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# 4.2 Modeling with linear functions  (Page 3/9)

 Page 3 / 9

## Using a linear model to investigate a town’s population

A town’s population has been growing linearly. In 2004, the population was 6,200. By 2009, the population had grown to 8,100. Assume this trend continues.

1. Predict the population in 2013.
2. Identify the year in which the population will reach 15,000.

The two changing quantities are the population size and time. While we could use the actual year value as the input quantity, doing so tends to lead to very cumbersome equations because the y -intercept would correspond to the year 0, more than 2000 years ago!

To make computation a little nicer, we will define our input as the number of years since 2004.

To predict the population in 2013 ( $t=9$ ), we would first need an equation for the population. Likewise, to find when the population would reach 15,000, we would need to solve for the input that would provide an output of 15,000. To write an equation, we need the initial value and the rate of change, or slope.

To determine the rate of change, we will use the change in output per change in input.

The problem gives us two input-output pairs. Converting them to match our defined variables, the year 2004 would correspond to $\text{\hspace{0.17em}}t=0,$ giving the point $\text{\hspace{0.17em}}\left(0,\text{6200}\right).\text{\hspace{0.17em}}$ Notice that through our clever choice of variable definition, we have “given” ourselves the y -intercept of the function. The year 2009 would correspond to $\text{\hspace{0.17em}}t=\text{5,}$ giving the point $\text{\hspace{0.17em}}\left(5,\text{8100}\right).$

The two coordinate pairs are $\text{\hspace{0.17em}}\left(0,\text{6200}\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(5,\text{8100}\right).\text{\hspace{0.17em}}$ Recall that we encountered examples in which we were provided two points earlier in the chapter. We can use these values to calculate the slope.

We already know the y -intercept of the line, so we can immediately write the equation:

$P\left(t\right)=380t+6200$

To predict the population in 2013, we evaluate our function at $\text{\hspace{0.17em}}t=9.$

$\begin{array}{ccc}\hfill P\left(9\right)& =& 380\left(9\right)+6,200\hfill \\ & =& 9,620\hfill \end{array}$

If the trend continues, our model predicts a population of 9,620 in 2013.

To find when the population will reach 15,000, we can set $\text{\hspace{0.17em}}P\left(t\right)=15000\text{\hspace{0.17em}}$ and solve for $\text{\hspace{0.17em}}t.$

$\begin{array}{ccc}\hfill 15000& =& 380t+6200\hfill \\ \hfill 8800& =& 380t\hfill \\ \hfill t& \approx & 23.158\hfill \end{array}$

Our model predicts the population will reach 15,000 in a little more than 23 years after 2004, or somewhere around the year 2027.

A company sells doughnuts. They incur a fixed cost of $25,000 for rent, insurance, and other expenses. It costs$0.25 to produce each doughnut.

1. Write a linear model to represent the cost $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ of the company as a function of $\text{\hspace{0.17em}}x,$ the number of doughnuts produced.
2. Find and interpret the y -intercept.

a. $\text{\hspace{0.17em}}C\left(x\right)=0.25x+25,000\text{\hspace{0.17em}}$ b. The y -intercept is $\text{\hspace{0.17em}}\left(0,25,000\right).\text{\hspace{0.17em}}$ If the company does not produce a single doughnut, they still incur a cost of \$25,000.

A city’s population has been growing linearly. In 2008, the population was 28,200. By 2012, the population was 36,800. Assume this trend continues.

1. Predict the population in 2014.
2. Identify the year in which the population will reach 54,000.
1. 41,100
2. 2020

## Using a diagram to build a model

It is useful for many real-world applications to draw a picture to gain a sense of how the variables representing the input and output may be used to answer a question. To draw the picture, first consider what the problem is asking for. Then, determine the input and the output. The diagram should relate the variables. Often, geometrical shapes or figures are drawn. Distances are often traced out. If a right triangle is sketched, the Pythagorean Theorem relates the sides. If a rectangle is sketched, labeling width and height is helpful.

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Rukmini
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zbMATH — the first resource for mathematics

Theorems on the uniqueness and thermal properties of stationary, nonsingular, quasifree states on spacetimes with a bifurcate Killing horizon. (English) Zbl 0861.53074
In this paper the authors study in some detail quasifree Hadamard states on spacetimes with bifurcate Killing horizons. These manifolds are globally hyperbolic spacetimes possessing the structure necessary to naturally generalize the ideas used in previous work by B. S. Kay [Commun. Math. Phys. 100, No. 1, 57-81 (1985; Zbl 0578.46062)] to study the thermal properties of quasifree states on Schwarzschild spacetimes.
After establishing necessary geometrical properties of this class of spacetimes in Chapter 2 and then necessary functional analytic properties of quasifree (KMS) states on the CCR-algebra (defined in the usual manner on these globally hyperbolic spacetimes) in Chapter 3 (where a rigorous definition of Hadamard states finally appears in the literature), the authors prove that any quasifree Hadamard state that is invariant under the flow determined by the Killing vector field must be unique and pure. Moreover, if the spacetime admits a certain discrete reflection symmetry (which obtains in analytic spacetimes) such states must be KMS with respect to the Killing flow at the Hawking temperature $$T=\kappa/2\pi$$, when restricted to the right or left wedge subalgebras of the global CCR-algebra. ($$\kappa$$ is the surface gravity of the horizon.) The authors also prove that Killing-flow-invariant Hadamard states do not exist on (the CCR-algebra on) Schwarzschild-de Sitter spacetimes and on the globally hyperbolic region of Kerr spacetimes. Indeed, for each of these two classes of spacetimes, two independent, rigorous arguments (which have natural generalizations to different classes of spacetimes) for the nonexistence of such states are given, providing additional insight into the underlying reasons for the absence of such states. The question of existence in the case of other classes of spacetimes with bifurcate Killing horizons is left open.
This paper is an important contribution to the small body of mathematically rigorous results known about quantum field theories on curved spacetimes.

MSC:
 53Z05 Applications of differential geometry to physics 81T20 Quantum field theory on curved space or space-time backgrounds 81T05 Axiomatic quantum field theory; operator algebras 83C47 Methods of quantum field theory in general relativity and gravitational theory
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I: functional analysis, () · Zbl 0517.47006 [58] Whiting, B.F., Mode stability of the Kerr black hole, J. math. phys., 30, 1301-1305, (1989) · Zbl 0689.53041 [59] Streater, R.F.; Wightman, A.S., Spin and statistics and all that, (1964), Benjamin New York, PCT · Zbl 0135.44305 [60] Dieudonné, J., () [61] Haag, R.; Kastler, D., An algebraic approach to quantum field theory, J. math. phys., 5, 848-861, (1964) · Zbl 0139.46003 [62] talk (see Workshop Chairman’s Report by A. Ashtekar, pp. 453-456). [63] Gonnella, G.; Kay, B.S., Can locally Hadamard quantum states have non-local singularities? classical quantum gravity, 6, 1445-1454, (1989) · Zbl 0678.53082 [64] Fredenhagen, K.; Haag, R., On the derivation of Hawking radiation associated with the formation of a black hole, Commun. math. phys., 127, 273-284, (1990) · Zbl 0692.53040 [65] Lüders, C.; Roberts, J.E., Local quasiequivalence and adiabatic vacuum states, Commun. math. phys., 134, 29-63, (1990) · Zbl 0749.46045 [66] R.M. Wald, Vacuum states in spacetimes with Killing horizons, in: Quantum Mechanics in Curved Space-Time, eds. J. Audretsch and V. de Sabbata (Plenum, New York), in press. [67] Frolov, V.P.; Thorne, K.S., Renormalized stress-energy tensor near the horizon of a slowly evolving, rotating black hole, Phys. rev. D, 39, 2125-2154, (1989) [68] Landsman, N.P.; van Weert, Ch.G., Real- and imaginary-time field theory at finite temperature and density, Phys. rep., 145, 141-249, (1987) [69] Gibbons, G.W.; Perry, M.J., Black holes in thermal equilibrium, Phys. rev. lett., 36, 985-987, (1976) [70] Gibbons, G.W.; Perry, M.J., Black holes and thermal Green functions, (), 467-494
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Consider two real valued signals, $x(t)$ band-limited to $[-500\:Hz,500\:Hz]$ and $y(t)$ band-limited to $[-1\:kHz,1\:kHz].$ For $z(t) = x(t)\cdot y(t),$ the Nyquist sampling frequency (in  $kHz$) is _______.


