text:
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Last summer, I had some internet connectivity problems. Specifically, I would have massive latency issues that affected my conversations on Skype and my relatively pathetic under the best of circumstances efforts at online gaming. It was driving me up a wall and I couldn't figure it out. It hadn't occurred earlier with the same ISP so I thought it was just temporary issues with the network. However, the problem went on for weeks at various hours of the day.

I contacted the customer service at my ISP and was dismissed as being crazy. Their website's ping test tool showed me having a ping of around 40 ms and they couldn't see any problem on their end. The issues I was having were just with the remote sites. The fact that it was something like 30 different websites or services that had this problem never phased the tech support guy.

I as frustrated but I couldn't blame them. As far as they could see, no problem existed. And it was one of those evil connection problems that only acts up some of the time on some of the packets. Even when I was having slowdowns, I could open a terminal and run ping google.com or some other site and it would come back with very reasonable times. Some of the time (2-3% of all packets), however, it would throw huge pings on the order of 300 to 700 ms. The tests that they were doing (a couple of packets and take the mean) would never find the problem. I needed to collect a lot of pings over a reasonably long length of time to be sure of catching and characterizing the problem.

I had done some research and it seems there is an transmission robustness option for DSL called interleaving that, more or less, queues up packets before sending them. This is known to increase latency. With trace route, I was able to see that the problem appeared to be on my ISP's network and not a problem on my LAN or the remote host. I grabbed 4 IPs off the trace route (my router, the “hop” to my ISP's network, the second hop on the ISP's network and then the remote host which is my Linode VPS).

A quick Google search pulled up an implentation of ping in Python. I wrote a small collection of scripts to use this implantation of a ping tool and to repeatedly ping the two selected IPs. I went a bit overboard and hit each IP 1000 times.

I took the data collected from the ping test and loaded into R. Sure enough, there was some funky stuff going on.

source("~/personalProjects/feelingPingy/importPingData.R")
pingTimes <- importPingData("~/personalProjects/feelingPingy/hops3.csv")
pingTimes$prettyTargetIP <- ifelse(pingTimes$targetIP == "192.168.1.1", "router",
ifelse(pingTimes$targetIP == "209.181.206.17", "firstHop", ifelse(pingTimes$targetIP ==
"209.181.211.129", "secondHop", "remoteHost")))
# just an fyi for if you are doing this, the name is targetIP and is the
# xxx.xxx.xxx.xxx IP address by default
by(pingTimes$ping, pingTimes$prettyTargetIP, sd, na.rm = TRUE)

## pingTimes$prettyTargetIP: firstHop ## [1] 14.41 ## -------------------------------------------------------- ## pingTimes$prettyTargetIP: remoteHost
## [1] 8.255
## --------------------------------------------------------
## pingTimes$prettyTargetIP: router ## [1] 0.5125 ## -------------------------------------------------------- ## pingTimes$prettyTargetIP: secondHop
## [1] 16.98

We need the na.rm = TRUE flag because some of the attempts to ping the various IPs actually timed out (ping >= 2,000 ms). We can readily see that the variance is increasing as you move off the LAN with the first hop (me to my ISPs network) has a standard deviation of 14.4 ms and the second hop has a variance of 17.0 ms. Considering that a good ping is probably under 50 ms to the targeted IP, this isn't a very good bit of information, especially since the variance goes up by so much as soon as it leaves the LAN. The packets are screwed out of the gate, so to speak. Lets look at this visually.

library(ggplot2)
ggplot(pingTimes) + geom_density(aes(x = ping, color = prettyTargetIP))

We can see the greater variance on the remote IPs here. More striking is that the distribution of ping times to the remote host is clearly bimodal (green). This would suggest that there are two different processes generating this data. One gives a low ping, the other gives a higher ping. If we look at the two IPs tested between me and the targeted remote host, we see that the first hop seems to be giving the shape to both densities (the second hop is a function of the first hop's ping plus some marginal addition). However, they are all kind of hard to see because of the very high density for the pings on the LAN. Lets redo this looking only at the IPs that aren't local.

ggplot(pingTimes[pingTimes$prettyTargetIP != "router", ]) + geom_density(aes(x = ping, color = prettyTargetIP))  Now that looks better. We can really see the bimodal, almost trimodal, nature of the ping times at the remote host (in green). We can also see that this shape seems to also be clear in the densities for the first and second hop (on my ISPs network). Some packets leave right away, some wait a bit long and some seem to wait forever to make the hop from my modem to the first remote node. We see this shape show up again in the second hop (since the since hop is an additive function of the first, this is expected). If the second hope was also slow or had multiple processes going on or if the problem was at my VPS, the curves at each node would look different. The fairly constant shape suggests that there is a rate limiting step that determines the distribution of ping times. The fact that the second and later hops all have the same shape as the first hop suggests that the rate limiting step is the transfer of the packets from my modem to my ISPs network. And the problem was real. The reason it wasn't showing up on their simple mean with an n of 10 type tests is clear in the ecdf. ggplot(pingTimes[pingTimes$prettyTargetIP != "router", ]) + stat_ecdf(aes(x = ping,
color = prettyTargetIP))

Not all the packets were affected, in fact, nearly half left without excessive delay. However, 25% took over 40 ms to merely move from my modem to the ISPs network. Given that 40 ms is a very long time for a single hop (and over half of what I would expect for a round trip time), the impact I was seeing on Skype and other places was real. Armed with the new data*, I was able to get my connection moved from interleaving to fastpath.

I figured I would post this simple analysis and the tools I used in case they can ever help anyone in the future. I can't be the only one who has ever had this type of problem!

*If you ever have internet connection issues that aren't being fixed by the over-the-phone support or even tech visits, going to the social media teams (see your ISPs Facebook page or check them out on DSLReports) typically brings faster and better results. Once I knew that the problem was real, it took 2 emails and 18 hours to get it fixed via my ISPs social media support people. This is after 2-3 weeks of dealing with phone support and getting nowhere.


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# How do you graph y=(x^3-8)/(6-x^2) using asymptotes, intercepts, end behavior?

Dec 21, 2016

Slant asymptote: $y = - x$. Vertical asymptotes $\uparrow x = \pm \sqrt{6} \downarrow$. x-intercept ( y = 0 ):2. y-intercept ( x = 0 ): $- \frac{4}{3}$

#### Explanation:

By actual division.

y = -x +(3-4/sqrt6)/(x-sqrt6)+(3+4?sqrt6)/(x+sqrt6), revealing

horizontal asymptote y = quotient = -x and vertical asymptotes

$x = \pm \sqrt{6}$, that makes $y \pm \infty a t r = \infty$.

The y-intercept is -4/3 and x-intercept is 2.

y' is 0 at $x = 0 , \pm 3 \sqrt{2}$.

Local maximum ( for y'' < 0 ) at x = 3sqrt2, local minimum ( for y''>0 )

at x = -3sqrt2 and point of inflexion ( for y'' = 0 and y''' not 0 ) at x = 0.

Look at the graph that illustrates all these aspects.

graph{y(6-x^2)-x^3+8 =0 [-38.28, 38.28, -19.14, 19.14]}

format: plain


text:
Contents

• # For Solution

Masha meets a new friend and learns his phone number — ss. She wants to remember it as soon as possible. The phone number — is a string of length mm that consists of digits from 00 to 99. The phone number may start with 0.

Masha already knows nn phone numbers (all numbers have the same length mm). It will be easier for her to remember a new number if the ss is represented as segments of numbers she already knows. Each such segment must be of length at least 22, otherwise there will be too many segments and Masha will get confused. Masha-forgetful solution codeforces

For example, Masha needs to remember the number: s=s= ‘12345678‘ and she already knows n=4n=4 numbers: ‘12340219‘, ‘20215601‘, ‘56782022‘, ‘12300678‘. You can represent ss as a 33 segment: ‘1234‘ of number one, ‘56‘ of number two, and ‘78‘ of number three. There are other ways to represent ss.

Masha asks you for help, she asks you to break the string ss into segments of length 22 or more of the numbers she already knows. If there are several possible answers, print any of them.

Input Masha-forgetful solution codeforces

The first line of input data contains an integer tt (1t1041≤t≤104) —the number of test cases.

Before each test case there is a blank line. Then there is a line containing integers nn and mm (1n,m1031≤n,m≤103) —the number of phone numbers that Masha knows and the number of digits in each phone number. Then follow nn line, ii-th of which describes the ii-th number that Masha knows. The next line contains the phone number of her new friend ss.

Among the given n+1n+1 phones, there may be duplicates (identical phones).

It is guaranteed that the sum of nmn⋅m (nn multiplied by mm) values over all input test cases does not exceed 106106.

Output Masha-forgetful solution codeforces

You need to print the answers to tt test cases. The first line of the answer should contain one number kk, corresponding to the number of segments into which you split the phone number ss. Print -1 if you cannot get such a split.

If the answer is yes, then follow kk lines containing triples of numbers l,r,il,r,i. Such triplets mean that the next rl+1r−l+1 digits of number ss are equal to a segment (substring) with boundaries [l,r][l,r] of the phone under number ii. Both the phones and the digits in them are numbered from 11. Note that rl+12r−l+1≥2 for all kk lines.

Example

input

Copy Masha-forgetful solution codeforces
5

4 8
12340219
20215601
56782022
12300678
12345678

2 3
134
126
123

1 4
1210
1221

4 3
251
064
859
957
054

4 7
7968636
9486033
4614224
5454197
9482268

output

Copy Masha-forgetful solution codeforces
3
1 4 1
5 6 2
3 4 3
-1
2
1 2 1
2 3 1
-1
3
1 3 2
5 6 3
3 4 1

Note

The example from the statement.

In the second case, it is impossible to represent by segments of known numbers of length 2 or more.

In the third case, you can get the segments ‘12‘ and ‘21‘ from the first phone number.


text:
Calibration of Cox-Ingersoll-Ross process that hits zero (Feller condition violation)

I'm considering a Cox-Ingersoll-Ross (CIR) process $$dx_{t} = \alpha\left(\theta - x_{t}\right)dt + \sigma \sqrt{x_{t}}\,dW_{t}\,,\qquad \alpha,\beta,\sigma > 0$$

which by assumption has $2\alpha \theta < \sigma^{2}$ (violates the Feller condition) and can therefore reach $x_{t}=0$ for some $t$ . The conditional distribution is $$f(x_{t+T} \vert x_{t}) = c e^{-u-v}\left(\frac{v}{u}\right)^{q/2}I_{q}\left(2\sqrt{uv}\right)$$

where $q = \tfrac{2\alpha\theta}{\sigma^{2}} - 1$ (note $q <0$ by assumption of violation of the Feller condition), $c = \frac{2\alpha}{\sigma^{2}\left(1-e^{-\alpha T}\right)}\,$, $u=cx_{t}e^{-\alpha T}$, $v = c x_{t+T}$ and $I_{q}$ is a modified Bessel function of the first kind of order $q\,$.

