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Abstract

Star clusters are often used as tracers of major star formation events in external galaxies as they can be studied up to much larger distances than individual stars. It is vital to understand their evolution if they are used to derive, for example, the star formation history of their host galaxy. More specifically, we want to know how cluster lifetimes depend on their environment and structural properties such as mass and radius. This review presents a theoretical overview of the early evolution of star clusters and the consequent long term survival chances. It is suggested that clusters forming with initial densities of survive the gas expulsion, or “infant mortality”, phase. At Myr they are bound and have densities of . After this time they are stable against expansion by stellar evolution, encounters with giant molecular clouds and will most likely survive for another Hubble time if they are in a moderate tidal field. Clusters with lower initial densities () will disperse into the field within a few 10s of Myrs. Some discussion is provided on how extra galactic star cluster populations and especially their age distributions can be used to gain insight in disruption.

Keywords. galaxies: star clusters, open clusters and associations: general

IAUS 266.  Star cluster disruption] Star cluster disruption Mark Gieles] Mark Gieles

1 Introduction

In the Milky Way the majority of star formation occurs in embedded clusters (Lada & Lada 2003). At the same time, only a small fraction of stars in the disc resides in (open) clusters indicating that most of the clusters/associations are not long lived and disruption on time-scales of the order of the ages of open clusters is important.

Historically, studies on the lifetimes of star clusters focussed on the open clusters in the Milky Way. In the end of the 50’s the scarcity of old (Gyr) open clusters was noted nearly simultaneously in three studies: van den Bergh (1957), von Hoerner (1958) and Oort (1958). It was already speculated then by these authors that this was due to their finite lifetimes. The first quantitative explanation for this came from Spitzer (1958) who showed that interactions with Giant Molecular Clouds (GMCs) are a plausible cause for the disruption of open clusters. In his model a cluster with a density of survives the periodic heating due to passing GMCs for about 200 Myr. Wielen (1971) constructed an age distribution of open clusters within 1 kpc of the sun and derived a median lifetime of 200 Myr. Because of this remarkable agreement with the theoretical predictions by Spitzer, GMC encounters have since then been held as the dominant disruption mechanism for open clusters. Another supporting argument for this idea came from the fact that the old open clusters are strongly concentrated towards the Galactic anti-centre, where the density of GMCs is low, and their survival chances high (van den Bergh & McClure 1980 and also Fig. 1).

Later, Elson & Fall (1985) and Hodge (1987) constructed age distributions for the LMC and SMC clusters, respectively. They both compared to Wielen’s result and showed that the age distributions in the Magellanic Clouds extend to older ages, and consequently concluded that these clusters survive longer than their Milky Way counterparts.

This review attempts to summarise work done on the disruption of star clusters since the seminal studies mentioned above. Three time spans are considered in which different destruction mechanisms are important:

• first few Myr: Expulsion of residual gas from the star formation process

• first few 100 Myr: Mass loss due to stellar evolution

• first Gyr: Interactions with GMCs and 2-body relaxation in a tidal field

These points will be discussed from a theoretical and observational viewpoint in § 2 and § 3, respectively.

2 Theoretical considerations

2.1 The first few Myrs: gas expulsion and “infant mortality” of clusters

A lot of attention has recently gone to the disruption of star clusters just after their formation due to the removal of residual gas from the star formation process by the feedback from hot stars. This was already considered by Tutukov (1978) and Lada & Lada (2003) argue it could be the explanation for the strong drop at a few Myrs in the age distribution of Galactic embedded and open clusters. The arguments why this is destructive are as follows: if the embedded stars are in virial equilibrium with their surrounding, then the stellar velocities are too high once the gas is removed. The cluster consequently expands or even complete dissolves. If we define the total mass of the system as , with the stellar mass and the gas mass, then virial equilibrium before gas expulsion can be written as

 v2rms=GMtot2rv. (2.0)

Here is the gravitational constant, is the root-mean square velocity of the stars and is the virial radius of the (embedded) cluster. The latter relates to the half-mass radius as , slightly dependent on the density profile. Since is three times the 1-dimensional velocity dispersion squared, , and is times the half-mass radius in projection, , Eq. 2.1 can also be written in observable quantities as

 σ21d=GMtotηreff, (2.0)

with . In most models it is assumed that the gas expulsion time-scale, , is much shorter than the crossing time of the stars, . The latter is defined as

 tcr = 2√2(r3vGM)1/2 (2.0) ≈ 0.65Myr(ρh103M⊙pc−3)−1/2. (2.0)

For the positions of the stars are static during the change of the potential and the cluster’s response can be calculated using the impulsive approximation.

In here the velocities of the stars right after gas removal are still defined by the total mass of the embedded cluster (Eq. 2.1) and hence the cluster will expand to find a new equilibrium. Lets take , such that is the star formation efficiency (SFE). The final radius can then be expressed in and the initial radius, , as (e.g. Hills 1980)

 rvrv(0)=ϵ2ϵ−1. (2.0)

From this it can be seen that for the cluster dissolves. Since this value is high for a SFE, gas expulsion is a plausible cause for the disruption of a large fraction of the embedded clusters. A value of , however, is for several reasons an upper limit for complete destruction of the cluster:

• Escape of unbound stars: For the system is globally unbound since the total energy (kinetic plus potential) is positive. However, most of the positive energy is carried away by escaping stars which have velocities much higher than . Almost all dynamical models find that for a fraction of the stars will remain bound (see Fig. 2).

• Sub-virial initial conditions: This was first discussed by Lada et al. (1984) who considered a collapsing cloud in which the stars are dynamically “cold” while the gas is removed (see also Proszkow et al. 2009). If the stellar velocities are only a fraction of the virial velocities then Eq. 2.1 becomes , i.e. the condition for complete disruption becomes , which for is (Goodwin 2008), much lower than the analytic estimate of for initially virialised embedded clusters. The combination of and the initial virial state, or the effective SFE (Goodwin & Bastian 2008), thus determines the survival chance.

• Clumpy initial conditions: Fellhauer et al. (2009) showed that if stars form in a clumpy fashion, and if these clumps merge quickly, then the survival chance of the merger remnant is enhanced.

• Slow gas removal: If the gas expulsion time-scale is longer than , the cluster will expand adiabatically and this can result in a bound cluster for as low as (e.g. Geyer & Burkert 2001; Baumgardt & Kroupa 2007 and Fig. 2).

Therefore, the importance of the gas removal phase in disrupting star clusters depends on many parameters: the initial density of the embedded cluster, the gas removal time-scale, the initial virial state of the embedded cluster, the star formation efficiency and the initial configuration of the stars (clumpy or smooth density profile). This has led to several different predictions. For example: most models assume that gas removal is instantaneous. If is the same for all clusters then the disruption rate is independent of cluster mass. Baumgardt et al. (2008) advocate, however, that gas expulsion less efficient in disrupting massive clusters because of their higher binding energy per unit mass. This makes the general prediction that the exposed cluster mass function should be shallower than the embedded cluster mass function (see Parmentier et al. 2008).

Since the embedded phase is short lasting (Myr) most of these parameters are poorly constrained from observations and the interpretation of the models is therefore sensitive to the assumptions that are made for the initial conditions. In § 3 possible observational signatures of this early disruption will be discussed.

2.2 The first few 100 Myrs: mass loss due to stellar evolution

Clusters that survive the gas expulsion phase continuously lose mass through the evolution of the member stars. For a Kroupa type IMF between and the cluster mass reduces roughly by 10%/20%/30% in the first 10/100/500 Myr. Since these stellar evolution time-scales are much longer than typical values of , the cluster will in most cases react by an adiabatic expansion. This expansion is analogous to the increase of the orbital separation of a binary when one of the members loses mass. It is far less sensitive to the fractional mass loss than in the impulsive approximation, since

 rv(t)rv(0) = M(0)M(t) (2.0) = 1μ. (2.0)

In the last step is used for the fraction of mass that remains and this can be compared to in the instantaneous mass loss case (Eq. 2.1). For the cluster will expand by a factor of two and not lose mass, whereas for impulsive mass loss the cluster nearly dissolves, as we have seen in § 2.1.

The situation is somewhat more complicated than this simple prediction. The expansion is more severe and can even result in complete disruption if the cluster is mass segregated before the bulk of the stellar evolution takes place (Vesperini et al. 2009). If the cluster is not primordially mass segregated, it will still segregate while the stars evolve (Applegate 1986) and this can eventually lead to an enhanced expansion at later times. Finally, the fate of the cluster also depends on the Jacobi (or tidal) radius (). The combined effect of mass loss by stellar evolution and dynamical evolution in a tidal field was considered by Fukushige & Heggie (1995). They show that when clusters expand to a radius of they lose equilibrium and most of their stars overflow in a few crossing times. As was the case with gas expulsion, a lot more observational constraints on the initial conditions of star clusters, such as the degree of primordial mass segregation and the initial radius with respect to the tidal radius, are needed before we can make general statements about the effect of stellar evolution on the evolution of star clusters.

2.3 Interaction with Giant Molecular Clouds

An external disruptive factor, already considered by Spitzer (1958), is cluster encounters with Giant Molecular Clouds (GMCs). Since GMCs are typically more massive than clusters, the cluster is more affected by an encounter than the cloud (Theuns 1991). The cluster lifetime due to periodic heating of passing clouds is inversely proportional to the volume density of molecular gas, , and proportional to the density of the cluster, a typical result for disruption by external tidal “shocks” (e.g. Ostriker et al. 1972)

 tGMCdis≈1Gyr(ρ⊙gasρgas)(ρh10M⊙pc−3). (2.0)

Here is the molecular gas density in the solar neighbourhood and the constant is taken from Gieles et al. (2006) and equals within a factor of two the original result by Spitzer (1958). The dependence of on means that lifetimes of star clusters also roughly scale inversely with the observable surface density of molecular gas, enabling us to make order of magnitude predictions for the lifetimes of clusters in other galaxies. In spiral galaxies disruption by GMC encounters is especially important in the early stages of evolution since clusters form in the thin gaseous disc where is high. Older clusters are typically more associated with the thick disc where is low and GMC encounters are less frequent. Since young (1 Gyr) clusters in spiral galaxies have a near constant radius (e.g. Larsen 2004), Gieles et al. (2006) argue that the dissolution time due to GMC encounters is longer for more massive clusters, due to their higher density. It is not clear whether this lack of a mass-radius relation is a universal property imprinted by formation, or the result of an evolutionary effect. This point will be addressed in § 3.