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# Mesh Lighting with .X format

This topic is 4150 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.

## Recommended Posts

Hi all, I've been using Milkshape to do simple models like planets mapped with a material and texture. Then I tried to use the DirectX Model Viewer to view the model, everything is ok in place. I also noticed that the normals are there. Next, I throw them into my own code using some of D3DX libraries. However, with a Directional Light, the "planet" mesh still doesn't get lighted up. This also happened to some of my borrowed models as well. Please note that I've ensured that the light parameters are correct and that its range is far enough to hit the planet. But somehow, the planet doesn't get lighted up. I've spoken to some of my friends who had experience using .X format. They told me that there is a part of the file I have to remove. Does anyone have any idea with that? Please help... Thanks, Thrawn80

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Sometimes you have to delete stuff depending on the exporter but thats probably not your issue because I export from MilkShape and it works fine.

Try the usual stuff. Make sure that lighting is enabled. Make sure that your material's ambient and diffuse are not the same. Make sure that you have that specific light enabled.

Is it completely black? That could be a different problem. You could post your code so we can check it out.

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Hi Programmer101,

May I ask, why can't I put the same values for Ambient and Diffuse? Please forgive my ignorance but I'm still relatively a newbie in 3D meshes...

On the other hand, the code should be fine because there are some of the models that actually react to the light settings.

It's a directional light, so I don't think attenuation, phi and theta has anything to do with it. Perhaps I'm wrong also :)

Thanks,
Thrawn80

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Hello, Have you tryed calling this
Device->SetRenderState(D3DRS_NORMALIZENORMALS, true);

this will renormalize the normals may help next time send some code so we don't need to guess.

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Hi ankhd,

Thanks for that. I'm going to try it out. Actually I also don't know what code to paste since my code is already working for some of the models.

Which is y I don't mention abt the codes. More rather, I'm inclined to believe that it's the exporter's fault....

I'm using Milkshape, but the exported is DirectX (JT).. so there might be some things that I hav to add or remove from the .X file.

Perhaps if there is anyone who has insights on this can provide some help? Please advise.

Regards,
Thrawn80

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You can't use the same ambient and diffuse values because the color will never be lighter than the ambient color and never darker than the diffuse so by making them the same you eliminate your lighting range.

Again, I've never had to delete anything from my JT exported models.

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Hi all,

I've managed to resolve the problem.

Here's how -
...// the material description of the .x file (ORIGINAL)Material Material{    0.886275;0.886275;0.886275;1.000000;;    0.000000;    0.698039;0.698039;0.698039;;    0.147059;0.147059;0.147059;;    TextureFileName    {        "Blackstone.jpg";    }}

Noticed that 0.000000; ? that's actually the Power of the material. After going thru some DirectX documentations on materials, I've decided to try changing this value.

Material Material{    0.886275;0.886275;0.886275;1.000000;;    80.000000;    0.698039;0.698039;0.698039;;    0.147059;0.147059;0.147059;;    TextureFileName    {        "Blackstone.jpg";    }}

Now the power is at 80, the lighting works along with the specular lighting. And yes, there is no need to remove anything from exporter for Milkshape.

Please let me know if you think that there are any errors or discreptancies in my conclusion.

Thanks a lot !

Thrawn80

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## Elementary Algebra

Published by Cengage Learning

# Chapter 2 - Real Numbers - 2.4 - Exponents - Problem Set 2.4: 100

#### Answer

$= -125$

#### Work Step by Step

We plug in 5 for x to obtain: $-x^{3}$ $= -5^{3}$ $= -(5 \times 5 \times 5)$ $= -(5 \times 25)$ $= -125$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


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~/1 POINTSSERCP11 21.4.P.021.resistor (R 9.00 102 0), capacitor (C 0.250 uF), and an inductor (L 2.00 H) are connected series

Question:

~/1 POINTS SERCP11 21.4.P.021. resistor (R 9.00 102 0), capacitor (C 0.250 uF), and an inductor (L 2.00 H) are connected series across 102-Hz AC source for which MVmax 1,50 102 V, Calculate the impedance of the circuit: (b) Calculate the maximum current delivered bv the source Calculate the phase angle between the current Jnd voltage: (d) Is the current leading or lagging behind the voltage? The current lcads tne voltage The current lags behind the voltage_ There pnase difference between the current and voltage: Need Help? RundIt

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Solve each system by elimination or substitution. $\left\{\begin{array}{l}{-x+5 y=3} \\ {2 x-10 y=4}\end{array}\right.$
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## Cryptology ePrint Archive: Report 2008/383

Hierarchical Identity Based Encryption with Polynomially Many Levels

Craig Gentry and Shai Halevi

Abstract: We present the first hierarchical identity based encryption (HIBE) system that has full security for more than a constant number of levels. In all prior HIBE systems in the literature, the security reductions suffered from exponential degradation in the depth of the hierarchy, so these systems were only proven fully secure for identity hierarchies of constant depth. (For deep hierarchies, previous work could only prove the weaker notion of selective-ID security.) In contrast, we offer a tight proof of security, regardless of the number of levels; hence our system is secure for polynomially many levels. Our result can very roughly be viewed as an application of Boyen's framework for constructing HIBE systems from exponent-inversion IBE systems to a (dramatically souped-up) version of Gentry's IBE system, which has a tight reduction. In more detail, we first describe a generic transformation from identity based broadcast encryption with key randomization" (KR-IBBE) to a HIBE, and then construct KR-IBBE by modifying a recent construction of IBBE of Gentry and Waters, which is itself an extension of Gentry's IBE system. Our hardness assumption is similar to that underlying Gentry's IBE system.

Category / Keywords: public-key cryptography /


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107 views

The strength of a salt solution is $p$% if $100$ ml of the solution contains $p$ grams of salt. If three salt solutions $A, B, C$ are mixed in the proportion $1:2:3$, then the resulting solution has strength $20$%. If instead the proportion is $3:2:1$, then the resulting solution has strength $30$%. A fourth solution, $D$, is produced by mixing $B$ and $C$ in the ratio $2:7$. The ratio of the strength of $D$ to that of $A$ is

1. $2:5$
2. $1:3$
3. $1:4$
4. $3:10$

1
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# Best configuration for 400W resistive heating device

Suppose I wanted to heat a large aluminum disk with joule heating. I would use one of its flat faces, which has a radius of about a foot (30cm, SA = 0.28 sq m) and enough resistors to dissipate 400W of energy.

If I use 12V across a circuit:

P = V^2/R

P = 320W

V = 12V

R = 0.45 ohms

and the current for such a circuit would be I = V/R = 12/0.45 =~ 27 A (which seems very high)

Is this correct? Is this kind of set up "safe"?

--

And if you are interested:

If you were to build this circuit, how would you arrange resistors of any R value while:

1. Distributing the heat across a large (1ft radius) surface
2. Remaining under 100W per resistor
3. Maintaining R=.45 ohm total circuit resistance as closely as possible

Here is one attempt of mine, using 10 parallel 4ohm resistors (.4 ohm circuit resistance):

simulate this circuit – Schematic created using CircuitLab

If you think it is safe/worthwhile, I am open to setups with higher voltage, but for some reason 12v seems to be the standard.

Also if I wanted to use PWM (DC) to control the output of this circuit, what would be the cheapest and most efficient way to set that up with Arduino?

My disk is about 10.8 kg with specific heat capacity (Al) = 0.9 J/gC

So I chose 320W because this would allow me to heat by 10 deg C every 5 min:

0.9*10800*10/320 =~ 300 sec = 5 min

• It'll work. Main safety issues come from the heat and the high current. If you're paranoid about safety or have difficulty finding 30A rated fuses, switches, etc, you could wire pairs of resistors as separate 6A circuits. Aug 7 '15 at 11:00

Personally, I would (if possible) have a groove milled into the disk and use a custom tubular heater that runs from the mains. I'd specify the heater at about 1.5x the power I think is required.

Tubular heaters are like the elements in a domestic oven- swaged with powdered ceramic inside- and are very long-lived and robust.

Easily controlled using an SSR- 500W would only be a few amperes.

