}+\left(-0.54\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\right)\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-6.5\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

The position and velocity at t = 10.0 s are, finally,

$\stackrel{\to }{r}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left(216.0\stackrel{^}{i}-88.0\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{m}$
$\stackrel{\to }{v}\left(10.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left(24.1\stackrel{^}{i}-6.5\stackrel{^}{j}\right)\text{m/s}.$

The magnitude of the velocity of the skier at 10.0 s is 25 m/s, which is 60 mi/h.

## Significance

It is useful to know that, given the initial conditions of position, velocity, and acceleration of an object, we can find the position, velocity, and acceleration at any later time.

With [link] through [link] we have completed the set of expressions for the position, velocity, and acceleration of an object moving in two or three dimensions. If the trajectories of the objects look something like the “Red Arrows” in the opening picture for the chapter, then the expressions for the position, velocity, and acceleration can be quite complicated. In the sections to follow we examine two special cases of motion in two and three dimensions by looking at projectile motion and circular motion.

At this University of Colorado Boulder website , you can explore the position velocity and acceleration of a ladybug with an interactive simulation that allows you to change these parameters.

## Summary

• In two and three dimensions, the acceleration vector can have an arbitrary direction and does not necessarily point along a given component of the velocity.
• The instantaneous acceleration is produced by a change in velocity taken over a very short (infinitesimal) time period. Instantaneous acceleration is a vector in two or three dimensions. It is found by taking the derivative of the velocity function with respect to time.
• In three dimensions, acceleration $\stackrel{\to }{a}\left(t\right)$ can be written as a vector sum of the one-dimensional accelerations ${a}_{x}\left(t\right),{a}_{y}\left(t\right),\text{and}\phantom{\rule{0.2em}{0ex}}{a}_{z}\left(t\right)$ along the x- , y -, and z- axes.
• The kinematic equations for constant acceleration can be written as the vector sum of the constant acceleration equations in the x , y , and z directions.

## Conceptual questions

If the position function of a particle is a linear function of time, what can be said about its acceleration?

If an object has a constant x -component of the velocity and suddenly experiences an acceleration in the y direction, does the x- component of its velocity change?

No, motions in perpendicular directions are independent.

If an object has a constant x- component of velocity and suddenly experiences an acceleration at an angle of $70\text{°}$ in the x direction, does the x- component of velocity change?

## Problems

The position of a particle is $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}+5.0\stackrel{^}{j}-6.0t\stackrel{^}{k}\right)\phantom{\rule{0.2em}{0ex}}\text{m}.$ (a) Determine its velocity and acceleration as functions of time. (b) What are its velocity and acceleration at time t = 0?

A particle’s acceleration is $\left(4.0\stackrel{^}{i}+3.0\stackrel{^}{j}\right)\text{m/}{\text{s}}^{2}.$ At t = 0, its position and velocity are zero. (a) What are the particle’s position and velocity as functions of time? (b) Find the equation of the path of the particle. Draw the x- and y- axes and sketch the trajectory of the particle.

a. $\stackrel{\to }{v}\left(t\right)=\left(4.0t\stackrel{^}{i}+3.0t\stackrel{^}{j}\right)\text{m/s},$ $\stackrel{\to }{r}\left(t\right)=\left(2.0{t}^{2}\stackrel{^}{i}+\frac{3}{2}{t}^{2}\stackrel{^}{j}\right)\phantom{\rule{0.2em}{0ex}}\text{m}$ ,
b. $x\left(t\right)=2.0{t}^{2}\text{m,}\phantom{\rule{0.2em}{0ex}}y\left(t\right)=\frac{3}{2}{t}^{2}\text{m,}\phantom{\rule{0.2em}{0ex}}{t}^{2}=\frac{x}{2}⇒y=\frac{3}{4}x$

A boat leaves the dock at t = 0 and heads out into a lake with an acceleration of $2.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}\stackrel{^}{i}.$ A strong wind is pushing the boat, giving it an additional velocity of $2.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\stackrel{^}{i}+1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}\stackrel{^}{j}.$ (a) What is the velocity of the boat at t = 10 s? (b) What is the position of the boat at t = 10s? Draw a sketch of the boat’s trajectory and position at t = 10 s, showing the x- and y -axes.

The position of a particle for t >0 is given by $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}-7.0{t}^{3}\stackrel{^}{j}-5.0{t}^{-2}\stackrel{^}{k}\right)\phantom{\rule{0.2em}{0ex}}\text{m}.$ (a) What is the velocity as a function of time? (b) What is the acceleration as a function of time? (c) What is the particle’s velocity at t = 2.0 s? (d) What is its speed at t = 1.0 s and t = 3.0 s? (e) What is the average velocity between t = 1.0 s and t = 2.0 s?