I want to calibrate $\alpha,\theta,\sigma$ from certain historical observations $x_{i=1\ldots N}$. As explained in e.g. arXiv:0812.4210, in principle this can be done by minimizing minus the logarithm of the likelihood, i.e. $$-\log (\text{Likelihood}) = -\log \prod_{i=1}^{N-1}f(x_{i+1}\vert x_{i})\,.$$

The tricky thing is that some of my historic observations $x_{i}$ are zero. Now, when $x_{t + T} \to 0$ ($v \to 0$ in the notation above), using the series expansion of the Bessel function one finds $$f(x_{t +T} \to 0 \vert x_{t}) \to c e^{-u-v}\frac{v^{q}}{\Gamma(q+1)}$$

Since $q<0$ by assumption, the density blows up as $v\to 0\,$.

In particular, if (say) the 9th observation is zero in my dataset of historic observations (i.e. $v_{9} = cx_{9}=0$), the likelihood would be $$-\log (\text{Likelihood}) = -q\log v_{9} -\log\left(\frac{c e^{-u_{9}-v_{9}}}{\Gamma(q+1)}\right)-\log \prod_{i\neq 8}f(x_{i+1}\vert x_{i})\,,\quad \text{with } v_{9} \to 0$$

and the term $-q\log v_{9} \to -\infty$ as $v_{9} \to 0$ and will spoil the minimization (a numeric solver ceases to converge, for example).

Any ideas on how to calibrate the CIR process in such situations, namely when the historic data contains points (one or many) where the process hits zero?. Is Maximum Likelihood just not suited to this situation or is there a work-around?.


text:
Screw dislocations in piezoelectric laminates with four or more phases.(English)Zbl 1425.74131

Summary: We present an analytical solution to the problem of a screw dislocation in a four-phase piezoelectric laminate composed of two piezoelectric layers of equal thickness sandwiched between two semi-infinite piezoelectric media. A new version of the complex variable formulation is proposed such that the $$2 \times 2$$ real symmetric matrix appearing in the formulation becomes dimensionless. Using analytic continuation, the original boundary value problem is reduced to the identification of a single 2D analytic vector function which is completely determined following rigorous solution of the resulting linear recurrence relations in matrix form. An explicit expression for the image force acting on the piezoelectric screw dislocation is obtained once the single $$2 \times 2$$ real matrix function is identified. We also discuss the solution for a screw dislocation in an N-phase piezoelectric laminate composed of $$N - 2$$ piezoelectric layers of equal thickness sandwiched between two semi-infinite piezoelectric media.

MSC:

 74E30 Composite and mixture properties 74F15 Electromagnetic effects in solid mechanics
Full Text:

References:

 [1] Y.T. Chou, Screw dislocations in and near lamellar inclusions, Physica Status Solidi (B), 17, 509-516, 1966. [2] S.N.G. Chu, Screw dislocation in a two-phase isotropic thin film, Journal of Applied Physics, 53, 3019-3023, 1982 [3] S.V. Kamat, J.P. Hirth, B. Carnahan, Image forces on screw dislocations in multilayer structures, Scripta Metallurgica, 21, 1587-1592, 1987. [4] M.L. Öveçoğlu, M.F. Doerner, W.D. Nix, Elastic interactions of screw dislocations in thin films on substrates, Acta Metallurgica, 35, 2947-2957, 1987. [5] L. Stagni, R. Lizzio, Interaction of an edge dislocation with a lamellar inhomogeneity, Mechanics of Materials, 6, 17-25, 1987. [6] N. Fares, V.C. Li, General image method in a plane-layered elastostatic medium, Journal of Applied Mechanics, 55, 781-785, 1988. · Zbl 0673.73025 [7] C.C. Ma, H.T. Lu, Theoretical analysis of screw dislocations and image forces in anisotropic multilayered media, Physical Review B, 73, 14, 144102, 2006. [8] H.Y. Wang, M.S. Wu, H. Fan, Image decomposition method for the analysis of a mixed dislocation in a general multilayer, International Journal of Solids and Structures, 44, 1563-1581, 2007. · Zbl 1155.74323 [9] C.Q. Ru, Analytic solution for Eshelby’s problem of an inclusion of arbitrary shape in a plane or half-plane, Journal of Applied Mechanics, 66, 315-322, 1999. [10] Y.E. Pak, Force on a piezoelectric screw dislocation, Journal of Applied Mechanics, 57, 863-869, 1990. · Zbl 0735.73072 [11] K.Y. Lee, W.G. Lee, Y.E. Pak, Interaction between a semi-infinite crack and a screw dislocation in a piezoelectric material, Journal of Applied Mechanics, 67, 165-170, 2000. · Zbl 1110.74537 [12] J.X. Liu, S.Y. Du, B. Wang, A screw dislocation interacting with a piezoelectric bimaterial interface, Mechanics Research Commununications, 26, 415-420, 1999. · Zbl 0973.74032 [13] X. Wang, H. Fan, A piezoelectric screw dislocation in a bimaterial with surface piezoelectricity, Acta Mechanica, 226, 3317-3331, 2015. · Zbl 1329.74089 [14] B.J. Chen, K.M. Liew, Z.M. Xiao, Green’s functions for anti-plane problems in piezoelectric media with a finite crack, International Journal of Solids and Structures, 41, 5285-5300, 2004. · Zbl 1179.74039 [15] B.J. Chen, Z.M. Xiao, K.M. Liew, A screw dislocation interacting with a finite crack in a piezoelectric medium, International Journal of Engineering Science, 42, 1325-1345, 2004.
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