2.4 2-body relaxation in the tidal field of the host galaxy

On somewhat longer time-scales stars are lost from clusters through 2-body relaxation in which stars get accelerated to the escape velocity, (e.g. Ambartsumian 1938; Spitzer 1940). The relevant time-scale of escape is, to first order, the half-mass relaxation time (Spitzer 1987)

 trh=0.138N1/2r3/2h√¯mGlnΛ, (2.0)

where is the number of stars, is the mean stellar mass and is the Coulomb logarithm: (Giersz & Heggie 1994). The relaxation time is approximately the time stars need to establish a Maxwellian velocity distribution. A fraction of the stars in the tail of the distribution have velocities larger than and consequently escape. Assuming that this high velocity tail is refilled every then the dissolution time-scale is . For isolated clusters . For a Maxwellian velocity distribution a fraction has and then . For tidally limited cluster is higher since is lower. For a typical cluster density profile , implying (Spitzer 1987). The escape fraction is often taken constant (e.g. Gnedin & Ostriker 1997). But depends on (through ) and on the strength of the tidal field, or (through ). Effectively, depends on the ratio (e.g. Spitzer & Chevalier 1973; Wielen 1988). Gieles & Baumgardt (2008) show that for (the tidal regime). Together with Eq. 2.4 we then find for clusters on circular orbits in the tidal regime that , apart from the slowly varying Coulomb logarithm. Here is the angular frequency in the galaxy and and are the galactocentric distance and the velocity around the galaxy centre, respectively. So for a flat rotation curve we find for a cluster of a given mass (e.g. Chernoff & Weinberg 1990; Vesperini & Heggie 1997). This linear dependence of on makes it difficult to explain the universality of the globular cluster mass function from dynamical evolution of a power-law initial cluster mass function (e.g. Vesperini et al. 2003), but this will not be discussed here. Note that can also be written as , where is the density within , which is how Lee & Ostriker (1987) present it.

Fukushige & Heggie (2000) show that the linear scaling of with is slightly affected by the finite time it takes stars to find the Lagrangian “exit” points. Baumgardt (2001) showed that is of importance and finds for equal mass clusters that . Baumgardt & Makino (2003) find that this scaling also holds for models of clusters with a mass spectrum, stellar evolution and for different types of orbits in a logarithmic potential. Their result for can be summarised as

 tdis≈2Myr(NlnΛ)3/4RGkpc(VG220kms−1)−1(1−ε), (2.0)

where is the eccentricity of the orbit. If the Coulomb logarithm is taken into account the scaling is something like (Lamers et al. 2005).

3 Observational signatures of cluster disruption

3.1 The velocity dispersion of young clusters

Ho & Filippenko (1996) were the first to demonstrate that one can empirically determine the dynamical mass, , of star clusters beyond the Local Group using their integrated light. They used high resolution spectroscopy to measure from the width of the spectral lines and high angular resolution (HST) imaging to measure . The dynamical mass then follows from the virial relation (Eq. 2.1). They did this for the  Myr old super star cluster “A” in the dwarf starburst galaxy NGC 1569 and found that its is comparable to the masses of Milky Way globular clusters, concluding that cluster “A” is a young globular cluster. Since then this analysis has been applied to several dozen more clusters with different ages and masses and in different galaxies (e.g. Larsen et al. 2004; Bastian et al. 2006; McCrady & Graham 2007; Mengel et al. 2008).

3.2 Super-virial and dispersing?

The empirically derived relies on the assumption that the cluster is in virial equilibrium. One way of verifying the validity if this assumption is by comparing to the photometric mass, . The latter can be derived from the cluster’s total luminosity and age and a comparison to an SSP model. Bastian et al. (2006) provide a compilation of and values of 19 clusters. They find that for the somewhat older clusters ( Myr) there is good agreement between and . For many of the young (Myr) star clusters . Goodwin & Bastian (2006) suggest that this is a signature of gas expulsion and the “infant mortality” scenario discussed in § 2.1, i.e. they conclude that these clusters are super-virial.

In the left panel of Fig. 3 an updated version of Fig. 5 in Bastian et al. (2006) is given. It shows the light to dynamical mass ratio of 24 young star clusters in different galaxies. The data are compiled from literature (McCrady et al. 2003 ; Bastian et al. 2006; Smith et al. 2006; McCrady & Graham 2007; Moll et al. 2007; Östlin et al. 2007; Larsen et al. 2008; Mengel et al. 2008) and will be discussed in more detail in Gieles et al. (2009) and Portegies Zwart et al. (2010). The observations taken in the visible are shown in blue and the ones taken in the NIR are shown in red. The photometric evolution in the and band from the Bruzual & Charlot (2003) SSP models using a Chabrier IMF is shown as a full (blue) line and a dashed (red) line, respectively. All clusters have an age close to Myr. This is because of observational constraints: younger clusters are more obscured and at 10 Myr the red supergiants phase sets in making the cluster brighter (especially in the NIR, see Fig. 3) and therefore easier to study.

It is important to realise what initial conditions are required for the gas expulsion scenario to work. The models discussed in § 2.1 show that after instantaneous gas removal it takes roughly 10 initial crossing times to either completely disrupt or expand to a new virial equilibrium. In the models of Goodwin & Bastian (2006) this corresponds to  Myr since their clusters have Myr in the embedded phase. Only with an initial of a few Myrs it is possible to “catch” an unbound expanding cluster at 10 Myr. This corresponds to an initial density (stars and gas) of (Eq. 2.1). Since the clusters considered here have densities of (see Fig. 4), gas expulsion can not be responsible for the high velocity dispersions. The densities in the embedded phase were at least a factor higher because of the gas, but probably even more since they have also expanded since then. A similar conclusion can be drawn from the measured velocities, which are of the order of . With such velocities an unbound cluster dissolves into the field in a few Myrs, incompatible with their radii being only a few pc. We conclude that the “infant mortality” phase of dense clusters only lasts a few Myrs. The clusters considered here have evolved for at least and are therefore probably bound objects (an argument also made by McCrady and collaborators).

3.3 Binaries

The point remains that the measured velocity dispersions are too high for some of the clusters. It could be that the IMF assumption is invalid. The virial velocities are higher when the IMF is more bottom heavy, but the differences between the measured velocities and the virial velocities are probably too large to be explained by IMF variations. An alternative is that the velocities are dominated by orbital motions of binaries in the cluster, not taken into account in the virial relation (Eq. 2.1). Kouwenhoven & de Grijs (2008) considered this and they show that the ratio can be elevated by several factors (see also de Grijs in this contribution). But they concluded that the contribution of binaries to the velocity dispersion is less then 5% for clusters with . So binaries can not be responsible for the observed velocities according to their model. But Kouwenhoven & de Grijs (2008) did not include the mass dependent stellar mass-to-light ratio in their models and also did not take into account stars more massive than . So the effect will be stronger for the clusters considered in Fig. 3, since the light is dominated by massive stars () for which (primordial) multiplicity is high and short periods and mass ratio close to unity are common.

Gieles et al. (2009) derive a simple analytic expression for the additional velocity dispersion due to orbital motions of stars in binaries, , relative to the dynamical dispersion of a virialised cluster,

 σ2binσ21d≈(f0.25)(q0.6)32(m115M⊙)23(103daysP)23(M/reff105M⊙\textuppc−1)−1. (3.0)

Here is the binary fraction, is the mass ratio, is the mass of the primary star (the contribution of secondary star is ignored) and is the orbital period. The normalisation constants are based on properties of massive binaries in the Milky Way and appropriate for an age of Myr. Since the cluster light is dominated by the most massive stars the measured velocity dispersion is affected by binaries according to this relation. The prediction for including the velocities of binaries () is shown in the right panel of Fig. 3 (full line). The dashed lines show the 1 and 2 spread in the model due to stochastics. The discrepancy between and can largely be explained by virialised clusters with a modest binary population. Assuming that all clusters host a similar binary population, the clusters with a low dynamical dispersion, for which we use as a proxy (Eq. 2.1), are affected most. This could be the reason why the data show a decreasing ratio for an increasing ratio.

This suggest that these clusters are survivors of the gas expulsion phase and are at the beginning of the stellar evolution phase. In the next Hubble time these clusters will adiabatically expand by a factor of due to mass loss by stellar evolution if they are not mass segregated (§ 2.2) and it remains to conclude that these clusters are good young globular cluster candidates if the tidal field they are in is not too strong.

3.4 The mass radius relation of star clusters

Several studies have pointed out that there is a very weak dependence of cluster radius on mass/luminosity (e.g. Zepf et al. 1999; Larsen 2004; Bastian et al. 2005). For the clusters in Fig. 4 this seems not to be the case. Although these are only 24 clusters, and it is far from a tight relation, these objects seem more consistent with a constant density of , or within a factor of two. The difference between this figure and the studies mentioned above is that the clusters considered here are confined to a narrow (young) age range. The mass-radius relation is of importance for disruption since (§ 2.3). So for a constant cluster density GMCs will destroy all clusters equally fast. If the mass-radius relation evolves in time, for example to a near constant radius as found by several authors, then will become mass dependent. In any case, the densities of these clusters imply that is much longer than a Hubble time, even when is 10 times higher than in the solar neighbourhood (Eq. 2.3), so GMC encounters will not play an important role in the disruption of these clusters at this stage.

In a solar neighbourhood type tidal field the clusters in Fig. 4 have , within a factor of two true for all of them. This implies that these clusters are probably also very stable against the expansion due to mass loss by stellar evolution, since in § 2.2 it was argued that rapid dissolution due to mass loss by stellar evolution only occurs when increases to . A similar idea was recently put forward by Pfalzner (2009) who discovered two evolutionary sequences in young Galactic star clusters. A dense group that starts with a density of at a few Myr containing the Arches cluster, NGC 3603 and Trumpler 14, to which they refer as “starburst clusters”. They seem to expand along a sequence of constant mass, where at Myr there are the red supergiant clusters (RSGC01 and RSGC02) with a density of . The second sequence also starts at a few Myr, but with much lower densities () and expands along a sequence to densities comparable to the field star density. These are referred to as “leaky clusters”. The clusters in Fig. 4 could be compared to the red supergiant clusters in the Milky Way, i.e. the end stage of the starburst sequence.

4 Discussion

Up to the 90s the open clusters in the Milky Way and the populous clusters in the Magellanic Clouds were the targets of studies on disruption. HST has enabled us to establish the properties of larger populations of clusters containing more massive clusters in quiescent spiral galaxies (e.g. Larsen & Richtler 2000), interacting galaxies (e.g. Bastian et al. 2005; Whitmore et al. 1999) and merger remnants (e.g. Miller et al. 1997).

The primary tool used for studies on the disruption of clusters is the age distribution. Different groups give varying weights to the various aspects described in § 2 in the interpretation of the results. There are two flavours of empirically based cluster disruption models: one is based on externally driven cluster disruption. In this model the dissolution time depends on the mass of the cluster and its environment. It was based on the age and mass distributions of luminosity limited cluster samples in different galaxies (Boutloukos & Lamers 2003). The variations in dissolution time-scales are explained by differences in the tidal field strength (Lamers et al. 2005) and the GMC density (Gieles et al. 2006). The second model assumes that internally driven cluster disruption is most important. In this model roughly 80-90% of the number of clusters is destroyed each age dex resulting in a (mass limited) age distribution that declines roughly as . In this model it is assumed that the infant mortality of clusters proceeds for a few 100 Myr and that the disruption rate is equally fast for all masses. This model is based on the cluster population of the Antennae galaxies (Fall et al. 2005; Whitmore et al. 2007) and it was argued by these authors that their model is universal and should be able to describe the age distributions of cluster populations in other galaxies as well.