a. $\stackrel{\to }{v}\left(t\right)=\left(6.0t\stackrel{^}{i}-21.0{t}^{2}\stackrel{^}{j}+10.0{t}^{-3}\stackrel{^}{k}\right)\text{m/s}$ ,
b. $\stackrel{\to }{a}\left(t\right)=\left(6.0\stackrel{^}{i}-42.0t\stackrel{^}{j}-30{t}^{-4}\stackrel{^}{k}\right)\text{m/}{\text{s}}^{2}$ ,
c. $\stackrel{\to }{v}\left(2.0s\right)=\left(12.0\stackrel{^}{i}-84.0\stackrel{^}{j}+1.25\stackrel{^}{k}\right)\text{m/s}$ ,
d. $\stackrel{\to }{v}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=6.0\stackrel{^}{i}-21.0\stackrel{^}{j}+10.0\stackrel{^}{k}\text{m/s},\phantom{\rule{0.2em}{0ex}}|\stackrel{\to }{v}\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=24.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
$\stackrel{\to }{v}\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=18.0\stackrel{^}{i}-189.0\stackrel{^}{j}+0.37\stackrel{^}{k}\text{m/s},$ $|\stackrel{\to }{v}\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=199.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$ ,
e. $\stackrel{\to }{r}\left(t\right)=\left(3.0{t}^{2}\stackrel{^}{i}-7.0{t}^{3}\stackrel{^}{j}-5.0{t}^{-2}\stackrel{^}{k}\right)\text{cm}$
$\begin{array}{}\\ \\ \hfill {\stackrel{\to }{v}}_{\text{avg}}& =9.0\stackrel{^}{i}-49.0\stackrel{^}{j}-6.3\stackrel{^}{k}\text{m/s}\hfill \end{array}$

The acceleration of a particle is a constant. At t = 0 the velocity of the particle is $\left(10\stackrel{^}{i}+20\stackrel{^}{j}\right)\text{m/s}.$ At t = 4 s the velocity is $10\stackrel{^}{j}\text{m/s}.$ (a) What is the particle’s acceleration? (b) How do the position and velocity vary with time? Assume the particle is initially at the origin.

A particle has a position function $\stackrel{\to }{r}\left(t\right)=\text{cos}\left(1.0t\right)\stackrel{^}{i}+\text{sin}\left(1.0t\right)\stackrel{^}{j}+t\stackrel{^}{k},$ where the arguments of the cosine and sine functions are in radians. (a) What is the velocity vector? (b) What is the acceleration vector?

a. $\stackrel{\to }{v}\left(t\right)=\text{−sin}\left(1.0t\right)\stackrel{^}{i}+\text{cos}\left(1.0t\right)\stackrel{^}{j}+\stackrel{^}{k}$ , b. $\stackrel{\to }{a}\left(t\right)=\text{−cos}\left(1.0t\right)\stackrel{^}{i}-\text{sin}\left(1.0t\right)\stackrel{^}{j}$

A Lockheed Martin F-35 II Lighting jet takes off from an aircraft carrier with a runway length of 90 m and a takeoff speed 70 m/s at the end of the runway. Jets are catapulted into airspace from the deck of an aircraft carrier with two sources of propulsion: the jet propulsion and the catapult. At the point of leaving the deck of the aircraft carrier, the F-35’s acceleration decreases to a constant acceleration of $5.0\phantom{\rule{0.2em}{0ex}}\text{m/}{\text{s}}^{2}$ at $30\text{°}$ with respect to the horizontal. (a) What is the initial acceleration of the F-35 on the deck of the aircraft carrier to make it airborne? (b) Write the position and velocity of the F-35 in unit vector notation from the point it leaves the deck of the aircraft carrier. (c) At what altitude is the fighter 5.0 s after it leaves the deck of the aircraft carrier? (d) What is its velocity and speed at this time? (e) How far has it traveled horizontally?