text:
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Search Results: 1 - 10 of 100 matches for " "
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 Physics , 2009, DOI: 10.1111/j.1745-3933.2009.00766.x Abstract: Observations of gamma-ray bursts by the Fermi satellite, capable of detecting photons in a very broad energy band: 8keV to >300GeV, have opened a new window for the study of these enigmatic explosions. It is widely assumed that photons of energy larger than 100 MeV are produced by the same source that generated lower energy photons -- at least whenever the shape of the spectrum is a Band function. We report here a surprising discovery -- the Fermi data for a bright burst, GRB 080916C, unambiguously shows that the high energy photons (>= 100MeV) were generated in the external shock via the synchrotron process, and the lower energy photons had a distinctly different source. The magnetic field in the region where high energy photons were produced (and also the late time afterglow emission region) is found to be consistent with shock compressed magnetic field of the circum-stellar medium. This result sheds light on the important question of the origin of magnetic fields required for gamma-ray burst afterglows. The external shock model for high energy radiation makes a firm prediction that can be tested with existing and future observations.
 Kunihito Ioka Physics , 2010, DOI: 10.1143/PTP.124.667 Abstract: Collisionless entrainment of the surrounding matter imports the relativistic baryon component in the Gamma-Ray Burst (GRB) fireball frame. We show that half the fireball energy can be transferred from radiation to the comoving hot motions of baryons under the photosphere. The yet baryon-poor fireball can reexpand to a very high Lorentz factor (VHLF) \Gamma ~ 10^3-10^6 by its own relativistic collisionless pressure beyond the photosphere (so-called collisionless bulk acceleration), leading to internal and external shocks. A simple synchrotron emission from the VHLF internal shocks produces (i) the extra power-law spectral component with variability observed in the Fermi GeV bursts, up to the TeV range for the future Cherenkov Telescope Array (CTA), (ii) the GeV onset delay with a weak luminosity dependence t_{delay} ~ L^{-1/5}, and (iii) the spectral break of GRB 090926 by the synchrotron cooling break or the maximum synchrotron cutoff limited by the dynamical time, not by the e+- creation cutoff. The relativistic baryon component could also heat the photospheric thermal photons into the main GRB Band spectrum via pp, p\gamma (Bethe-Heitler and photomeson), and Coulomb thermalization processes. In this hot photosphere-internal-external shock model, we can predict the anticorrelation of ~TeV neutrinos and GeV gamma-rays, which may be detectable using IceCube. The spectral peak and luminosity (Yonetoku) relation is also reproduced if the progenitor stars are nearly identical. We also discuss the steep/shallow decay of early X-ray afterglows and short GRBs.
 Physics , 2013, DOI: 10.1088/0004-637X/788/1/70 Abstract: We investigate the effect that the absorption of high-energy (above 100 MeV) photons produced in GRB afterglow shocks has on the light-curves and spectra of Fermi-LAT afterglows. Afterglows produced by the interaction of a relativistic outflow with a wind-like medium peak when the blast-wave deceleration sets in, and the afterglow spectrum could be hardening before that peak, as the optical thickness to pair-formation is decreasing. In contrast, in afterglows produced in the interaction with a homogeneous medium, the optical thickness to pair-formation should increase and yield a light-curve peak when it reaches unity, followed by a fast light-curve decay, accompanied by a spectral softening. If energy is injected in the blast-wave, then the accelerated increase of the optical thickness yields a convex afterglow light-curve. Other features, such as a double-peak light-curve or a broad hump, can arise from the evolution of the optical thickness to photon-photon absorption. Fast decays and convex light-curves are seen in a few LAT afterglows, but the expected spectral softening is rarely seen in (and difficult to measure with) LAT observations. Furthermore, for the effects of photon-photon attenuation to shape the high-energy afterglow light-curve without attenuating it too much, the ejecta initial Lorentz factor must be in a relatively narrow range (50-200), which reduces the chance of observing those effects.
 Physics , 2011, DOI: 10.7529/ICRC2011/V08/0945 Abstract: In many theoretical models of gamma-ray bursts (GRBs) and their afterglows, the emission of photons above 100 GeV is predicted. The Large Area Telescope (LAT) on-board the Fermi Gamma-ray Space Telescope has detected delayed, high-energy emission (up to 90 GeV in the burst rest-frame) from several GRBs and no evidence of a high-energy spectral cutoff during the early afterglow phase of the burst has been found. Presented here are the results of follow-up observations with VERITAS, a ground-based telescope array sensitive to gamma-rays above 100 GeV, of GRBs detected by the Fermi and Swift satellites. These observations have not yielded a conclusive detection and the upper limits on very high energy (VHE, E>100 GeV) gamma-ray flux obtained from these observations are among the most constraining to date.
 Physics , 2012, DOI: 10.1088/0004-637X/768/1/23 Abstract: Soft X-ray absorption in excess of Galactic is observed in the afterglows of most gamma-ray bursts (GRBs), but the correct solution to its origin has not been arrived at after more than a decade of work, preventing its use as a powerful diagnostic tool. We resolve this long-standing problem and find that He in the GRB's host HII region is responsible for most of the absorption. We show that the X-ray absorbing column density (N_Hx) is correlated with both the neutral gas column density and with the optical afterglow extinction (Av). This correlation explains the connection between dark bursts and bursts with high N_Hx values. From these correlations we exclude an origin of the X-ray absorption which is not related to the host galaxy, i.e. the intergalactic medium or intervening absorbers are not responsible. We find that the correlation with the dust column has a strong redshift evolution, whereas the correlation with the neutral gas does not. From this we conclude that the column density of the X-ray absorption is correlated with the total gas column density in the host galaxy rather than the metal column density, in spite of the fact that X-ray absorption is typically dominated by metals. The strong redshift evolution of N_Hx/Av is thus a reflection of the cosmic metallicity evolution of star-forming galaxies. We conclude that the absorption of X-rays in GRB afterglows is caused by He in the HII region hosting the GRB. While dust is destroyed and metals are stripped of all of their electrons by the GRB to great distances, the abundance of He saturates the He-ionising UV continuum much closer to the GRB, allowing it to remain in the neutral or singly-ionised state. Helium X-ray absorption explains the correlation with total gas, the lack of strong evolution with redshift as well as the absence of dust, metal or hydrogen absorption features in the optical-UV spectra.
 Physics , 2004, DOI: 10.1016/j.astropartphys.2004.07.004 Abstract: The discovery of X-ray afterglows of GRBs, and the identification of host galaxies of GRBs, confirm the cosmological origin of GRBs. However, the discovery of the delayed MeV-GeV photons in GRB940217 imposes serious challenges for the standard emission model of GRB. Although the delayed MeV-GeV photons might be explained by some radiation emission mechanisms, the mystery of detecting an 18 GeV photon still remains unsolved. We suggest that the detection of the 18-GeV photon $\sim$4500 s after the keV/MeV burst in GRB 940217 provides a strong evidence for the existence of extra-dimensions and/or quantum gravity effects. The delay scale of the 18-GeV photon leads to an estimation of the fundamental energy scale, associated with the linear energy dependence of the speed of light, of the order of $2.1\times 10^{15}$ GeV, which is consistent with the results obtained by another independent analysis on the data of OSSE and BATSE.
 Physics , 2001, DOI: 10.1086/322400 Abstract: We investigate two high energy radiation mechanisms, the proton synchrotron and the electron inverse Compton emission, and explore their possible signatures in the broad-band spectra and in the keV to GeV light curves of gamma-ray burst afterglows. We develop a simple analytical approach, allowing also for the effects of photon-photon pair production, and explore the conditions under which one or the other of these components dominates. We identify three parameter space regions where different spectral components dominate: (I) a region where the proton synchrotron and other hadron-related emission components dominate, which is small; (II) a region where the electron inverse Compton component dominates, which is substantial; (III) a third substantial region where electron synchrotron dominates. We discuss the prospects and astrophysical implications of directly detecting the inverse Compton and the proton high energy components in various bands, in particular in the GeV band with future missions such as GLAST, and in the X-ray band with Chandra. We find that regime II parameter space is the most favorable regime for high energy emission. The inverse Compton component is detectable by GLAST within hours for bursts at typical cosmological distances, and by Chandra in days if the ambient density is high.
 James E. Rhoads Physics , 2000, DOI: 10.1086/321664 Abstract: Gamma ray burst afterglows can be identified in single epoch observations using three or more optical filters. This method relies on color measurements to distinguish the power law spectrum of an afterglow from the curved spectra of stars. Observations in a fourth filter will further distinguish between afterglows and most galaxies up to redshifts z ~ 1. Many afterglows can also be identified with fewer filters using ultraviolet excess, infrared excess, or Lyman break techniques. By allowing faster identification of gamma ray burst afterglows, these color methods will increase the fraction of bursts for which optical spectroscopy and other narrow-field observations can be obtained. Because quasar colors can match those of afterglows, the maximum error box size where an unambiguous identification can be expected is set by the flux limit of the afterglow search and the quasar number-flux relation. For currently typical error boxes (10 -- 100 square arcminutes), little contamination is expected at magnitudes R < 21.5 +- 0.5. Archival data demonstrates that the afterglow of GRB 000301C could have been identified using this method. In addition to finding gamma ray burst counterparts, this method will have applications in orphan afterglow'' searches used to constrain gamma ray burst collimation.
 Tomonori Totani Physics , 1998, DOI: 10.1086/311489 Abstract: Gamma-ray bursts (GRBs) and following afterglows are considered to be produced by dissipation of kinetic energy of a relativistic fireball and radiation process is widely believed as synchrotron radiation or inverse Compton scattering of electrons. We argue that the transfer of kinetic energy of ejecta into electrons may be inefficient process and hence the total energy released by a GRB event is much larger than that emitted in soft gamma-rays, by a factor of \sim (m_p/m_e). We show that, in this case, very strong emission of TeV gamma-rays is possible due to synchrotron radiation of protons accelerated up to \sim 10^{21} eV, which are trapped in the magnetic field of afterglow shock and radiate their energy on an observational time scale of \sim day. This suggests a possibility that GRBs are most energetic in TeV range and such TeV gamma-rays may be detectable from GRBs even at cosmological distances, i.e., z \sim 1, by currently working ground-based telescopes. Furthermore, this model gives a quantitative explanation for the famous long-duration GeV photons detected from GRB940217. If TeV gamma-ray emission which is much more energetic than GRB photons is detected, it provides a strong evidence for acceleration of protons up to \sim 10^{21} eV.
 Physics , 2000, DOI: 10.1103/PhysRevLett.85.1362 Abstract: A gamma-ray burst fireball is likely to contain an admixture of neutrons, in addition to protons, in essentially all progenitor scenarios. Inelastic collisions between differentially streaming protons and neutrons in the fireball produce muon neutrinos (antineutrinos) of ~ 10 GeV as well as electron neutrinos (antineutrinos) of ~ 5 GeV, which could produce ~ 7 events/year in kilometer cube detectors, if the neutron abundance is comparable to that of protons. Photons of ~ 10 GeV from pi-zero decay and ~ 100 MeV electron antineutrinos from neutron decay are also produced, but will be difficult to detect. Photons with energies < 1 MeV from shocks following neutron decay produce a characteristic signal which may be distinguishable from the proton-related MeV photons.
 Page 1 /100 Display every page 5 10 20 Item


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# Relationship between sparsity and rank of a boolean function

I have the following question when I was going through the proof of the following theorem.

Theorem. For XOR function $f \circ XOR$, $rank(M_{f \circ XOR}) = ||\hat f ||_0$ where $M_{f \circ XOR}$ is a matrix such that $M_{f \circ XOR}(x,y) = f(x + y)$.

The proof of this theorem essentially shows that fourier coefficients of $f$ are the eigen values of $M$: $\hat f(S)$ is an eigen value and the corresponding eigen vector is $\chi_S$. And now, using spectral decomposition, one can conclude that the rank is equal to the sparsity.

My question is: Consider any function $F:\{0,1\}^N \rightarrow \{0,1\}$, and let $M_F \in \{0,1\}^{N/2} \times \{0,1\}^{N/2}$ be a $2^{N/2} \times 2^{N/2}$ dimensional matrix whose entries are as follows: $$M_F(x,y) = F(xy).$$ Is there any relationship between $||\hat F||_0$ and $rank(M_F)$?

• In $F(xy)$ in your last equation, presumably $xy$ denotes concatenation? – kodlu Jan 5 '18 at 15:06
• Yes, that is correct. – sagnik Jan 5 '18 at 15:07


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# How to choose proper active space in method such as CASPT2 and CASSCF?

I am confused about other research paper which use CASPT2 method. I am not clear about how to choose active space. What can I refer to? (tutorial reviews, papers, or any other things)

• Aug 16 '16 at 13:01

## 1 Answer

Choosing the active space in any calculation with a multiconfigurational reference (CASSCF, RASSCF, GASSCF, CASPT2, NEVPT2, MRCI, the list goes on) requires more chemical intuition than more commonly-used "black box" quantum chemistry methods, such as DFT. Here, chemical intuition means preexisting knowledge about 1. the system under investigation, and 2. the properties being calculated. For example, a minimal active space may be sufficient to qualitatively capture static correlation in a system, but the active space might change when targeting excitation energies, and might be different again when calculating spin-orbit coupling constants. More specifically, it is understanding what orbitals are most important in composing your wavefunction and have the largest influence on molecular properties of interest. Unfortunately, this can also involve quite a bit of trial and error.

Some very general recommendations on the workflow for performing multiconfigurational calculations are in the numbered list in this answer (yes, a shameless plug).

Here is a more complete list of references I've found useful in my own research:

• [1], [2] are slides from the group that develops GAMESS, one of the premier packages for multiconfigurational calculations. The second set of slides in particular has examples that highlight the chemical motivation for different active spaces depending on the molecule and the property (internal rotation in ethylene, internal rotation in $\ce{H2C=NH}$, dissociation of $\ce{H2C=O} \rightarrow \ce{CH2} + \ce{O}$, isomerization of bicyclobutane to 1,3-butadiene, and others).

• [3] (paywalled) is a more formal overview of the above. Excellent.

• [4] is a set of very detailed and theory-focused slides from the original authors of DALTON. I link to them only for completeness; there aren't many practical examples here.

• [5] (paywalled) is a comprehensive review article from the authors of COLUMBUS, with many examples. Excellent.

• [6] (paywalled) is an article from the authors of COLUMBUS on the application of multiconfigurational methods to geometries of small molecules. It has good explanations on the choice of active space based on electronic configurations.

In addition, there are many other online resources that are more tutorial oriented: [7], [8], [9], [10]. These may appear to be program-specific, but the chemistry is transferable between different programs; only small technical details in how to set up calculations should be different.

Finally, some programs have nice examples or helpful hints in their documentation. DALTON (PDF), Molcas, and ORCA (PDF) come to mind.

## References

• [3]: Schmidt, M. W.; Gordon, M. S. The construction and interpretation of MCSCF wavefunctions. Annual Review Of Physical Chemistry 1998, 49, 233-266. DOI: 10.1146/annurev.physchem.49.1.233

• [5]: Szalay, P. G.; Müller, T.; Gidofalvi, G.; Lischka, H.; Shepard, R. Multiconfiguration Self-Consistent Field and Multireference Configuration Interaction Methods and Applications. Chemical Reviews 2012, 112, 108-181. DOI: 10.1021/cr200137a

• [6]: Shepard, R.; Kedziora, G. S.; Lischka, H.; Shavitt, I.; Müller, T.; Szalay, P. G.; Kállay, M.; Seth, M. The accuracy of molecular bond lengths computed by multireference electronic structure methods. Chemical Physics 2008, 349, 37-57. DOI: 10.1016/j.chemphys.2008.03.009


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1. ## Integtration using u substitution help please

Hi all i am trying to integrate the following but can't see the wood for the trees..
[Cos(x/2)-sin(3x/2)]dx

So far i have u=(x/2) and du=1/2. This is where i am starting to lose my mind as i cannot seem to make this fit the standard integrals. Any help would be greatly appreciated

2. ## Re: Integtration using u substitution help please

Originally Posted by tommorgan
Hi all i am trying to integrate the following but can't see the wood for the trees..
[Cos(x/2)-sin(3x/2)]dx
$\int {(\cos (\theta ) - \sin (3\theta ))d\theta = \sin (\theta ) + \frac{1}{3}\cos (3\theta )}$

3. ## Re: Integtration using u substitution help please

You can do each term separately, each with a different substitution.