• This would be a nice solution, but probably more expensive than I have funding for. They wouldn't happen to fabricate these for $100 or less would they? – Ryan Aug 7 '15 at 4:32 • I've been looking around for hours and the resistors are one of the cheapest and easiest to assemble, but at the cost of safety I suppose. Another solution I'm looking at is this: protovision.com/2011/09/25/milled-reprap-heated-bed-pcb But I don't have access to a CNC mill at the moment. – Ryan Aug 7 '15 at 4:40 • What country are you in? Electric skillet looks like a goodish fit. 10 Watt ceramic resistors would also be easy enough and no great safety hazard and moderate cost. | Also possible with some insulating head scratching needed is NiChrome wire of suitable rating. This can be obtained in a range of thicknesses to suit current and power needed. Aug 7 '15 at 11:54 • Custom calrod elements were not expensive last time I bought them, should be well under$100 for a couple. Aug 7 '15 at 12:23
• Ok thanks for the help everyone. I will try out a prototype with just a couple resistors and see how it goes, but since you mention the custom heating element, I may switch plans if I don't like the resistors. (btw I am in the US)
– Ryan
Aug 7 '15 at 21:17

What Spehro says re mains encapsulated element BUT it's called an electric skillet (electric frypan in NZ) - obtained at garage sales / yard sales (what country are you in) for a few dollars or somebody's attic or dumpster for free. Or Amazon / Walmart new from about $30. Typically 1000 Watt. Some are 1500 Watt, maybe more. PWM to suit. Heat distribution evenness may be an issue but a copper disk on pan base should help. Get friendly machine shop to remove sides OR use a hacksaw, file and effort if necessary. This gives you a metal heated base with sealed embedded element able to meet UL & other consumer safety specs. Amazon selection Amazon cheapie -$12.99!!! but at 7" square may not easily spread heat evenly enough.

____________________________

Walmart $low to$ high

• @Ryan But it still gets zero votes :-). Low or mo cost, robust, waterproof, available, minimal effort, ... :-). Aug 8 '15 at 14:57


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Here's a brilliant puzzle we stumbled across...

You spin a wheel and it randomly lands of $£1$, $£2$ or END. If you land on $£1$ or $£2$, you bank the money and spin the wheel again. You keep spinning until you land on END, at which point you win the money you have banked. On average, how much do you expect to win?

For an extra challenge...

What if the wheel now has values $£1$, $£2$, ..., $£\text{n}$ as well as END and and you play the same game? How much do you expect to win now?

Post solutions below!

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$$\require{cancel}$$

# 26.3: Anti-derivatives and integrals

In the previous section, we were concerned with determining the derivative of a function $$f(x)$$. The derivative is useful because it tells us how the function $$f(x)$$ varies as a function of $$x$$. In physics, we often know how a function varies, but we do not know the actual function. In other words, we often have the opposite problem: we are given the derivative of a function, and wish to determine the actual function. For this case, we will limit our discussion to functions of a single independent variable.

Suppose that we are given a function $$f(x)$$ and we know that this is the derivative of some other function, $$F(x)$$, which we do not know. We call $$F(x)$$ the anti-derivative of $$f(x)$$. The anti-derivative of a function $$f(x)$$, written $$F(x)$$, thus satisfies the property: \begin{aligned} \frac{dF}{dx}=f(x)\end{aligned} Since we have a symbol for indicating that we take the derivative with respect to $$x$$ ($$\frac{d}{dx}$$), we also have a symbol, $$\int dx$$, for indicating that we take the anti-derivative with respect to $$x$$: \begin{aligned} \int f(x) dx &= F(x) \\ \therefore \frac{d}{dx}\left(\int f(x) dx\right) &= \frac{dF}{dx}=f(x)\end{aligned} Earlier, we justified the symbol for the derivative by pointing out that it is like $$\frac{\Delta f}{\Delta x}$$ but for the case when $$\Delta x\to 0$$. Similarly, we will justify the anti-derivative sign, $$\int f(x) dx$$, by showing that it is related to a sum of $$f(x)\Delta x$$, in the limit $$\Delta x\to 0$$. The $$\int$$ sign looks like an “S” for sum.

While it is possible to exactly determine the derivative of a function $$f(x)$$, the anti-derivative can only be determined up to a constant. Consider for example a different function, $$\tilde F(x)=F(x)+C$$, where $$C$$ is a constant. The derivative of $$\tilde F(x)$$ with respect to $$x$$ is given by: \begin{aligned} \frac{d\tilde{F}}{dx}&=\frac{d}{dx}\left(F(x)+C\right)\\ &=\frac{dF}{dx}+\frac{dC}{dx}\\ &=\frac{dF}{dx}+0\\ &=f(x)\end{aligned} Hence, the function $$\tilde F(x)=F(x)+C$$ is also an anti-derivative of $$f(x)$$. The constant $$C$$ can often be determined using additional information (sometimes called “initial conditions”). Recall the function, $$f(x)=x^2$$, shown in Figure A2.2.1 (left panel). If you imagine shifting the whole function up or down, the derivative would not change. In other words, if the origin of the axes were not drawn on the left panel, you would still be able to determine the derivative of the function (how steep it is). Adding a constant, $$C$$, to a function is exactly the same as shifting the function up or down, which does not change its derivative. Thus, when you know the derivative, you cannot know the value of $$C$$, unless you are also told that the function must go through a specific point (a so-called initial condition).

In order to determine the derivative of a function, we used Equation A2.2.1. We now need to derive an equivalent prescription for determining the anti-derivative. Suppose that we have the two pieces of information required to determine $$F(x)$$ completely, namely:

1. the function $$f(x)=\frac{dF}{dx}$$ (its derivative).
2. the condition that $$F(x)$$ must pass through a specific point, $$F(x_0)=F_0$$.

The procedure for determining the anti-derivative $$F(x)$$ is illustrated above in Figure A2.3.1. We start by drawing the point that we know the function $$F(x)$$ must go through, $$(x_0,F_0)$$. We then choose a value of $$\Delta x$$ and use the derivative, $$f(x)$$, to calculate $$\Delta F_0$$, the amount by which $$F(x)$$ changes when $$x$$ changes by $$\Delta x$$. Using the derivative $$f(x)$$ evaluated at $$x_0$$, we have: \begin{aligned} \frac{\Delta F_0}{\Delta x} &\approx f(x_0)\;\;\;\; (\text{in the limit} \Delta x\to 0 )\\ \therefore \Delta F_0 &= f(x_0) \Delta x\end{aligned} We can then estimate the value of the function $$F_1=F(x_1)$$ at the next point, $$x_1=x_0+\Delta x$$, as illustrated by the black arrow in Figure A2.3.1 \begin{aligned} F_1&=F(x_1)\\ &=F(x+\Delta x) \\ &\approx F_0 + \Delta F_0\\ &\approx F_0+f(x_0)\Delta x\end{aligned} Now that we have determined the value of the function $$F(x)$$ at $$x=x_1$$, we can repeat the procedure to determine the value of the function $$F(x)$$ at the next point, $$x_2=x_1+\Delta x$$. Again, we use the derivative evaluated at $$x_1$$, $$f(x_1)$$, to determine $$\Delta F_1$$, and add that to $$F_1$$ to get $$F_2=F(x_2)$$, as illustrated by the grey arrow in Figure A2.3.1: \begin{aligned} F_2&=F(x_1+\Delta x) \\ &\approx F_1+\Delta F_1\\ &\approx F_1+f(x_1)\Delta x\\ &\approx F_0+f(x_0)\Delta x+f(x_1)\Delta x\end{aligned} Using the summation notation, we can generalize the result and write the function $$F(x)$$ evaluated at any point, $$x_N=x_0+N\Delta x$$: \begin{aligned} F(x_N) \approx F_0+\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x\end{aligned} The result above will become exactly correct in the limit $$\Delta x\to 0$$:

$F(x_N) = F(x_0)+\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x$

Let us take a closer look at the sum. Each term in the sum is of the form $$f(x_{i-1})\Delta x$$, and is illustrated in Figure A2.3.2 for the same case as in Figure A2.3.1 (that is, Figure A2.3.2 shows $$f(x)$$ that we know, and Figure A2.3.1 shows $$F(x)$$ that we are trying to find).

