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# Find a segment dividing a triangle in equal areas.

In Euclid's geometry context, I have the following problem: Let ABC a triangle and P laying in AB. We need to find a point Q in AC or BC such that the triangle APQ has the half of the area of ABC. Let M be the medium point of AB. Construct a parallel segment to PC passing by M, and call Q it's intersection with AC (or BC). Prove that Q is the point we find.

-

The median $CM$ divides $\triangle ABC$ into two triangles of equal area. If $P=M$ we are finished. We can suppose without loss of generality that $P$ lies between $A$ and $M$.
Note that the area of $\triangle CMB$ is the area of $\triangle MBQ$ plus the area of triangle $MQC$.
But the area of $\triangle MQC$ is the same as the area of $\triangle MQP$, since they have the same base $MQ$, and by parallelism the same height.
Thus the area of $\triangle PBQ$ is the same as the area of $\triangle CMB$, that is, half the area of $\triangle ABC$. This completes the proof.


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# Calculus

These are all my mathematics notes related to Calculus. More articles that are more complicated will come in the future, but I want to finish all the easy stuff.

## Prerequisites

Knowledge of functions and operations between functions such as composition, addition, subtraction, multiplication, etc... are needed in order to understand most of this. Knowing the shape of a graph for polynomials, exponential, and trigonometric functions is also very helpful. Knowing some trigonometry is essential for stuff later on.

format: plain


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Question

In NPN transistors, ${{10}^{10}}$ electrons enter the emitter region in ${{10}^{-6}}s$. If 2% electrons are lost in base region then collector current and current amplification factor (β) respectively are \begin{align} & A.1.57mA,49 \\ & B.1.92mA,70 \\ & C.2mA,25 \\ & D.2.25mA,100 \\ \end{align}

Verified
130.5k+ views
Hint: First of all find out the emitter current. After that find out the collector current using the formula
${{I}_{c}}=\alpha {{I}_{e}}$
$\alpha$ is given by how much current from the emitter reaches the collector. So using this $\alpha$ amplification factor can be found out.

Complete step by step answer:
Current amplification factor in a bi junction transistor is described as the ratio of output current to its input current. In the common base configuration, current amplification factor is given as the ratio of collector current to the emitter current. The low current travels from the voltage source into the transistor base. The current at the base will turn on the transistor. Then the current is amplified and travels from the emitter of the transistor to the collector. This is how current amplification takes place in a bi junction transistor. The amplifier is a very complex circuit which is using properties of a transistor.
Here let us first write what all are given in the question.
Number of electrons entering emitter,
${{n}_{e}}={{10}^{10}}$
Time taken=${{10}^{-6}}s$
Therefore the current in the emitter is
${{I}_{e}}=\dfrac{{{q}_{e}}\times {{n}_{e}}}{t}$
${{I}_{e}}=\dfrac{{{10}^{10}}\times 1.6\times {{10}^{-19}}}{{{10}^{-6}}}$
$=1.6mA$
Since only 2% current is lost, therefore 98% will reach the collector
So
$\alpha =0.98$
And also we know that
${{I}_{c}}=\alpha {{I}_{e}}$
Therefore the collector current will be,
\begin{align} & =\alpha \times {{I}_{e}} \\ & =0.98\times 1.6 \\ & {{I}_{c}}\approx 1.57mA \\ \end{align}
Also the current amplification factor will be,
$\beta =\dfrac{\alpha }{1-\alpha }=49$

Hence the correct answer is given by the option A.

Note:
MOSFETs are controlled by voltage not on their own. It is used to amplify current using a MOSFET by passing the input current through a resistor there by converting the current signal to a voltage. This signal voltage is controlling the MOSFET, which gives an output signal current.


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# Exam C Practice Problem 17 – Estimating Claim Frequency

Both Problems 17-A and 17-B use the following information.

An insurance portfolio consists of independent risks.

For each risk in this portfolio, the number of claims in a year has a Poisson distribution with mean $\theta$. The parameter $\theta$ follows a Gamma distribution.

A risk is randomly selected from this portfolio. Prior to obtaining any claim experience, the number of claims in a year for this risk has a distribution with mean 0.6 and variance 0.72.

After observing this risk for one year, insurance company records indicate that there are 2 claims for this risk.

___________________________________________________________________________________

Problem 17-A

After knowing the insurance company records, what is the expected number of claims per year for this risk?

$\text{ }$

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.60$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.83$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0.91$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 1.25$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.00$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

___________________________________________________________________________________

Problem 17-B

After knowing the insurance company records, what is the variance of the number of claims per year for this risk?

$\text{ }$

$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{26}{36}$

$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{32}{36}$

$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{35}{36}$

$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{42}{36}$

$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{45}{36}$

___________________________________________________________________________________

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

$\text{ }$

___________________________________________________________________________________

___________________________________________________________________________________

$\copyright \ 2013 \ \ \text{Dan Ma}$


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# Quantum mechanics/Atomic spectra

 Subject classification: this is a chemistry resource.
 Type classification: this is a lesson resource.

## Combination of two orbital angular momenta (different orbitals)

In classical physics, the total angular momentum is the (vector) sum of the angular momenta of the particles present in the system; the resulting total depends on the orientation of the components. In quantum mechanics, if a system has two particles with angular momentum L1 and L2 and angular momentum quantum numbers l1 and l2, we can define a total angular momentum L=L1+L2. L behaves as any angular momentum and the eigenvalues of ${\displaystyle {\hat {\mathbf {L} }}^{2}}$  are ħ²L(L+1), i.e.

${\displaystyle {\hat {\mathbf {L} }}^{2}\psi =\hbar ^{2}L(L+1)\psi }$  (Eq. 1)

L is the total angular momentum quantum number and can take values |l1 - l2|, |l1 - l2|+1, ..., l1 + l2.

Example 1. Consider an atom with one electron in orbital 2p (l1=1) and one electron in orbital 3d (l2=2). The total angular momentum quantum number for this system can be L=1,2,3. States with different total angular momentum have different energy so that there are (at least) 3 different states deriving from a configuration 2p3d.

Example 2. Consider the excited He atom with configuration 1s2p. The allowed quantum number for this configuration is just L=1.

The component along the z axis of the angular momentum is ${\displaystyle {\hat {L}}_{z}}$  and its eigenvalues can take the values ħmL with mL = -L, -L+1, ..., L.

Wavefunctions with L=0, 1, 2, 3,... are indicated with the capital letters S, P, D, F,...

## Combination of two spin angular momenta (different orbitals)

In a system with two electrons in different orbitals, you can define that total spin as the sum of the individual spins: ${\displaystyle {\hat {\mathbf {S} }}={\hat {\mathbf {S} }}_{1}+{\hat {\mathbf {S} }}_{2}}$ . The total spin behaves as any angular momentum, and the eigenvalues of ${\displaystyle {\hat {\mathbf {S} }}^{2}}$  are ħ²S(S+1), i.e.

${\displaystyle {\hat {\mathbf {S} }}^{2}\psi =\hbar ^{2}S(S+1)\psi }$  (Eq. 2)

S is the total spin quantum number and takes the value between |s1 - s2| and s1 + s2, i.e. between 0 and 1 in the case of 2 electrons, because the individual spin quantum numbers are each equal to ½.

The component along the z axis of the total spin is ${\displaystyle {\hat {S}}_{z}}$  and its eigenvalues can take the values ħmS with mS = -S, -S+1, ..., S. There are in total 2S+1 eigenvalues. 2S+1 is called spin multiplicity. States with multiplicity 1, 2, 3, 4,... are called singlet, doublet, triple, quartet, etc.

In the two examples in the section above, the total spin can take the value of 0 and 1 and so the configuration gives rise to a singlet and a triplet. The energy of a wavefunction depends on the spin multiplicity (or S).

## Spectroscopic term

The electronic configuration is not enough to describe the wavefunction, because for each configuration there are different ways to combine the orbital and spin angular momenta of the electrons (different combinations have different energies). The labels of a wavefunction which indicate the total angular momentum (S, P, D,...) and the spin multiplicity (singlet, doublet, triplet) are called spectroscopic terms.

The spectroscopic terms originating from the configuration 2p3d of the example 1 above are 1P, 1D, 1F, 3P, 3D, 3F.

The spectroscopic terms originating from the configuration 1s2p of the excited helium atom are 1P and 3P.

You can neglect the electrons in filled sub-shells, so that the spectroscopic term of the ground state of lithium 1s22s is 2S. The spectroscopic terms of the excited configuration of beryllium 1s22s2p are 1P and 3P.

## Electrons in equivalent orbitals

In sections 1 and 2 we considered electrons in different orbitals. If they are in the same orbital the calculation of the allowed spectroscopic terms is slightly more laborious. Consider atomic carbon with 2 electrons in 2p. You have to arrange the electrons in all possible ways allowed by the Pauli principle and then identify by inspection the spectroscopic terms on the basis of the values of mS and mL The spectroscopic terms of atomic carbon are 1D, 3P, 1S.