The universal cluster dissolution scenario has led to some controversy. Chandar et al. (2006) show that the age distribution of SMC clusters declines as , almost the same as in the Antennae. Gieles et al. (2007) used the same data set and show that the age distribution of massive clusters () is flat in the first few 100 Myr. They reproduce the result for the full sample and conclude that the decline is caused by detection incompleteness, confirmed by de Grijs & Goodwin (2008) using similar arguments. Boutloukos & Lamers (2003) show that for a constant formation history and no disruption, the age distribution of a luminosity limited sample declines as , where is the index of the cluster initial mass function. The index describes the cluster fading with age: , with being the flux at wavelength of a cluster with constant mass. For the -band , resulting in , with due to fading alone when . If the cluster mass function is a Schechter function (Gieles 2009; Larsen 2009), the age distribution of a luminosity limited sample not affected by disruption is as steep as (see also Konstantopoulos in this contribution). Both models rely on the assumption that the cluster formation history has been constant in the age range considered. Bastian et al. (2009) recently pointed out that this is invalid for interacting galaxies like the Antennae galaxies and that a recent burst can be confused by mass independent cluster disruption.

Another issue, especially important for distant cluster populations, is what we actually define as a star cluster. Fig. 5 shows the age distribution of clusters and associations in the central region of the SMC (Chiosi et al. 2006). The sample is split in two sub-samples of clusters, based on their size. The age distribution of large clusters falls off much quicker than the one of the compact group. The median radius of both samples are 4.5 pc and 9 pc, respectively. At a distance of 20 Mpc it will be very hard to tell these two groups apart. A speculative suggestion is that more of the short-lived “associations” or “leaky clusters” are included in cluster samples in distant galaxies, that would not be considered as genuine star clusters if they could be resolved.

Acknowledgement

I thank Brad Whitmore and Iraklis Konstantopoulos for an interesting discussion over lunch at Copacabana beach during this conference and the organisers of IAUS 266 for an interesting meeting!

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text:
Calculus Volume 1

# Chapter 4

### Checkpoint

4.1

$172πcm/sec,172πcm/sec,$ or approximately 0.0044 cm/sec

4.2

$500ft/sec500ft/sec$

4.3

$110rad/sec110rad/sec$

4.4

$−0.61ft/sec−0.61ft/sec$

4.5

$L(x)=2+112(x−8);L(x)=2+112(x−8);$ 2.00833

4.6

$L(x)=−x+π2L(x)=−x+π2$

4.7

$L(x)=1+4xL(x)=1+4x$

4.8

$dy=2xex2dxdy=2xex2dx$

4.9

$dy=1.6,dy=1.6,$ $Δy=1.64Δy=1.64$

4.10

The volume measurement is accurate to within $21.6cm3.21.6cm3.$

4.11

7.6%

4.12

$x=−23,x=−23,$ $x=1x=1$

4.13

The absolute maximum is $33$ and it occurs at $x=4.x=4.$ The absolute minimum is $−1−1$ and it occurs at $x=2.x=2.$

4.14

$c=2c=2$

4.15

$522522$ sec

4.16

$ff$ has a local minimum at $−2−2$ and a local maximum at $3.3.$

4.17

$ff$ has no local extrema because $f′f′$ does not change sign at $x=1.x=1.$

4.18

$ff$ is concave up over the interval $(−∞,12)(−∞,12)$ and concave down over the interval $(12,∞)(12,∞)$

4.19

$ff$ has a local maximum at $−2−2$ and a local minimum at $3.3.$

4.20

Both limits are $3.3.$ The line $y=3y=3$ is a horizontal asymptote.

4.21

Let $ε>0.ε>0.$ Let $N=1ε.N=1ε.$ Therefore, for all $x>N,x>N,$ we have

$|3−1x2−3|=1x2<1N2=ε|3−1x2−3|=1x2<1N2=ε$

Therefore, $limx→∞(3−1/x2)=3.limx→∞(3−1/x2)=3.$

4.22

Let $M>0.M>0.$ Let $N=M3.N=M3.$ Then, for all $x>N,x>N,$ we have

$3x2>3N2=3(M3)22=3M3=M3x2>3N2=3(M3)22=3M3=M$

4.23

$−∞−∞$

4.24

$3535$

4.25

$±3±3$

4.26

$limx→∞f(x)=35,limx→∞f(x)=35,$ $limx→−∞f(x)=−2limx→−∞f(x)=−2$

4.29

$y=32xy=32x$

4.30

The function $ff$ has a cusp at $(0,5)(0,5)$ $limx→0−f′(x)=∞,limx→0−f′(x)=∞,$ $limx→0+f′(x)=−∞.limx→0+f′(x)=−∞.$ For end behavior, $limx→±∞f(x)=−∞.limx→±∞f(x)=−∞.$

4.31

The maximum area is $5000ft2.5000ft2.$

4.32

$V(x)=x(20−2x)(30−2x).V(x)=x(20−2x)(30−2x).$ The domain is $[0,10].[0,10].$

4.33

$T(x)=x6+(15−x)2+12.5T(x)=x6+(15−x)2+12.5$

4.34

The company should charge $7575$ per car per day.

4.35

$A(x)=4x1−x2.A(x)=4x1−x2.$ The domain of consideration is $[0,1].[0,1].$

4.36

$c(x)=259.2x+0.2x2c(x)=259.2x+0.2x2$ dollars

4.37

$11$

4.38

$00$

4.39

$limx→0+cosx=1.limx→0+cosx=1.$ Therefore, we cannot apply L’Hôpital’s rule. The limit of the quotient is $∞∞$

4.40

$11$

4.41

$00$

4.42

$ee$

4.43

$11$

4.44

The function $2x2x$ grows faster than $x100.x100.$

4.45

$x1≈0.33333333,x2≈0.347222222x1≈0.33333333,x2≈0.347222222$

4.46

$x1=2,x2=1.75x1=2,x2=1.75$

4.47

$x1≈−1.842105263,x2≈−1.772826920x1≈−1.842105263,x2≈−1.772826920$

4.48

$x1=6,x2=8,x3=263,x4=809,x5=24227;x*=9x1=6,x2=8,x3=263,x4=809,x5=24227;x*=9$

4.49

$−cosx+C−cosx+C$

4.50

$ddx(xsinx+cosx+C)=sinx+xcosx−sinx=xcosxddx(xsinx+cosx+C)=sinx+xcosx−sinx=xcosx$

4.51

$x4−53x3+12x2−7x+Cx4−53x3+12x2−7x+C$

4.52

$y=−3x+5y=−3x+5$

4.53

$2.93sec,64.5ft2.93sec,64.5ft$

### Section 4.1 Exercises

1.

$88$

3.

$±1310±1310$

5.

$2323$ ft/sec

7.

The distance is decreasing at $390mi/h.390mi/h.$

9.

The distance between them shrinks at a rate of $132013≈101.5mph.132013≈101.5mph.$

11.

$9292$ ft/sec

13.

It grows at a rate $4949$ ft/sec

15.

The distance is increasing at $(13526)26(13526)26$ ft/sec

17.

$−56−56$ m/sec

19.

$240π240π$ m2/sec

21.

$12π12π$ cm

23.

The area is increasing at a rate $(33)8ft2/sec.(33)8ft2/sec.$

25.

The depth of the water decreases at $128125π128125π$ ft/min.

27.

The volume is decreasing at a rate of $(25π)16ft3/min.(25π)16ft3/min.$

29.

The water flows out at rate $(2π)5m3/min.(2π)5m3/min.$

31.

$3232$ m/sec

33.

$2519π2519π$ ft/min

35.

$245π245π$ ft/min

37.

The angle decreases at $4001681rad/sec.4001681rad/sec.$

39.

$100π mi/min100π mi/min$

41.

The angle is changing at a rate of $1125rad/sec.1125rad/sec.$

43.

The distance is increasing at a rate of $62.5062.50$ ft/sec.

45.

The distance is decreasing at a rate of $11.9911.99$ ft/sec.

### Section 4.2 Exercises

47.

$f′(a)=0f′(a)=0$

49.

The linear approximation exact when $y=f(x)y=f(x)$ is linear or constant.

51.

$L(x)=12−14(x−2)L(x)=12−14(x−2)$

53.

$L(x)=1L(x)=1$

55.

$L(x)=0L(x)=0$

57.

0.02

59.

$1.99968751.9996875$

61.

$0.0015930.001593$

63.

$1;1;$ error, $~0.00005~0.00005$

65.

$0.97;0.97;$ error, $~0.0006~0.0006$

67.

$3−1600;3−1600;$ error, $~4.632×10−7~4.632×10−7$

69.

$dy=(cosx−xsinx)dxdy=(cosx−xsinx)dx$

71.

$dy=(x2−2x−2(x−1)2)dxdy=(x2−2x−2(x−1)2)dx$

73.

$dy=−1(x+1)2dx,dy=−1(x+1)2dx,$ $−116−116$

75.

$dy=9x2+12x−22(x+1)3/2dx,dy=9x2+12x−22(x+1)3/2dx,$ $−0.1−0.1$

77.

$dy=(3x2+2−1x2)dx,dy=(3x2+2−1x2)dx,$ $0.20.2$

79.

$12xdx12xdx$

81.

$4πr2dr4πr2dr$

83.

$−1.2πcm3−1.2πcm3$

85.

$−100−100$ ft3

### Section 4.3 Exercises

91.

93.

95.

97.

Since the absolute maximum is the function (output) value rather than the x value, the answer is no; answers will vary

99.

When $a=0a=0$

101.

Absolute minimum at 3; Absolute maximum at −2.2; local minima at −2, 1; local maxima at −1, 2

103.

Absolute minima at −2, 2; absolute maxima at −2.5, 2.5; local minimum at 0; local maxima at −1, 1

105.

107.

109.

$x=1x=1$

111.

None

113.

$x=0;x=±2x=0;x=±2$

115.

None

117.

$x=−1,1x=−1,1$

119.

Absolute maximum: $x=4,x=4,$ $y=332;y=332;$ absolute minimum: $x=1,x=1,$ $y=3y=3$

121.

Absolute minimum: $x=12,x=12,$ $y=4y=4$

123.

Absolute maximum: $x=2π,x=2π,$ $y=2π;y=2π;$ absolute minimum: $x=0,x=0,$ $y=0y=0$

125.

Absolute maximum: $x=−3;x=−3;$ absolute minimum: $−1≤x≤1,−1≤x≤1,$ $y=2y=2$

127.

Absolute maximum: $x=π4,x=π4,$ $y=2;y=2;$ absolute minimum: $x=5π4,x=5π4,$ $y=−2y=−2$

129.