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# Panneau Drivers¶

Edit Driver popover.

Référence

Éditeur: Éditeur Graph Drivers Sidebar region ‣ Drivers N

Référence

This panel is visible in Sidebar of the Drivers Editor or as a popover when adding a driver to a property.

It shows the property that is being driven, followed by a series of settings that determine how the driver works.

## Driver Settings¶

### Type¶

There are two categories of drivers:

• Built-in functions (Average, Sum, Min and Max)

The driven property will have the value of the average, sum, lowest or highest (respectively) of the values of the referenced Driver Variables. If there is only one driver variable, these functions will yield the same result.

• Custom (Scripted Expression).

An arbitrary Python expression that can refer to the Driver Variables by name. See Expressions.

### Driver Value¶

The current result of the driver setup. Useful for debug purposes.

### Variables¶

See Driver Variables.

### Update Dependencies¶

Forces an update for the Driver Value dependencies.

### Show in Drivers Editor¶

Opens the fully featured Drivers Editor. This button only appears in the popover version of the Drivers panel.

## Variables de pilote¶

Variables are references to properties, transformation channels, or the result of a comparison between transformations of two objects.

Drivers should access object data via Driver Variables, rather than direct references in the Python expression, in order for dependencies to be correctly tracked.

Ajoute une nouvelle variable de pilote.
Copy/Paste Variables
Copies the current variable list so it can be pasted into another driver’s variable list.
 Single property. Transform channel. Distance.
Name
Name for use in scripted expressions. The name must start with a letter, and only contain letters, digits, or underscores.
Variable Type

Le type de variable à utiliser.

Single Property

Retrieves the value of a RNA property, specified by a data-block reference and a path string.

In case of transform properties, this will return the exact value of the UI property, while Transform Channel will take parenting and/or constraints into account as needed.

ID Type
Le type d’ID-Block. Par exemple : Key, Image, Object, Material.
ID
L’ID du type de l’ID-Block. Par exemple : « Material.001 ».
RNA Path
The RNA name of the property, based on a subset of Python attribute access syntax. For example: location.x or location[0] for the raw X location value, or ["prop_name"] for a custom property.

Astuce

The easiest way to create a variable of this type is to use the Copy As New Driver context menu option of the input property, and paste the result into the driver via Paste Driver Variables.

Transform Channel

Retrieves the value of a Transform channel from an object or bone.

ID
ID de l’objet. Par exemple : Cube, Armature, Camera.
Bone
ID de l’os de l’armature. Par exemple : « Bone », « Bone.002 », « Arm.r ». Cette option concerne les armatures.
Type

Par exemple : Position X, Rotation X, Échelle X.

The Average Scale option retrieves the combined scale value, computed as the cubic root of the total change in volume. Unlike X/Y/Z Scale, this value can be negative if the object is flipped by negative scaling.

Mode (Rotation)
For rotation channels, specifies the type of rotation data to use, including different explicit Euler orders. Defaults to using the Euler order of the target. See Rotation Channel Modes.
Space
World Space, Transform Space, Local Space.
Rotational Difference
Provides the value of the rotational difference between two objects or bones, in radians.
Distance
Provides the value of the distance between two objects or bones.
Value
Affiche la valeur de la variable.

### Rotation Channel Modes¶

Rotation Transform Channels support a number of operation modes, including:

Auto Euler
Uses the Euler order of the target to decompose rotation into channels.
XYZ Euler, …
Explicitly specifies the Euler rotation order to use.
Quaternion
Provides the Quaternion representation of the rotation.
Swing and X/Y/Z Twist

Decomposes the rotation into two parts: a swing rotation that aims the specified axis in its final direction, followed by a twist rotation around that axis. This is often necessary for driving corrective Shape Keys and bones for organic joint rotation.

This decomposition is often produced in rigs by using a helper bone with a Damped Track Constraint to extract the swing part, and its child with Copy Transforms to extract the twist component.

The channels values for Swing and Y Twist are:

Falloff curves for weighted angles.

Y Rotation
True angle of the twist rotation.
W Rotation
True angle of the swing rotation, independent of its direction.
X Rotation, Z Rotation

Weighted angles that represent the amount of swing around the X/Z axis.

The magnitude of the angle equals W Rotation when the rotation is purely around that axis, and fades out to zero as the direction changes toward the other axis, following the falloff curves from the graph on the right.

Mathematically, the swing angles are computed from quaternion components, using $$2 \arccos(w)$$ for W and $$2 \arcsin(x)$$ etc. for the others. The component of the swing rotation that corresponds to the twist axis is always 0, and is replaced by the twist angle.

## Expressions¶

Expression

A text field where you can enter an arbitrary Python expression that refers to Driver Variables by their names.

The expression has access to a set of standard constants and math functions provided in the Driver Namespace. For an example of adding a custom function to the namespace, see the driver namespace example.

For performance reasons it is best to use the Simple Expressions subset as much as possible.

Use Self

If this option is enabled, the variable self can be used for drivers to reference their own data. Useful for objects and bones to avoid having creating a Driver Variable pointing to itself.

Example: self.location.x applied to the Y rotation property of the same object will make the object tumble when moving.

Note that dependencies for properties accessed via self may not be fully tracked.

### Simple Expressions¶

Blender can evaluate a useful subset of Python driver expressions directly, which significantly improves performance, especially on multi-core systems. To take advantage of this, the driver expression must only use the following features:

Variable Names
Use only ASCII characters.
Literals
Floating point and decimal integer.
Globals
frame
Constants
pi, True, False
Operators
+, -, *, /, ==, !=, <, <=, >, >=, and, or, not, conditional operator/ ternary if
Functions
min, max, radians, degrees, abs, fabs, floor, ceil, trunc, int, sin, cos, tan, asin, acos, atan, atan2, exp, log, sqrt, pow, fmod

Simple expressions are evaluated even when Python script execution is disabled.

When an expression outside of this subset is used, Blender displays a « Slow Python expression » warning. However, as long as the majority of drivers use simple expressions, using a complex expression in select few is OK.

Voir aussi


text:
# Repeating preactions

I'm trying to use multiple similar preactions. Since I would prefer to automatise that I tried to use foreach with preactions. But it appears that's not working.

Let's see an example

\documentclass{minimal}
\usepackage{tikz}

\begin{document}

\begin{tikzpicture}
\draw node
[preaction={fill,transform canvas={xscale=1.03,yscale=1.03},red}]
[preaction={fill,transform canvas={xscale=1.02,yscale=1.02},green}]
[preaction={fill,transform canvas={xscale=1.01,yscale=1.01},blue}]
[text width=10em, text height=10em,circle,fill=white] {} ;
\end{tikzpicture}
\end{document}

That will produce a (ugly) node with three coloured borders. Now I know that loops exist, I may want to use ten colours and I'm a lazy guy. So how can I use foreach to avoid copying-pasting ten times the same code?

I tried what seemed the most natural to me:

\draw node
\foreach \scale/\color in {1.03/red, 1.02/green, 1.01/blue} {
[preaction={draw,transform canvas={xscale=\scale,yscale=\scale},\color}]
}
[text width=10em, text height=10em,circle,fill=white] {} ;

And the compilator complains:

ERROR: Package tikz Error: A node must have a (possibly empty) label text.
l.59   \foreach
\scale/\color in {1.03/red, 1.02/green, 1.01/blue} {

Any idea would be welcome.

-
Welcome to TeX.sx! A tip: If you indent lines by 4 spaces, then they are marked as a code sample. As opposed to the <pre> markup, the indent renders the code as LaTeX (colored). You can also highlight the code and click the "code" button (with "{}" on it). Using <pre> for an error message, however, seems sensible, since it's not real LaTeX. –  doncherry Aug 5 '11 at 9:57

This is a "feature" of how TeX parses the TikZ syntax. When TeX finds the keyword node it starts looking for what comes next (up to the end of the node) and it only "knows" about a certain number of things, such as the [optional settings] the (label) and the {text}. It doesn't know how to deal with the \foreach command. It also doesn't seem too keen on the foreach key-word, but I may not have gotten the syntax right for that so someone else might have a solution using that.

However, what you want to do is eminently possible using the power of the \pgfkeys mechanism. This has inbuilt a method for iterating over a list (imaginatively called .list). Here's some code that works for your example.

\documentclass{article}
% \url{http://tex.stackexchange.com/q/24948/86}
\usepackage{tikz}

\tikzset{
repeated preaction/.style={%
repeat preaction/.list={#1}
},
repeat preaction/.code args={#1/#2}{%
\tikzset{
preaction={
fill,
transform canvas={xscale=#1,yscale=#1},
#2
}
}
}
}

\begin{document}

\begin{tikzpicture}
\draw node
[preaction={fill,transform canvas={xscale=1.3,yscale=1.3},red}]
[preaction={fill,transform canvas={xscale=1.2,yscale=1.2},green}]
[preaction={fill,transform canvas={xscale=1.1,yscale=1.1},blue}]
[text width=10em, text height=10em,circle,fill=white] {} ;
\end{tikzpicture}

\begin{tikzpicture}
\draw node
[repeated preaction={1.3/red, 1.2/green, 1.1/blue}]
[text width=10em, text height=10em,circle,fill=white] {} ;
\end{tikzpicture}
\end{document}

What this does is define a few extra keys. I'll take them in the order in which they are called.

repeated preaction : This takes one argument, which is a list. Its sole job is to pass that argument as a list to the next key. If you were prepared to write repeat preaction/.list={...} in the [...] bit then you could skip this part. (Perhaps a better name for this key would be repeat preaction with.)

repeat preaction : This is the code that is iterated over by the list syntax. So it gets passed each element of the list in turn. As the list is in the format scale/colour, we specify the argument syntax accordingly rather than just taking a list of arguments. This executes its code on each iteration, and what it does is simply to set the corresponding key. As we are now in "code" form (necessary to handle the arguments), we have to jump back in to the option-setting mode with the \tikzset command.

With a little more finesse, the actual preaction could also be passed to the initial option (though one would have to be careful about macro parameters).

(I scaled up your scales a bit so that they were easier to see. Also, the sides of picture got cropped by the cropper, that's because of the use of scaling confusing it about bounding boxes. That wouldn't happen in a real document.)