## Hund's empirical rule

It is useful to know what is the lowest energy spectroscopic term originating from a given configuration. Hund's rule states that:

 The level with the largest multiplicity has the lowest energy. For a given multiplicity, the term with the largest L has the lowest energy.

The ground state of atomic carbon is 3P (which has the largest multiplicity). This is why in elementary textbooks, electrons in partially filled shells have parallel spin.

## The last complication: spin-orbit coupling

Everything said so far implied that the spin does not appear in the Hamiltonian and its only effect is to change the structure of the spatial wavefunction. It is true that this is the main role of spin, but there is a very small term in the Hamiltonian that depends on the spin and this is called the spin-orbit coupling (which will not be written explicitly). Spin orbit coupling is important only for heavy atoms or when very accurate energy levels are needed (spectroscopy). The effect of spin orbit coupling is to split the spectroscopic terms in energy levels. The number of energy levels originating from a spectroscopic term is given by the possible values of the total angular momentum J = L + S (angular plus spin) with eigenfunction

${\displaystyle {\hat {\mathbf {J} }}^{2}\psi =\hbar ^{2}J(J+1)\psi }$

The quantum number J can take the values |L - S|, |L - S|+1, ..., L + S.

The spectroscopic term 3P (L=1, S=1) can have J=0,1,2 and therefore 3 finely spaced energy levels indicated as 3P0, 3P1, 3P2.

## Summary

Lessons 12-13 give an example of how quantum mechanics is used when the problem is too difficult to be solved exactly. You have to proceed by successive approximations, each time more refined. The aim is to understand the experiment (in this case, how many energy levels are present in an atom) and not to predict the exact value of the energy levels.

1. The first level of approximation is the independent electron approximation. Each electron is in an orbital and the energy is determined only by the orbital occupation or electronic configuration.
2. The second level of approximation is considering that the energy is different if the total orbital angular momentum and spin angular momentum are different. We differentiate the energy in spectroscopic terms depending on the values of L and S.
3. The most accurate level takes into account the spin orbit coupling. The energy levels are identified by the total angular momentum J.

Exercise
1. Find the spectroscopic terms originating from the following configurations (two electrons in two non-equivalent orbitals):
• 1s2s
• 1s3s
• 2p3p
• 2s3d
2. Find the spectroscopic term originating from the ground states configuration of the sodium atom and the boron atom.
3. Find the spin-orbit energy levels of hydrogen with electron in 2p and 3d.

Next: Lesson 13 - Introduction to molecular orbital theory


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# How to find the limit of cos(2x)/x as x approaches zero [duplicate]

Basicly the title. I know it's easy but I can't figure it out.  Not a duplicate has a 2x instead of x.

• what is $\cos 0?$ – Will Jagy Nov 15 '15 at 21:51
• It is when $x\to 0^+$? (i.e., only by positive values?) As asked, the limit does not exist. – Clement C. Nov 15 '15 at 21:54
• "Basically the title" is not a question. Please make your question's body self contained. If your entire question fits into the title, it is not a good question. – Asaf Karagila Nov 15 '15 at 22:39
• If its a short question then it can fit into the title. Doesn't mean its a bad question. – Earl Smith Nov 16 '15 at 23:53


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# Math Help - laplace transforms to solve IVPs

1. ## laplace transforms to solve IVPs

Use Laplace Transforms to solve the following Initial Value Problems:

(i)

$\dot{y_1} = y_{2}$
$\dot{y_2} = -y_{1} + e^t$

with $y_{1}(0) = 0, y_{2}(0) = 0$

(ii)

$\dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $y(0) = 0$, $\dot{y}(0) = 0$ and $\ddot{y}(0) = 0$

2. Originally Posted by wik_chick88
Use Laplace Transforms to solve the following Initial Value Problems:

(i)

$\dot{y_1} = y_{2}$
$\dot{y_2} = -y_{1} + e^t$

with $y_{1}(0) = 0, y_{2}(0) = 0$
There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

$\ddot{y}_1=\dot{y}_2=-y_{1} + e^t$

Then solving the new equation for \dot y:

$\ddot{y}_1=-y_{1} + e^t$

Laplace transforming:

$s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1$

Applying the initial conditions $y_1(0)=0$, $\dot{y}_1=0$ this becomes:

$Y_1(s)=-\frac{1}{(s-1)^2(s+1)}$

which you invert by the use of partial fractions and a table of inverse Laplace transforms.

The second method involves writing this as a vector ODE:

$\dot{ {\bold{y}} }={\bold{A}}{\bold{y}} + (0,e^t)^T$

where:

${\bold{A}}=\left[\begin{array}{rr}0&1\\-1&0\end{array}\right]$

and: ${\bold{y}}=(y_1,y_2)^T$

Now take the Laplace transform:

$s{ {\bold{Y}} }(s)-{\bold{y}}(0)={\bold{A}}{\bold{Y}}(s) + (0,-1/(s-1))^T$

Then (as : ${\bold{y}}(0)=(0,0)^T$)

${\bold{Y}}(s)=({\bold{A}}-s{\bold{I}})^{-1}(0,1/(s-1))^T$

Now with some matrix algebra you can find $Y_1(s)$ and $Y_2(s)$ and proceed as before.

CB

3. Originally Posted by wik_chick88

(ii)

$\dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $y(0) = 0$, $\dot{y}(0) = 0$ and $\ddot{y}(0) = 0$
Here you have:

$\dddot{y} =u(t-3) t$

where u(.) is the Heavyside unit step function, and:

$\mathcal{L}\dddot{y}=s^3Y(s)-s^2 \ddot{y}(0)-s\dot{y}(0)-y(0)$.

The rest is routine.

CB

4. ... use Laplace Transforms to solve the following Initial Value Problems:

$\dddot{y} = \begin{cases} t & \ t < 3 \\ 0 & \ t \geq 3 \end{cases}$ with $y(0) = 0$, $\dot{y}(0) = 0$ and $\ddot{y}(0) = 0$

If we intend to apply the LT for resolving a DE, then we have to impose that $t \ge 0$. So the proposed third order differrential equation must be written as…

$\dddot{y} = \begin{cases} t & \ 0 \le t < 3 \\ 0 & \ t \geq 3 \end{cases} =$ $t\cdot u(t) - (3+t) \cdot u(t-3)$

... where $u(*)$ is the Heavyside unit step function...

Kind regards

$\chi$ $\sigma$

5. Originally Posted by chisigma
If we intend to apply the LT for resolving a DE, then we have to impose that $t \ge 0$. So the proposed third order differrential equation must be written as…
No we don't, the default LT in use is the one sided LT and so what the function does at negative arguments is irrelevant to the forward transform.

But the inverse transform is only valid for $t\ge 0$, since any function that agree on the non-negative half real axis have the same (one-sided) LT.

So the solution obtained using the LT to a IVP is only valid for non-negative times.

CB

6. Originally Posted by CaptainBlack
There are two ways (at least) of doing this the first involves differentiating one of the equation and then substituting the other into the new equation:

$\ddot{y}_1=\dot{y}_2=-y_{1} + e^t$

Then solving the new equation for \dot y:

$\ddot{y}_1=-y_{1} + e^t$

Laplace transforming:

$s^2Y_1(s)-s y_1(0)-\dot{y}_1(0)=Y_1(s)-\frac{1}{s-1}, \ \ \ \ s>1$

Applying the initial conditions $y_1(0)=0$, $\dot{y}_1=0$ this becomes:

$Y_1(s)=-\frac{1}{(s-1)^2(s+1)}$

which you invert by the use of partial fractions and a table of inverse Laplace transforms.
hi thanks so much! i do have a question though...
when you laplace transformed it, why did the $-y_{1} + e^t$ get laplace transformed to $Y_{1}(s) - \frac{1}{s-1}$ wouldn't it be $-Y_{1}(s) + \frac{1}{s-1}$??


text:
almost 4 years ago

系統：

Mac OS X Yosemite 10.10.1
Ruby 2.1.5
Rails 4.1.8

相關連結

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原因

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Yes, Slim is speedy! Slim was developed right from the start with performance in mind.
Benchmarks are done for every commit at http://travis-ci.org/slim-template/slim.
Don't trust the numbers? That's as it should be. Please try the benchmark rake task yourself!

However in our opionion you should use Slim because of its features and syntax. We just ensure that Slim doesn't have a negative impact on the performance of your application.


text:
RUS  ENG JOURNALS   PEOPLE   ORGANISATIONS   CONFERENCES   SEMINARS   VIDEO LIBRARY   PACKAGE AMSBIB
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 Mat. Sb.: Year: Volume: Issue: Page: Find

 Mat. Sb. (N.S.), 1984, Volume 124(166), Number 3(7), Pages 335–352 (Mi msb2056)

Estimates of the rate of convergence for certain minimization algorithms for strongly convex functions

P. A. Vitushkin

Abstract: The convergence of certain minimization algorithms for strongly convex functions is investigated. Namely, convergence with the rate of a geometric progression is proved for the method of coordinatewise descent and one variant of the method of feasible directions. An estimate of the ratio of the progression in dependence on the number of variables is given for the method of coordinatewise descent.
Bibliography: 3 titles.