As you can see, each term in the sum corresponds to the area of a rectangle between the function $$f(x)$$ and the $$x$$ axis (with a piece missing). In the limit where $$\Delta x\to 0$$, the missing pieces (shown by the hashed areas in Figure A2.3.2) will vanish and $$f(x_i)\Delta x$$ will become exactly the area between $$f(x)$$ and the $$x$$ axis over a length $$\Delta x$$. The sum of the rectangular areas will thus approach the area between $$f(x)$$ and the $$x$$ axis between $$x_0$$ and $$x_N$$: \begin{aligned} \lim_{\Delta x\to 0}\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x=\text{Area between f(x) and x axis from x_0 to x_N}\end{aligned}

Re-arranging Equation A2.3.1 gives us a prescription for determining the anti-derivative: \begin{aligned} F(x_N) - F(x_0)&=\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x\end{aligned} We see that if we determine the area between $$f(x)$$ and the $$x$$ axis from $$x_0$$ to $$x_N$$, we can obtain the difference between the anti-derivative at two points, $$F(x_N)-F(x_0)$$

The difference between the anti-derivative, $$F(x)$$, evaluated at two different values of $$x$$ is called the integral of $$f(x)$$ and has the following notation:

$\int_{x_0}^{x_N}f(x) dx=F(x_N) - F(x_0)=\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x$

As you can see, the integral has labels that specify the range over which we calculate the area between $$f(x)$$ and the $$x$$ axis. A common notation to express the difference $$F(x_N) - F(x_0)$$ is to use brackets: \begin{aligned} \int_{x_0}^{x_N}f(x) dx=F(x_N) - F(x_0) =\big [ F(x) \big]_{x_0}^{x_N}\end{aligned}

Recall that we wrote the anti-derivative with the same $$\int$$ symbol earlier: \begin{aligned} \int f(x) dx = F(x)\end{aligned} The symbol $$\int f(x) dx$$ without the limits is called the indefinite integral. You can also see that when you take the (definite) integral (i.e. the difference between $$F(x)$$ evaluated at two points), any constant that is added to $$F(x)$$ will cancel. Physical quantities are always based on definite integrals, so when we write the constant $$C$$ it is primarily for completeness and to emphasize that we have an indefinite integral.

As an example, let us determine the integral of $$f(x)=2x$$ between $$x=1$$ and $$x=4$$, as well as the indefinite integral of $$f(x)$$, which is the case that we illustrated in Figures A2.3.1 and A2.3.2. Using Equation A2.3.2, we have: \begin{aligned} \int_{x_0}^{x_N}f(x) dx&=\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} f(x_{i-1}) \Delta x \\ &=\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} 2x_{i-1} \Delta x \end{aligned} where we have: \begin{aligned} x_0 &=1 \\ x_N &=4 \\ \Delta x &= \frac{x_N-x_0}{N}\end{aligned} Note that $$N$$ is the number of times we have $$\Delta x$$ in the interval between $$x_0$$ and $$x_N$$. Thus, taking the limit of $$\Delta x\to 0$$ is the same as taking the limit $$N\to\infty$$. Let us illustrate the sum for the case where $$N=3$$, and thus when $$\Delta x=1$$, corresponding to the illustration in Figure A2.3.2: \begin{aligned} \sum_{i=1}^{i=N=3} 2x_{i-1} \Delta x &=2x_0\Delta x+2x_1\Delta x+2x_2\Delta x\\ &=2\Delta x (x_0+x_1+x_2) \\ &=2 \frac{x_3-x_0}{N}(x_0+x_1+x_2) \\ &=2 \frac{(4)-(1)}{(3)}(1+2+3) \\ &=12\end{aligned} where in the second line, we noticed that we could factor out the $$2\Delta x$$ because it appears in each term. Since we only used 4 points, this is a pretty coarse approximation of the integral, and we expect it to be an underestimate (as the missing area represented by the hashed lines in Figure A2.3.2 is quite large).

If we repeat this for a larger value of N, $$N=6$$ ($$\Delta x = 0.5$$), we should obtain a more accurate answer: \begin{aligned} \sum_{i=1}^{i=6} 2x_{i-1} \Delta x &=2 \frac{x_6-x_0}{N}(x_0+x_1+x_2+x_3+x_4+x_5)\\ &=2\frac{4-1}{6} (1+1.5+2+2.5+3+3.5)\\ &=13.5\end{aligned}

Writing this out again for the general case so that we can take the limit $$N\to\infty$$, and factoring out the $$2\Delta x$$: \begin{aligned} \sum_{i=1}^{i=N} 2x_{i-1} \Delta x &=2 \Delta x\sum_{i=1}^{i=N}x_{i-1}\\ &=2 \frac{x_N-x_0}{N}\sum_{i=1}^{i=N}x_{i-1}\end{aligned} Now, consider the combination: \begin{aligned} \frac{1}{N}\sum_{i=1}^{i=N}x_{i-1}\end{aligned} that appears above. This corresponds to the arithmetic average of the values from $$x_0$$ to $$x_{N-1}$$ (sum the values and divide by the number of values). In the limit where $$N\to \infty$$, then the value $$x_{N-1}\approx x_N$$. The average value of $$x$$ in the interval between $$x_0$$ and $$x_N$$ is simply given by the value of $$x$$ at the midpoint of the interval: \begin{aligned} \lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{i=N}x_{i-1}=\frac{1}{2}(x_N+x_0)\end{aligned} Putting everything together: \begin{aligned} \lim_{N\to\infty}\sum_{i=1}^{i=N} 2x_{i-1} \Delta x &=2 (x_N+x_0)\lim_{N\to\infty}\frac{1}{N}\sum_{i=1}^{i=N}x_{i-1}\\ &=2 (x_N-x_0)\frac{1}{2}(x_N+x_0)\\ &=x_N^2 - x_0^2\\ &=(4)^2 - (1)^2 = 15\end{aligned} where in the last line, we substituted in the values of $$x_0=1$$ and $$x_N=4$$. Writing this as the integral: \begin{aligned} \int_{x_0}^{x_N}2x dx=F(x_N) - F(x_0)=x_N^2 - x_0^2\end{aligned} we can immediately identify the anti-derivative and the indefinite integral: \begin{aligned} F(x) &= x^2 +C \\ \int 2xdx&=x^2 +C\end{aligned} This is of course the result that we expected, and we can check our answer by taking the derivative of $$F(x)$$: \begin{aligned} \frac{dF}{dx}=\frac{d}{dx}(x^2+C) = 2x\end{aligned} We have thus confirmed that $$F(x)=x^2+C$$ is the anti-derivative of $$f(x)=2x$$.

Exercise $$\PageIndex{1}$$

The quantity $$\int_{a}^{b}f(t)dt$$ is equal to

1. the area between the function $$f(t)$$ and the $$f$$ axis between $$t=a$$ and $$t=b$$
2. the sum of $$f(t)\Delta t$$ in the limit $$\Delta t\to 0$$ between $$t=a$$ and $$t=b$$
3. the difference $$f(b) - f(a)$$.

## Common anti-derivative and properties

Table A2.3.1 below gives the anti-derivatives (indefinite integrals) for common functions. In all cases, $$x,$$ is the independent variable, and all other variables should be thought of as constants:

Function, $$f(x)$$ Anti-derivative, $$F(x)$$
$$f(x)=a$$ $$F(x)=ax+C$$
$$f(x)=x^n$$ $$F(x)=\frac{1}{n+1}x^{n+1}+C$$
$$f(x)=\frac{1}{x}$$ $$F(x)=\ln(|x|)+C$$
$$f(x)=\sin(x)$$ $$F(x)=-\cos(x)+C$$
$$f(x)=\cos(x)$$ $$F(x)=\sin(x)+C$$
$$f(x)=\tan(x)$$ $$F(x)=-\ln(|\cos(x)|)+C$$
$$f(x)=e^x$$ $$F(x)=e^x+C$$
$$f(x)=\ln(x)$$ $$F(x)=x\ln(x)-x+C$$

Table A2.3.1: Common indefinite integrals of functions.

Note that, in general, it is much more difficult to obtain the anti-derivative of a function than it is to take its derivative. A few common properties to help evaluate indefinite integrals are shown in Table A2.3.2 below.

Anti-derivative Equivalent anti-derivative
$$\int (f(x)+g(x)) dx$$ $$\int f(x)dx+\int g(x) dx$$ (sum)
$$\int (f(x)-g(x)) dx$$ $$\int f(x)dx-\int g(x) dx$$ (subtraction)
$$\int af(x) dx$$ $$a\int f(x)dx$$ (multiplication by constant)
$$\int f'(x)g(x) dx$$ $$f(x)g(x)-\int f(x)g'(x) dx$$ (integration by parts)

Table A2.3.2: Some properties of indefinite integrals.