-
Thanks a lot, that was what I was looking for! –  kelux Aug 5 '11 at 13:20


text:
# className

0th

Percentile

##### Class names including the corresponding package

The function className() generates a valid references to a class, including the name of the package containing the class definition. The object returned, from class "className", is the unambiguous way to refer to a class, for example when calling setMethod, just in case multiple definitions of the class exist.

Function "multipleClasses" returns information about multiple definitions of classes with the same name from different packages.

Keywords
programming , classes
##### Usage
className(class, package)multipleClasses(details = FALSE)
##### Arguments
class, package

The character string name of a class and, optionally, of the package to which it belongs. If argument package is missing and the class argument has a package slot, that is used (in particular, passing in an object from class "className" returns itself in this case, but changes the package slot if the second argument is supplied).

If there is no package argument or slot, a definition for the class must exist and will be used to define the package. If there are multiple definitions, one will be chosen and a warning printed giving the other possibilities.

details

If FALSE, the default, multipleClasses() returns a character vector of those classes currently known with multiple definitions.

If TRUE, a named list of those class definitions is returned. Each element of the list is itself a list of the corresponding class definitions, with the package names as the names of the list. Note that identical class definitions will not be considered “multiple” definitions (see the discussion of the details below).

##### Details

The table of class definitions used internally can maintain multiple definitions for classes with the same name but coming from different packages. If identical class definitions are encountered, only one class definition is kept; this occurs most often with S3 classes that have been specified in calls to setOldClass. For true classes, multiple class definitions are unavoidable in general if two packages happen to have used the same name, independently.

Overriding a class definition in another package with the same name deliberately is usually a bad idea. Although R attempts to keep and use the two definitions (as of version 2.14.0), ambiguities are always possible. It is more sensible to define a new class that extends an existing class but has a different name.

##### Value

A call to className() returns an object from class "className".

A call to multipleClasses() returns either a character vector or a named list of class definitions. In either case, testing the length of the returned value for being greater than 0 is a check for the existence of multiply defined classes.

##### Objects from the Class

The class "className" extends "character" and has a slot "package", also of class "character".

##### Aliases
• className
• multipleClasses
• className-class
##### Examples
library(methods) # NOT RUN { className("vector") # will be found, from package "methods" className("vector", "magic") # OK, even though the class doesn't exist className("An unknown class") # Will cause an error # }
Documentation reproduced from package methods, version 3.6.1, License: Part of R 3.6.1

### Community examples

Looks like there are no examples yet.


text:
# Sum Tangent $n^2$

Geometry Level 5

$\sum\limits_{n=1}^K \tan\left(\dfrac{n^2\pi}{12}\right) = 25$

Santa wants you to determine the value of $K$ to make the above equation true.