Full text: PDF file (773 kB)
References: PDF file   HTML file

English version:
Mathematics of the USSR-Sbornik, 1985, 52:2, 331–346

Bibliographic databases:

UDC: 519.615.7
MSC: Primary 26B25; Secondary 41A60, 49D07, 49D10, 65D15

Citation: P. A. Vitushkin, “Estimates of the rate of convergence for certain minimization algorithms for strongly convex functions”, Mat. Sb. (N.S.), 124(166):3(7) (1984), 335–352; Math. USSR-Sb., 52:2 (1985), 331–346

Citation in format AMSBIB
\Bibitem{Vit84} \by P.~A.~Vitushkin \paper Estimates of the rate of convergence for certain minimization algorithms for strongly convex functions \jour Mat. Sb. (N.S.) \yr 1984 \vol 124(166) \issue 3(7) \pages 335--352 \mathnet{http://mi.mathnet.ru/msb2056} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=752225} \zmath{https://zbmath.org/?q=an:0583.90079|0562.90071} \transl \jour Math. USSR-Sb. \yr 1985 \vol 52 \issue 2 \pages 331--346 \crossref{https://doi.org/10.1070/SM1985v052n02ABEH002894}


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# Wydawnictwa

## Wydawnictwa

• ### Preprinty z --wszystkie-- 2020 2019 2017 2016 2015 2013 2012 2011 2010 2009 2008 2007 2006 2005 2004 2003 2002 2001

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757 On regular periodic solutions to the Navier-Stokes equations. Case B, March 2020
756 Global regular periodic solutions to equations of weakly compressible barotropic fluid motions. Case C, March 2020
755 Global regular periodic solutions to equations of weakly compressible barotropic fluid motions. Case B, March 2020
754 CHARACTERIZATION OF RAPIDLY DECREASING DISTRIBUTIONS ON R^n BY CONVOLUTIONS WITH TEST FUNCTIONS, March 2019
753 TOPOLOGIES IN THE SET OF RAPIDLY DECREASING DISTRIBUTIONS, February 2019
752 ON RAPIDLY DECREASING DISTRIBUTIONS, May 2017
751 ON THE EXCHANGE BETWEEN CONVOLUTION AND MULTIPLICATION VIA THE FOURIER TRANSFORMATION, February 2017
750 ON THE LOCALLY CONVEX SPACE OF RAPIDLY DECREASING DISTRIBUTIONS, June 2016
749 BIFURCATION SETS OF FAMILIES OF REFLEXIONS ON SURFACES IN R 3, September 2015
748 Poisson algebras and singular symplectic forms associated to Ak type singularities., July 2015
747 Singularities of affine equidistants: Extrinsic geometry of surfaces in 4-space. July 2015, July 2015
746 Tests for first-order stochastic dominance, December 2013
745 Simple examples of affine manifolds with infinitely many exotic models, July 2013
744 Singularities of solvable Hamiltonian mappings, April 2013
743 The pullbacks of principal coactions, March 2013
742 Discrete ill-posed problems: two-parameter regularization, November 2012
741 Combinatorial cycles of tetrahedral chains, November 2012
740 "A note on the paper ""Stability and stabilizability of infinite dimensional systems"" by A. J. Pritchard and J. Zabczyk, SIAM Review, 23 (1981), 25-52", October 2012
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738 The K-theory of the triple-Toeplitz deformation of the complex projective plane, July 2012
737 The cocycle condition for multi-pullbacks of algebras, July 2012
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735 On congruences for the sums $\sum_{i=1}^{[n//r]}{\chi_n(i)\over i^k}$ of E. Lehmer's type, July 2012
734 An easy proof of Gowers' FINk theorem, December 2011
733 The K-theory of Heegaard quantum lens spaces, October 2011
732 Piccard-Vessiot theory for real fields, October 2011
731 Regularization methods for mathematical model of laser beams, October 2011
730 On convergence of an alternating method for a Cauchy problem for the Helmholtz equation, June 2011
729 The Petrovski? condition!!and!!rapidly decreasing distritutions, January 2011
728 Vector fields with distributions and invariants of ODEs, December 2010
727 Finding two almost homoclinic solutions for a class of second order perturbed Hamiltonian systems, December 2010
726 Estimating AR(1) with Entropy Loss Function, August 2010
725 On the Tarski-Seidenberg Theorem, June 2010
724 Computing Gromov-Witten Invariants of some Fano Varieties, May 2010
723 Division with Remainder in Algebras with Valuation, May 2010
722 On Splitting Polynomials in Noncommutative Extensions, May 2010
721 Maximal Commutative Subalgebras, Poisson Geometry and Hochschild Homology, May 2010
720 Distributive Lattices and Cohomology, May 2010
719 On Actions of of Compact Quantum Group, May 2010
718 Examples of Non-compact Quantum Group Actions, May 2010
717 Proceedings of the 8-th Research Seminar of Difference and Differential Operators, 7-14 July 2007, April 2010
716 On the Russell Problem, March 2010
715 On Constructing Optimal Families of Pairing-friendly Elliptic Curves, March 2010
714 Regularization of linear operators, Hilbert space methods, March 2010
713 Shifts and Periodicity in Algebraic Analysis, February 2010
712 Convolution Semigroups of Rapidly Decreasing Distributions, February 2010
711 Finite Closed Coverings!!of Compact Quantum Spaces, February 2010
710 Tridendriform Algebras and Combinatorial Hopf Algebras, December 2009
709 Generic Singularities of Symplectic and Quasi-symplectic Immersions, December 2009
708 Secret Sharing Matrices, September 2009
707 Admissible Tracks in Shamir's Scheme, September 2009
706 Conditional stability estimates and regularization with applications to Cauchy problems for the Helmholtz equation, May 2009
705 Remarks of uniform convergence of random variables and statistics, May 2009
704 Symplectic invariants of parametric singularities, May 2009
703 Regularity of Solutions to Nonlinear Nondiagonal Parabolic System, February 2009
702 On a Random Number of Disorders, February 2009
701 Cauchy's problem for system of PDE with constant coefficients and semigroups of operators, January 2009
700 Optimal stopping of a risk process with disruption and interest rates, January 2009
699 Completeness of bond market driven by Lévy process, December 2008
698 F-Doubly Stochastic Markov Chains - A New Class of Processes for Modeling Credit Rating , December 2008
697 Defaultable bonds with an infinite number of Lévy factors, December 2008
696 Optimal detection of homogeneous segment of observations in stochastic sequence, December 2008
695 Homoclinic solutions for a class of autonomous second order Hamiltonian systems with a superquadratic potential, November 2008
694 Regularity of Ornstein-Uhlenbeck processes driven by a Lévy white noise, November 2008
693 Fourier-like methods for equations with separable variables, October 2008
692 Structual properties of semilinear SPDEs driven by cylindrical stable processes, October 2008
691 Nonlinear separable equations in linear spaces and commutative Leibniz algebras, September 2008
690 On incompleteness of bond markets with infinite number of random factors, September 2008
689 A note on small branching fluctuation limits of catalytic superprocesses with immigration, September 2008
688 Möj mistrz i promotor Witold Pogorzelski (1895-1963), August 2008
687 Conley index in Hilbert spaces and the Leray-Schauder degree, March 2008
686 Equivalence of measures corresponding to the Hilbert space valued Lévy processes, December 2007
685 Positivity of crredit spreads, August 2007
684 Direct Numerical Simulation of turbulence using divergence-free wavelets, June 2007
683 How to define convex functions on differentiable manifolds, June 2007
682 Approximations for solutions of Lévy-type stochastic differential equations, May 2007
681 Contact Equivalence and characteristic Cones of Ordinary Differential Equations, May 2007
680 Orthogonal Helmholtz decomposition in arbitrary dimension using divergence-free and curl-free wavelets, May 2007
679 On the Cancellation Problem, April 2007
678 Diffeomorphisms that are Symplectomorphisms, March 2007
677 Heath-Jarrow-Morton-Musiela Equation of Bond Market, February 2007
676 Shannon wavelet approximations of linear differential operators, January 2007
675 Characteristic vector fields of generic distributions of corank 2, December 2006
674 A note on completeness of bond market in infinite dimension, December 2006
673 Exponential moments for HJM models with jumps, November 2006
672 Symplectic Singularities of Varieties: The Method of Algebraic Restrictions, October 2006
671 Manifolds with a Unique Embedding, September 2006
670 On Thom polynomials for A4 (-) via Schur functions, September 2006
669 Invariant measures for forward rate HJM model with Lévy noise, September 2006
668 Uniform asymptotic normality for the Bernoulli scheme, June 2006
667 Invariant measures for a class of stochastic evolution equations, March 2006
666 On Fourier transforms of distribution semigroups, March 2006
665 Data driven score tests of fit for a semiparametric homoscedastic linear regression model, January 2006
664 The complex symplectic moduli spaces of uni-model parametric plane curve singularities, January 2006
662 Wavelet moment method for Cauchy problem fo the Helmhotz equation, December 2005
661 Gamma-minimax prediction in exponential families with quadratic variance function, November 2005
660 Approximate solution of a Cauchy problem for the Helmholtz equation, September 2005
659 Symplectic packing of a torus, July 2005
658 Lévy modelling of defaultable bonds, July 2005
657 L-statistics as nonparametric Quantile Estimators, June 2005
656 Estimating quantiles with Linex Loss Function. Approximation to VaR Estimation, April 2005
655 On a pairing between super Lie and periodic homology, March 2005
654 Differential operators of k[x]/(xn+1), December 2004
653 Optimal quantile estimators. Small sample approach, November 2004
652 The K-Theiry of Heegaard-Type Quantum 3-Spheres, September 2004
651 HJM condition for models with Levy noise, September 2004
650 Twistor spaces of complex grassmannians, April 2004
649 Portfolio diversification Markovian prices, April 2004
648 Historical remarks on bounded sets, April 2004
647 Stable anti-Yetter-Drinfeld modules, March 2004
646 A note on Stochastic Burgers' System of Equations, December 2003
645 Relative Poincare lemma, contractibility, quasi-homogeneity and vector fields tangent to a singular variety, October 2003
644 Representations of the convolution algebra D(R) and distrubution groups, July 2003
643 Hopf-cyclic homology and cohomology with coefficients, July 2003
642 The Chern-Galois character, July 2003
641 On the Łojasiewicz exponent and effective nullstellensatz, June 2003
640 Unbiased estimation of survival probability for censored data with known censoring times, April 2003
639 Chern numbers for two families of noncommutative Hopf fibrations, March 2003
638 Effective WLLN, SLLN and CLN in statistical models, March 2003
637 Nonanalycity in time of solutions of KdV equation, February 2003
636 On new geometric conditions for some weakly lower semicontinous functionals with application to the rank-one conjecture of Morrey (to apperar in Proc. Roy. Soc. Edingburgh), January 2003
635 Between the classical theorem of Young and Convergence Teorem in Set-Valued Analysis, January 2003
634 Graph C*-algebras and Z2-quotients of Quantum Spheres, November 2002
633 Haagerup inequality and contractivity properties of Ornstein-Uhlenbeck semigroup for general commutation relations, October 2002
632 Property Gamma for von Neumann algebras connected with general commutation relations, October 2002
631 Hypoeliptic diffusions and cyclic cohomology, October 2002
630 Regularizatin of discrete ill-posed problems, September 2002
629 Abstract convex analysis in metric spaces, August 2002
628 Frequency moments inverse problem and maximum (Shannon vs. Renyi-Tsallis) entropy, August 2002
627 Divisibility properties of generalized Vandermonde and Cauchy determinants,, August 2002
626 Constructing median-unbiased estimators in one-parameter families of distributions via optimal nonparametric estimation and stochastic ordering,, July 2002
625 Generalized Sturm separation theorem,, April 2002
624 Locally trivial quantum Hopf fibration,, March 2002
623 Joint replenishment multiproduct inventory problem with continous production and discrete demands, February 2002
622 Survey on the Orthogonal Franklin System, February 2002
kis The Wider spaces, representations of the convolution algebra L1(R+) and one parameter semigroups of operators,, January 2001
621b On ramification values of a polynomial mapping, December 2001
621a On generalized critical values of a polynomial mapping, December 2001
620 On an inequality for a free boundary problem for equations of a viscous compressible heat conducting capilary fluid, September 2001
619 The most stable estimator of location under integrable contaminants, September 2001
618 On Tableau consistency in weak theories, June 2001
617 Pseudoresolvents and distributions semigroups, July 2001
616 On global special solutions for Navier-Stokes equations with boundary slip conditions in a cylindrical domain. Existence, June 2001
615 On some presentation of graphs, May 2001
614b On assymptotic critical values of a complex polynomial, April 2001
614a On biffurcation points of a complex polynomial, April 2001
613 Holomorphic Characterizations of Certains Classes of Distributions, January 2001
612 Almost sure and moment stability of stochastic partial differential equations, January 2001
611 Pricing Option for Markovian Models, January 2001