how can I convert mile to meter per hour
1 mile * 1609m
Boon
hey can someone show me how to solve the - "Hanging from the ceiling over a baby bed ...." question
i wanted to know the steps
Shrushti
sorry shrushti..
Rashid
which question please write it briefly
Asutosh
Olympus Mons on Mars is the largest volcano in the solar system, at a height of 25 km and with a radius of 312 km. If you are standing on the summit, with what initial velocity would you have to fire a projectile from a cannon horizontally to clear the volcano and land on the surface of Mars? Note that Mars has an acceleration of gravity of 3.7 m/s2 .
what is summit
Asutosh
highest point on earth
Ngeh
पृथवी को इसके अक्ष पर कितने कोणीय चाल से घूमाऐ कि भूमधय पे आदमी का भार इसके वासतविक भार से 3/5अधिक हो
best
Murari
At a post office, a parcel that is a 20.0-kg box slides down a ramp inclined at 30.0° 30.0° with the horizontal. The coefficient of kinetic friction between the box and plane is 0.0300. (a) Find the acceleration of the box. (b) Find the velocity of the box as it reaches the end of the plane, if the length of the plane is 2 m and the box starts at rest.
As an IT student must I take physics seriously?
yh
Bernice
hii
Raja
IT came from physics and maths so I don't see why you wouldn't
conditions for pure rolling
Md
the time period of jupiter is 11.6 yrs. how far is jupiter from the sun. distance of earth from rhe sun is 1.5*10 to the power 11 meter.
lists 5 drawing instruments and their uses
that is a question you can find on Google, anyway of top of my head, compass, ruler, protractor, try square, triangles.
Rongfang
A force F is needed to break a copper wire having radius R. The force needed to break a copper wire of radius 2R will be
2F
Jacob
it will be doubled
kelvin
double
Devesh
The difference between vector and scaler quantity
vector has both magnitude & direction but scalar has only magnitude
Manash
my marunong ba dto mag prove ng geometry
ron
how do I find resultant of four forces at a point
Inusah
use the socatoa rule
kingsley
draw force diagram, then work out the direction of force.
Rongfang
In a closed system of forces... Summation of forces in any direction or plane is zero... Resolve if there is a need to then add forces in a particular plane or direction.. Say the x direction... Equate it tk zero
Jacob
define moment of inertia
it is the tendency for a body to continue in motion if is or continue to be at rest if it is.
prince
what is Euler s theorem
what is thermocouple?
joining of two wire of different material forming two junctions. If one is hot and another is cold the it will produce emf...
joining of two metal of different materials to form a junction in one is hot & another is cold
Manash
define dimensional analysis
mathematical derivation?
Hira
explain what Newtonian mechanics is.
a system of mechanics based of Newton laws motion this is easy difenation of mean...
Arzoodan
what is the meaning of single term,mechanics?
jyotirmayee
mechanics is the science related to the behavior of physical bodies when some external force is applied to them
Lalita
SO ASK What is Newtonian mechanics in physics? Newtonian physics, also calledNewtonian or classical mechanics, is the description of mechanical events—those that involve forces acting on matter—using the laws of motion and gravitation formulated in the late seventeenth century by English physicist
Suleiman
can any one send me the best reference book for physics?
Prema
concept of physics by HC verma, Fundamentals of Physics, university of physics
tq u.
Prema
these are the best physics books one can fond both theory and applications.
can any one suggest best book for maths with lot of Tricks?
Vivek
what is the water height in barometer?
SUNEELL
13.5*76 cm. because Mercury is 13.5 times dense than Mercury
LOVE
water is 13.5 times dense than the Mercury
LOVE
plz tell me frnds the best reference book for physics along with the names of authors.
Prema
i recomended the reference book for physics from library University of Dublin or library Trinity college
Arzoodan
A little help here... . 1. Newton's laws of Motion, are they applicable to motions of all speeds? 2.state the speeds which are applicable to Newtons laws of Motion
Derek
mechanics which follows Newtons law
Manash
The definition of axial and polar vector .
Arpita
polar vector which have a starting point or pt. of applications is,force,displacement
jyotirmayee
axial vector represent rotational effect and act along the axis of rotation b
jyotirmayee
prove Newton's first law of motion
prince
Hello frnds what is physics in general?
Ngeh
A block of mass m is attached to a spring with spring constant k and free to slide along a horizontal frictionless surface. At t=0, the block spring system is stretched on amount x>0 from the equilibrium position and is released from rest Vx = 0 What is the period of oscillation of the block? What
Ella
What is the velocity of the block when it first comes back to the equilibrium position?
Ella


text:
Hope for the best, but prepare for the worst.

— English Proverb

Be prepared.

— The Scout Motto

By failing to prepare, you are preparing to fail.

— Benjamin Franklin

## The survival essentials

We've seen all throughout history, and as recently as Hurricane Katrina, that for a wide variety of reasons civilization breaks down from time to time.  These disruptions can last days, weeks, or even years.  Preparation often determines who survives.

Here are some things that preppers typically stock, in order of importance:

1. Guns and ammo.  (Without a means of self defense, you cannot count on keeping any other supplies.)

2. Water, food, and shelter.

3. Medical supplies, emergency power, communication equipment.

4. Heirloom seeds, farming equipment, etc.

Of course, just having the equipment is not enough.  You have to know how to use them.  The time to prepare is before, not during, the disaster

## Survival communication

Disaster communications serve two critical functions:

1. Coordinates the activities of family, friends, and neighbors.

2. Tells you what's happening around the country and around the world to help you better plan your own actions.

Amateur radio, also known as ham radio, can help with both.  The entry level ("Technician class") ham radio license lets you operate on the VHF and UHF frequencies, which are most useful for local communications.   For example, you could post a lookout at the top of the local hill and communicate with him via ham radio.

The second level ("General class") ham radio license also allows you to operate on the HF frequencies, which means you can talk all around the world without relying on the grid, satellites, or other infrastructure, simply by bouncing signals off the ionosphere.  If you don't want to rely on the mainstream media for accurate news, you will want to talk directly to people outside your own area.  Unfortunately, operating HF radios is not as simple as picking up a cell phone and dialing a number.  You need to know about feed lines, antennas, SWR, the characteristics of the various HF bands, MUF, procedural signals, etc.  You will learn about these topics while studying to pass the license exams, and you will pick up even more by practicing with the radio once you are licensed.

## How to get started in ham radio

Here's a step-by-step guide to getting ready to use ham radio for survival communications:

1. Decide which level of communications capability you need:

• A Technician class license is useful mostly for local communications.

• A General class allows you to talk all around the world.

• An Extra class license has a few minor additional privileges, but nothing important for survival.

2. Choose a target exam date.  Assume you will need about 10 hours to study for the Technician exam.  To get the General class license, you must pass an additional exam, so tack on an additional 20 hours to study for the General class exam.  If you want the Extra class license, add on another 30 hours to study for the Extra class exam.  The good news is that you no longer have to learn Morse code for any class of ham radio license.

It sounds like a lot of work, but all of this studying is not a waste of time, as you are learning information that could someday save your life.