Absolute minimum: $x=−2,x=−2,$ $y=1y=1$

131.

Absolute minimum: $x=−3,x=−3,$ $y=−135;y=−135;$ local maximum: $x=0,x=0,$ $y=0;y=0;$ local minimum: $x=1,x=1,$ $y=−7y=−7$

133.

Local maximum: $x=1−22,x=1−22,$ $y=3−42;y=3−42;$ local minimum: $x=1+22,x=1+22,$ $y=3+42y=3+42$

135.

Absolute maximum: $x=22,x=22,$ $y=32;y=32;$ absolute minimum: $x=−22,x=−22,$ $y=−32y=−32$

137.

Local maximum: $x=−2,x=−2,$ $y=59;y=59;$ local minimum: $x=1,x=1,$ $y=−130y=−130$

139.

Absolute maximum: $x=0,x=0,$ $y=1;y=1;$ absolute minimum: $x=−2,2,x=−2,2,$ $y=0y=0$

141.

$h=924549m,h=924549m,$ $t=30049st=30049s$

143.

The absolute minimum was in 1848, when no gold was produced.

145.

Absolute minima: $x=0,x=0,$ $x=2,x=2,$ $y=1;y=1;$ local maximum at $x=1,x=1,$ $y=2y=2$

147.

No maxima/minima if $aa$ is odd, minimum at $x=1x=1$ if $aa$ is even

### Section 4.4 Exercises

149.

One example is $f(x)=|x|+3,−2≤x≤2f(x)=|x|+3,−2≤x≤2$

151.

Yes, but the Mean Value Theorem still does not apply

153.

$(−∞,0),(0,∞)(−∞,0),(0,∞)$

155.

$(−∞,−2),(2,∞)(−∞,−2),(2,∞)$

157.

2 points

159.

5 points

161.

$c=233c=233$

163.

$c=12,1,32c=12,1,32$

165.

$c=1c=1$

167.

Not differentiable

169.

Not differentiable

171.

Yes

173.

The Mean Value Theorem does not apply since the function is discontinuous at $x=14,34,54,74.x=14,34,54,74.$

175.

Yes

177.

The Mean Value Theorem does not apply; discontinuous at $x=0.x=0.$

179.

Yes

181.

The Mean Value Theorem does not apply; not differentiable at $x=0.x=0.$

183.

$b=±2cb=±2c$

185.

$c=±1πcos−1(π2),c=±1πcos−1(π2),$ $c=±0.1533c=±0.1533$

187.

The Mean Value Theorem does not apply.

189.

$12c+1−2c3=5212880;12c+1−2c3=5212880;$ $c=3.133,5.867c=3.133,5.867$

191.

Yes

193.

It is constant.

### Section 4.5 Exercises

195.

It is not a local maximum/minimum because $f′f′$ does not change sign

197.

No

199.

False; for example, $y=x.y=x.$

201.

Increasing for $−2 and $x>2;x>2;$ decreasing for $x<−2x<−2$ and $−1

203.

Decreasing for $x<1,x<1,$ increasing for $x>1x>1$

205.

Decreasing for $−2 and $1 increasing for $−1 and $x<−2x<−2$ and $x>2x>2$

207.

a. Increasing over $−22,−22,$ decreasing over $x<−2,x<−2,$ $−1 b. maxima at $x=−1x=−1$ and $x=1,x=1,$ minima at $x=−2x=−2$ and $x=0x=0$ and $x=2x=2$

209.

a. Increasing over $x>0,x>0,$ decreasing over $x<0;x<0;$ b. Minimum at $x=0x=0$

211.

Concave up on all $x,x,$ no inflection points

213.

Concave up on all $x,x,$ no inflection points

215.

Concave up for $x<0x<0$ and $x>1,x>1,$ concave down for $0 inflection points at $x=0x=0$ and $x=1x=1$

217.

219.

221.

a. Increasing over $−π2 decreasing over $x<−π2,x>π2x<−π2,x>π2$ b. Local maximum at $x=π2;x=π2;$ local minimum at $x=−π2x=−π2$

223.

a. Concave up for $x>43,x>43,$ concave down for $x<43x<43$ b. Inflection point at $x=43x=43$

225.

a. Increasing over $x<0x<0$ and $x>4,x>4,$ decreasing over $0 b. Maximum at $x=0,x=0,$ minimum at $x=4x=4$ c. Concave up for $x>2,x>2,$ concave down for $x<2x<2$ d. Infection point at $x=2x=2$

227.

a. Increasing over $x<0x<0$ and $x>6011,x>6011,$ decreasing over $0 b. Minimum at $x=6011x=6011$ c. Concave down for $x<5411,x<5411,$ concave up for $x>5411x>5411$ d. Inflection point at $x=5411x=5411$

229.

a. Increasing over $x>−12,x>−12,$ decreasing over $x<−12x<−12$ b. Minimum at $x=−12x=−12$ c. Concave up for all $xx$ d. No inflection points

231.

a. Increases over $−14 decreases over $x>34x>34$ and $x<−14x<−14$ b. Minimum at $x=−14,x=−14,$ maximum at $x=34x=34$ c. Concave up for $−34 concave down for $x<−34x<−34$ and $x>14x>14$ d. Inflection points at $x=−34,x=14x=−34,x=14$

233.

a. Increasing for all $xx$ b. No local minimum or maximum c. Concave up for $x>0,x>0,$ concave down for $x<0x<0$ d. Inflection point at $x=0x=0$

235.

a. Increasing for all $xx$ where defined b. No local minima or maxima c. Concave up for $x<1;x<1;$ concave down for $x>1x>1$ d. No inflection points in domain

237.

a. Increasing over $−π4 decreasing over $x>3π4,x<−π4x>3π4,x<−π4$ b. Minimum at $x=−π4,x=−π4,$ maximum at $x=3π4x=3π4$ c. Concave up for $−π2 concave down for $x<−π2,x>π2x<−π2,x>π2$ d. Infection points at $x=±π2x=±π2$

239.

a. Increasing over $x>4,x>4,$ decreasing over $0 b. Minimum at $x=4x=4$ c. Concave up for $0 concave down for $x>823x>823$ d. Inflection point at $x=823x=823$

241.

$f>0,f′>0,f″<0f>0,f′>0,f″<0$

243.

$f>0,f′<0,f″<0f>0,f′<0,f″<0$

245.

$f>0,f′>0,f″>0f>0,f′>0,f″>0$

247.

True, by the Mean Value Theorem

249.

True, examine derivative

### Section 4.6 Exercises

251.

$x=1x=1$

253.

$x=−1,x=2x=−1,x=2$

255.

$x=0x=0$

257.

Yes, there is a vertical asymptote

259.

Yes, there is vertical asymptote

261.

$00$

263.

$∞∞$

265.

$−17−17$

267.

$−2−2$

269.

$−4−4$

271.

Horizontal: none, vertical: $x=0x=0$

273.

Horizontal: none, vertical: $x=±2x=±2$

275.

Horizontal: none, vertical: none

277.

Horizontal: $y=0,y=0,$ vertical: $x=±1x=±1$

279.

Horizontal: $y=0,y=0,$ vertical: $x=0x=0$ and $x=−1x=−1$

281.

Horizontal: $y=1,y=1,$ vertical: $x=1x=1$

283.

Horizontal: none, vertical: none

285.

Answers will vary, for example: $y=2xx−1y=2xx−1$

287.

Answers will vary, for example: $y=4xx+1y=4xx+1$

289.

$y=0y=0$

291.

$∞∞$

293.

$y=3y=3$

295.

297.

299.

301.

303.

305.

307.

309.

$limx→1−f(x)=-∞andlimx→1−g(x)=∞limx→1−f(x)=-∞andlimx→1−g(x)=∞$

### Section 4.7 Exercises

311.

The critical points can be the minima, maxima, or neither.

313.

False; $y=−x2y=−x2$ has a minimum only

315.

$h=623h=623$ in.

317.

$11$

319.

$100ft by100ft100ft by100ft$

321.

$40ft by40ft40ft by40ft$

323.

$19.73ft.19.73ft.$

325.

$84bpm84bpm$

327.

$T(θ)=40θ3v+40cosθvT(θ)=40θ3v+40cosθv$

329.

$v=bav=ba$

331.

approximately $34.02mph34.02mph$

333.

$44$

335.

$00$

337.

Maximal: $x=5,y=5;x=5,y=5;$ minimal: $x=0,y=10x=0,y=10$ and $y=0,x=10y=0,x=10$

339.

Maximal: $x=1,y=9;x=1,y=9;$ minimal: none

341.

$4π334π33$

343.

$66$

345.

$r=2,h=4r=2,h=4$

347.

$(2,1)(2,1)$

349.

$(0.8351,0.6974)(0.8351,0.6974)$

351.

$A=20r−2r2−12πr2A=20r−2r2−12πr2$

353.

$C(x)=5x2+32xC(x)=5x2+32x$ Differentiating, setting the derivative equal to zero and solving, we obtain $x=253x=253$ and $h=2543h=2543$.

355.

$P(x)=(50−x)(800+25x−50)P(x)=(50−x)(800+25x−50)$

### Section 4.8 Exercises

357.

$∞∞$

359.

$12a12a$

361.

$1nan−11nan−1$

363.

Cannot apply directly; use logarithms

365.

Cannot apply directly; rewrite as $limx→0x3limx→0x3$

367.

$66$

369.

$−2−2$

371.

$−1−1$

373.

$nn$

375.

$−12−12$

377.

$1212$

379.

$11$

381.

$1616$

383.

$11$

385.

$00$

387.

$00$

389.

$−1−1$

391.

$∞∞$

393.

$00$

395.

$1e1e$

397.

$00$

399.

$11$

401.

$00$

403.

$tan(1)tan(1)$

405.

$22$

### Section 4.9 Exercises

407.

$F(xn)=xn−xn3+2xn+13xn2+2F(xn)=xn−xn3+2xn+13xn2+2$

409.

$F(xn)=xn−exnexnF(xn)=xn−exnexn$

411.

$|c|>0.5|c|>0.5$ fails, $|c|≤0.5|c|≤0.5$ works

413.

$c=1f′(xn)c=1f′(xn)$

415.

a. $x1=1225,x2=312625;x1=1225,x2=312625;$ b. $x1=−4,x2=−40x1=−4,x2=−40$

417.

a. $x1=1.291,x2=0.8801;x1=1.291,x2=0.8801;$ b. $x1=0.7071,x2=1.189x1=0.7071,x2=1.189$

419.

a. $x1=−2625,x2=−1224625;x1=−2625,x2=−1224625;$ b. $x1=4,x2=18x1=4,x2=18$

421.

a. $x1=610,x2=610;x1=610,x2=610;$ b. $x1=2,x2=2x1=2,x2=2$

423.

$3.1623or−3.16233.1623or−3.1623$

425.