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text:
# [R] survival::survfit,plot.survfit

Bernhard Reinhardt bernhard.reinhardt at dlr.de
Thu Feb 26 10:03:24 CET 2009

Jeff Xu wrote:
> I am confused when trying the function survfit.
> my question is:  what does the survival curve given by plot.survfit mean?
> is it the survival curve with different covariates at different points?
> or just the baseline survival curve?
>
> for example, I run the following code and get the survival curve
>
> ####
> library(survival)
> fit<-coxph(Surv(futime,fustat)~resid.ds+rx+ecog.ps,data=ovarian)
> plot(survfit(fit,type="breslow"))
> summary(survfit(fit,type="breslow"))
> ####
>
> for the first two failure points, we have s(59|x1)=0.971, s(115|x2)=0.942
> how can we guarantee that s(59|x1) is always greater than s(115|x2)?
> since s(59|x1)=s_0(59)^exp(\beta'x1) and s(115|x2)=s_0(115)^exp(\beta'x2),
> we can manipulate covariates to make s(59|x1) < s(115|x2), right?
> do I miss anything?

In advance: I´m a beginner in survival analysis, too. But I think I can

plot(survfit(fit)) should plot the survival-function for x=0 or
equivalently beta'=0. This curve is independent of any covariates.

If you want to see the impact of residual-status=2 you could add
something like:

attach(ovarian)
ovarian_new <- data.frame(resid.ds=2,
rx=(mean(rx)),ecog.ps=mean(ecog.ps))
detach()

plot(survfit(fit), newdata=ovarian_new)

This should give you the survival-function for an average patient with
residual-status 2.

Regards

Bernhard


text:
# PETSc Krylov Subspace and nullspace

How is the nullspace correction implemented in the PETSc conjugate gradient solver? I searched the source code and documentation, but could not find references on the actual implementation.

For background, I am solving linear systems arising from a bilinear form which is elliptic in a quotient space $\mathbb{R}^n\big/\ker{\mathbf{Z}}$, where $\mathbf{Z}$ is an $m \times n$ matrix having non-trivial null-space.

I compute SVD of $\mathbf{Z}$ and feed the right singular vectors corresponding to "big" singular values as nullspace vectors to the PETSc conjugate gradient solver. It works like a charm, except when $\mathbf{Z}$ is not of academic size.

It is just a projection $1 - ZZ^T$. If you are digging around in the implementation, look at KSP_PCApplyBAorAB() and related functions.


text:
sin 3x = 3 sin x - 4 sin³ x. cos 3x = 4 cos³ x  I'm assuming the problem you are trying to solve is to prove that sin2x * sinx = cosx. I believe you wrote the problem incorrectly, as simplifying you get 2sin^2x  16 May 2019 Sin2x is 2sinxcosx from the identity so just substitute. sin2x - cosx = 2sinxcosx - cosx = cosx(2sinx - 1). Upvote • 0 Downvote. Add comment.

2*sin x*cos x-cos x=0, разложим левую часть на  26 Jul 2011 sin 2x = 2 sin x cos x. cos 2x = cos² x - sin² x = 2 cos² x - 1 = 1 - 2 sin² x. tan 2x = (2 tan x) / (1 - tan² x). sin 3x = 3 sin x - 4 sin³ x. cos 3x = 4 cos³ x  I'm assuming the problem you are trying to solve is to prove that sin2x * sinx = cosx.

Solve the trig equation for x from 0 to 2pi. Simplify (sin(2x))/(cos(x)) Apply the sine double-angle identity. Cancel the common factor of . Tap for more steps Cancel the common factor.

$$= -\cos x\int { \sin x \sin^2(x) } + \sin x\int { \cos x \sin^2(x) } .$$ You should finish evaluating the above integrals. Share. Cite. Follow edited May 1 '16 at 17:47.

i though jennifer s answer s is the best answer.. 0 0. Still have questions? Get your cos(2x) = cos 2 (x) – sin 2 (x) = 1 – 2 sin 2 (x) = 2 cos 2 (x) – 1 Half-Angle Identities The above identities can be re-stated by squaring each side and doubling all of the angle measures. Solve the trig equation for x from 0 to 2pi.
Maria skolan malmo

Summary of Double-Angles. • Sine: sin 2x = 2 sin x cos x. • Cosine  a) 2 sin2 x − 5 sinx + 2 = 0; b) sin2 2x − sin 2x = 0; c) sin2 x − sinx +6=0.

cos(2x) = cos 2 (x) - sin 2 (x) = 2 cos 2 (x) - 1 = 1 - 2 sin 2 (x).
Hitta lagfarter på nätet

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text:
# Water in a Rectangular Reservoir Having Base 80 M by 60 M I S 6.5 M Deep. - Mathematics

Water in a rectangular reservoir having base 80 m by 60 m i s 6.5 m deep. In what time can the water be emptied by a pipe ôf which the cross-section is a square of side 20 cm, if the water runs through the pipe at the rate of 15 km/hr.