## Common uses of integrals in Physics - from a sum to an integral

Integrals are extremely useful in physics because they are related to sums. If we assume that our mathematician friends (or computers) can determine anti-derivatives for us, using integrals is not that complicated.

The key idea in physics is that integrals are a tool to easily performing sums. As we saw above, integrals correspond to the area underneath a curve, which is found by summing the (different) areas of an infinite number of infinitely small rectangles. In physics, it is often the case that we need to take the sum of an infinite number of small things that keep varying, just as the areas of the rectangles.

Consider, for example, a rod of length, $$L$$, and total mass $$M$$, as shown in Figure A2.3.3. If the rod is uniform in density, then if we cut it into, say, two equal pieces, those two pieces will weigh the same. We can define a “linear mass density”, $$\mu$$, for the rod, as the mass per unit length of the rod: \begin{aligned} \mu = \frac{M}{L}\end{aligned} The linear mass density has dimensions of mass over length and can be used to find the mass of any length of rod. For example, if the rod has a mass of $$M=5\text{kg}$$ and a length of $$L=2\text{m}$$, then the mass density is: \begin{aligned} \mu=\frac{M}{L}=\frac{(5\text{kg})}{(2\text{m})}=2.5\text{kg/m}\end{aligned} Knowing the mass density, we can now easily find the mass, $$m$$, of a piece of rod that has a length of, say, $$l=10\text{cm}$$. Using the mass density, the mass of the $$10\text{cm}$$ rod is given by: \begin{aligned} m=\mu l=(2.5\text{kg/m})(0.1\text{m})=0.25\text{kg}\end{aligned} Now suppose that we have a rod of length $$L$$ that is not uniform, as in Figure A2.3.3, and that does not have a constant linear mass density. Perhaps the rod gets wider and wider, or it has a holes in it that make it not uniform. Imagine that the mass density of the rod is instead given by a function, $$\mu(x)$$, that depends on the position along the rod, where $$x$$ is the distance measured from one side of the rod.

Now, we cannot simply determine the mass of the rod by multiplying $$\mu(x)$$ and $$L$$, since we do not know which value of $$x$$ to use. In fact, we have to use all of the values of $$x$$, between $$x=0$$ and $$x=L$$.

The strategy is to divide the rod up into $$N$$ pieces of length $$\Delta x$$. If we label our pieces of rod with an index $$i$$, we can say that the piece that is at position $$x_i$$ has a tiny mass, $$\Delta m_i$$. We assume that $$\Delta x$$ is small enough so that $$\mu(x)$$ can be taken as constant over the length of that tiny piece of rod. Then, the tiny piece of rod at $$x=x_i$$, has a mass, $$\Delta m_i$$, given by: \begin{aligned} \Delta m_i = \mu(x_i) \Delta x\end{aligned} where $$\mu(x_i)$$ is evaluated at the position, $$x_i$$, of our tiny piece of rod. The total mass, $$M$$, of the rod is then the sum of the masses of the tiny rods, in the limit where $$\Delta x\to 0$$: \begin{aligned} M &= \lim_{\Delta x\to 0}\sum_{i=1}^{i=N}\Delta m_i \\ &= \lim_{\Delta x\to 0}\sum_{i=1}^{i=N} \mu(x_i) \Delta x\end{aligned} But this is precisely the definition of the integral (Equation A2.3.1), which we can easily evaluate with an anti-derivative: \begin{aligned} M &=\lim_{\Delta x\to 0}\sum_{i=1}^{i=N} \mu(x_i) \Delta x \\ &= \int_0^L \mu(x) dx \\ &= G(L) - G(0)\end{aligned} where $$G(x)$$ is the anti-derivative of $$\mu(x)$$.

Suppose that the mass density is given by the function: \begin{aligned} \mu(x)=ax^3\end{aligned} with anti-derivative (Table A2.3.1): \begin{aligned} G(x)=a\frac{1}{4}x^4 + C\end{aligned} Let $$a=5\text{kg/m}^{4}$$ and let’s say that the length of the rod is $$L=0.5\text{m}$$. The total mass of the rod is then: \begin{aligned} M&=\int_0^L \mu(x) dx \\ &=\int_0^L ax^3 dx \\ &= G(L)-G(0)\\ &=\left[ a\frac{1}{4}L^4 \right] - \left[ a\frac{1}{4}0^4 \right]\\ &=5\text{kg/m}^{4}\frac{1}{4}(0.5\text{m})^4 \\ &=78\text{g}\\\end{aligned}

With a little practice, you can solve this type of problem without writing out the sum explicitly. Picture an infinitesimal piece of the rod of length $$dx$$ at position $$x$$. It will have an infinitesimal mass, $$dm$$, given by: \begin{aligned} dm = \mu(x) dx\end{aligned} The total mass of the rod is the then the sum (i.e. the integral) of the mass elements \begin{aligned} M = \int dm\end{aligned} and we really can think of the $$\int$$ sign as a sum, when the things being summed are infinitesimally small. In the above equation, we still have not specified the range in $$x$$ over which we want to take the sum; that is, we need some sort of index for the mass elements to make this a meaningful definite integral. Since we already know how to express $$dm$$ in terms of $$dx$$, we can substitute our expression for $$dm$$ using one with $$dx$$: \begin{aligned} M = \int dm = \int_0^L \mu(x) dx\end{aligned} where we have made the integral definite by specifying the range over which to sum, since we can use $$x$$ to “label” the mass elements.

One should note that coming up with the above integral is physics. Solving it is math. We will worry much more about writing out the integral than evaluating its value. Evaluating the integral can always be done by a mathematician friend or a computer, but determining which integral to write down is the physicist’s job!


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Physics Course/Motion/Periodic Motion/Wave

Water WaveEdit

Concentric CirclesEdit

Concentric circles travels in a direction when there is a cobble drops on the water surface

${\displaystyle x^{2}+y^{2}=r^{2}}$
${\displaystyle Tan\theta ={\frac {y}{x}}}$

Which can be expressed as

${\displaystyle y(\theta )=rSin\theta }$
${\displaystyle x(\theta )=rCos\theta }$

High Low WaveEdit

Water wave moves in a direction that changes its amplitude sinusoidally

${\displaystyle y(t)=rSin\omega t}$
${\displaystyle \omega ={\frac {\lambda }{t}}=\lambda f}$

With

${\displaystyle \lambda }$ : Wavelength . The distance between 2 crests (highest point)
${\displaystyle f}$ : Frequency . frequency of time to complete a cycle


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# Improving Power Spectral Estimation using Multitapering: Precise asteroseismic modeling of stars, exoplanets, and beyond

29 Sep 2022  ·  Aarya A. Patil, Gwendolyn M. Eadie, Joshua S. Speagle, David J. Thomson ·

Asteroseismic time-series data have imprints of stellar oscillation modes, whose detection and characterization through time-series analysis allows us to probe stellar interiors physics. Such analyses usually occur in the Fourier domain by computing the Lomb-Scargle (LS) periodogram, an estimator of the \textit{power spectrum} underlying unevenly-sampled time-series data. However, the LS periodogram suffers from the statistical problems of (1) inconsistency (or noise) and (2) bias due to high spectral leakage. In addition, it is designed to detect strictly periodic signals but is unsuitable for non-sinusoidal periodic or quasi-periodic signals. Here, we develop a multitaper spectral estimation method that tackles the inconsistency and bias problems of the LS periodogram. We combine this multitaper method with the Non-Uniform Fast Fourier Transform (\texttt{mtNUFFT}) to more precisely estimate the frequencies of asteroseismic signals that are non-sinusoidal periodic (e.g., exoplanet transits) or quasi-periodic (e.g., pressure modes). We illustrate this using a simulated and the Kepler-91 red giant light curve. Particularly, we detect the Kepler-91b exoplanet and precisely estimate its period, $6.246 \pm 0.002$ days, in the frequency domain using the multitaper F-test alone. We also integrate \texttt{mtNUFFT} into the \texttt{PBjam} package to obtain a Kepler-91 age estimate of $3.96 \pm 0.48$ Gyr. This $36$\% improvement in age precision relative to the $4.27 \pm 0.75$ Gyr APOKASC-2 (uncorrected) estimate illustrates that \texttt{mtNUFFT} has promising implications for Galactic archaeology, in addition to stellar interiors and exoplanet studies. Our frequency analysis method generally applies to time-domain astronomy and is implemented in the public Python package \texttt{tapify}, available at \url{https://github.com/aaryapatil/tapify}.