$0,−1or10,−1or1$

427.

$00$

429.

$0.5188or−1.29060.5188or−1.2906$

431.

$00$

433.

$4.4934.493$

435.

$0.159,3.1460.159,3.146$

437.

We need $ff$ to be twice continuously differentiable.

439.

$x=0x=0$

441.

$x=−1x=−1$

443.

$x=5.619x=5.619$

445.

$x=−1.326x=−1.326$

447.

There is no solution to the equation.

449.

It enters a cycle.

451.

$00$

453.

$−0.3513−0.3513$

455.

Newton: $1111$ iterations, secant: $1616$ iterations

457.

Newton: three iterations, secant: six iterations

459.

Newton: five iterations, secant: eight iterations

461.

$E=4.071E=4.071$

463.

$4.394%4.394%$

### Section 4.10 Exercises

465.

$F′(x)=15x2+4x+3F′(x)=15x2+4x+3$

467.

$F′(x)=2xex+x2exF′(x)=2xex+x2ex$

469.

$F′(x)=exF′(x)=ex$

471.

$F(x)=ex−x3−cos(x)+CF(x)=ex−x3−cos(x)+C$

473.

$F(x)=x22−x−2cos(2x)+CF(x)=x22−x−2cos(2x)+C$

475.

$F(x)=12x2+4x3+CF(x)=12x2+4x3+C$

477.

$F(x)=25(x)5+CF(x)=25(x)5+C$

479.

$F(x)=32x2/3+CF(x)=32x2/3+C$

481.

$F(x)=x+tan(x)+CF(x)=x+tan(x)+C$

483.

$F(x)=13sin3(x)+CF(x)=13sin3(x)+C$

485.

$F(x)=−12cot(x)−1x+CF(x)=−12cot(x)−1x+C$

487.

$F(x)=−secx−4cscx+CF(x)=−secx−4cscx+C$

489.

$F(x)=−18e−4x−cosx+CF(x)=−18e−4x−cosx+C$

491.

$−cosx+C−cosx+C$

493.

$3x−2x+C3x−2x+C$

495.

$83x3/2+45x5/4+C83x3/2+45x5/4+C$

497.

$14x−2x−12x2+C14x−2x−12x2+C$

499.

$f(x)=−12x2+32f(x)=−12x2+32$

501.

$f(x)=sinx+tanx+1f(x)=sinx+tanx+1$

503.

$f(x)=−16x3−2x+136f(x)=−16x3−2x+136$

505.

Answers may vary; one possible answer is $f(x)=e−xf(x)=e−x$

507.

Answers may vary; one possible answer is $f(x)=−sinxf(x)=−sinx$

509.

$5.8675.867$ sec

511.

$7.3337.333$ sec

513.

$13.7513.75$ ft/sec2

515.

$F(x)=13x3+2xF(x)=13x3+2x$

517.

$F(x)=x2−cosx+1F(x)=x2−cosx+1$

519.

$F(x)=−1(x+1)+1F(x)=−1(x+1)+1$

521.

True

523.

False

### Review Exercises

525.

True, by Mean Value Theorem

527.

True

529.

Increasing: $(−2,0)∪(4,∞),(−2,0)∪(4,∞),$ decreasing: $(−∞,−2)∪(0,4)(−∞,−2)∪(0,4)$

531.

$L(x)=1716+12(1+4π)(x−14)L(x)=1716+12(1+4π)(x−14)$

533.

Critical point: $x=3π4,x=3π4,$ absolute minimum: $x=0,x=0,$ absolute maximum: $x=πx=π$

535.

Increasing: $(−1,0)∪(3,∞),(−1,0)∪(3,∞),$ decreasing: $(−∞,−1)∪(0,3),(−∞,−1)∪(0,3),$ concave up: $(−∞,13(2−13))∪(13(2+13),∞),(−∞,13(2−13))∪(13(2+13),∞),$ concave down: $(13(2−13),13(2+13))(13(2−13),13(2+13))$

537.

Increasing: $(14,∞),(14,∞),$ decreasing: $(0,14),(0,14),$ concave up: $(0,∞),(0,∞),$ concave down: nowhere

539.

$33$

541.

$−1π−1π$

543.

$x1=−1,x2=−1x1=−1,x2=−1$

545.

$F(x)=2x3/23+1x+CF(x)=2x3/23+1x+C$

547.

Inflection points: none; critical points: $x=−13;x=−13;$ zeros: none; vertical asymptotes: $x=−1,x=−1,$ $x=0;x=0;$ horizontal asymptote: $y=0y=0$

549.

The height is decreasing at a rate of $0.1250.125$ m/sec

551.

$x=abx=ab$ feet

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# If the number of square units in the area of circle С is twice the num

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If the number of square units in the area of circle С is twice the num  [#permalink]

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If the number of square units in the area of circle С is twice the number of linear units in the circumference of C, what is the number of square units in the area?

(A) 4
(B) 8
(C) 4pi
(D) 8pi
(E) 16pi
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Posts: 51067
Re: If the number of square units in the area of circle С is twice the num  [#permalink]

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09 Mar 2011, 07:52
banksy wrote:
252. If the number of square units in the area of circle С is twice the number of linear units in the circumference of C, what is the number of square units in the area?
(A) 4
(B) 8
(C) 4pi
(D) 8pi
(E) 16pi

Given: $$area=\pi{r^2}=x$$ and $$circumference=2\pi{r}=\frac{x}{2}$$. Question: $$x=?$$

$$\pi{r^2}=2*2\pi{r}$$ --> $$r=4$$ --> $$area=\pi{r^2}=16\pi=x$$.

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# calculus homework help... sketching a graph of f

• Nov 12th 2012, 06:00 PM
asilvester635
calculus homework help... sketching a graph of f
The figure shows the graph of f" . Sketch agraph of f. (The answer is not unique.
Attachment 25679
• Nov 12th 2012, 06:09 PM
jakncoke
Re: calculus homework help... sketching a graph of f
How about $f(x) = 3x^2 - \frac{x^3}{3}$

$f'(x) = 6x - x^2$

$f''(x) = 6 - 2x$ which is the graph in ur picture
• Nov 12th 2012, 06:13 PM
MarkFL
Re: calculus homework help... sketching a graph of f
Quote:

Originally Posted by jakncoke
How about $f(x) = 3x^2 - \frac{x^3}{3}$

$f'(x) = 6x - x^2$

$f''(x) = 6 - 2x$ which is the graph in ur picture

I don't think the OP is supposed to use integration of obtain a member of the family of curves, but rather to use the implications of the given graph. Also, this topic has been posted twice.
• Nov 17th 2012, 05:41 PM
asilvester635
Re: calculus homework help... sketching a graph of f
jakncoke how did you get f(x) = 3x^2 - x^3/3?


text:
# How to increase the line thickness of a Seaborn Line?

MatplotlibPythonData Visualization

To increase the line thickness of a Seaborn line, we can take the following steps

## Steps

• Set the figure size and adjust the padding between and around the subplots.

• Create a dataframe, df, of two-dimensional, size-mutable, potentially heterogeneous tabular data.

• Create a Seaborn line plot with linewidth value in the argument. Here we have set linewidth=7.

• Rotate the tick params, i.e., labels by 45 degrees.

• To display the figure, use Show() method.

## Example

import seaborn as sns
from matplotlib import pyplot as plt
import pandas as pd
import numpy as np

plt.rcParams["figure.figsize"] = [7.00, 3.50]
plt.rcParams["figure.autolayout"] = True

df = pd.DataFrame(
dict(
time=list(pd.date_range("2021-01-01 12:00:00", periods=10)),
speed=np.linspace(1, 10, 10)
)
)

ax = sns.lineplot(x="time", y="speed", data=df, lw=7)

ax.tick_params(rotation=45)

plt.show()

## Output

It will produce the following output −

Published on 09-Oct-2021 09:32:57


text:
## Future value formula annuity due

12 Apr 2019 An annuity due is an annuity in which the cash flows occur at the start of each period. Due to the advance nature of cash flows, each cash flow  The future value of annuity due formula is used to calculate the ending value of a series of payments or cash flows where the first payment is received

12 Apr 2019 An annuity due is an annuity in which the cash flows occur at the start of each period. Due to the advance nature of cash flows, each cash flow  The future value of annuity due formula is used to calculate the ending value of a series of payments or cash flows where the first payment is received  5 Feb 2020 The future value of an annuity due formula is used to predict the end result of a series of payments made over time, including the income that is  annuity-due is (1 + i) times the present value of the corresponding payment in an It is possible to derive algebraic formulas to compute the present and future. Annuity due is the equal payment made at the beginning of the year. is known as annuity due and its future value is calculated by using the following formula:. g = growth rate of payment. n = number of years. m = number of compounding periods per year. A formula for the future value of a growing annuity due can be. formula for the present value of an increasing annuity, as well as the special case in an annuity due, payments or receipts occur at the beginning of each.

## formula for the present value of an increasing annuity, as well as the special case in an annuity due, payments or receipts occur at the beginning of each.

4 Oct 2019 FV of an Annuity Due formula – How the Future Value of an Annuity Due is calculated. \text{Future Value} = \text{Annuity Payment} \times \. “  Future Worth of $1 Per Period (FW$1/P); Sinking Fund Factor (SFF); Present All of the formulas and factors in AH 505 pertain to ordinary annuities only. Annuity Due. An annuity due is an annuity in which the cash flows, or payments, occur at   Then to solve I need to transform compounding period to payment period or patment period to compounding period. I'am using the next formulas: Compounding  Calculating the present value of annuity due is a simple 2 step procedure: First, you calculate the future value as a regular annuity; Secondly, you compound the   Future value of annuity calculator is designed to help you to estimate the value of a series of Annuity due: Payments are made at the beginning of each period - rental lease payments, life The two basic annuity formulas are as follows:. Formula. Formula Sheet Download. future value of annuity formula. FVAn = Future value of ordinary annuity for n years.

### Future Value - Amount to which an investment will grow after Time until CF Cash flow Present value Formula in Column C. 0 Present value of an annuity due.

5 Feb 2020 The future value of an annuity due formula is used to predict the end result of a series of payments made over time, including the income that is  annuity-due is (1 + i) times the present value of the corresponding payment in an It is possible to derive algebraic formulas to compute the present and future. Annuity due is the equal payment made at the beginning of the year. is known as annuity due and its future value is calculated by using the following formula:. g = growth rate of payment. n = number of years. m = number of compounding periods per year. A formula for the future value of a growing annuity due can be. formula for the present value of an increasing annuity, as well as the special case in an annuity due, payments or receipts occur at the beginning of each. The present value annuity due calculation formula is as follows: Present Value Annuity Due Formula. Where: PVAD = present value annuity due. C = amount of   Formula Method for Annuity-due: Present Value: 1 + νk + ν2k + ν3k + ททท + νn−k . = (1 - (νk )(n/k)). 1 - νk by SGS. Accumulated Value at time t = n is: (1 + i)n an|i.