#### Solution

Given that,
Flow of water = 15km / hr
=15000m/hr.
Volume of water coming out of the pipe in one hour
20/100xx20/100xx± 5000=600m^3

"Volume of the tank"= 80xx60xx6.5

=31200m^3
∴Time taken to empty the tank
="Volume of tank"/"volume of water coming out of the pipe in one hour"
=31200m^3

=52  hours.

Concept: Volume of a Cuboid
Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 18 Surface Areas and Volume of a Cuboid and Cube
Exercise 18.2 | Q 24 | Page 31

Share


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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 7.8: Using Trigonometric Ratios

Difficulty Level: At Grade Created by: CK-12

## Introduction

The Ramp Dilemma (Part 2)

Here is the problem from the last lesson. Now that you know the math needed, think about this problem and the sine ratio.

Mr. Watson’s homeroom decided to do a community service repairing the ramp outside the shed. The fresh coat of paint shone in the bright sunlight and Mr. Watson walked across the grass with all of his students to look at the ramp outside the door of the shed.

“Has that always been there?” asked Dan.

“No, in fact it was just brought out here yesterday,” Mr. Watson explained.

“Well if it’s brand new, then why do we need to fix it?” Emily asked.

“Because it doesn’t fit neatly under the door.”

Sure enough, the students looked and could see that the back of the ramp was too tall and the students would need to fix the back to make it fit beneath the doorway. Fixing this ramp would help everyone because it would make it easy to push or pull the cart with the athletic equipment out onto the field and since the equipment was used by local kid’s teams as well as the school, this was a definite way to give back to the community.

“What do we need to do?” Dan asked.

They looked at the ramp. Mr. Watson drew the following sketch.

“Well that’s not very helpful,” Dan commented.

“Sure it is,” Emily said.

This lesson is about using trigonometric ratios. By the end of the lesson, you will know how to figure out the height of the ramp.

What You Will Learn

By the end of this lesson, you will be able to complete the following skills.

• Use calculator or tables to find sine, cosine and tangent ratios for each of the acute angles of right triangles with given side lengths.
• Determine and use the sine ratio to find the length of an opposite side of a right triangle, given the angle and the hypotenuse.
• Determine and use the cosine ratio to find the length of an adjacent side of a right triangle, given the angle and hypotenuse.
• Determine and use the tangent ratio to find the height of a right triangle, given the angle and the base.
• Use calculators or tables to solve real – world problems involving right triangles and trigonometric ratios.

Teaching Time

I. Use Calculator or Tables to Find Sine, Cosine and Tangent Ratios for Each of the Acute Angles of Right Triangles with Given Side Lengths

Now that you have a basic understanding of trigonometric ratios and how they are formed, it is time to learn more about how they can be used. You now know that a trigonometric ratio for a specific angle will remain constant no matter how large or small the triangle is. The idea is that the sides will always be in proportion to each other. So, if you know the measure of an angle (and can therefore identify the value of a trigonometric ratio) and the value of one side, you can use trigonometry to calculate the lengths of other sides. The trick is to use good algebra technique, and make sure that every time you set up a ratio, you are putting the values and variables in the correct places.

You can find trigonometric ratios by using your calculator.

You understand trigonometric ratios and have had a chance to practice reading specific values out of a table. You can find the ratio for any trigonometric value using your calculator. Take a moment to locate the buttons for sine, cosine, and tangent on the calculator. Keep in mind that usually, sine is abbreviated as sin, cosine is usually abbreviated as cos, and tangent is usually abbreviated as tan. Press the key of the ratio you want to find, and enter the angle in question. If you hit enter, or calculate, the calculator will show you the value of that specific ratio.

Example

Use your calculator to find the values of the following ratios.

\begin{align*}\text{sine} \ 47^{\circ}\end{align*}

\begin{align*}\text{cosine} \ 23^{\circ}\end{align*}

\begin{align*}\text{tangent} \ 82^{\circ}\end{align*}

You can find the values for each ratio using your calculator. When dealing with large decimals values, it is usually best to round the numbers to the nearest thousandth. It gives you a reasonably accurate value without being too long of a number to work with.

The sine of \begin{align*}47^{\circ}\end{align*} is 0.73135370161917..., or about 0.731.

The cosine of \begin{align*}23^{\circ}\end{align*} is 0.92050485345244..., or about 0.921.

The tangent of \begin{align*}82^{\circ}\end{align*} is 7.11536972238419..., or about 7.115.

Notice that because we are using a calculator, that we don’t need to know the side lengths. The calculator figures out the trigonometric ratio based on proportions and the given angle. Figuring out sine, cosine and tangents using a calculator is as easy as pressing a button!

II. Determine and Use the Sine Ratio to Find the Length of an Opposite Side of a Right Triangle, Given the Angle and the Hypotenuse

You have already learned how to write and determine the sine ratio, now we are going to use what we have learned in problem solving.

As you recall, the ratio of sine is \begin{align*}\frac{opposite}{hypotenuse}\end{align*}. If you know the sine value of the angle in question, and the length of the hypotenuse, you can find the measure of the opposite side. Look at the algebraic manipulation below.

\begin{align*}\sin e \angle X&=\frac{opposite}{hypotenuse}\\ \sin e \angle X \times hypotenuse&=\frac{opposite}{hypotenuse} \times hypotenuse\\ \sin e \angle X \times hypotenuse&=opposite\end{align*}

If you multiply the sine of any angle \begin{align*}X\end{align*} and the length of the hypotenuse, the result is the length of the opposite side.

Write this statement down in your notebook. Be sure that you write that it is connected to the sine ratio.

Now let’s apply this to an example.

Example

What is the length of side \begin{align*}AC\end{align*} in the triangle below?

Use the following equation to find the length of the side opposite angle \begin{align*}B\end{align*}. Notice that we are going to multiply the sine of Angle \begin{align*}B\end{align*} by the length of the hypotenuse to find the other side. Therefore, you will have to find the sine of angle \begin{align*}B\end{align*} first. Then you multiply that times the length of the hypotenuse. Notice that we are finding the length of the opposite side with sine.

\begin{align*} \sin e \angle B \times hypotenuse&=opposite\\ \sin e 14.5 \times 5&=opposite\\ 0.250 \times 5&=opposite\\ 1.25 &=opposite\end{align*}

The length of side \begin{align*}AC\end{align*} is 1.25 units.

We can apply a similar rule when we work with cosines. Let’s take a look.

III. Determine and Use the Cosine Ratio to Find the Length of an Adjacent Side of a Right Triangle, given the Angle and the Hypotenuse

Now as we work with cosines, we will be using given information to find the length of the adjacent side of a right triangle. Remember that the adjacent side is the side next to the angle that we are working with. As you recall, the ratio of cosine is \begin{align*}\frac{adjacent}{hypotenuse}\end{align*}. If you know the cosine value of the angle in question, and the length of the hypotenuse, you can find the measure of the adjacent side. Look at the algebraic manipulation below.

\begin{align*}\cos ine \angle X&=\frac{adjacent}{hypotenuse}\\ \cos ine \angle X \times hypotenuse&=\frac{adjacent}{hypotenuse} \times hypotenuse\\ \cos ine \angle X \times hypotenuse&=adjacent\end{align*}

If you multiply the cosine of any angle \begin{align*}X\end{align*} and the length of the hypotenuse, the result is the length of the adjacent side.

Write this statement in your notebook. Be sure to include that it is for cosines.

Example

What is the length of side \begin{align*}BC\end{align*} in the triangle below?

Use the following equation to find the length of the side adjacent to angle \begin{align*}B\end{align*}. Notice that to find the length of the adjacent side that you will first need to find the cosine for angle \begin{align*}B\end{align*}. Then you can multiply that answer with the length of the hypotenuse. This will give you the measurement of the side next to or adjacent to the angle.

\begin{align*}\cos ine \angle B \times hypotenuse&=adjacent\\ \cos ine 14.5 \times 5&=adjacent\\ 0.968 \times 5 &=adjacent\\ 4.84 &=adjacent\end{align*}

The length of side \begin{align*}BC\end{align*} is 4.84 units.

Finally, let’s look at tangents.

IV. Determine and Use the Tangent Ratio to Find the Height of a Right Triangle, Given the Angle and the Base

Now we can use the tangent ratio to find the length of the height of a right triangle. We can see that the height in a right triangle looks like side \begin{align*}a\end{align*}. This is because of the type of triangle that a right triangle is. As you recall, the ratio of tangent is \begin{align*}\frac{opposite}{adjacent}\end{align*}. If you know the tangent value of the angle in question, and the length of the adjacent side, you can find the measure of the opposite side. Look at the algebraic manipulation below.

\begin{align*}\tan gent \angle X&=\frac{opposite}{adjacent}\\ \tan gent \angle X \times adjacent &= \frac{opposite}{adjacent} \times adjacent\\ \tan gent \angle X \times adjacent &=opposite\end{align*}

If you multiply the tangent of any angle \begin{align*}X\end{align*} and the length of the adjacent side, the result is the length of the opposite side.