I recommend that you try to average at least one hour every day, and more if possible.  So decide how much free time you can devote to studying, then figure out when you could take the exam.

Look up exam sessions in your area by zip code at http://www.arrl.org/find-an-amateur-radio-license-exam-session.  (If you don't see any sessions in your immediate area, expand your search to a wider area using the dropdown box to the right of the zip code.)  Picking your exam date up front keeps you focused on studying.

3. Create an account on this website.  The HamTestOnline website is the most efficient way to get a ham radio license.  It integrates study materials with question drill using the actual exam questions.  So you are learning about the aspects of ham radio you need to understand, and at the same time preparing to get a high score on the exam.

4. Subscribe to the desired courses.  Everyone starts with the Technician exam, and most survivalists take the General class exam, as well.

5. Stay in Study mode until your score bar reads at least 85% on each course.  That will ensure that you are ready to pass the exams.

6. Take the exams at a local exam session.  Don't be afraid to take multiple exams in a single session.  We've had many students pass two exams in a single sitting.  In fact, we've had at least a hundred students go from zero to Extra in a single session.  Since they only charge a single exam fee (typically $15) no matter how many exams you take, taking multiple exams at once saves both time and money. 7. Buy one or more radios. For example, as of this writing the BaoFeng UV-5R Dual-Band Ham Radio VHF/UHF portable (handheld) radio costs just$32!  The Elecraft KX3 is an ultra-compact HF radio excellent for hunkering down or bugging out.  Depending on the options you select, it costs around $1000. If that's too much for your budget, you could buy a less feature-rich, used HF radio at local hamfest for around$200.  To operate HF you're also going to need some kind of antenna.  If bugging out is an option, you might prefer something portable like the Buddipole or Ventenna HFp Vertical.

8. Find a solution for emergency power — emergency generator, solar panels, etc.  If you live in a sunny area, solar panels will provide free power for decades.  Without power, a radio is as useful as a rock.

9. Use your radios until you are completely comfortable with them.  Participate in your local ARRL Field Day and other ham radio contests.  It's amazing how much fun you can have while learning essential skills!

10. When you're not using the equipment, you might consider investing in EMP bags or a Faraday box for safe storage.

## See you on the air

In a The End Of The World As We Know It (TEOTWAWKI) scenario, most people are not going to be prepared.  But the ham radio frequencies are going to be filled with survivalists and preppers.  I'll talk to you then!


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# Distance between two points in a picture

I have this picture:

I'de like to obtain the distance between two of these points. The black scale bar is 3$$\mu m$$.

To do so, I import the image to Mathematica and call it pic. Then by finding dominant colors, I define red as

red = RGBColor[0.827134595933227, 0.00309700865836607,
0.08256587730260687, 1.];

And then I obtain the coordinates of the centers of the dots as

  centroid[pic_,color_] := ComponentMeasurements[
Binarize@ColorDetect[pic,color],
"Centroid"
][[All,2]]

coordinates = centroid[pic,red]

So

            {{5.5, 39.5}, {13.9828, 37.5345}, {34.2308, 37.8846}, {23.6034,
34.8103}, {5.38571, 32.5571}, {15., 29.}, {1.19231, 24.8077}}

Now, let's say we want to obtain the distance between two points of coordinates: {15., 29.} and {23.6034, 34.8103} which is 10.3816.

Now, my question is that how can I also determine the size of scale and obtain the real distance between these two points?

• You can try using the function copyCurve [here][1]. It was aimed for other purposes, but it can also easily do what you need. [1]: mathematica.stackexchange.com/questions/44355/… Feb 10 '21 at 8:55
• This can be done with the copyCurve function. As well as the distance definition. Feb 10 '21 at 13:05

First, let's get the scale bar out of the image. There are a few ways to do this (MorphologicalBinarize and friends), but I went with the more eye-bally approach of using ColorReplace. Here, we're replacing black pixels with white, and all other pixels with black, to get a mask. Note that ColorReplace has a third argument d, which you could use to fine-tune what it sees as "black".

i = Import["https://i.stack.imgur.com/yR6UY.gif"]

c = ColorReplace[i, {Black -> White, _ -> Black}]

Compare with a simple MorphologicalBinarize - the choice is yours:

Now we can get the bounding box of our rectangle, and simply take the difference:

box = ComponentMeasurements[c, "BoundingBox"][[1, 2]]

Differences /@ Transpose@box

{{38.}, {5.}}

So the longest dimension of our bounding box is 38 pixels long, and the short side is 5 pixels long.

Just to check, I counted the pixels another way, using SequenceCases on our replaced image, and each line of 1s, the pixels of the scale bar, had a length of 38.

SequenceCases[
Flatten@MorphologicalComponents[c], {p : Repeated[1]} :> Length@{p}]

{38, 38, 38, 38, 38}

Either way, we get about 38 pixels, so now we know the length of the bar, and thus we can get the size of a pixel (assuming pixels are square here).

pixelLength = Quantity[3, "Microns"] / 38

Quantity[0.0789474, "Microns"]

and now if we get the Euclidean distance between your points and multiply it by our pixel length, we get the distance in microns.