PDF Abstract

## Categories

Instrumentation and Methods for Astrophysics Earth and Planetary Astrophysics Solar and Stellar Astrophysics Applications


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Home > Standard Error > How To Find Standard Error Of The Mean In Spss

# How To Find Standard Error Of The Mean In Spss

## Contents

n. The .597 is the numerical description of how tightly around the imaginary line the points lie. Sign in 2 Loading... In quotes, you need to specify where the data file is located on your computer. have a peek at this web-site

h. Why don't we just discard the variance in favor of the standard deviation (or reversely)? Appease Your Google Overlords: Draw the "G" Logo Why can't we use the toilet when the train isn't moving? In the syntax below, the get file command is used to load the data into SPSS.

## Standard Error Of Measurement Spss

For large sample sizes, however, the two formulas have virtually identical outcomes. mean, or more simply as SEM. Each variable that was listed on the variables= statement will have its own line in this part of the output.

Watch QueueQueueWatch QueueQueue Remove allDisconnect Loading... Suppose the mean number of bedsores was 0.02 in a sample of 500 subjects, meaning 10 subjects developed bedsores. They have neither the time nor the money. Standard Error Of Measurement Calculator Loading...

Descriptive statistics a. How To Calculate Standard Error Of Measurement In Excel Available at: http://www.scc.upenn.edu/čAllison4.html. These are the ratios of the mean of the differences to the standard errors of the difference under the two different assumptions: (-4.86995 / 1.30419) = -3.734, (-4.86995/1.33189) = -3.656. f. 75 - This is the 75% percentile, also know as the third quartile.

The Standard Error of the estimate is the other standard error statistic most commonly used by researchers. Standard Error Of Measurement Example On the other hand, with small variability, the difference is more clear as in the third graph. f. Instead, we use standard deviation.

## How To Calculate Standard Error Of Measurement In Excel

Skewness - Skewness measures the degree and direction of asymmetry. However, many statistical results obtained from a computer statistical package (such as SAS, STATA, or SPSS) do not automatically provide an effect size statistic. Standard Error Of Measurement Spss It is given by where s is the sample deviation of the observations and N is the number of valid observations. Spss Output Interpretation Amplitude of a Sinus, Simple question Cohomology of function spaces Why does the state remain unchanged in the small-step operational semantics of a while loop?

The variance is the squared standard deviation. Check This Out m. Case processing summary a. Comparing Means Comparing means allows us to look at differences between groups of participants Choose Statistics, Compare Means, Means Continuous variables go in Dependent List Grouping variable goes in Independent List P Value Spss

Std. - Standard deviation is the square root of the variance. The independent samples t-test compares the difference in the means from the two groups to a given value (usually 0). So we conclude that the mean forwrite is different from 50. Source The median splits the distribution such that half of all values are above this value, and half are below.

Sign in to make your opinion count. Standard Error Of Measurement Definition In principle, it's awkward that to different statistics basically express the same property of a set of numbers. The SPSS Output Viewer will appear with your results in it.

## RStatsInstitute 299,865 views 10:19 Effect Size - Duration: 5:46.

Then you calculate SEM as follows: $$SEM= SD*(\sqrt{1-ICC})$$ share|improve this answer edited Dec 13 '13 at 13:47 Ming-Chih Kao 683518 answered Feb 17 '13 at 3:14 amin 111 Note that the standard error difference is calculated differently under the two different assumptions. In Harry Potter book 7, why didn't the Order flee Britain after Harry turned seventeen? Paired Samples T Test Mean - These are the respective means of the variables.

DrKKHewitt 16,164 views 4:31 SPSS Video #1: Entering & Defining Variables In SPSS - Duration: 4:14. c. Taken together with such measures as effect size, p-value and sample size, the effect size can be a very useful tool to the researcher who seeks to understand the reliability and have a peek here Generated Mon, 17 Oct 2016 15:10:02 GMT by s_wx1094 (squid/3.5.20)

A confidence interval for the mean specifies a range of values within which the unknown population parameter, in this case the mean, may lie. Therefore, the difference may well come by chance. SPSS computes SE for the mean, the kurtosis, and the skewness A small value indicates a greater stability or smaller sampling err Measures of the shape of the distribution(measures h.

It should be noted that this formula is not restricted to the use of an estimate of ICC; in fact, you can plug in any "valid" measure of reliability (most of Obsessed or Obsessive? It is given by where s is the sample deviation of the observations and N is the number of valid observations. If you're data contain only a sample from your target population, see below.

Loading... In other words, it tests whether the difference in the means is 0. Standard Deviation and Histogram Right, let's make things a bit more visual. Each variable that was listed on the variables= statement in the above code will have its own line in this part of the output.


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# ECE - Study Guide: Physical Memory addresses how many bytes?

Attempting to do this study guide, but i'm not finding too much help online. Any help is welcome, and a explanation would be nice, as my text book does not describe how to answer a problem given the information.

The figure: ][2]

Questions:

1. How much physical memory can the microprocessor address?

2. How many words does the memory IC contain?

3. What is the address range for the memory? That is, if one wanted to access this memory with a program, what range of addresses would "talk" to this memory chip?

• You should be able to quickly find answers to this type of question on the web. For example, I searched for address range of 24 bits and found Wikipedia "24 bit". You have been fortunate that anyone answered, we usually expect more evidence of effort. – gbulmer Dec 18 '15 at 9:38
• @gbulmer well that's not quite what I was looking for nor is that the correct cpu/memory design – StingRay21 Dec 18 '15 at 14:36
• I think the wikipedia link answers part 1 of your question, "How much physical memory can the microprocessor address?". Similarly part 2 of your question is answered by the Wikipedia 18-bit address range of the memory. If that is not the case, then IMHO you need to update your question to explain why, because the answer you have accepted is that same information. – gbulmer Dec 18 '15 at 14:41
• @gbulmer ahh I see now. I didn't see the correlation when I very first searched and looked through those wiki pages. Thank you very much :) – StingRay21 Dec 18 '15 at 15:09

3. What value do the upper bits of the address bus (those connected to the address decode module) have to be in order for the select_memory signal to be asserted? (make a lookup table for the circuit of the decode described). Now you know the starting address (upper bits select the memory, lower bits are 0), and the ending address (upper bits still selecting the memory, lower bits all 1). So convert both those binary numbers to decimal and you have your range of addresses.