Write this statement in your notebook. Be sure that it is connected to tangents.

Example

What is the length of side \begin{align*}QR\end{align*} in the triangle below?

Use the following equation to find the length of the side opposite angle \begin{align*}P\end{align*}. Notice that to find the measure of the opposite side we need to first find the tangent of the angle we are working with. Then we can take that measure and multiply it with the adjacent side.

\begin{align*}\tan gent \angle P \times adjacent &=opposite\\ \tan gent \ 72 \times 8 &=opposite\\ 3.078 \times 8 &= opposite\\ 24.624 &= opposite\end{align*}

The length of side \begin{align*}QR\end{align*} is 24.624 units.

Now let’s look at how we can apply this information to real – life problems.

V. Use Calculators or Tables to Solve Real – World Problems Involving Right Triangles and Trigonometric Ratios

All of the ratios that you have been using are connected to Trigonometry. We can use trigonometry in real – world situations to solve real problems. It is important to think of trigonometry as a tool you have to solve different types of problems. Just as you know how to use the Pythagorean to find the third side of a right triangle when you know two, use trigonometry to find missing sides when you know specific angle measurements in a triangle. As you are given values in a problem, either create or add to a diagram to represent the information in the clearest possible way.

Write this statement down in your notebook.

Now let’s look at an example.

Example

Craig launched a model rocket and wanted to calculate how high it went. He stood 10 feet away from the launch site and used a tool to calculate the angle between the ground and his line of sight to the rocket at its highest point. His data is shown in the diagram below.

Calculate the greatest height the model rocket reached.

In this problem, you have a lot of information, but the only important data is the shape of the triangle, the length of the base, and the measurement of the angle. The base is the side adjacent to where Craig is standing, and you want to find the value of the side opposite where Craig was standing. For these purposes, the hypotenuse of the triangle is irrelevant. Since you have the angle, and the adjacent side, you can find a tangent to discover the height of the triangle.

\begin{align*}\tan gent &=\frac{opposite}{adjacent}\\ \tan 65^{\circ} &= \frac{opposite}{10 \ feet}\\ 2.145 &=\frac{opposite}{10 \ feet}\\ 2.145 \times 10 \ feet &= \frac{opposite}{10 \ feet} \times 10 \ feet\\ 21.45 \ feet &=opposite\end{align*}

Craig’s rocket reached it’s highest height at 21.45 feet, almost 21 and \begin{align*}\frac{1}{2}\end{align*} feet high!

## Real-Life Example Completed

The Ramp Dilemma (Part 2)

Here is the original problem once again, be sure to reread it one last time.

Mr. Watson’s homeroom decided to do a community service repairing the ramp outside the shed. Tache fresh coat of paint shone in the bright sunlight and Mr. Watson walked across the grass with all of his students to look at the ramp outside the door of the shed.

“Has that always been there?” asked Dan.

“No, in fact it was just brought out here yesterday,” Mr. Watson explained.

“Well if it’s brand new, then why do we need to fix it?” Emily asked.

“Because it doesn’t fit neatly under the door.”

Sure enough, the students looked and could see that the back of the ramp was too tall and the students would need to fix the back to make it fit beneath the doorway. Fixing this ramp would help everyone because it would make it easy to push or pull the cart with the athletic equipment out onto the field and since the equipment was used by local kid’s teams as well as the school, this was a definite way to give back to the community.

“What do we need to do?” Dan asked.

They looked at the ramp. Mr. Watson drew the following sketch.

“Well that’s not very helpful,” Dan commented.

“Sure it is,” Emily said.

Now apply the sine ratio and figure out the height of the ramp.

Solution to Real – Life Example

First, we take the measurements and use the sine ratio.

\begin{align*}\text{Sine} \ 15^{\circ} = \frac{Opposite}{Hypotenuse}\end{align*}

The opposite in this example is the missing side. We use \begin{align*}x\end{align*} to represent this unknown measure. This is the measure that we are searching for.

\begin{align*}\text{Sine} \ 15^{\circ} = \frac{opposite}{hypotenuse}=\frac{x}{4 \ ft}\end{align*}

Now we can multiply both sides by 4 to find the measure.

\begin{align*}4 \ \text{Sine} \ 15 &= x\\ 2.6 & = x\end{align*}

The height of the ramp is 2.6 feet

## Vocabulary

Here are the vocabulary words that are found in this lesson.

Sine
a ratio between the opposite side and the hypotenuse of a given angle.
Cosine
a ratio between the adjacent side and the hypotenuse of a given angle.
Tangent
a ratio between the opposite side and the adjacent side of a given angle.
Trigonometric Ratio
used to find missing side lengths of right triangles when angle measures have been given.

## Time to Practice

1. True or false. There are three trigonometric ratios.
2. True or false. The sine ratio is the opposite compared to the adjacent.
3. True or false. The tangent ratio is the opposite compared to the adjacent.
4. True or false. The sine ratio is the opposite compared to the hypotenuse.
5. True or false. The cosine ratio is the adjacent compared to the hypotenuse.

Directions: Use the information given and what you have learned about trigonometric ratios to figure out the measure of each missing side.

1. Tangent angle \begin{align*}A \ 45^{\circ}\end{align*}, adjacent side length – 5, what is the length of the opposite side?
2. Tangent angle \begin{align*}B \ 12^{\circ}\end{align*}, adjacent side length – 6, what is the length of the opposite side?
3. Tangent angle \begin{align*}C \ 20^{\circ}\end{align*}, adjacent side length – 15, what is the length of the opposite side?
4. Sine angle \begin{align*}D \ 2^{\circ}\end{align*}, hypotenuse – 12, what is the length of the opposite side?
5. Sine angle \begin{align*}E \ 65^{\circ}\end{align*}, hypotenuse – 8, what is the length of the opposite side?
6. Sine angle \begin{align*}F \ 45^{\circ}\end{align*}, hypotenuse – 2, what is the length of the opposite side?

Directions: Answer each of the following questions.

1. What is the tangent ratio?
2. What is the cosine ratio?
3. What is the sine ratio?

Directions: Draw a diagram for each problem and then solve each problem using trigonometric ratios. Each problem has two parts, a diagram and an answer.

1. Mary wants to find out how tall a tree is. She knows if she measures the angle to from the ground to the tip of the tree 8 feet away, it is \begin{align*}85^{\circ}\end{align*}. How tall is the tree?
2. Barry ties one end of string to the top of his apartment building, drops it to the ground, and then pulls it out from the building. If the angle the string makes with the ground is \begin{align*}75^{\circ}\end{align*} and the length of the string is 125 feet, how tall is the building?
3. A suspension bridge has a large metal pipe connecting the roadway to the top of the tower. The angle the pipe makes with the road is \begin{align*}40^{\circ}\end{align*} and the length of the pipe is 75 yards. What is the distance between the place where the roadway and pipe intersect and the base of the tower?

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# Popular depictions of electromagnetic wave: is there an error?

Here are some depictions of electromagnetic wave, similar to the depictions in other places:

Isn't there an error? It is logical to presume that the electric field should have maximum when magnetic field is at zero and vise versa, so that there is no moment when the both vectors are zero at the same time. Otherwise one comes to a conclusion that the total energy of the system becomes zero, then grows to maximum, then becomes zero again which contradicts the conservation law.

The depictions you're seeing are correct, the electric and magnetic fields both reach their amplitudes and zeroes in the same locations. Rafael's answer and certain comments on it are completely correct; energy conservation does not require that the energy density be the same at every point on the electromagnetic wave. The points where there is no field do not carry any energy. But there is never a time when the fields go to zero everywhere. In fact, the wave always maintains the same shape of peaks and valleys (for an ideal single-frequency wave in a perfect classical vacuum), so the same amount of energy is always there. It just moves.

To add to Rafael's excellent answer, here's an explicit example. Consider a sinusoidal electromagnetic wave propagating in the $z$ direction. It will have an electric field given by

$$\mathbf{E}(\mathbf{r},t) = E_0\hat{\mathbf{x}}\sin(kz - \omega t)$$

Take the curl of this and you get

$$\nabla\times\mathbf{E}(\mathbf{r},t) = \left(\hat{\mathbf{z}}\frac{\partial}{\partial y} - \hat{\mathbf{y}}\frac{\partial}{\partial z}\right)E_0\sin(kz - \omega t) = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$

Using one of Maxwell's equations, $\nabla\times\mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t}$, you get

$$-\frac{\partial\mathbf{B}(\mathbf{r},t)}{\partial t} = -E_0 k\hat{\mathbf{y}}\cos(kz - \omega t)$$

Integrate this with respect to time to find the magnetic field,

$$\mathbf{B}(\mathbf{r},t) = -\frac{E_0 k}{\omega}\hat{\mathbf{y}}\sin(kz - \omega t)$$

Comparing this with the expression for $\mathbf{E}(\mathbf{r},t)$, you find that $\mathbf{B}$ is directly proportional to $\mathbf{E}$. When and where one is zero, the other will also be zero; when and where one reaches its maximum/minimum, so does the other.