EuclideanDistance[{5.5, 39.5}, {13.9828, 37.5345}] * pixelLength

Quantity[0.687437, "Microns"]

This answer will not hold up well if the bar is not always aligned perfectly to the image border, as it is in your test image. You could consider ComponentMeasurements[c, "Length"] or "CaliperLength" to get the length of the bar instead in that case. However, it may take the distances between opposite corners of the bar, which may not be what you want.

• I did not have my coffee before this answer - I would wait until a better-caffeinated person comes by and gives a better response. I suspect my approach is not scientific enough and potentially just plain wrong. Feb 10 '21 at 9:43
• Not that I'm aware of, but you could simply create a function called pixelDistanceInMicrons[p1_, p2_, pixelLength_] := EuclideanDistance[p1, p2] * pixelLength and call it like pixelDistanceInMicrons[{5.5, 39.5}, {13.9828, 37.5345}, Quantity[0.0789474, "Microns"]]. Feb 10 '21 at 10:36
• You could also count the pixels by hand, I suppose. Feb 11 '21 at 9:02
• If the bar is diagonal, counting pixels in the way I have done it here will not work, as I am counting them left to right. As I mention, I think "CaliperLength" will do a reasonable job in this case. I am hopeful someone else will post a more scientific answer as I'm not convinced this method would be enough for Real Science and I'm not sure I can help you further because of that. Hope my answer at least helps you find your way to a more useful result :) Feb 11 '21 at 12:06


text:
Mathematics Calculus Derivative - Formal Definition | Limit | Continuity and...

# Derivative – Formal Definition | Limit | Continuity and Differentiability

-

In simple sentence, derivative is the slope of a tangent line. Slope means change in y direction over change in x direction $(\frac{Δy}{Δx})$. However, in this article we will try to understand the in fundamental concept of derivative in calculus. But to understand derivative we also need to have a basic understanding on what is limit. So we will start our discussion on formal definition of derivative with the basic concept of limit.

### Limit of a Function:

Say, a function $f(x) = \frac{x^{2}-4}{x-2}$. In this function, we get an output for every possible input of x except +2. If we substitute x=2, then the function become $f(2)=\frac{0}{0}$, which is an indeterminate form. So we can say that at the point, $x=2$ we can not define the function. And therefore this brings the concept of limit. Because we are unable of determining the value of the function when $x=2$, therefore we can determine the value of the function when x is almost 2, or the value of x near about 2, or approaching to 2 from both left-side and right-side of 2.

Therefore, if x approaches to 2 from its left-side of number line, the function tends to become: x=1.9, f(1.9)= 3.9. when x=1.99, f(x)=3.99. Similarly if x approaches to 2 from the right-side of the number line, the value of the function tends to become: x=2.1, f(x)=4.1, when x=2.11, f(x)=4.11 and so on. Here we can can see that as x approaches to 2, f(x) approaches to 4. So, we can say, f(x)→4 as x→2. This is the concept of limit.

Definition: A function f(x) is defined when x is near the number a, but not a. It means  f has a value on some open interval that contains a, except possibly at a itself. Then we write:

$$\lim_{x\to a}f(x)=L,\quad\text{f(x)→L, as x→a.}$$

So, this means that we can get f(x) “as close as we want” to L by making x sufficiently close to a. Here L is the limit of the function. Now we can find out the value of function $f(x)=\frac{x^{2}-4}{x-2}$ when x→2 using limit:

$$\lim_{x\to 2}f(x)=\frac{(x+2)(x-2)}{(x-2)}\\ \lim_{x\to 2}f(x)=(x-2)=2+2=4,\\ \quad\text{bcoz, x→2, but x≠2}$$

When does the Limit Exist: limit of a function f(x) exists, only when for some value of  x=a, $f(a)=\frac{0}{0}$. Also if for any value x=a, $f(a)≠\frac{0}{0}$, then we can say that the limit does not exists. For example, for the function $f(x)=\frac{|x|}{x},$ limit does not exists. before going to definition of derivative, let get to know a little about continuity.

### Continuity:

A function is said to be continuous if we able to graph the function without taking off our pen, or pencil. So the definition of continuity refers that A function is continuous at a point x=a, if

$$\lim_{x\to a} f(x)= f(a)$$

as x approaches to a, the y value of x, f(x) will be equal to y value of a, f(a). Which means at the point a, f(x)=f(a), when x→a. And to find f(x) = f(a), we have to approach x→a from both left-side of a, x→$a^{-}$ and right-side of a, x→$a^{+}$. Therefore,we can define that a function is continuous if:

$$\lim_{x\to a^{-}} f(x)=\lim_{x\to a^{+}} f(x)=f(a)$$

## The formal definition of Derivative

At the beginning we have said that derivative is the slope of tangent line. Now we will see what that really means. We know, a slope (m) is the rate of change of vertical variable (y), with respect to horizontal variable (x). So, $m=\frac{Δy}{Δx}$.