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The angles made by this line with the +ve direactions of the coordinate axes: θx, θy, and θz are used to find the direction cosines of the line: cos θx, cos θy, and cos θz. answered Aug 22, 2018 by SunilJakhar (89.0k points) selected Aug 22, 2018 by Vikash Kumar . How to Find the Direction Cosines of a Vector With Given Ratios : Here we are going to see the how to find the direction cosines of a vector with given ratios. v = v x e x + v y e y + v z e z , {\displaystyle \mathbf {v} =v_ {x}\mathbf {e} _ {x}+v_ {y}\mathbf {e} _ {y}+v_ {z}\mathbf {e} _ {z},} where ex, ey, ez are the standard basis in Cartesian notation, then the direction cosines are. How do you find the direction cosines and direction angles of the vector? Transcript. Direction cosines and direction ratios of a vector : Consider a vector as shown below on the x-y-z plane. In this explainer, we will learn how to find direction angles and direction cosines for a given vector in space. Direction cosines : (x/r, y/r, z/r) x/r = 3/ √89. We know that in three-dimensional space, we have the -, -, and - or -axis. We know that the vector equation of a line passing through a point with position vector vec a and parallel to the vector vec b is   $\overrightarrow{r} = \overrightarrow{a} + \lambda \overrightarrow{b}$  Here, $\overrightarrow{a} = 4 \hat{i} + \hat{k}$, $\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$, $\overrightarrow{r} = \left( 4 \hat{i} + 0 \hat{j}+ \hat{k} \right) + \lambda \left( - 2 \hat{i} + 6 \hat{j} - 3 \hat{k} \right)$, $\text{ Here } , \lambda \text{ is a parameter } . Example, 3 Find the direction cosines of the line passing through the two points (– 2, 4, – 5) and (1, 2, 3). of a vector (line) are the cosines of the angles made by the line with the + ve directions of x, y & z axes respectively. (Give the direction angles correct to the nearest degree.) Apart from the stuff given in "How to Find the Direction Cosines of a Vector With Given Ratios", if you need any other stuff in math, please use our google custom search here. d. or d and is the distance between and Px yz11 11 ,, Px yz22 22 ,,. These direction numbers are represented by a, b and c. Ex 10.2, 12 Find the direction cosines of the vector  + 2  + 3  . Hence direction cosines are ( 3/ √89, -4/ √89, 8 / √89) Direction ratios : Direction ratios are (3, -4, 8). Precalculus Vectors in the Plane Direction Angles. It it some times denoted by letters l, m, n.If a = a i + b j + c j be a vector with its modulus r = sqrt (a^2 + b^2 + c^2) then its d.cs. The cartesian equation of the given line is, \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}$, $\frac{x - 4}{- 2} = \frac{y - 0}{6} = \frac{z - 1}{- 3}$, This shows that the given line passes through the point (4,0,1) and its direction ratios are proportional to -2,6,-3, $\frac{- 2}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{6}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}, \frac{- 3}{\sqrt{\left( - 2 \right)^2 + 6^2 + \left( - 3 \right)^2}}$, $= \frac{- 2}{7}, \frac{6}{7}, \frac{- 3}{7}$  Thus, the given line passes through the point having position vector  $\overrightarrow{a} = 4 \hat{i} + \hat{k}$  and is parallel to the vector $\overrightarrow{b} = - 2 \hat{i} + 6 \hat{j} - 3 \hat{k}$. © Copyright 2017, Neha Agrawal. Solution : x = 3, y = 1 and z = 1 |r vector| = r = √(x 2 + y 2 + z 2) = √3 2 + 1 2 + 1 2) = √(9+1+1) = √11. z/r = 8/ √89. Then ∠ PRO = ∠ PSO = ∠ PTO = 90º. Direction cosines (d.cs.) Find the direction cosines of the line  $\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .$  Also, reduce it to vector form. (7, 3, -4) cos(a) =… Find the direction cosines and direction angles of the vector determining the norm of a vector in space, vector operations in space, evaluating simple trigonometric expressions. One such property of the direction cosine is that the addition of the squares of … Direction Cosines and Direction Ratios. View Answer Find the direction cosines of the vector 6 i ^ + 2 j ^ − 3 k ^ . The direction cosines are not independent of each other, they are related by the equation x 2 + y 2 + z 2 = 1, so direction cosines only have two degrees of freedom and can only represent direction and not orientation. Question 1 : If We will begin by considering the three-dimensional coordinate grid. The direction cosine of the vector can be determined by dividing the corresponding coordinate of a vector by the vector length. |r vector|  =  r  =  √(x2 + y2 + z2)   =  √(32 + (-4)2 + 82), Hence direction cosines are ( 3/√89, -4/√89, 8/√89), |r vector|  =  r  =  √(x2 + y2 + z2)   =  √32 + 12 + 12), Hence direction cosines are ( 3/√11, 1/√11, 1/√11), |r vector|  =  r  =  √(x2 + y2 + z2)   =  √02 + 12 + 02), |r vector|  =  r  =  √(x2 + y2 + z2)   =  √52 + (-3)2 + (-48)2, |r vector|  =  r  =  √(x2 + y2 + z2)   =  √32 + 42 + (-3)2, |r vector|  =  r  =  √(x2 + y2 + z2)   =  √12 + 02 + (-1)2. Property of direction cosines. find direction cosines of a vector in space either given in component form or represented graphically. The sum of the squares of the direction cosines is equal to one. Ex 10.2, 13 Find the direction cosines of the vector joining the points A (1, 2,−3) and B (−1,−2,1), directed from A to B. 1 Answer. If the position vectors of P and Q are i + 2 j − 7 k and 5 i − 3 j + 4 k respectively then the cosine of the angle between P Q and z-axis is View solution Find the direction cosines of the vector a = i ^ + j ^ − 2 k ^ . In this video, we will learn how to find direction angles and direction cosines for a given vector in space. vectors; Share It On Facebook Twitter Email. z^^)/(|v|). Let us assume a line OP passes through the origin in the three-dimensional space. How to Find the Direction Cosines of a Vector With Given Ratios". 12.1 Direction Angles and Direction Cosines. (3) From these definitions, it follows that alpha^2+beta^2+gamma^2=1. (ii) 3i vector + j vector + k vector. After having gone through the stuff given above, we hope that the students would have understood, "How to Find the Direction Cosines of a Vector With Given Ratios". Then, the line will make an angle each with the x-axis, y-axis, and z-axis respectively.The cosines of each of these angles that the line makes with the x-axis, y-axis, and z-axis respectively are called direction cosines of the line in three-dimensional geometry. To find the direction cosines of the vector a is need to divided the corresponding coordinate of vector by the length of the vector. Find the direction cosines of a vector which is equally inclined to the x-axis, y-axis and z-axis. The magnitude of vector d is denoted by . Given a vector (a,b,c) in three-space, the direction cosines of this vector are Here the direction angles, , are the angles that the vector makes with the positive x-, y- and z-axes, respectively.In formulas, it is usually the direction cosines that occur, rather than the direction angles. Ex 11.1, 2 Find the direction cosines of a line which makes equal angles with the coordinate axes. Find the Magnitude and Direction Cosines of Given Vectors : Here we are going to see how to find the magnitude and direction cosines of given vectors. How to Find the Direction Cosines of a Vector With Given Ratios". For example, take a look at the vector in the image. Entering data into the vector direction cosines calculator. Any number proportional to the direction cosine is known as the direction ratio of a line. 22 d dxx yy zz21 2 1 2 1. Geospatial Science RMIT THE DISTANCE d BETWEEN TWO POINTS IN SPACE . Students should already be familiar with. Solution for Find the direction cosines and direction angles of the vector. Therefore, we can say that cosines of direction angles of a vector r are the coefficients of the unit vectors, and when the unit vector is resolved in terms of its rectangular components. In three-dimensional geometry, we have three axes: namely, the x, y, and z-axis. So direction cosines of the line = 2/√41, 6/√41, -1/√41. A( 1, 2 , −3) B(−1, −2, 1) () ⃗ = (−1 − 1) ̂ + (−2 − 2) ̂ + (1−(−3)) ̂ = –2 ̂ – 4 ̂ + 4 ̂ Directions ratios are a = – 2, b = –4, & c = 4 Magnitude The unit vector coordinates is equal to the direction cosine. To find the direction cosines of a vector: Select the vector dimension and the vector form of representation; Type the coordinates of the vector; Press the button "Calculate direction cosines of a vector" and you will have a detailed step-by-step solution. Best answer. if you need any other stuff in math, please use our google custom search here. 2 (2) DIRECTION COSINES OF A LINE BETWEEN TWO POINTS IN SPACE How to Find a Vector’s Magnitude and Direction. Lesson Video . Direction cosines of a line making, with x – axis, with y – axis, and with z – axis are l, m, n l = cos , m = cos , n = cos Given the line makes equal angles with the coordinate axes. By Steven Holzner . If you’re given the vector components, such as (3, 4), you can convert it easily to the magnitude/angle way of expressing vectors using trigonometry. Example: Find the direction cosines of the line joining the points (2,1,2) and (4,2,0). Click hereto get an answer to your question  Find the direction ratios and the direction cosines of the vector a = (5î - 3ĵ + 4k̂). All rights reserved.What are Direction cosines and Direction ratios of a vector? What this means is that direction cosines do not define how much an object is rotated around the axis of the vector. Find the Magnitude and Direction Cosines of Given Vectors - Practice Question. Find the Direction Cosines of the Line 4 − X 2 = Y 6 = 1 − Z 3 . y/r = -4/ √89. Prerequisites. We know, in three-dimensional coordinate space, we have the -, -, and -axes.These are perpendicular to one another as seen in the diagram below. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Direction cosines : (x/r, y/r, z/r) x/r = 3/ √11 The coordinates of the unit vector is equal to its direction cosines. are … Find the direction cosines and direction ratios of the following vectors. Also, Reduce It to Vector Form. My Vectors course: https://www.kristakingmath.com/vectors-courseLearn how to find the direction cosines and direction angles of a vector. \], Chapter 28: Straight Line in Space - Exercise 28.1 [Page 10], CBSE Previous Year Question Paper With Solution for Class 12 Arts, CBSE Previous Year Question Paper With Solution for Class 12 Commerce, CBSE Previous Year Question Paper With Solution for Class 12 Science, CBSE Previous Year Question Paper With Solution for Class 10, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Arts, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Commerce, Maharashtra State Board Previous Year Question Paper With Solution for Class 12 Science, Maharashtra State Board Previous Year Question Paper With Solution for Class 10, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Arts, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Commerce, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 12 Science, CISCE ICSE / ISC Board Previous Year Question Paper With Solution for Class 10, PUC Karnataka Science Class 12 Department of Pre-University Education, Karnataka. Let R, S and T be the foots of the perpendiculars drawn from P to the x, y and z axes respectively. Find the direction cosines of a vector 2i – 3j + k . 0 votes . 1 Answer A. S. Adikesavan Jul 1, 2016 ... How do I find the direction angle of vector #<-sqrt3, -1>#? 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The origin Give the direction cosines is equal to one vector is equal to one the =. By dividing the corresponding coordinate of a line which makes equal angles with the coordinate axes P to the cosines... Angles with the coordinate axes SunilJakhar ( 89.0k points ) selected Aug 22, 2018 by SunilJakhar ( points! Or d and is the distance d BETWEEN TWO points in space coordinate grid j +! Look at the vector in space cosine is that direction cosines for a given vector in the three-dimensional space Kumar! Direction Ratios of the line 4 − x 2 = y 6 = 1 − z 3 find direction... Rotated around the axis of the vector use our google custom search here z/r! ) selected Aug 22, 2018 by Vikash Kumar answered Aug 22, 2018 by SunilJakhar 89.0k... Let r, s and T be the foots of the following Vectors or graphically! ) selected Aug 22, 2018 by SunilJakhar ( 89.0k points ) selected Aug 22, by... The foots of the line joining the points ( 2,1,2 ) and of distance r from origin. The x, y and z axes respectively 4 − x 2 = y 6 = 1 z. Ratio of a line OP passes through the origin in the three-dimensional space y/r, z/r ) =. J vector + j vector + k line which makes equal angles with the coordinate axes lesson in., -1/√41 of the vector length given Ratios '' cosines and direction angles correct to the direction cosines a! The perpendiculars drawn from P to the x, y and z respectively... Shown below on the x-y-z plane in math, please use our google custom search here direction... 6/√41, -1/√41 − 3 k ^ to one = 1 − z 3, and... ( 4,2,0 ): ( x/r, y/r, z/r ) x/r = 3/ √89 ( points... Evaluating simple trigonometric expressions, Px yz22 22,, points ( 2,1,2 ) (! The foots of the squares of … direction cosines of a vector in the image in component form represented...
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# Indefinite Integration