For an electromagnetic wave in free space, conservation of energy is expressed by Poynting's theorem,

$$\frac{\partial u}{\partial t} = -\nabla\cdot\mathbf{S}$$

The left side of this gives you the rate of change of energy density in time, where

$$u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right)$$

and the right side tells you the electromagnetic energy flux density, in terms of the Poynting vector,

$$\mathbf{S} = \frac{1}{\mu_0}\mathbf{E}\times\mathbf{B}$$

Poynting's theorem just says that the rate at which the energy density at a point changes is the opposite of the rate at which energy density flows away from that point.

If you plug in the explicit expressions for the wave in my example, after a bit of algebra you find

$$\frac{\partial u}{\partial t} = -\omega E_0^2\left(\epsilon_0 + \frac{k^2}{\mu_0\omega^2}\right)\sin(kz - \omega t)\cos(kz - \omega t) = -\epsilon_0\omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$

(using $c = \omega/k$) and

$$\nabla\cdot\mathbf{S} = \frac{2}{\mu_0}\frac{k^2}{\omega}E^2 \sin(kz - \omega t)\cos(kz - \omega t) = \epsilon_0 \omega E_0^2 \sin\bigl(2(kz - \omega t)\bigr)$$

thus confirming that the equality in Poynting's theorem holds, and therefore that EM energy is conserved.

Notice that the expressions for both sides of the equation include the factor $\sin\bigl(2(kz - \omega t)\bigr)$ - they're not constant. This mathematically shows you the structure of the energy in an EM wave. It's not just a uniform "column of energy;" the amount of energy contained in the wave varies sinusoidally from point to point ($S$ tells you that), and as the wave passes a particular point in space, the amount of energy it has at that point varies sinusoidally in time ($u$ tells you that). But those changes in energy with respect to space and time don't just come out of nowhere. They're precisely synchronized in the manner specified by Poynting's theorem, so that the changes in energy at a point are accounted for by the flux to and from neighboring points.

There is no contradiction with conservation law, you are just applying it wrongly. To do it right, you have to consider a small closed region and check whether the variation of energy inside this domain is the same as (minus) the energy flux through its boundaries. In the case of a electromagnetic wave, when the energy density decreases inside the region, it just means that the energy has left the region you are considering.

• On the other hand, if the phases of B and E are shifted, the total energy is constant because $(\sin t)^2+(\cos t)^2$=1 which is what is expected. – Anixx Jan 28 '11 at 18:08
• Dear Anixx, please listen to Rafael, he is completely right. There is absolutely no reason why the total energy density should be constant. – Luboš Motl Jan 28 '11 at 18:11
• Energy in electromagnetic wave moves from magnetic field to magnetic field and vise versa like in a spring it moves from kinetic form to potential. The total energy remains constant. When energy of electric field is at maximum, the energy of magnetic is zero. Energy cannot apper from nothing. – Anixx Jan 28 '11 at 18:37
• Energy of magnetic field is $c_1B^2$ and energy of electric field is $c_2E^2$. They cannot both be at maximum and then come to a minimum. – Anixx Jan 28 '11 at 18:40
• And also look at the Maxwell equations. $\operatorname{rot} \mathbf{E} = -\frac{\partial \mathbf{B}} {\partial t}$ This equation clearly states that electric field E is proportional to minus derivative of B by time and the fourth Maxwell equation says that B is proportional to derivative of E by time. So if E is sine, B is cosine. – Anixx Jan 28 '11 at 18:49

Look at the original, Heinrich Hertz, 1889! If E=B=0 on a plane, S=0 => no energy Transport through that plane. I agree with Annix: "The pictures may be depicting a static wave, but certainly, not a propagating wave."

In a (small) vicinity of any given point the instantaneous total energy $u=u(t)$ varies in time. The average total energy $\langle u\rangle=\frac{E_0^2}{2c \mu_0}$ does not:

$$\frac{d \langle u\rangle}{dt}=0.$$

Thus, energy is conserved for any small vicinity for any point.

Without the detailed mathematics you need to use is Maxwell's equations, for example

$$\nabla \times \mathbf E = -\dfrac{\partial \mathbf B}{\partial t}$$

and a sketch of an electromagnetic wave travelling in the positive z-direction.

In simple terms the equation tells you that the rate of change of the electric field with position in the z-direction is equal to the rate of change of flux with time.

Consider position $L$ with the electric field and the magnetic field being at maximum positive values.
The gradient of the electric field in the z-direction is zero.
The sketch shows the magnetic field at a slightly later time with the magnetic field hardly changing hardly at all and in the limit not at all.
So you have $0$ (gradient of the electric field equal to $0$ (minus rate of magnetic field).
At this position the magnetic field is at a maximum but it is the rate of change of the magnetic field which is important.

Consider position $M$ where the gradient of the electric field in the z-direction is at a maximum magnitude but negative.
This is where the magnetic field is changing most rapidly with time.
Even though the magnetic field/flux is zero at this position it is the rate of change of magnetic field which is important and note that it is a positive increase. With the rate of change of magnetic field a maximum multiplying it by minus one equates it to the maximum negative value of the rate of change with position in the z-direction of electric field.

A similar analysis will show that the sketch graph is consistent with another of Maxwell's equations.

$$\nabla \times \mathbf B = \mu_0 \epsilon_0 \dfrac{\partial \mathbf E}{\partial t}$$

## protected by ACuriousMind♦Apr 19 '15 at 17:16

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).


text:
1
JEE Main 2021 (Online) 26th August Morning Shift
Numerical
+4
-1
Let the line L be the projection of the line $${{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 4} \over 2}$$ in the plane x $$-$$ 2y $$-$$ z = 3. If d is the distance of the point (0, 0, 6) from L, then d2 is equal to _______________.
2
JEE Main 2021 (Online) 27th July Evening Shift
Numerical
+4
-1
The distance of the point P(3, 4, 4) from the point of intersection of the line joining the points. Q(3, $$-$$4, $$-$$5) and R(2, $$-$$3, 1) and the plane 2x + y + z = 7, is equal to ______________.
3
JEE Main 2021 (Online) 27th July Morning Shift
Numerical
+4
-1
Let a plane P pass through the point (3, 7, $$-$$7) and contain the line, $${{x - 2} \over { - 3}} = {{y - 3} \over 2} = {{z + 2} \over 1}$$. If distance of the plane P from the origin is d, then d2 is equal to ______________.
4
JEE Main 2021 (Online) 25th July Evening Shift
Numerical
+4
-1
If the lines $${{x - k} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$ and
$${{x + 1} \over 3} = {{y + 2} \over 2} = {{z + 3} \over 1}$$ are co-planar, then the value of k is _____________.
JEE Main Subjects
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text:
The name describes the general “learner” that is used without narrowing it down to of the type such as “classif” or “regr” and the name. For example the name “classif.svm” is “svm”.

getLearnerName(learner)

## Arguments

learner [Learner] Learner

## Value

[character(1)]


text:
# IQ can be measured by which of the following formulae?

Free Practice With Testbook Mock Tests

## Options:

1. $$\frac {Age}{Mental \; Age} \times 100$$

2. $$\frac {Mental\;Age}{Chronological \; Age} \times 100$$

3. $$\frac {Mental\;Age}{Chronological \; Age} \times 10$$

4. None of the above

### Correct Answer: Option 2 (Solution Below)

This question was previously asked in

Official Paper 4: Tripura TET 2018 Paper 1

## Solution:

Intelligence Quotient is commonly known as IQ refers to the score of a standardized test that assesses and measures human intelligence.

• The first test to measure intelligence was developed by Binet and Simon in 1905.
• Terman in 1916 revised the test and devised the concept of Intelligence Quotient.
• An individual’s Intelligence Quotient (IQ) is determined by the formula given below,

$${\bf{Intelligence}}\;{\bf{Quotient}}\;\left( {{\bf{IQ}}} \right) = \frac{{{\bf{Mental}}\;{\bf{Age}}\;\left( {{\bf{MA}}} \right)\;}}{{{\bf{Chronological}}\;{\bf{Age}}\;\left( {{\bf{CA}}} \right)}} \times 100$$

• Where, Mental Age is based on intellectual development.
• Chronological Age is the number of years a person has lived.

Hence, it could be concluded that IQ can be measured by $$\frac {Mental\;Age}{Chronological \; Age} \times 100$$.