### Use this calculator to determine the future value of an annuity due which is a series of equal The future value is computed using the following formula:.

12 Apr 2019 An annuity due is an annuity in which the cash flows occur at the start of each period. Due to the advance nature of cash flows, each cash flow  The future value of annuity due formula is used to calculate the ending value of a series of payments or cash flows where the first payment is received  5 Feb 2020 The future value of an annuity due formula is used to predict the end result of a series of payments made over time, including the income that is  annuity-due is (1 + i) times the present value of the corresponding payment in an It is possible to derive algebraic formulas to compute the present and future. Annuity due is the equal payment made at the beginning of the year. is known as annuity due and its future value is calculated by using the following formula:.

## Calculating the present value of annuity due is a simple 2 step procedure: First, you calculate the future value as a regular annuity; Secondly, you compound the

14 Feb 2019 Your mother gives you \$100 cash for a birthday present, and says, “Spend it wisely. The bank could use formulas, future value tables, a financial calculator, or a spreadsheet Type = 0 for regular annuity, 1 for annuity due. 10 Jan 2011 Learn how to calculate the future value of an annuity due with your TI BA II Plus or HP 12c Financial calculator.

Future Value Annuity Due Calculator - Given the interest rate per time period, number of time periods and present value of an annuity you can calculate its future  31 Dec 2019 Therefore, the formula for the future value of an annuity due refers to the value on a specific future date of a series of periodic payments, where  12 Apr 2019 An annuity due is an annuity in which the cash flows occur at the start of each period. Due to the advance nature of cash flows, each cash flow


text:
# .

In mathematics, the Duistermaat–Heckman formula, due to Duistermaat and Heckman (1982), states that the pushforward of the canonical (Liouville) measure on a symplectic manifold under the moment map is a piecewise polynomial measure. Equivalently, the Fourier transform of the canonical measure is given exactly by the stationary phase approximation.

Berline & Vergne (1982) and, independently, Atiyah & Bott (1984) showed how to deduce the Duistermaat–Heckman formula from a localization theorem for equivariant cohomology.
References

Berline, Nicole; Vergne, Michele (1982), "Classes caracteristiques equivariantes. Formule de localisation en cohomologie equivariante", Comptes rendus de l'Académie des sciences
Atiyah, Michael Francis; Bott, Raoul (1984), "The moment map and equivariant cohomology", Topology. an International Journal of Mathematics 23 (1): 1–28, doi:10.1016/0040-9383(84)90021-1, MR 721448
Duistermaat, J. J.; Heckman, G. J. (1982), "On the variation in the cohomology of the symplectic form of the reduced phase space", Inventiones Mathematicae 69 (2): 259–268, doi:10.1007/BF01399506, MR 674406

Mathematics Encyclopedia


text:
# In one Tech region the number of students in Math 136 is 20% more than the number of students in Math 123. If there are 480 students in Maths 136, how many students are in Math 123?

In one Tech region the number of students in Math 136 is 20% more than the number of students in Math 123. If there are 480 students in Maths 136, how many students are in Math 123?
You can still ask an expert for help

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Nathaniel Kramer

Step 1
Acording to the given information,
Let the number of students in math123 be ‘x’
So, according to question in math136 there are 20% more students than math 123.
Number of students in math $136=480$
$480=\frac{20}{100}x+x$
Step 2
Now solve the above equation.
$480=\frac{20}{100}x+x$
$480=\frac{120x}{100}$
$\frac{480\left(100\right)}{120}=x$
$400=x$
Therefore, number of students in math123 are 400.


text:
03 Oct 2013

# Munin, mysql and semaphore: how to deal with the "identifier removed" error

History / Edit / PDF / EPUB / BIB / 6 min read (~1004 words)

The following depicts how I “solved” a problem I recently had regarding munin, its mysql plugins and the shared memory cache library used by the plugins (written in perl and using IPC::ShareLite).

First off, let’s begin with a description of the problem. I posted the following on serverfault.com in hope I’d get help from someone more experienced than I am.

I’ve recently setup a munin-node on a CentOS server. All was working fine until I tried to add the apache plugin (which works fine).

For some odd reason, the mysql plugins for munin that used to work ceased to work… I’m now getting a weird error whenever I’m running the plugin with munin-run. For instance

munin-run mysql_files_tables

returns me

IPC::ShareLite store() error: Identifier removed at /usr/lib/perl5/vendor_perl/5.8.8/Cache/SharedMemoryBackend.pm line 156

but sometimes it will also return

table_open_cache.value 64
Open_files.value 58
Open_tables.value 64
Opened_tables.value 19341

but after a while it will revert to the previous error.

I do not have any knowledge about the IPC or the ShareLite library so I don’t really know were to start looking. Since it is a module related to shared memory, I tried tracking down shared memory segments with ipcs without much success.

I haven’t yet rebooted the machine as it is used for many projects (I’d obviously like to be able to diagnose the problem without requiring a restart if it was possible).

Has anyone faced this problem? (a quick search on google didn’t present any relevant help)

Thanks for the help!

Obviously, one can see quickly that this is a quite specific question that not many may have actually encountered. Thus, I didn’t expect to receive much help out of it (and I didn’t).

I had left this issue on the side for a couple of days hoping to come back to it at some point. Munin and the mysql plugins were installed on two servers and it was working fine on both of them (and a third one as master node). After a minor change, one of two client nodes stopped working correctly while the other was still fine. After a couple of days though the second server also decided to exhibit a similar issue…

Tonight I remembered about strace, which is pretty awesome in circumstances like this one. I went ahead and launched strace munin-run mysql_files_tables which outputted a lot of stuff and then stopped at the following point:

...
ioctl(4, SNDCTL_TMR_TIMEBASE or TCGETS, 0x7fff13da8e30) = -1 ENOTTY (Inappropriate ioctl for device)
lseek(4, 0, SEEK_CUR)                   = 0
read(4, "# Carp::Heavy uses some variable"..., 4096) = 4096
brk(0x163e7000)                         = 0x163e7000
read(4, "\n    redo if $Internal{$caller};"..., 4096) = 1737
close(4)                                = 0
write(2, "IPC::ShareLite store() error: Id"..., 123IPC::ShareLite store() error: Identifier removed at /usr/lib/perl5/vendor_perl/5.8.8/Cache/SharedMemoryBackend.pm line 156
) = 123
semop(14581770, 0x2ab08bb67cf0, 3

and when it is actually fixed, the application would end instead (outputting a bunch of stuff such as the following)

...
stat("/usr/lib64/perl5/auto/Storable/_freeze.al", {st_mode=S_IFREG|0644, st_size=706, ...}) = 0
stat("/usr/lib64/perl5/auto/Storable/_freeze.al", {st_mode=S_IFREG|0644, st_size=706, ...}) = 0
open("/usr/lib64/perl5/auto/Storable/_freeze.al", O_RDONLY) = 4
ioctl(4, SNDCTL_TMR_TIMEBASE or TCGETS, 0x7fffe7223570) = -1 ENOTTY (Inappropriate ioctl for device)
lseek(4, 0, SEEK_CUR)                   = 0
read(4, "# NOTE: Derived from ../../lib/S"..., 4096) = 706
close(4)                                = 0
semop(917514, {{1, 0, 0}, {2, 0, 0}, {2, 1, SEM_UNDO}}, 3) = 0
semop(917514, {{2, -1, SEM_UNDO|IPC_NOWAIT}}, 1) = 0
semop(917514, {{1, 0, 0}, {2, 0, 0}, {2, 1, SEM_UNDO}}, 3) = 0
shmdt(0x7fc30021f000)                   = 0
semop(917514, {{2, -1, SEM_UNDO|IPC_NOWAIT}}, 1) = 0
...

#### How to “fix” the problem

What you can see in the first output above is pretty interesting. The semop call gives you the semid the process is trying to obtain (the semaphore used to synchronize different processes using the same shared memory). The signature of the semop function is as follow:

int semop(int semid, struct sembuf *sops, unsigned nsops);

where
semid: semaphore id
sops: pointer to a sembuf struct

struct sembuf {
u_short sem_num; /* semaphore # */
short   sem_op;  /* semaphore operation */
short   sem_flg; /* operation flags */
};

nsops: the length of sops

Upon first inspection, you can see that the sembuf in the first case seems to be invalid if you compare it with the working version where it is actually resolved (strace displays something such as {{2, -1, SEM_UNDO|IPC_NOWAIT}} instead of 0x2ab08bb67cf0. But that is not helping me much.

With that semid you can do two things: first, you can check if it is still alive by calling ipcs, second, you can remove it with ipcrm -s semid.

In my case the “fix” itself was to remove the semaphore that the plugin wasn’t able to obtain (the reason of this still elude me though). After the removal of the semaphore, it is possible again to run munin correctly and the identifier removed error is gone.

I will have to do more research as to how/why this issue occurs as I’ve seen it happen only on CentOS machines so far (the master server is a Debian machine).

18 Sep 2013

# Accepting self-signed (or unknown) certificates with TortoiseGit

History / Edit / PDF / EPUB / BIB / 2 min read (~322 words)

For some odd reason it seems that the system designed into TortoiseGit doesn't allow the user to interact with git when it requires user interaction. For instance, accepting self-signed certificated is not possible, which gives you the known issue 318.

As of TortoiseGit 1.8.5.0, it is still not possible to accept certificates through the GUI. But it is possible to get your git repository to work with TortoiseGit (and work with the required certificate).

You will need to have msysgit installed and available in your PATH for the following to work.

The first step is to run some git command, for instance git clone https://myserver.com/depot via a command line so that git may auto-accept (or ask you to accept) the certificate. This step is crucial to get the certificate details saved onto your machine.

What you will want to do next is go to C:\Program Files (x86)\Git\.subversion and copy everything into %USERPROFILE%\.subversion. Basically, this should copy the certicates that were accepted by msysgit so they can be used by TortoiseGit.

Another, and possibly better solution, is to create a symbolic link so that those 2 folders are in fact a single one. For instance, you could do something such as

move %USERPROFILE%\.subversion %USERPROFILE%\.subversion_backup
mklink /D %USERPROFILE%\.subversion "C:\Program Files (x86)\Git\.subversion"

which will make %USERPROFILE%\.subversion point to your C:\Program Files (x86)\Git\.subversion folder. This has the benefit that any future certificate will work both for msysgit and TortoiseGit.

Thanks to Mexx’ C# Corner for pointing out the solution.

05 Jan 2013

# 2012 ManicTime review

History / Edit / PDF / EPUB / BIB / 8 min read (~1572 words)

As 2012 ended, I wanted to take a look back at this year and review my computer usage/consumption in order to reduce time wasting activity. That time should be channeled into more meaningful activities like learning a new language, improving my current skills, practicing piano and more.

The following data has been collected from August 11, 2012 to December 31, 2012. There is about 14 days which do not have any data (application was closed).