• Oct 27th 2010, 08:00 PM
Glitch
Indefinite Integration
The question:

$\int \frac{dx}{4 + x^2}$

I'm fairly sure I have to use some sort of substitution for this question. But I can't seem to get it to work. >_<
• Oct 27th 2010, 08:07 PM
pickslides
No substitution required, use the fact that

$\int \frac{a}{a^2 + x^2}~dx = Tan^{-1}\left(\frac{x}{a}\right)+C$
• Oct 27th 2010, 09:45 PM
Glitch
Quote:

Originally Posted by pickslides
No substitution required, use the fact that

$\int \frac{a}{a^2 + x^2}~dx = Tan^{-1}\left(\frac{x}{a}\right)+C$

What is this relationship called? I'm a little confused.
• Oct 27th 2010, 09:51 PM
TheEmptySet
This is just a formula for the indefinite integral

here is where it comes from

$\displaystyle \int \frac{a}{x^2+a^2}dx$

let $x=a\tan(\theta) \implies dx=a\sec^2{\theta}d\theta$ this gives

$\displaystyle \int \frac{a}{a^2\tan^2(\theta)+a^2}a\sec^2{\theta}d\th eta=\int d\theta =\theta=\tan^{-1}\left(\frac{x}{a} \right)+c$
• Oct 27th 2010, 09:58 PM
Glitch
Ahh, ok. Is there other similar formulas I should be aware of?

Also, I don't understand how the numerator is considered 'a' in my question when the denominator has '4', which is not $a^2$
• Oct 28th 2010, 01:52 AM
TheCoffeeMachine
Or $\displaystyle \int \frac{1}{x^2+a^2}\;{dx} = \frac{1}{a}\tan^{-1}{\frac{x}{a}}+k$ (same thing).

Or letting $x = 2\tan{x}$ in the OP's integral (same thing again).
• Oct 28th 2010, 01:08 PM
pickslides
Quote:

Originally Posted by Glitch
Ahh, ok. Is there other similar formulas I should be aware of?

There are quite a few.

Here's some that are similar

$\int \frac{1}{\sqrt{a^2-x^2}}~dx = Sin^{-1}\left(\frac{x}{a}\right)+C$

$\int \frac{-1}{\sqrt{a^2-x^2}}~dx = Cos^{-1}\left(\frac{x}{a}\right)+C$
• Oct 28th 2010, 04:21 PM
TheCoffeeMachine
Quote:

Originally Posted by Glitch
Also, I don't understand how the numerator is considered 'a' in my question when the denominator has '4', which is not $a^2$

It doesn't matter since we can write any number $k$ as $(\sqrt{k})^2$. For example, $4 = (\sqrt{4})^2 = 2^2$.


text:
# zbMATH — the first resource for mathematics

Superintegrability and time-dependent integrals. (English) Zbl 07144745
Summary: While looking for additional integrals of motion of several minimally superintegrable systems in static electric and magnetic fields, we have realized that in some cases Lie point symmetries of Euler-Lagrange equations imply existence of explicitly time-dependent integrals of motion through Noether’s theorem. These integrals can be combined to get an additional time-independent integral for some values of the parameters of the considered systems, thus implying maximal superintegrability. Even for values of the parameters for which the systems don’t exhibit maximal superintegrability in the usual sense they allow a completely algebraic determination of the trajectories (including their time dependence).
##### MSC:
 37J15 Symmetries, invariants, invariant manifolds, momentum maps, reduction (MSC2010) 37J35 Completely integrable finite-dimensional Hamiltonian systems, integration methods, integrability tests
Full Text:
##### References:
 [1] Crampin, M., Constants of the motion in Lagrangian mechanics, Internat. J. Theoret. Phys. 16 (10) (1977), 741-754 [2] Gubbiotti, G.; Nucci, M. C., Are all classical superintegrable systems in two-dimensional space linearizable?, J. Math. Phys. 58 (1) (2017), 14 pp., 012902 [3] Jovanović, B., Symmetries of line bundles and Noether theorem for time-dependent nonholonomic systems, J. Geom. Mech. 10 (2) (2018), 173-187 [4] Lie, S., Theorie der Transformationsgruppen, Teil I-III, Leipzig: Teubner, 1888, 1890, 1893 [5] López, C.; Martínez, E.; Rañada, M. F., Dynamical symmetries, non-Cartan symmetries and superintegrability of the $$n$$-dimensional harmonic oscillator, J. Phys. A 32 (7) (1999), 1241-1249 [6] Marchesiello, A.; Šnobl, L., Superintegrable 3D systems in a magnetic field corresponding to Cartesian separation of variables, J. Phys. A 50 (24) (2017), 24 pp., 245202 [7] Marchesiello, A.; Šnobl, L., An infinite family of maximally superintegrable systems in a magnetic field with higher order integrals, SIGMA Symmetry Integrability Geom. Methods Appl. 14 (092) (2018), 11 pp [8] Mariwalla, K. H., A complete set of integrals in nonrelativistic mechanics, J. Phys. A 13 (9) (1980), 289-293 [9] Nucci, M. C.; Leach, P. G.L., The harmony in the Kepler and related problems, J. Math. Phys. 42 (2) (2001), 746-764 [10] Olver, P. J., Applications of Lie groups to differential equations, Graduate Texts in Mathematics, vol. 107, Springer-Verlag, New York, 1993, second edition [11] Prince, G., Toward a classification of dynamical symmetries in Lagrangian systems, Proceedings of the IUTAM-ISIMM symposium on modern developments in analytical mechanics, Vol. II (Torino, 1982, vol. 117, 1983, pp. 687-691 [12] Sarlet, W.; Cantrijn, F., Generalizations of Noether’s theorem in classical mechanics, SIAM Rev. 23 (4) (1981), 467-494 [13] Sarlet, W.; Cantrijn, F., Higher-order Noether symmetries and constants of the motion, J. Phys. A 14 (2) (1981), 479-492
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