This is how we can find the slope for straight line. Change of y variable, $Δy=y_{2}-y_{1}$, with respect to change of x variable $Δx=x_{2}-x_{1}$. But for any curve y=f(x), slope just touches its surface. Let slope touches the curve at the point $p(x, f(x))$. Therefore, to find slope at p, we have to imagine another point $p_{1}((x+h), f(x+h)),$ closer to p, such that h→0, but h≠0. Because the nearer the point $p_{1}$ to p, the accurate the value of slope will be. So in this case the slope of the curve at point p:

$$Slope=\frac{f(x+h)-f(x)}{(x+h)-x}$$

But when h=0, the above slope become the form $\frac{0}{0}$, So h≠0 but h→0, therefore we have to include limit to define slope of the curve:

$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{(x+h)-x}$$

This is called the derivative of the function f(x) at the point $p(x, f(x))$. So,

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{(x+h)-x}$$

The line that touches the curve y=f(x) at p, refers to as tangent line. As above slope we derive from the tangent line, therefore the derivative of a function defined as slope of tangent line. And this definition of derivative also refers to as formal definition of derivative.

Derivative of a function y originally expressed as $\frac{dy}{dx}$. because it is similar to $\frac{Δy}{Δx}$, but in this case Δx→0, also Δy→0.

Which means, $\frac{dy}{dx}$ indicates the instantaneous change of the function, or infinitesimal change in y with respect to infinitesimal change in x. Also indicate how fast that change happen.

### Differentiability

if (x+h) indicates a point c on x-axis, then (x+h)=c, and y=f(x+h)=f(c), such that x→c. Then we can define the derivative of the function f(x) at the point x=c:

$$f'(c)=\lim_{x\to c}\frac{f(x)-f(c)}{x-c}$$

Here, we can define that if f is differentiable at x=c, then f is continuous at x=c. So if we know that f is differentiable then we can find the limit at x=c exists, and also the function is continuous at x=c. We also can say that if f is not continuous at x=c, then f is not differentiable at x=c. A function can be continuous but not differentiable, see here. Therefore we can say that a function to be differentiable at the point x=c, its limit must exist at that point. know more.

Read more article on calculus, here

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This article we will talk about the Macro. It is a function in microsoft excel which basically...


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# Is it possible to have an HTML canvas element in the background of my page?

In this article we are going to learn about is it possible to have an HTML canvas element in the background of my page.

You might try adding a CSS style with a position: fixed (or absolute, if applicable) to the canvas so that any material that comes after it will sit on top of it.

To get better understanding on is it possible to have an HTML canvas element in the background of my page, let’s look into the following examples…

## Example 1

In the following example we are using the position: absolute to apply CSS style to the canvas.

<!DOCTYPE html>

<canvas id="mytutorial" width="450" height="500" style="border:1px solid #ABEBC6;"></canvas>

Welcome To Tutorialspoint

On running the above script, the output window pops up, displaying the canvas as the background with the text “welcome to tutorialspoint” on the webpage.

## Example 2

Let’s consider the another example for having an HTML canvas element in the background of my page.

<!DOCTYPE html>

<canvas id='tutorial'></canvas>

On running the above script, it will generate an output consisting of stroke text filled on the canvas displayed on the webpage, acting as a background to the page.


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# Proof of the divisibility rule of 17.

Rule: Subtract 5 times the last digit from the rest of the number, if the result is divisible by 17 then the number is also divisible by 17.

How does this rule work? Please give the proof.

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Is this homework? The imperative at the end sounds like it. –  celtschk Jul 23 '12 at 10:36
@celtschk: I don't think so, the question probably comes from math.stackexchange.com/questions/174114/… –  Generic Human Jul 23 '12 at 10:44
@celtschk no this is not a homework.This question comes from math.stackexchange.com/questions/174114/… (same link as provided by @generichuman). –  abhinav8 Jul 23 '12 at 10:52
@celtschk if he wanted to mask it as non homework, i'm sure he is clever enough. why would it matter? –  IIIIIllllllllIlllllIIIIIIIIlll Jul 23 '12 at 17:31
@АртёмЦарионов: On one hand, it would matter because it should then get the [homework] tag, on the other hand it would mean that the tone of the last sentence shouldn't bother you because it was just copied from a context where it was appropriate. –  celtschk Jul 23 '12 at 17:41

Write your number $10a+b$.

Then because 10 and 17 are relatively prime, $$17\mid a-5b \iff 17\mid 10a-50b \iff 17\mid 10a+b$$ The last equivalence is because $10a+b-(10a-50b) = 51b$ is always a multiple of 17.

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Let $$n=\sum_{k=0}^N 10^k a_k$$ be the number we want to test for divisibility, where the $a_k$s are the digits in the decimal expansion of $n$. We form the second number $m$ by the process you describe, $$m = \sum_{k=1}^N 10^{k-1} a_k - 5 a_0 = \frac{n-a_0}{10}- 5 a_0$$ Now suppose $17|m$. Then there exists a natural number $b$ such that $17b = m$. We then have $$17 b = \frac{n-a_0}{10}- 5 a_0$$ $$10 * 17 b = n-a_0- 50 a_0 \implies n= 17(10b + 3a_0)$$ and so $n$ is divisible by 17.

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10x+y will be divisible by an odd prime p iff a.p.x ~ b(10x+y) is divisible by p.

Now by Euclid's GCD algorithm, we can find integers a,b such that a.p ~ b.10=1 where (p,10)=1.

If p=17, by observation 3.17 - 5.10=1.

So, 3.17.x-5(10x+y)=x-5y.

If this is divisible by 17, so will be 10x+y.