The data covers my computer usage with over 966.21h of active usage. During the period for which I collected data, the computer was also powered off for 1686.30h and left unused (away) for approximately 58.04h.

If we account 2 months of 31 days + 2 months of 30 days + 20 days (August) - 14 days without data = 128 days of data. This would average to 7.55h/day of active computer usage. The way it is currently "structured" however is that computer usage during the week is about 4-5h/day while on the week-end, it is about 12h/day.

This sounds a bit high, but there's a reason to this. I'm not ACTIVELY using the computer for all that time. In ManicTime, the computer is considered active if the computer is being used at least once within a 60 minutes time frame. This means I could potentially be using the computer for 1 minute (or less) every hour and it would count as an active usage of 1h. But for the sake of this review, I'll consider myself as a computer addict (which I am) and will count every minute as an active minute.

The following top 10 items accounts for 876.18h out of the 966.21h of active usage of the computer.

 Application Hours Google Chrome 457.76 Remote Desktop Connection 179.11 League of Legends (TM) Client 72.69 VLC media player 62.9 HexChat 35.81 Free Alarm Clock 15.02 mRemote 14.9 Sublime Text 2 14.56 Windows Explorer 12.13 Torchlight II 11.3

On first sight we can see that I spend a lot of time browsing the web. I do various things on there and since it is the biggest chunk of my time, it is worth looking at what I do exactly on the web. The following table is the top 10 websites I've spent time on.

#### Web activity

This covers 259.15h out of 457.76h (56.6%) spent in Chrome. This means that I have a long-tail (a list of many different websites which I visit for a brief period) of 198.61h. The major time consumer here is www.reddit.com, which accounts for 36.2% of my time browsing. Even though reddit is a news/media website (useful for staying up to date with world events, not sure I do that...), it also contains a lot of content which I would categorize as time wasters: funny pictures, pictures of cats, videos as well as discussions about topics of interests (computer science, software engineer, robotics, electronics, etc.). I spent about 1.3h/day during the 128 days for which I collected data, which I find to be quite a lot.

As for the other sites, here's a couple of notes:
www.jolteon.net: This is a bug tracker I use to track new features/bugs where I work. It serves as a personal system for me to track these issues. I use it frequently to update task statuses as well as enter anything that I may have forgotten to add during the day. I also like to review it frequently to remind myself of what is left to work on (and let my mind figure that out while I sleep)
www.youtube.com: I often get on youtube because of reddit. I enjoy watching documentaries which last from 30-45 minutes on average.
docs.google.com: I've spent some time writing documents in Google Docs simply because it allowed me to share them with others so they could review my work.
-confidential-: This is a website I use to manage "things" for work. I generally go there every day and it takes me from 5-15 minutes on average.
www.twitch.tv: Watching streamers of Starcraft 2 and LoL for a while.
www.google.ca: Looks like I spend a lot of time searching...
en.wikipedia.org: Whenever I don't know something about a subject of interest, wiki is a good source (generally...)
www.facebook.com: Checking that everyone I know is still alive

#### Back to the apps

If we go back to the applications I use, the next in the list is Remote Desktop Connection. I use Remote Desktop Connection to connect to my PC at work so that I can do some work from home. As you can see, I have spent almost 1.44h/day working remotely. Considering that I am not a "work at home" employee, I find this to be outrageously high. I would like to see this be as close as possible to 0h/day.

Next is League of Legends (TM) Client. I've recently been interested in the game and started to play it on a more regular basis. I would like to keep this at around 1h/day or lower.

I've used VLC media player to watch series as well as movie on my PC. Series are 20-45 minutes while movies varies from 1h40 to 3h40. I'd say that about 225h/year looks like an acceptable amount of time spent on this.

I've stopped using MSN to chat with friends. My main communication channel is now through IRC, for which I use the HexChat client. I want to spend a maximum of 1h/day on communication though.

I'm not too sure why Free Alarm Clock is part of the top 10. I believe this has to do with the note I wrote at the beginning mentioning that ManicTime would consider a program active if there was some movement on the screen in the last 60 minutes. Since Free Alarm Clock was set to show up (and take focus) every hour, it is quite possible that it simply appeared from time to time while I was away and "sucked" the time out of whatever was running in the background.

mRemote is another application I used briefly to do remote desktop. Since it doesn't support multiple monitor remoting, I've stopped using it.

Sublime Text 2 is my text editor of choice. I haven't spent a lot of time in it since August mainly because I haven't been doing any coding at all in the past few months.

Windows Explorer Some time spent searching for files on my PC!

I've played through the whole campaign of Torchlight II, which was pretty awesome! I'd be really happy to try multiplayer with some willing friends to see what the end content is like (single player end of game content was pretty funny, but playing it alone wasn't very satisfying for me).

#### Conclusion

The important part of this process, other than reviewing what time was spent on this year, is to decide new objectives going forward. This means deciding what should be cut down, reduced, increased or added. For each application I've already determined what was my goal/limits thus I simply need to make sure I follow them. A monthly review should be sufficient.

#### Recommendations

• Reduce reddit usage below 1h/day
• Reduce/cut time spend doing remoting for work
• Redistribute free time on learning activities and skills improvement
05 Jan 2013

# Short SSD setup guide

History / Edit / PDF / EPUB / BIB / 1 min read (~143 words)

Backup everything. First rename, then remove when you confirmed it worked. Better safe than sorry.

REM Set the new root location
REM Ex. NEW_ROOT=D:
REM     NEW_ROOT=D:\Data
REM     NEW_ROOT=D:\MyStuff\MyComputer
set NEW_ROOT=D:

REM (Make sure that C:\Users doesn't exist anymore)
move C:\Users C:\Users.Backup

REM DO NOT DO THE FOLLOWING
REM (Rename C:\ProgramData before creating the link)
move C:\ProgramData C:\ProgramData.Backup

REM Update the TEMP folder at system level
setx /m TEMP %NEW_ROOT%\Windows\Temp
setx /m TMP %NEW_ROOT%\Windows\Temp

move C:\Windows\Temp C:\Windows\Temp.Backup
REM END DO NOT DO
05 Aug 2012

# PHPUnit - No coverage report

History / Edit / PDF / EPUB / BIB / 2 min read (~387 words)

I was trying to get PHPUnit to give me some coverage report for a project I had not worked on for a long time. I had received a github pull request from someone and I wanted to see what the coverage was on the project to see if the submitter had done a good job of covering his code, but I couldn't get PHPUnit to generate a report that contained any data. All I would get was a couple of empty directory folders pages, which was useless.

I had code coverage work on another project in the same environment I was in, so I was pretty sure that my problem had to do either with how I had setup PHPUnit for that particular project, or that something else was interfering with the report generation.

I tried a couple of things, starting by calling phpunit from the command line using different arguments:

--coverage-html report test\symbol_test.php

Would generate some report with data in it, good!

-c test\phpunit.xml (logging set in phpunit.xml)

Would generate an empty report, not good...

--coverage-html report -c test\phpunit.xml

Would generate an empty report, not good...

So at that point I saw that it was working correctly and that something was definitely wrong with my phpunit.xml configuration file. I went back to the phpunit.de manual, specifically on the configuration page, and tried to figure out my problem.

For code coverage to be included in your report, you have to add a filter, be it a blacklist or a whitelist, but you have to have a filter.

So I quickly added a filter such as

<filter>
<whitelist>
<directory>../</directory>
</whitelist>
</filter>

which would whitelist everything that is in the project (my project root is one level above test). Ran phpunit in the test folder and I finally got a report with data!


text:
217 views

A car $\text{A}$ starts from a point $\text{P}$ towards another point $\text{Q}$. Another car $\text{B}$ starts (also from $\text{P})\; 1$ hour after the first car and overtakes it after covering $30\%$ of the distance $\text{PQ}$. After that, the cars continue. On reaching $\text{Q}$, car $\text{B}$ reverses and meets car $\text{A}$, after covering $23\frac{1}{3}$ of the distance $\text{QP}$. Find the time taken by car $\text{B}$ to cover the distance $\text{PQ}$ (in hours).

1. $3$
2. $4$
3. $5$
4. $3\frac{1}{3}$

1


text:
length assortment co to jest
ABAP definicja length of the assortment. Co znaczy for every plant and material group that can, for.
Znaczenie length assortment definicja.

Czy przydatne?

# Definicja length of the assortment

Co znaczy:

Dimensionless factor defined for every plant and material group that can, for example, be calculated from planning figures or sales statistics.

Assortment length is taken into consideration for stock allocation. The relative length of each plant's assortment can be used to determine the quotas allocated to each plant.

Słownik i definicje SAPa na .


text:
Question

# A chord of a circle of radius $12cm$ subtends an angle of $120°$ at the centre. Find the area of the corresponding segment of the circle. (Use $\pi =$$3.14$and $\surd 3=1.73$)

Open in App
Solution

## Step 1: Find the area of minor sector of the circleRadius, $r=12cm$Draw a perpendicular $OD$on chord $AB$and it will bisect chord $AB$.So, $AD=DB$Radius $=r=12cm$Area of triangle $=\frac{1}{2}×base×height$Area of segment=Area of the sector -the area of a triangleThe area of the minor sector $=\frac{\theta }{360}×\pi {r}^{2}$$=\frac{120°}{360°}×3.14×{12}^{2}$$=150.72c{m}^{2}$Step 2: Find the area of $\Delta AOB$From the $\Delta AOB$,$\angle OAB=\frac{180°-\left(120°\right)}{2}=30°\left[\because OA=OB=r,\text{andintriangletheangleoppositetoequalsidesareequal}\right]$Now, $C\mathrm{os}30°=\frac{AD}{OA}$$=\frac{\sqrt{3}}{2}=\frac{AD}{12}$Or, $AD=6\surd 3cm$We know that $OD$bisects $AB$. So,$AB=2×AD=12\surd 3cm$Now, $S\mathrm{in}30°=\frac{OD}{OA}$$=\frac{1}{2}=\frac{OD}{12}$$\therefore OD=6cm$So, the area of $\Delta AOB=\frac{1}{2}×base×height$Here, base $AB=12\sqrt{3}$andHeight$=OD=6$So, area of $\Delta AOB=\frac{1}{2}×12\sqrt{3}×6=36\sqrt{3}cm=62.28c{m}^{2}$Step 3: Find the area of Minor segment$\text{∴AreaofthecorrespondingMinorsegment=AreaoftheMinorsector–AreaofΔAOB}$ $=150.72c{m}^{2}–62.28c{m}^{2}$ $=88.44c{m}^{2}$Step 4: Find the area of major segmentArea of major segment $=$Area of circle $-$Area of minor segment $={\mathrm{\pi r}}^{2}-88.44\phantom{\rule{0ex}{0ex}}=3.14×{12}^{2}-88.44\phantom{\rule{0ex}{0ex}}=452.16-88.44\phantom{\rule{0ex}{0ex}}=363.72{\mathrm{cm}}^{2}$Hence, the area of the corresponding segment of the circle are $88.44c{m}^{2}$ and $363.72c{m}^{2}$