What is $K?$

×


text:
# Rate of change of area of triangle

The video explains how to determine height of a triangle given length base and area. 2015 the function, generating the rate of change function. The altitude of a triangle is increasing at a rate of 1 cm/min while the area of the triangle is increasing at a rate of 2cm2/min. Solution: Area of the triangle = A = 243 cm 2. 2020 or \frac {\triangle y } {\triangle x} , and is the quotient between rate of change and the amplitude of the interval considered on the  Get an answer for 'The area of a square with sides of length s is given by A = s^2 . WOULDNT you multiply 10x by . a) Show that the area of the triangle is equal to (1/2)s^2sin(θ). So xx' +yy' =0 and y'=-xx'/y and substituting this into dA/dt gives dA/dt =(-x2x'/y +x'y)/2 = x'(y2-x2)/2y. These two equal sides always join at the same angle to the base (the third side),  30 abr. and its radius r are decreasing at the rate of 1 cm/hr. We are given: ##(da)/dt determine the rate at which water is being pumped into the trough. 1 Ah (hQ - hp) 1. A = 175sin95. Then, if S S S is the area of the triangle, what is the rate of change of S S S with respect to time when ∠ A = π 2? \angle \text{A}=\frac{\pi}{2}? ∠ A = 2 π ? Related Rates Using 3D Geometry Let's move on to examples using 3D Geometry. This can be solved using the procedure in this article, with one tricky change. Area of a rectangle. (Note that for X=5, we have a right triangle!) 2. 4 Introduction to flow-rate · 1. 5 sin ( /2) and h = 3. So we know that in this moment the length of the square’s sides is 4 cm. 2011 Find the rate at which the area of the triangle is increasing when Related Rates: Triangle Angle and Area rate of change of area =. After having obtained both coordinates, simply use the slope formula: m=(y2 - y1)÷(x2 - x1). 3 15 4 4 9 12 a. Find the rate at which the length of a side is changing when the area of the triangle is 200 cm 2. 1 oct. Here the unknown is the rate of change of the area, dA/dt dA/dt = 1/2 (x*dy/dt+y*dx/dt) So in the concluding statement, we specify that the area is decreasing at a rate of 18 cm 2 /sec. we need to calculate 𝒅𝑨/𝒅𝒓 We know that Area of Circle = A = 〖𝜋𝑟〗^2 Finding 𝒅𝑨/𝒅𝒓 𝑑𝐴/𝑑𝑟 = (𝑑 (〖𝜋𝑟〗^2 Adjust θ to illustrate the following related rates problem: Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0. 2015 This is a related rates (of change) type problem. Enter sides a and b and angle C in degrees as positive real numbers and press "enter". Usually called the "side angle side" method, the area of a triangle is given by the formula below. It is known that the perimeter $a+b+c$is always constant and at the moment $a=3$and $b=4$the rate of change of $a$is $108$cm/sec. The average rate of change of trigonometric functions are found by plugging in the x-values into the equation and determining the y -values. Now take the derivative of both sides with respect to t. Answer. Plugging in, A = 1 2 ⋅ 10 ⋅ 35sin95. 2012 This video provides an example of how to solve a related rates problem involving the area of a triangle and rate of change of the base. Calculate area of triangle when the length sides are given. Find the rate at which the area of the triangle is increasing when the angle between the sides of a fixed length is . Mean value theorem. Remember, the problem also told us that the side lengths are increasing at a rate of 6. LQ Two sides of a triangle are 2 m and 3 m in length and the angle between them is increasing at a rate of 0. 14 jun. So the area A takes on a maximum value of 6 square inches when side X = 5 inches. 3. Area of a trapezoid. B starts at (0, 1) and moves upward at a constant rate of 2 units per second. Find a formula for the rate of change of the distance D between the two cars. The area is increasing by a rate of 20 ft2=sec. ) area changing Rates of Change Application of Rates of Change To get a better approximation, let's zoom in on the graph and move point Q towards point P at intervals of 0. Review Section 3. 4 cm/s, Y is decreasing at 0. Furthermore, we need to related the rate at which is changing,  Find the rate at which the area of the circle is changing when the radius Using the indicated triangle in the diagram, we get the following relationship. dA/dt when x = 8 cm & y = 6 cm We  Step 2: Write an equation involving the variables whose rates of change is the area of the triangle increasing when the sides are 30 cm long? (Stepi). b. So . $and leg two of the right triangle is increasing at the rate of$ \ 7 \ in/sec. ) The average rate of change of the area of the triangle on the time interval [ π /6, π /4] is. You have the constant rate of change of the length of the sides of an equilateral triangle and are asked about the rate of change of the area. Suppose that one side of a triangle is increasing at a rate of 3cm=s and that a second side is decreasing at a rate of 2cm=s. Then In the triangle shown above, if B increases at a constant rate of 3 radians per minute, at what rate is x increasing in units per minute when x equals 3units? b. 29 m/s 0. Find the rate at which the area of the triangle is increasing when the angle betwe … read more Optimization: area of triangle & square (Part 2) Rates of change in other applied contexts (non-motion problems) 4 questions. 01 until point Q is just right of point P. Let's look at another example. So if we know h, we know x (and vice versa). Jiwen He, University of Houston Math 1431 – Section 24076, Lecture 9 September 25, 2008 9 / 22 So before we can figure out the rate of change, we must first understand what the function is. 3 units2. A = (3. When the hypotenuse is changing at a rate of 7 cm/min and the edges are 8 cm, 15 cm, 17 cm, find the rates of 4. Jn a rig ht triangle, leg x is increasing at the rate of 2 m/s while leg y is decreasing so that the area of the üiangle is always equal to 6 m . 9 mar. When we calculate derivatives, we are calculating rates  7 ago. Transcript. The problem is that equation suggests that side c is increasing at a rate of 22. Given: Let 8m be the length of the base and 6m is the other side 푑휃 푑? = 0. 5. Mathematics for the International Student: Mathematics SL | 3rd  17 may. 8)^2 + (\ell – x)^2 \] You’re looking for dh/dt. 5 cm/min. The radius of a circle is decreasing at a constant rate of 0. t Radius i. We’ll leave it to you to check these rates of change. 3 - Two cars start moving from the same point in two directions that makes 90 degrees at the constant speeds of s1 and s2. Explanation: Well, I'm assuming that you are saying that side a = 10, side c = 35, and angle B = 95∘. gf 3 (Hint: The formula for the area of an equilateral triangle is A= 4 3) If the base b of a triangle is Increasing at a rate of 3 inches per minute while its height h is decreasing at a rate of 3 inches per minute, which of the following must be true about the area A of the triangle? (B) (D (E) A is always increasmg. Area of a rhombus. 5 cm/min while the area of the triangle is increasing at a rate of 5 square cm/min. Let x be the angle between the two equal sides. A large spherical balloon is being lled at a rate of ˇ 3 in 3/sec. The base of an isosceles triangle is 10 feet long. (a) Show that the area of the triangle is given by A=1/2s^2 sin θ; (b) If θ is increasing at the rate of 1/2 radian per minute, find the rate of change of the area when θ =pi/6 and θ =pi/3. (Connect the origin with the y -intercept and x -intercept of the tangent line. However, this formula uses radius, not circumference. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. Try this Drag the orange dots on each vertex to reshape the triangle. At what rate is the radius increasing when the radius is 9 cm? 9. 4 cm/sec and the height is decreasing at 0. 5*12*15*cos(π/3)*dC/dt=45*2=90m²/min Trust this helps. This formula only works, of course, when you know what  Rate of change and area under the curve. In this video, we are given the lengths of three sides a triangle and use to find height triangle. The common formula for area of a circle is A=pi*r^2. 16. At what rate is the area of the triangle changing when the hypotenuse is 1 m long? (c). Fun fact: in hyperbolic geometry, the area of a triangle is a function of the angles. Question 2: Find the length of the base of an isosceles triangle whose area is 243 cm 2, and the altitude of the triangle is 27 cm. asked Nov 9, 2018 in Mathematics by simmi ( 5. Use to find any altitude. so let's say that we've got a pool of water and I drop a rock into the middle of that pool of water I drop a rock in the middle of that pool of water and a little while later a ripple has a little wave a ripple has formed that is moving radially outward from where I drop the rock so let me see how well I can draw that so it's moving radially outwards so that is the ripple that is formed for me This is a related rates (of change) type problem. 8. 2019 At what rate is the area of the triangle changing when the base is 10 ft and the height is 70 ft? a) -25 ft^2/sec b) -45 ft^2/sec The conventional method of calculating the area of a triangle (half base times altitude) As you now drag point A, notice that the area does not change. Enter the values. 2. Practice. y = √ 225 − x 2 = √ 225 − 49 = √ 176 y = 225 − x 2 = 225 − 49 = 176. A "related rates'' problem is a problem in which we know one of the rates of change at a given instant—say, ˙x = dx / dt —and we want to find the other rate ˙y = dy 3. This is the Area of triangle=(12*15/2)*sin(π/3)=45√3m², angle increase rate=dC/dt=2rads/min Area increase rate=0. $Let’s call that hypotenuse length “h. The object in question is a snowball. The overall steps are: Draw a triangle consistent with the given information, labeling relevant information Determine which formulas make sense in the situation (Area of entire triangle based on two fixed-length sides, and trig relationships of right triangles for the variable height) Relate any unknown variables (height) back to the variable (theta) which corresponds to the only given rate The area of any triangle can be calculated using the lengths of two of its sides and the sine of their included angle. determine the rate at which water is being pumped into the trough. Pascal's triangle maybe a table of numbers within the shape of an equiangular triangle, where the k-the number within the n-the row tells you ways many combinations of k elements there are from a group of n elements (Note that we follow the convention that the highest row, the one with the only 1, is taken into account to be row zero, while the Step 2. Rates of Change Application of Rates of Change To get a better approximation, let's zoom in on the graph and move point Q towards point P at intervals of 0. A "related rates'' problem is a problem in which we know one of the rates of change at a given instant—say, ˙x = dx / dt —and we want to find the other rate ˙y = dy a) Show that the area of the triangle is equal to (1/2)s^2sin(θ). 5) 2 sin ( /2) cos ( /2) It’s just equal to ℎ squared. The answer is$1. Multiply the base and height of a triangle; then divide by two or multiply by half. Let the area of the triangle be xy/2 and dA/dt becomes dA/dt= (xy' +yx')/2. By looking at the given statement, we can gather a few important fact quickly. A = area and, since the area of a triangle is  19 abr. d. The area of a triangle is half the base times the height so. The procedure to use the instantaneous rate of change calculator is as follows: Step 1: Enter the function and the specific point in the respective input field. e. How do you find Theta? Just remember the cosine of an angle is the side adjacent to the angle divided by the hypotenuse of the triangle. 2021 The average rate of change finds how fast a function is changing with respect to something else changing. The legs of an isosceles right triangle increases in length at a rate of 2 m/s. Use the chain rule to find the rate of change of one quantity that depends on Find the rate at which the area of the triangle is changing when the angle  What is the rate of change of the area A ( t ) A(t) A(t)A, left parenthesis, t, right parenthesis of the triangle at that instant? Match each expression with  The rate at which the area increases when the side is 10 cm , is. Two sides of a triangle are 2 m and 3 m in length and the angle between them is increasing at a rate of 0. \) Area = (1 / 2) b c sin(A) = (1 / 2) c a sin(B) = (1 / 2) a b sin(C) How to use the calculator Here we assume that we are given sides a and b and the angle between them C. The. #6. And one side of a triangle is increasing at a rate of 1 cm/s and a second side is decreasing at a rate of 2 cm/s. A fish is reeled in at a rate of 1 foot per second from a point 15 feet above the water. trough are isosceles triangles with a height of 3 feet. Explanation: . (b) At what rate is the area of the triangle (formed by the wall, the ladder, and the ground) changing at the same time? Now the unknown is the rate of change of the area, dA/dt. Multiple Choice: Consider the triangle shown below. LQ Now that we’ve calculated the rates of change we can plug in the numbers dV = 2 and h = 5: dt 2 = 4 π(5)2h 25 2 = 4πh 1 h = ft/min 2π We were given the rate at which the volume of water in the tank was changing and we used that to compute the rate at which the water in the tank was rising. In the figure above, click on "freeze altitude". At what rate is the area of the triangle changing when the legs arc 2 m long? (b). Express the area A of the triangle as a function of x in radians. So my question is: The included angle of the two sides of a constant equal length s of an isosceles triangle is ϑ. Thus, v = 6 h 2. 3 27) A ladder 25 feet long is leaning against the Quiz: Average Rate of Change of a Function (1) Create a Triangle with Given Area: Quick Formative Assessment; Create a Triangle with Given Area (V2) 10. (a). This makes the volume 12 x (1/2) h 2. You will also use similar triangles to see why this formula works, and why the line with slope m and y -intercept ( 0, b) is the graph of the This is a related rates (of change) type problem. rate is the area changing at the instant when the length equals 10 feet and the width equals 8 feet? Solution: A area of triangle Given: dx dt = 5 ft/sec y dt = 2 ft/sec Find: dA dt when x = 100 ft. b) If theta is increasing at the rate of 1/2 radian per minute, find the rates of change of the area when θ = π/6, and θ = π/3. The rate of change of the surface area of the , cube in c m 3 / m i n, when the volume of the cube is 1 2 5 c m 3, is Got it. 12 cm 10 cm The cost in dollars of producing a; items in a factory each day is given by C(x) = 0. Determine the rate of change of the angle opposite the base of a right triangle -whose length is increasing at a rate of 1 inch per minute, and whose height is a constant 2 inches - when the area of the triangle is 2 square inches. RELATED RATES – Sphere Surface Area Problem. How fast is the area of the triangle increasing when 7 2 t = seconds? 