 IQ Range Descriptive Label Above 130 Very Superior Intelligence (gifted) 120 to 129 Superior Intelligence 110 to 119 High Average Intelligence 90 to 109 Average Intelligence 80 to 89 Low Average Intelligence 71 to 79 Borderline Intellectual Functioning 55 to 70 Mild Intellectual Disability 40 to 54 Moderate Intellectual Disability 25 to 39 Severe Intellectual Disability Below 25 Profound Intellectual Disability


text:
# Adjust column width of a table

I am working on it:

\documentclass{article}
\newcommand\htext[1]{\texttt{0x\MakeLowercase{#1}}}

\begin{document}
\begin{table}[h]
\centering
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
\htext{FFFFFFFFFFFFFFFFF} & \htext{FFFFFFFFFFFFFFFFF} & \htext{FFFFFFFFFFFFFFFFF} &  &  &  &  \\
\hline
\end{tabular}
\end{table}
\end{document}

The empty 4 columns will have something very similar. Now, notice that, an Overfull \hbox error occurs.

How can I adjust width of each column so that each \htext{...} entry is split in to more than one lines (I don't want to reduce font size)? Something like setting p{2in} did not help (as it just created one cell entry overlapping with other).

-
Do you any suggestions or requirements for how the string FFFFFFFFFFFFFFFFF might be split across two lines? Maybe unsurprisingly, TeX does not appear to have a hyphenation rule for this string, so you need to provide such information explicitly. Separately, are you really looking to typeset tables containing strings of FFFFFFFFFFFFFFFFF (or ffffffffffffff? Must make for really fascinating reading... :-) – Mico Aug 10 '14 at 21:08
@Mico Just splitting across two lines is sufficient - no hyphenation needed. For your 2nd query, I need to write hex strings of 17 to 20 characters (not all are equal to FFFFFFFFFFFFFFFFF, of course). I have used FFFFFFFFFFFFFFFFF just to illustrate the scenario. – pushpen.paul Aug 10 '14 at 21:13
You probably want to use a variation on tabular{|p{40cm}|... or tabularx{|X|...} with the tabularx package. – Andrew Aug 10 '14 at 21:15
@Andrew - What the OP is encountering is that TeX does not "know" how and where to split up these rather long hex strings. – Mico Aug 10 '14 at 21:26
Splitting a 22 character string into two will not work, because seven columns would still be too wide. – egreg Aug 10 '14 at 21:57

Here is a solution, adapted from a macro I found on Stack Overflow: I define a \hyphenatestring macro, that takes string, converts it to lower case and makes it ‘hyphenatable anywhere’. This macro can be used inside font-changing commands:

\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{tabularx}

\makeatletter
\def\hyphenatestring#1{\xHyphen@te#1$\unskip} \def\xHyphen@te{\@ifnextchar${\@gobble}{\sw@p{\hskip 0pt plus 1pt\xHyphen@te}}}
\def\sw@p#1#2{\MakeLowercase{{#2}}#1}%\texttt
\makeatother

\newcommand\htexttt[1]{\texttt{\hyphenatestring{0x#1}}}
\newcommand\htextsf[1]{\textsf{\hyphenatestring{0x#1}}}

\begin{document}

\begin{table}[h]
\centering
\begin{tabularx}{\linewidth}{|*{7}{X|}}
\hline
\htexttt{AAAAAAAAAAAAAAAAA}&  & \textbf{\htextsf{CCCCCCCCCCCCCCCCC}} &  & \textit{\htexttt{FFFFFFFFFFFFFFFFF}}{} &  &  \\
\hline
\end{tabularx}
\end{table}

\end{document}

-

Borrowing from the answer to this, you can define a command to insert a hyphenation break after each character.

\usepackage{xparse}

\NewDocumentCommand{\splitlist}{>{\SplitList{}}m}

After that, you can redefine your command to make use of this:

\newcommand\htext[1]{\texttt{0x\MakeLowercase{\splitlist{#1}}}}

And then play with the column widths until it fits your needs:

\documentclass{article}
\usepackage{xparse}

\NewDocumentCommand{\splitlist}{>{\SplitList{}}m}

\newcommand\htext[1]{\texttt{0x\MakeLowercase{\splitlist{#1}}}}

\begin{document}
\begin{table}[h]
\centering
\begin{tabular}{|p{0.11\textwidth}|p{0.11\textwidth}|p{0.11\textwidth}|%
p{0.11\textwidth}|p{0.11\textwidth}|p{0.11\textwidth}|p{0.11\textwidth}|}
\hline
\htext{FFFFFFFFFFFFFFFFF} & \htext{FFFFFFFFFFFFFFFFF} &
\htext{FFFFFFFFFFFFFFFFF} & \htext{FFFFFFFFFFFFFFFFF} &
\htext{FFFFFFFFFFFFFFFFF} & \htext{FFFFFFFFFFFFFFFFF} &
\htext{FFFFFFFFFFFFFFFFF} \\
\hline
\end{tabular}
\end{table}
\end{document}

Result:

-
Overfull \hbox still remains. Anyway, I appreciate your attempt. – pushpen.paul Aug 10 '14 at 23:01
Are you sure it was overfull? I got underfull on this example (probably because of that last 'f'), but like I said, you can fiddle with the column widths. I didn't test it until the error was gone because it will depend on your paper size, font size, etc, so there is no one-size-fits-all solution, I think. – Zé Loff Aug 10 '14 at 23:06
Why \-? You can add \hspace{0pt} and no hyphen will show. – egreg Aug 10 '14 at 23:07
@egreg Perfect. I didn't know that – Zé Loff Aug 10 '14 at 23:09
@ZéLoff Here goes the error: line 13: Overfull \hbox (4.65147pt too wide) in paragraph. Before that, there are 14 Underfull \hboxes also. – pushpen.paul Aug 11 '14 at 4:13


text:
## Friday, 21 February 2014

### Probability distributions

The chart I showed you, with the distributions we are studying marked in yellow, can be downloaded here. It is from the paper Univariate Distribution Relationships, L. M. Leemis and J. T. McQueston. In that paper you can find explanations of the various relationships that are indicated by the arrows.

Here also is a simpler chart about Relationships between probability distributions. This one shows some relations that the above chart misses out, such as how the binomial can come from the Poisson as $X \mid X+Y$ (see Examples sheet 2, #5).

Wikipedia has a nice List of probability distributions.

In 16.5 of the notes I mention that if $X$ and $Y$ are i.i.d. exponential r.vs then $X/(X+Y)$ is a uniformly distributed r.v. You will prove this in Examples sheet 4, #8. A general method will be given in Lecture 20. But it can also be done this way:

$P(X/(X+Y) \leq t)=P(X(1-t)/t\leq Y)=\int_{x=0}^\infty\left(\int_{y=x(1-t)/t}^{\infty}\lambda e^{-\lambda y}dy\right)\lambda e^{-\lambda x}dx = t$

which is the c.d.f. of $U[0,1]$.


text:
# relationship between coefficients of a differential equation when product of solutions is constant

Suppose $y_1 y_2$ is the product of the two solutions of the differential equation $y'' + p_1 y' + p_2y = 0$ and it is a constant. How can the requirement for this be $2p_1p_2 + p_2' = 0$? I have tried several manipulations but cannot get the desired requirement. Any hints?

Thanks

-
You didn't try enough. You should use the fact that constants have a zero derivative, differentiating over and over again, and using the differential equation to always get expressions involving only $y_1,y_2$ and their first derivatives. –  Yuval Filmus Jun 30 '11 at 4:15
Let $y_1 y_2 = C \neq 0$. Differentiate once: $$y'_1 y_2 + y_1 y'_2 = 0.$$ Differentiate again: $$y''_1 y_2 + 2y'_1y'_2 + y_1 y''_2 = 0.$$ Substitute the differential equation: $$2y'_1 y'_2 = p_1 y'_1 y_2 + p_2 y_1 y_2 + p_1 y_1 y'_2 + p_2 y_1 y_2 = 2Cp_2.$$ Differentiate this: $$2y''_1 y'_2 + 2y'_1 y''_2 = 2Cp'_2.$$ Substitute the differential equation: $$0 = 2Cp'_2 + 2p_1y'_1y'_2 + 2p_2y_1y'_2 + 2p_1y'_1y'_2 + 2p_2y'_1y_2 = 2Cp'_2 + 4p_1y'_1y'_2 = 2Cp'_2 + 4Cp_1p_2.$$ Since $C \neq 0$, we deduce that $$p'_2 + 2p_1 p_2 = 0.$$


text:
# Levi umbilical surfaces in complex space

created by monti on 08 Jul 2005
modified on 26 Jul 2006

[BibTeX]

Accepted Paper

Inserted: 8 jul 2005
Last Updated: 26 jul 2006

Journal: Journal fuer die Reine und Angewandte Mathematik
Year: 2006

Abstract:

We define a complex connection on a real hypersurface of $\C^{n+1}$ which is naturally inherited from the ambient space. Using a system of Codazzi-type equations, we classify connected real hypersurfaces in $\C^{n+1}$, $n\ge 2$, which are Levi umbilical and have non zero constant Levi curvature. It turns out that such surfaces are contained either in a sphere or in the boundary of a complex tube domain with spherical section.