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text:
Corpus ID: 117545836

# Backgrounds To Higgs Boson Searches from Asymmetric Internal Conversion

@article{Gray2011BackgroundsTH,
title={Backgrounds To Higgs Boson Searches from Asymmetric Internal Conversion},
author={R. Gray and C. Kilic and Michael Park and S. Somalwar and S. Thomas},
journal={arXiv: High Energy Physics - Phenomenology},
year={2011}
}
• R. Gray, +2 authors S. Thomas
• Published 2011
• Physics
• arXiv: High Energy Physics - Phenomenology
• A class of potential backgrounds for Higgs boson searches in the W W to di-lepton channel at both the Tevatron and Large Hadron Collider is presented. Backgrounds from W photon production with external conversion of the on-shell photon in detector material to an asymmetric electron-positron pair, with loss of the trailing electron or positron has been treated adequately in Higgs searches. Here we consider analogous backgrounds from the production of a W with an off-shell photon and with the… CONTINUE READING
22 Citations

#### References

SHOWING 1-10 OF 11 REFERENCES
A brief introduction to PYTHIA 8.1
• Physics, Computer Science
• Comput. Phys. Commun.
• 2008
• 4,224
• PDF


text:
Norm of $||(x-x^*,y-y^*,z-z^*) ||\leq ||\nabla J(x,y,z)||$

I want to prove that

$$||(x-x^*,y-y^*,z-z^*) ||\leq ||\nabla J(x,y,z)||,$$

where $$J(x,y,z)=\exp(\frac{x+y+z}{2016})+\frac{x^2+2y^2+3z^2}{2}$$ and $$(x^*,y^*,z^*)$$ is the minimum of $$J$$.

Also, I have $$||(x-x^*,y-y^*,z-z^*) ||^2\leq U^t H U$$ with $$H$$ the Hessian matrix of $$J$$ and $$U=(x-x^*,y-y^*,z-z^*).$$

Could you give me some clue to complete the proof?

Thanks!


text:
# How to put the
…
 tag very first para in the environment like theorem

I am trying to create a one environment in tex4ht, in the html file
...
 tag is missing in the very first para of the environment

I am using the below tex coding

\documentclass{book}
\newenvironment{exe}{Exercise: }{}
\begin{document}
This version uses the powerful Uniscribe library in order to better handle Unicode and complex language processing.

\begin{exe}
This Theorem explain how it works in the example

This Theorem explain how it works in the example
\end{exe}
This version uses the powerful Uniscribe library in order to better handle Unicode and complex language processing (including support for bidirectional text).

\end{document}

and I used the below coding in the cfg file

\ConfigureEnv{exe}
{\ifvmode\IgnorePar\fi\EndP\HCode{
}}
{\ifvmode\IgnorePar\fi\EndP\HCode{
}}
{}
{}

The output in the html file is:

This version uses the powerful Uniscribe library in order

Exercise:  This Theorem explain how it works in the example

This Theorem explain how it works in the example

This version uses the powerful Uniscribe library in order to better handle
Unicode and complex language processing (including support for bidirectional
text).

I compile the file with the following arguments

htlatex sample "sample.cfg,xhtml"

Please suggest how to insert the para tag
...
 in the opening para of the new environment?

You need to insert explicit paragraphs with \par command in this case:

\ConfigureEnv{exe}
{\ifvmode\IgnorePar\fi\EndP\HCode{
}\par}
{\ifvmode\IgnorePar\fi\EndP\HCode{
}\par}
{}
{}

The result:

   Exercise:  This Theorem explain how it works in the example

   This Theorem explain how it works in the example

   This version uses the powerful Uniscribe library in order to better handle
Unicode and complex language processing (including support for bidirectional
text).

• Its working now thank you for your timely help. I think Inserting paragraph tag is more tricky in TeX4HT – Venkatesan Ramachandiran Feb 7 '17 at 10:15
• sorry another one doubt, why the closing para tag
 coming with newline in the html file, how to run on the tag with the line. – Venkatesan Ramachandiran Feb 7 '17 at 10:17
• @VenkatesanRamachandiran the newlines are inserted everywhere, where the line in broken in the DVI file, so every paragraph inserts new line. – michal.h21 Feb 7 '17 at 10:25
• @VenkatesanRamachandiran but you can use something like HTML tidy to reformat the HTML file. – michal.h21 Feb 7 '17 at 10:35
• I want to generate xml structures from TeX4ht!!! – Venkatesan Ramachandiran Feb 7 '17 at 10:41


text:
## Thursday, 28 January 2021

### A clear 28° halo on snow surface

Sunshine has been a rare phenomenon in Finland this winter. On 16th January, the sky was finally clear and temperature had dropped to -25°C, so I headed to a nearby lakeshore to observe possible surface halos. There were indeed 22° and 46° halos visible on the snow, so I took a set of 335 photos by moving back and forth of a 50 m stretch of a road along the lake. The mirrored stack revealed that this was an odd radius display with 18°, 20°, 23° and 24° halos present. An interesting feature was that 22° and 23° halos were equally strong and well defined in the stack.

Two days later I got another chance to photograph surface halos in the same place. The temperature was -8°C. Only a modest 22° halo with some bright glints outside it were visible to naked eye, so I did not expect much from the stack. Nevertheless, I took 147 photos. This time the mirrored stack looked even better than two days ago: a strong 22° halo accompanied by several odd radius halos, including a clear 28° ring.

A comparison with the previous display (limited by blue frame in the photograph below) provided another surprise: the prominent ”22° halo” was in fact a 23° halo and 22° halo was missing. So this was an odd radius display with 18°, 20°, 23°, 24°, 28°, 35° and 46° halos. Although there are not many observations of 28° halo on surface yet, it seems probable that it is not as rare as has been thought. This gives some hope that a crystal sample can be obtained one day.

The strong 23° halo and the absence of 22° halo in this surface display is interesting. It means that the usual ”22° halo” on snow may in some rare instances be something else than it looks, but it is not easy to detect such cases from the abundance of common 22° halos. For example, the first observation of surface 28° halo on 7 April 2012 by Jari Luomanen and Marko Riikonen looks rather similar as the current one, and it may perhaps contain 23° halo instead of 22° halo (yellow frame below, photo Jari Luomanen). Although all photos in this collage were taken with Samyang 8 mm fisheye lens, the result should be viewed with caution due to the robust alignment.


text:
# Changes between Version 10 and Version 11 of udg/ecoms/RPackage/biascorrection

Ignore:
Timestamp:
May 11, 2016 4:35:40 PM (6 years ago)
Comment:

--

### Legend:

Unmodified
 v10 prd.bc <- biasCorrection(obs, prd, prd, method = "eqm", multi.member = TRUE, window = 30) window = c(30,10)) plotMeanGrid(obs) plotMeanGrid(prd, multi.member = FALSE) The '''netcdf4 file''' resulting from this example can be downloaded here: http://www.meteo.unican.es/work/datasets/tas_qqmap_System4_WFDEI_2001_2010.nc4 {{{#!comment In order to preserve the inter member variability of seasonal forecasts, the bias correction methods should be jointly calibrated using the different ensemble members (considering the CDF of the joined series for the correction). This is the procedure followed by the biasCorrection function when using the option '''multi.member=TRUE'''. Note that multi.member=FALSE will independently calibrate each member and, therefore, inter-member variability will be destroyed. }}} Moreover, in order to take into account the model drift (the change of the model bias as a function of the lead time), the bias correction methods are applied considering the lead month of the predictions as an extra dimension. This is implemented in the biasCorrection function by considering a moving time window (one month in the above example: '''window = 30''', in days). The default recommended value is one-month (window=30), although some tests are being conducted in order to determine the optimum window to correct the available seasonal forecasts (further information will follow). In order to take into account the model drift (the change of the model bias as a function of the lead time), the bias correction methods are applied considering the lead month of the predictions as an extra dimension. This is implemented in the biasCorrection function by considering a moving time window and a time step for (one month in the above example and a time step of 10 days: window = c(30,10), in days). The default recommended value for window is one month, although some tests are being conducted in order to determine the optimum window to correct the available seasonal forecasts (further information will follow). = Multi-variable bias correction  = # Bias correction parameters interpolationMethod <- "nearest" # Both observation and forecast should be define on the same grid. Options: "nearest" and "bilinear" method <- "eqm" \ Empirical quantile mapping multi.member <- FALSE # Should members be adjusted sepparately (TRUE, default), or jointly (FALSE)? method <- "eqm" # Empirical quantile mapping pr.threshold <- 1 # The minimum value that is considered as a non-zero precipitation. window <- 30 # Numeric value specifying the time window width used to calibrate. The window is centered on the target day. Default to \code{NULL}, which considers the whole period available. window <- c(30,7) # Integer vector specifying the time window width and the time step used to calibrate. The window is centered on the target time frame. Default to \code{NULL}, which considers the whole period available. }}} leadMonth = leadMonth) prd <- interpGrid(prd, new.coordinates = getGrid(obs), prd <- interpGrid(prd, new.coordinates = getGrid(obs), method = interpolationMethod) prd <- if ("tp" %in% obs$Variable$varName) { biasCorrection(obs, prd, prd, pr.threshold = pr.threshold, method = method, multi.member = multi.member, biasCorrection(obs, prd, prd, pr.threshold = pr.threshold, method = method, window = window) } else { biasCorrection(obs, prd, prd, method = method, multi.member = multi.member, window = window) biasCorrection(obs, prd, prd, method = method, window = window) } # Exporting to netcdf4


text:
Below:

Image text transcribed for accessibility: Given the difference equation Derive (as demonstrated in the class) the control canonical form simulation diagram (where state variables are taken as outputs of time delay elements). Be sure to show how the simulation diagram is determined. From the diagram, derive the corresponding state space representation, that is, find A, B, C and D in For parts (c) and (d) assume A unit step input r(k) = Initial state vector x(0) = Write and execute a MATLAB program using the state and output equations of part (b) to recursively determine the state vector x(k) and output y(k) at successive time instants k = 1, 2, 3, 4, 5. Submit both the MATLAB code and the resulting output. Analytically determine the steady-state value (as k rightarrow infinity) of the state vector x(k) output y(k) Given the transfer function Find the discrete transfer function G(z) by table look-up. Express it as a ratio of polynomials with positive powers of z. Show your work in detail Repeat (a) using MATLAB. Let T=1, 0.5, and 0.1 sec. Find G(z) for each. Use MATLAB to generate the Bode diagrams of magnitude and phase for the given sample times. For each of the block diagrams above, using the procedure shown in class, give a step-by-step derivation of the expression for C*(s). The unit impulse response of a hold device is shown. Find the corresponding transfer function of the hold device. Show your work in detail. For the input shown below, sketch the output y(t) of a first order hold device over the time interval 0 le t le 5. Note that part (b) is independent of part (a).