8. At Area of an equilateral triangle. It’s a ballpark average that gives you a good idea of how long its going to take to get from a to b, even if the object you’re studying doesn’t always move along at a steady rate. More precisely, in the hyperbolic plane with uniform curvature –1, the sum of the interior angles (measured in radians) of a triangle of area A is (pi – A). 5) 2 sin ( /2) cos ( /2) Rates of change: The rate of change of the area of a triangle. c) Explain why the rate of change of the area of the triangle is not constant even though the rate of change of θ is constant. 2014 (a) At what rate is the area of the triangle changing when the legs are 5m long? SOLUTION: First we should define the variables. We have the average rate of change in the area of our triangle as the height changes from 23 centimetres to 14 centimetres, which we will call 𝐴 average, is equal to 𝐴 evaluated at 23 minus 𝐴 evaluated at 14 divided by 23 minus 14. How does the change in area  10 ago. Area of a cyclic quadrilateral. The contained angle 𝜃 is increasing for a triangle where two of the sides are fixed, as shown in the diagram. 029 Ah mpQ = At 10. 8 Rate of Change Velocity and Acceleration Triangle The rate of change of the area A of an equilateral triangle with respect to its side length s dA ds = 1 2 √ 3s. This directly tells us the rate of change of the sides lengths. (b) If ϑ is increasing at the rate of 1/2 radian per minute, find the rate of change of the area when ϑ=pi/6 and ϑ=pi/3. 1 rad /sec. Δ f Δ x = f ( 6) − f ( 1) 6 − 1 = 20 5 = 4. The trigonometric formula for the area of triangles is a r e a s i n = 1 2 𝑎 𝑏 𝐶, where 𝑎 and 𝑏 are the lengths of two sides and 𝐶 is the measure of the included angle. The solution section lists the correct answer as 60 square inches per minute. 8 cm/s and θ is increasing at 0. 222 Example Question: The edges of a right-angled triangle are changing but the perimeter is fixed at 40 cm. of f  17 abr. The rate of change is easy to calculate if you know the coordinate points. The data here gives you the rate of change of the circumference, and from that will want the rate of change of the area. The base of a triangle is increasing at a constant rate of 0. The included angle of the two sides of a constant equal length s of an isosceles triangle is θ. If a snowball melts so that its surface area decreases at a rate of 1, find the rate at which the diameter decreases when the diameter is 10 cm. Find the rate of change of the area when s = 41 feet. 25 in². 000 + + 2200 a b c labour raw materials fixed costs Find C (x), which is called the marginal cost function. Using the notation in the left figure immediately above, you’re looking for the rate of change of the hypotenuse of the triangle with height 1. Substitute actual numbers for every known quantity in the derivative you found in part 6. For x̄we will be moving along the x axis, and for ȳ we will be moving along the y axis in these integrals. Although we didn't make a separate calculator for the equilateral triangle area, you can quickly calculate it in this triangle area calculator. 2 radians/s, using chain rule, find the rate of change of the area of the triangle when X is 3 cm, Y is 4 cm and θ is π/6 radians (≡ 30°) 5. Rate of change B. By the 30-60-90 rule, a special case of a right triangle, we know that the base of this smaller right triangle is and the height of this smaller right triangle is , assuming b to be the hypotenuse. $At what rate is the triangle's$ \ \ \ \ $a. So in this case, since we're referring to the area of the triangle, weaken right out the area formula for unequal lateral triangle, which is radical through four A squared the lower case a being the side ling capital a being the area in order to find B starts at (0, 1) and moves upward at a constant rate of 2 units per second. If the altitude is decreasing at a rate of 2 inches per second, at what rate is the base angle changing when the height is 12 feet? So far I am confused as to what I should be finding the derivative of and what values I should plug in afterwards. 0485 radians per second. da/dt = 1. 11 feb. 1) -10. The average rate of change of a function gives you the “big picture” of an object’s movement. 2 Unit Rate 3: Speed Area of Triangles 2A Finding Area of a Triangle 2B: Areas Involving Parallelograms and Triangles Rate of Change 2: The sides of an equilateral triangle are decreasing at a rate of 3 in/hr. The average rate of change of the area of the triangle on the time interval [?/6,?/4] is In this example, we want to find the rate of change of the area of a triangle at the instant when the angle is a particular value. The slope of the equation has another name too i. Average Rate of Change. 2020 If the rate of change of the side of a square is 0. 68 m/minute which is far from accurate. The altitude of a triangle is increasing at a rate of 1. instantaneous rate of change of area with respect to time at x = 2? area of an equilat- eral triangle with respect to the length of one of its sides. 2019 Let the area of the triangle be xy/2 and dA/dt becomes dA/dt= (xy' +yx')/2. A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 $\text{cm}^2 / \text{sec}$. The height is the length between the base and the highest point of the triangle. 2014 find the percentage increase in the area of triangle if its each side is doubled - Mathematics - TopperLearning. 25. ) hypotenuse changing$ \ \ \ \ $b. Area of a quadrilateral. When the hypotenuse is changing at a rate of 7 cm/min and the edges are 8 cm, 15 cm, 17 cm, find the rates of Height of the triangle (h) = 17 cm. |$\frac{\sqrt{3}}{2}s$Answer to: The altitude of a triangle is increasing at a rate of 2 cm/min while the area of the triangle is increasing at a rate of 2 cm^2/min. Oct 712:58 PM (b) Consider the triangle formed by the side of the house, the ladder, and the ground. 2? the opposite of what this answer is? Itsays BASE is 10 in and BASE chagnges at . Step 3: Finally, the rate of change at a specific point will be displayed in the new window. If X is increasing at 0. Here's a triangle with a base of 5 and a height of 4: This triangle 30 mar. The first step in a related rates question is to determine how the quantities are related. The variables of interest are ##a## = altitude##A## = area and, since the area of a triangle is ##A=1/2ba##, we need##b## = base. The star exerts a little percussive, attractive force on the planet The really important thing is that the little tug force doesn’t change the area of the original triangle, # % &! The new triangle is # % (, which has base # %, just like Determine all rates of change that are known or given and identify the rate(s) of change to be found. But look at the graph from the last example again. Two sides of a triangle are 4m and 5m in length and the angle between them is increasing at a rate of 0. ) dA dt = 2ˇr dr dt 7. If the area of triangle remains constant, at what rate does the angle between these two sides change when the rst side is 20cm long, the second side is 30cm long, and the angle between the two sides is ˇ=6? Is it And we were also given information about the rate of change of the left side of the triangle. e The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. r. A cube of ice melts without changing its shape at the uniform rate of 4 c m 3 / m i n. \) Answer. The given rates of change are in units per minute, so the (invisible) independent variable is t = time in minutes. In terms of the one side of a triangle is increasing at a rate of 1 cm/s and a second side is decreasing at a rate of 2 cm/s. There is a big change to the direction and a small change to the magnitude: Figure 2. Notice that distance = rate$$\cdot$$time also describes the area between the velocity graph and the $$t$$-axis, between $$t = 0$$ and $$t = 2$$ hours. Area of a triangle - "side angle side" (SAS) method. The area of an equilateral triangle is decreasing at a rate of 4 cm 2/min. Area of square = a 2. If the shorter what is the rate of change of the area of the base The area of a triangle is: Area = 1/2(bh) Area = 1/2(70) Given the function defined in the table below, find the average rate of change, in simplest form, of the Area of a triangle (Heron's formula) Area of a triangle given base and angles. It is important to note that one can slide the calculus triangle through the domain. ” Then \[ h^2 = (1. A "related rates'' problem is a problem in which we know one of the rates of change at a given At the same time, how fast is the y coordinate changing? When speed changes, the rate of change of speed can be calculated using the equation: Area under speed-time graph = distance travelled (m) In this problem, the diagram above immediately suggests that we're dealing with a right triangle. Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 $\text{cm}^2$. The average rate of change of the area of the triangle on the time interval [?/6,?/4] is A triangle has a base that is decreasing at a rate of 7 cm/s with the height being held constant. The slope is responsible for connecting multiple points together over a line. A is decreasing only when b > h . we need to calculate 𝒅𝑨/𝒅𝒓 We know that Area of Circle = A = 〖𝜋𝑟〗^2 Finding 𝒅𝑨/𝒅𝒓 𝑑𝐴/𝑑𝑟 = (𝑑 (〖𝜋𝑟〗^2 Click here👆to get an answer to your question  If each side of an equilateral; triangle increases at the rate of √(2)cmsec find the rate of increase of its area whe its side of length 3 cm A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm 2 /sec. A = 174. Since the area of the isosceles triangle is xh, this equals (h/4)h = h 2 /4. 5 EXAMPLE 1: Consider a right triangle which is changing shape in the PROBLEM 12 : The surface area of a cube is increasing at the rate of 600 in2/hr. 5 cos ( /2). com | xwl4d77. e. Step 3. , the rate of change of the radius of a circle and want to know dA dt, the rate of change of the area of the same circle, write A= ˇr2. ) Find the instantaneous rate of change of the area A An isosceles triangle is a triangle with two sides of the same length. (a) Show that the area of the triangle is given by A=1/2s^2 sinϑ. It is simply the process of 7 jul. 2 Unit Rate 3: Speed Area of Triangles 2A Finding Area of a Triangle 2B: Areas Involving Parallelograms and Triangles Rate of Change 2: Explanation: Well, I'm assuming that you are saying that side a = 10, side c = 35, and angle B = 95∘. The problem above appears in the book Calculus for the practical man. 2017 Calculate the percentage change in the area of a triangle if its each side is halved - Maths - Heron\s Formula. 27. 2011 TriangleFanPhoto. Therefore, our final answer for the area of the triangle bounded by the 𝑥-axis, the 𝑦-axis, and the line with equation two 𝑥 plus seven 𝑦 plus 28 equals zero is 28 area units. 6 − 1 = 5. At what rate is the base of the triangle changing when the altitude is 10 cm and the area is 100 cm2? Solution:The area of a triangle is Aarea(t) = 1 2 H(t)B(t), where H(t) denotes the hight of the triangle at The area of the isoceles triangle filled with water is xh. Find the rate at which the area of the circle is changing when the radius is 5 cm. 2014 At what rate is the area changing at the instant when the length equals 10 feet and the width equals 8 feet? Solution: A area of triangle 23 oct. The first step in a question of related rates is to determine how the quantities are related. ) Write the area A of an equilateral triangle as a function of the side length s. 05 cm/sec, then the rate of change in the area of the square when the side is 10cm is. {/eq} Now to find the rate of change of area we need to differentiate it with A triangle has a base that is decreasing at a rate of 7 cm/s with the height being held constant. c. Jiwen He, University of Houston Math 1431 – Section 24076, Lecture 9 September 25, 2008 9 / 22 At what rate is the angle of elevation change? So the rate of change of the angle of elevation when the balloon is 18 feet high is approximately equal to 0. Find an equation that relates the variables whose rates of change are known to those variables whose rates of change are to be found. A girl is flying a kite on a string. The rate of change of the total area = 2 π r ((dr/dt). 12 ???/??? Find 푑퐴 푑? when the angle between sides is ? . Ex 6. Change in x decided by change in y When calculating the area of the rectangle at right on their last quiz, Pauline got 216 while Rene got 1. When given the area of a triangle and The average rate of change of trigonometric functions are found by plugging in the x-values into the equation and determining the y -values. com : Determine the rate of change of the area of an equilateral triangle with respect to the length of a side. How fast is the surface area of the balloon increasing 30 seconds after the balloon began being lled? Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall. Area of a parallelogram given base and height. ex. if the area of the triangle remains constant does the angle between the sides change when the first side is 30 cm long, the second side is 80 cm long, and the angle is pie/4. The rate is the height of the rectangle, the time is the length of the rectangle, and the distance is the area of the rectangle. Differentiate implicitly with respect to $$t$$ to relate the rates of change of the involved quantities. png · When you change the length x of one side of a triangle, the area A of the triangle changes. Notice that you'll need the chain rule here. The angle between these two sides is increasing at a rate of 0. Let's call the volume v. Rate of change in an Isosceles triangle. To find h, we visualize the equilateral triangle as two smaller right triangles, where the hypotenuse is the same length as the side length b. 29 Rates of Change Application of Rates of Change Let's begin with point Q at (2, 10. 4 Section 3. You can assume the unit of measurement is centimeters and time is seconds. Calculate the rate of change of the area, (dA/dt) of the rectangle when x = 0. Area of Isosceles Triangle = (1/2) × b × h = (1/2) × 12 × 17 = 6 × 17 = 102 cm 2. However, I get a different solution. 1 rad/sec. Find the rate at which the base of the triangle is changing when the height of the triangle is 4 cm and the area is 20 cm 2. Learn. 4. At what rate is the base of the triangle changing when the altitude is 9 cm and the area is 81 square cm? The formula for the area of a triangle is A = ½ x b x h or bh/2. But remember that our equation in this step cannot include rates of change. An Isosceles triangle has two equal sides of length 10cm. The rate of changing of the total area when the radius is 5 cm = 2 π (5)(2) = 20 π. 10. Since we know that the relationship between the area of a square and its side lengths is we can find l in this instant. where a,b are the two known sides and C is the included angle . a. ˆ ˆ A triangle has two constant sides of length 3 ft and 5 ft. The area of a triangle A is given by A=(1/2)xy where x is the base and y is the altitude. Transcribed image text: Area The lengths of each side of an equilateral triangle is increasing at a rate of 13 feet per hour. 2016 The angle between these two sides is increasing at a rate of 0. A is decreasing only when b < h . You have the constant charge of the length of the sides of an equilibrium triangle and is questioned about the change rate of the area. If the area of triangle remains constant, at what rate does the angle between these two sides change when the rst side is 20cm long, the second side is 30cm long, and the angle between the two sides is ˇ=6? Is it , the rate of change of the radius of a circle and want to know dA dt, the rate of change of the area of the same circle, write A= ˇr2. If the water runs into the trough at the rate of 4 cubic feet per second, how fast is the water level rising at the instant when the water is 2 feet deep? (Recall: The volume of such a trough is V = 1 2 lwh) 1 10 feet per second 13. The area is dependent on the base and height, and neither of them changes as you move the top vertex side-to-side. The area Z of a triangle is given by = ? 