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$17 = 2^3 + 3^2$

the sum of the exponents of $2$ and $3$ is $(3 + 2)$

i.e. the sum of the exponents is $5$

Now a number can be expressed as $abcde$

i.e. $abcde = (10^4)a + (10^3)b + (10^2)c + (10^1)d + e$

remove the units digit from $abcde$ to get $$abcd = (10^3)a + (10^2)b + (10^1)c + d$$

Now, let $A = abcd - (3 + 2)(e)$

Assume $A$ is divisible by $17$

i.e. $A = abcd - (5)e = (17)k$

i.e. $abcde = (10)(abcd) + e$

i.e. $abcde = (10)((17)k + (5)e) + e$

i.e. $abcde = (170)k + (50)e + e$

i.e. $abcde = (17)(10)k + (51)e$

i.e. $abcde = (17)(10)k + (17)(3)e$

i.e. $abcde = (17)((10)k + (3)e)$

i.e. $abcde$ is indeed divisible by $17$

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How are you using $\, 17 = 2^3+3^2\,$ to compute $\: 10^{-1}\equiv -5\pmod{17}$ ? $\ \$ –  Bill Dubuque Jul 23 '12 at 13:56
rajesh do you have an answer for the above question? –  IIIIIllllllllIlllllIIIIIIIIlll Jul 23 '12 at 17:29
The modulus of a congruence a=b (mod m) is the number m. It is the "base" with respect to which a congruence is computed (i.e., m gives the number of multiples of a that are "thrown out"). For example, when computing the time of day using a 12-hour clock obtained by adding four hours to 9:00, the answer, 1:00, is obtained by taking 9+4=1 (mod 12) (i.e., adding the hours with modulus 12). –  Rajesh K Singh Jul 27 '12 at 5:58
please clarify what is $10^{-1}$ –  Rajesh K Singh Jul 29 '12 at 10:38
is it $\frac{1}{10}$ –  Rajesh K Singh Jul 29 '12 at 10:39

Write the original number as 10x+y to separate out the last digit (y) from the number without the last digit (x). Recognize that divisibility by a 17 means you can write the number as 17n for a positive integer n. So our statement is if x-5y=17n then 10x + y = 17m, where x,y,m,n are positive integers.

Assume x - 5y = 17n. Now we try to get the left-hand side to look like our original number. First multiply both sides by 10. (10x - 50y = 170n), and then add 51y to both sides. (10x+ y = 170n + 51y). Factor 17 from the right hand side. (10x+y=17(10n+3y)). Recognize that 10n+3y is a positive integer as n and y are integers, let's define it as m, proving that 10x+y=17m meaning that the original number is divisible by 17 if x-5y was divisible by 17.

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Technical Article

# Dynamic Braking Resistors for Motor Control Applications

April 21, 2022 by David Peterson

## Large DC and AC motor drives often provide terminals for installing a braking resistor. What are these resistors, and how do they slow down a machine? What hazards and cautions must be considered?

### Mechanical Energy

Any machine in motion possesses kinetic energy. This energy is the result of some stored potential energy being ‘dumped’ into a motor or actuator, causing the energy to be converted into accomplished work. While the machine is in motion, two factors help to determine just how much energy must be dissipated to bring it to a stop: mass and velocity.

High kinetic energy can result from a large mass system or from a high velocity, or both. When the system has a large mass (a large flywheel, heavy drum roller), we refer to this as a high inertia system, meaning that motion will want to continue unless acted on by a significant external force (Newton's second law of motion).

When the kinetic energy and the inertia are both high, as is the case for many kinds of industrial equipment, the process of slowing or stopping the system can be a difficult task.

### Methods of Stopping Free Motion

To stop motion, you only have a couple of options. ‘Stopping’ in a more technical sense means to convert the kinetic energy of the system into other by-products that will lead to a gradual decrease until kinetic energy is zero.

If a device is left spinning, there are a handful of things that can stop it:

• Friction from bearings and contact points (if the bearings are good, this can take a very long time to slow down)
• Air resistance can be surprisingly significant if the geometry of the moving object interacts with the air around it
• If the device is driven from an electrical or fluid system, the machine may find itself trying to reverse-drive the fluid or current. This acts as a form of frictional resistance

In all three scenarios, we must ask one critical question about the kinetic energy. You cannot simply delete energy from a rotating object, so where does that energy actually go?

In all three cases, heat will be generated at the point of energy conversion. If the bearings or the air around the object are the source of friction, that is where the heat will be found. More friction (such as a worn bearing) will produce more heat. This is exactly why predictive maintenance examines machinery for problematic hot points to indicate excessive wear and friction.

In the case of a reverse driving of fluid and current, this can also generate large amounts of heat. Hydraulic fluid, being very viscous, can stop motion much faster and dissipate the heat more quickly, so it tends to be less of a problem. However, in the case of electricity, driving current back into the drive circuitry can be instantly catastrophic for the control system.

### Resistive Braking

One of the most common passive electronic components is the resistor. The simple function of this device is to convert electrical energy (current) into heat. This can accomplish a variety of tasks, such as dividing voltage in a series circuit, limiting current in a solid-state circuit, or simply creating heat in a baseboard heater.

In the scenario with the large load spinning, it is crucial for the kinetic energy to be safely dissipated as quickly as possible, as a long slowdown may mean halted production for many minutes - or even hours. The resistor is an obvious choice for this task, as long as the resistor is properly chosen for the task.

The energy to be lost is tremendous. No small prototype circuit board resistor will be suitable for this job. Those resistors have power ratings between ⅛ watt up to only 5-10 watts for the very largest.

To get an idea of what kind of power rating this resistor might need for the task of braking, imagine a steel flywheel with a diameter of 3.3 feet (1 m) and a mass of 330 lbs (150 kg), rotating at 250 rpm. We wish to slow this down to a stop in 3 seconds.

For rotational motion, the kinetic energy equation follows:

$$Energy = \frac{1}{2} \times Moment~of~inertia \times Angular~velocity^2$$

Moment of inertia for a cylinder:

$$Moment~of~inertia = \frac{1}{2} \times Mass \times Radius^2$$

$$Moment~of~inertia = 18.75~kg \cdot m^2$$

The angular velocity of this wheel:

$$Angular~velocity = \frac{2 \times \pi \times RPM}{60}$$

$$Angular~velocity \approx 26 \frac{radians}{second}$$

Therefore, plugging into the first energy formula, this system has 6337 joules of energy.

This unit, joules, is not a particularly common unit, but when energy is divided by time (the three-second stop time target), we will see the maximum power dissipated in watts.

$$Power = \frac{6337}{3}=2112~watts$$

When examining power braking resistor specification sheets, we should therefore expect to see power ratings in the hundreds to many thousands of watts.

### Braking Resistor Connections

You cannot simply install a braking resistor across the motor leads, since this would route power from the drive unit through the resistor. Instead, the drive units are outfitted with brake resistor terminals specifically for this purpose.

##### Figure 3. Some braking resistors do not need to be quite as large - smaller mass, velocity, and stipping rate can make a difference. This one is ceramic. Image used courtesy of Automation Direct

Many VFD and DC motor drive units will have clearly marked resistor terminals.

When a controlled stop is necessary, the drive unit automatically switches internal contacts from the driving source over to the output resistor. Suddenly, rather than receiving electricity from the source, the motor power is re-routed through the resistor. The motor is now a generator, and the generator is driving a resistor to dissipate energy faster.

### What Does ‘Dynamic’ Mean?

The power dissipated by the resistor is not constant. When the machine is at full speed and the stop is applied, the kinetic energy is at its maximum, so the power dissipated by the resistor will be maximum as well. As the load slows down, so does the energy dissipated.

In our example, it would be erroneous to say the resistor is consuming 2112 watts at all times during the entire slow-down cycle, but it will be at this value for a moment, so the resistor must be sized properly.

Eventually, as the load gets close to stopping, the energy dissipation would be quite small. This total energy consumption is predicted with an integration from calculus (total sum over time when the situation is changing).

In a physical sense, this would appear to the viewer as a large change in speed, and massive heat output from the resistor at the beginning, but would taper off over time.

In modern drive units, the amount of current released into the braking resistor is tuned carefully to control the power dissipation, and therefore the braking effect, so that the dynamic control is made even more useful for a wide variety of cases.

### Summary

Braking resistors are useful for large loads. The drive units can take advantage of these devices for more rapid slow-down cycles. In some cases, the energy may not be converted into heat, but perhaps into a battery chagrin circuit. In these cases, we call this ‘regenerative’ braking, used in electric vehicles, train locomotives, and more.


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# Aktuelles

## Einladung zum Vortrag (Zoom) von Herrn Prof. Dr. Gregory Berkolaiko am 24.03.2021 - 17:00 Uhr im Rahmen des Forschungsseminars Analysis

[24.03.2021]

Achtung neu: Der Vortrag findet erst um 17:00 Uhr statt!

Im Rahmen des Forschungsseminars Analysis hält am Mittwoch, den 24. März 2021, um 17:00 Uhr,

Herr Prof. Dr. Gregory Berkolaiko (Texas A&M University)

einen Vortrag zum Thema

Lateral variation principle and extrema of dispersion relation of periodic graphs.

Der Vortrag findet im Rahmen eines Zoom-Meetings statt:

https://fernuni-hagen.zoom.us/j/85219043827?pwd=eldtMXY2Uzk0ek1XVlo3c0lEcmUzQT09

Meeting ID: 852 1904 3827
Passcode: 93571026

Abstract: The first step in the proofs of several spectral geometry theorems is perturbing the operator "along" a given eigenfunction $f$, i.e. adding a perturbation $P$ that vanishes on $f$ and therefore leaves the corresponding eigenvalue $\lambda_0$ in its place.

But such perturbation may still affect the sequential number of $\lambda_0$ in the spectrum, creating a spectral shift. We will discuss a general theorem that recovers the value of the spectral shift by looking at the stability of $\lambda_0$ with respect to small variations of the perturbation $P$.

As an application of this result, we show that a large family of tight-binding models have a curious property: there is a local condition akin to second derivative test that detects if a critical point is a global (sic!) extremum. With some additional assumptions (time-reversal invariance and dimension 3 or less), we show that any local extremum of a given sheet of the dispersion relation is in fact the global extremum.

Based on joint work with Y. Canzani, G. Cox, P. Kuchment and J. Marzuola.

mathinf.webteam | 12.08.2021