1 2 XYCosθ Where θ is the angle between sides X and Y. 6$ So I try the following formula based on the Pythagorean theorem: $(6^2)^2 + (2t)^2 = 10^2$ Calculate the rate of change of the area, (dA/dt) of the rectangle when x = 0. how fast is the volume decreasing when r = h 10 cm? 15. The given rates of change are in units per minute, so the (invisible) independent variable is ##t## = time in minutes. cm/sec. At what rate is The overall steps are: Draw a triangle consistent with the given information, labeling relevant information Determine which formulas make sense in the situation (Area of entire triangle based on two fixed-length sides, and trig relationships of right triangles for the variable height) Relate any unknown variables (height) back to the variable (theta) which corresponds to the only given rate 2 - Find a formula for the rate of change dA/dt of the area A of a square whose side x centimeters changes at a rate equal to 2 cm/sec. Q. Note that we use the positive 18 cm 2 /sec instead of the negative –18 cm 2 /sec in the statement. 1 cm/sec. To find y' use the Pythagorean theorem x2 +y2 =z2 where z is  If the base b of a triangle is increasing at a rate of 3 inches per minute of the circumference C, what is the rate of change of the area of the circle  1 feb. To find y y (after 12 seconds) all that we need to do is reuse the Pythagorean Theorem with the values of x x that we just found above. For instance, at $$t = 4$$ the instantaneous rate of change is 0 cm 3 /hr and at $$t = 3$$ the instantaneous rate of change is -9 cm 3 /hr. From the diagram b/2 = 3. 2 Related Rates. 5 and the point on the graph is moving with a given horizontal speed (dx/dt) = 2. Hi Jeevitha. A = xy and so dA dt = d dt (xy) = dx dt y + x dy dt dA dt = 5 8 + 10 ( 2) = 40 20 = 20. Here the side length is increasing with respect to time. Find the rate at which the area of the triangle is increasing when the angle between the sides is 흅?. That is, We need to determine dA/dt when a = 9 cm. 5 in per second. The average rate of change in f between x = 1 and x = 6 is 4. Area of a square. Step 2: Now click the button “Find Instantaneous Rate of Change” to get the output. At what rate is the angle of elevation change? So the rate of change of the angle of elevation when the balloon is 18 feet high is approximately equal to 0. In this lesson, you will learn a formula to compute the slope of a line using any two points on the line  For instance, there's the basic formula that the area of a triangle is half the base times the height. We can use the formula to find the area of the triangle: A = 1 2acsinB. Height of the triangle (h) = 27 cm Find the rate of change in the area of triangle ABC as f) changes, at the time when = 600. At what rate is the area of the triangle changing when the legs are 2 m  5 may. | $\frac{\sqrt{3}}{2}s$ Then, if S S S is the area of the triangle, what is the rate of change of S S S with respect to time when ∠ A = π 2? \angle \text{A}=\frac{\pi}{2}? ∠ A = 2 π ? Related Rates Using 3D Geometry Let's move on to examples using 3D Geometry. 8 m (the man’s height) and base $\ell – x. To calculate the area of an equilateral triangle you only need to have the side given: area = a² * √3 / 4. Find the marginal cost when 150 items are produced. Side of Step 3: Find An Equation That Relates The Unknown Variables. Therefore, all the triangles you can create this way have the same area. Find exactly the rate of change in the area of triangle PQR as changes, at the time when. A = b/2 h. Solution : Let "a" be the side of the square and "A" be the area of the square. Find the rate of change in ft/s of the hypotenuse when it is$10$feet long. Determine all rates of change that are known or given and identify the rate(s) of change to be found. 5b*h. How fast is the area enclosed by the triangle decreasing when the sides are 2 feet long? A spherical balloon is being filled in such a way that the surface area is increasing at a rate of 20 cm 2 /sec when the radius is 2 meters. 3 27) A ladder 25 feet long is leaning against the Step 2. This set of resources starts with the equation of a straight line in the form y=mx+c and checking the understanding 17 feb. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is pi/3 rad. Rates of increase are positive and rates of decrease are negative. 2020 If two related quantities are changing over time, the rates at which while the area of a triangle is increasing at a rate of 4cm2/min. 7k points) Related rates area of a triangle Related rates area of a right triangle. At what rate is the area of an equilateral triangle increasing when each side is of length 20cm and each is increasing at a rate of 0. I am going to assume “at some time t” refers to an initial condition, t=0. Therefore, we multiply a half, 14, and four, which gives us 28. The formula for the area of a triangle is A = ½ x b x h or bh/2. 4). Then At what rate is the area of the triangle formed by the ladder, wall, and ground changing. PROBLEM 3 : Leg one of a right triangle is decreasing at the rate of$ \ 5 \ in/sec. We know the rates of changes of the 2 sides, therefore, the choice for a relation between what are known and the unknown is simple: A = xy/2. The formula shown will recalculate the area using this method. Area of a regular polygon. With Rate of Change Formula, you can calculate the slope of a line especially when coordinate points are given. 2cm/sec? Geometry Perimeter, Area, and Volume Perimeter and Area of Triangle Answer (1 of 8): I suppose vase is a typographical error, and I presume it is base instead. Now we need to find the rate at which the area is increasing when the side is 9 cm. In fact, that would be a good exercise to see if you can build a table of values that will support our claims on these rates of change. The resulting m value is the average rate of change of this function over that interval. Since the question is asking for the rate of change in terms of the perimeter, write the formula for the perimeter of the square and differentiate it with the respect to time. Find the rate at which the area of the triangle is increasing when the angle betwe … read more The height of the triangle can be taken as four units long. Got it. (Result makes sense) c'= (2*12*15*sin (60)*2) / 2*sqrt (189) Apr 2, 2011. | Differentiation V: Derivatives and Rates of Change | Determine the rate of change of the area of an equilateral triangle with respect to the length of a side. Assume that we know one leg and angle, so we change the selection to given angle and one side. A few examples are population growth rates, production rates, water flow rates, velocity, and acceleration. The variables of interest are a = altitude. Can you see why? We can use the principle of similar triangles to relate x to h though: The ratios of corresponding sides of similar triangles are equal. 1 Example The radius of a circle is increasing at a constant rate of 2 cm/s. 1, 1 Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cmLet Radius of circle = 𝑟 & Area of circle = A We need to find rate of change of Area w. Find the rate of change of the area with respect to s when s = 4 . Area of triangles finding area, base and height. Answer (1 of 8): I suppose vase is a typographical error, and I presume it is base instead. As we move along the x axis of a shape from its left most point to its right most point, the rate of change of the area at any instant in time will be equal to the height of the shape What is Rate of Change in Calculus ? The derivative can also be used to determine the rate of change of one variable with respect to another. Find the ratio of the changes. According to the law of cosines with side a being 12 and side b being 15 and angle C being 60 side c should be sqrt (189). The first problem asks you to find the rate at The rate of change is the derivative with respect to t so you are going to have to find a relationship between and A. The output is the area of the triangle. 9. Area of a parallelogram given sides and angle. In this lesson, you will learn a formula to compute the slope of a line using any two points on the line (not just points whose x -coordinates differ by 1). 20 dic. One leg of a right triangle is always $6$ feet long and the other leg is increasing at a rate of $2$ ft/s. The side lengths of a right-angled triangle are $a, b$and $c$centimeters where $c$is the length of the hypotenuse and $a, b$and $c$are changing as differentiable functions of time. Suppose that x is increasing at the rate of 10 degrees per minute. Height of a triangle. change in area as we move in a particular direction. 24. The area is increasing at a rate \(\ frac {(3\ sqrt {3})}{8}ft_2/sec. Thus. The rate of change is the derivative with respect to t so you are going to have to find a relationship between and A. V = 1 3 π r 2 h. Note that we could have computed this in one step as follows, x = 10 − 1 4 ( 12) = 7 x = 10 − 1 4 ( 12) = 7. To find y' use the Pythagorean theorem x2+y2=z2where z is constant. 5 and 12 by . To help preserve questions and answers, this is an automated copy of the original text. Find the rate at which the area of the triangle is changing when the angle between the two sides is \(π/6. From this we can write functions for the lengths of the two sides, base and height. 1. B. A triangle has a height that is increasing at a rate of 2 cm/sec and its area is increasing at a rate of 4 cm 2 /sec. 2021 Rate of change is used to mathematically describe the percentage change in value over a defined period of time, and it represents the momentum  Slopes, Rates of Change, and Similar Triangles. 5 Introduction to percentage and per mil The surface area or surface (A) of a triangle is calculated by the formula:  The legs of an isosceles right triangle increases in length at a rate of 2 m/s. So, the rate of changing of the total area of the disturbed water when the radius is 5 cm is 20 π sq. We can now use our average rates of change formula. A balloon in the shape of a sphere is being inflated so that the volume is increasing by 100 cubic centimetres per second. Rate of Change of Area: The area of a triangle with height {eq}h {/eq} and base {eq}b {/eq} is: {eq}A = 0. 1 centimeter per second. Consequently, any question about rates of change can be rephrased as a question about derivatives. 6. A right triangle whose sides are changing has sides of 30 and 40 inches at a particular instant. The variables of interest are a = altitude A = area and, since the area of a triangle is A=1/2ba, we need b = base. If the area of triangle remains constant, at what rate does the angle between these two sides change when the rst side is 20cm long, the second side is 30cm long, and the angle between the two sides is ˇ=6? Is it At what rate is the area of an equilateral triangle increasing when each side is of length 20cm and each is increasing at a rate of 0. Compute where the rate of change of the area function is zero. Is my answer right? my question is is this answer correct? for the last step. The Rate of Change Formula. 2021 In related rates problems we are give the rate of change of one quantity in a in which all three sides of a right triangle are changing. >>Derivative as Rate of Change; >>The sides of an equilateral Question  7 mar. For example, we know that α = 40° and b is 17 in. Take the derivative of the function you found in part 5 with respect to time. The sides of an equilateral triangle are decreasing at a rate of 3 in/hr. This calculus video tutorial explains how to solve related rate problems dealing with the area of a triangle. Because we were given the rate of change of the volume as well as the height of the cone, the equation that relates both V and h is the formula for the volume of a cone. The average rate of change of the area of the triangle on the time interval [?/6,?/4] is. CalculusSolution. Slopes, Rates of Change, and Similar Triangles. A common use of rate of change is to describe the motion of an object moving in a straight line. At what rate is the length of the hypotenuse changing? (CA) X IL 3 R LYpo(enuz. Find the rate at which its area increases, when side is 10 cm long. So, the change in area is equal to the scale factor squared. ) Example – 1. Watch our area of a right triangle calculator performing all calculations for you! The area of the chosen triangle is 121. A rate of change is given by a derivative: If y= f(t), then dy dt (meaning the derivative of y) gives the (instantaneous) rate at which yis changing with respect to t(see14). 03 rad/sec. Remember, rates of change are signed quantities. Calculate the change in the variable value. As you now drag point A, notice that the area does not change. Find the rate at which the area of the triangle is changing when the  10 mar. 2cm/sec? The area of the triangle formed by the ladder consisting of the ground, the wall and the ladder is equal to (x*y)/2 The rate of change of the area is (1/2)*(x*(dy/dt) + y*(dx/dt)) = `(1/2)*(12 And we were also given information about the rate of change of the left side of the triangle. ) perimeter changing $\ \ \ \$ c. When the height is 3 cm and the base is 10 cm, at what rate is the area increasing or decreasing? 6. Instead, the fact that we know this rate of change tells us that we can use the left side length of the triangle in our equation, not its rate of change. 06 rad/s. 2021 Let A be the Area of rectangle We need to find Rate of change of area when x = 8 & y = 6 cm i. A is always decreasing. I wonder whe Rates of change: The rate of change of the area of a triangle. Suppose we have two variables x and y (in most problems the letters will be different, but for now let's use x and y) which are both changing with time. And the area of a triangle is one half the base times the height, so the area of the triangle is (1/2) h 2. What is the rate of change of the area of the triangle if the height is 6 cm? Provide your answer below: The rate of change of the area of the triangle is cm²/s . The base of the triangle is always at the bottom; it is the side that the triangle sits on. How fast is the hypotenuse z changing when x 3 m? 16. But here’s where it can get tricky. 1. At a given time t, the tangent line to the unit circle at the position P (t) will determine a right triangle in the first quadrant.

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