text:
# BMI Calculator

Use this tool to calculate your BMI based on your height and weight.

## How is BMI interpreted ?

This table is valid for an adult male or female (see 'BMI index Limits' below).

BMI Interpretation
less than 16.5 Malnutrition
from 16.5 to 18.5 Underweight
from 18.5 to 25 Normal weight
from 25 to 30 Overweight
from 30 to 35 Moderate obesity
from 35 to 40 Severe obesity
greater than 40 Morbid obesity

## What is BMI?

BMI or Body Mass Index provides information on an individual's height-weight ratio. It is calculated with a relatively simple mathematical formula,

\text{BMI} = \text{Weight in kg} /(\text{(Height in cm)} * \text{(Height in cm)})

BMI is used to find out if your weight is suitable for your height. For a given weight and height of an individual, it is possible to statue quickly whether he is overweight, obese, lean, or normal build.

## What is a BMI too high or too low risk ?

Overly high BMI is often linked to risks of diseases and malfunctions including (non-exhaustive list),
- high blood pressure
- myocardial infarction
- type 2 diabetes
- rheumatological problems
- sleep apnea

A too low BMI may be related to the following risks (non-exhaustive list),
- malnutrition, nutrient deficiency (vitamins, minerals ...)
- weakened immune system (immunosuppression)
- osteoporosis

As a result, too low or too high BMI lead to the risk of serious illnesses and can significantly affect the person's quality of life. If you are in this case, medical follow-up is recommended in order to regain a more adequate body. A rebalancing of the diet, possibly associated with the practice of a suitable sports activity, can help to regain good physical fitness and normal BMI.

## BMI Index Limits

The BMI index as calculated above is not adequate for children, adolescents, seniors over 65 years of age, pregnant women, very sporty people, amputees, giant or dwarf people.


text:
# cartesian equation of a line in the space

In $\mathbb R^3$ we write the equation of a plane as $ax+by+cz=d$. Do we have a similar equation of a line in $\mathbb R^3$? In my knowledge we don't have such equation. All we know is that given a point on a line $L$ and a vector parallel to $L$ we can derive a vector equation and a parametric equation but we don't have an equation in the form $f(x,y,z)=0$ is this correct?

-
Depends on what you allow $f$ to be. – Qiaochu Yuan Jun 2 '12 at 15:27
$f$ is a map $\mathbb R^3\to \mathbb R$ – palio Jun 2 '12 at 15:35
Then take $f(x, y, z) = (ax + by + cz - d)^2 + (ex + fy + gz - h)^2$. – Qiaochu Yuan Jun 2 '12 at 15:40
this is a polynomial!! what about a line being an algebraic curve so it is given by two equations $f_i(x,y,z)=0$? – palio Jun 2 '12 at 17:07
I don't understand your question. You said "map." $f$ is a map. Yes, lines in $\mathbb{R}^3$ are given by two (linear) equations. They can also be given by one quadratic equation. (This is specific to $\mathbb{R}$.) – Qiaochu Yuan Jun 2 '12 at 17:14

I believe the nearest thing would be $$\frac{x-a}{d} = \frac{y-b}{e} = \frac{z-c}{f},$$ which is not quite of the form $f(x,y,z)=0$, but it is standard and about as close as you will get to what you want.
and we can rewrite the above in the form of two linearly independant equations $f_i(x,y,z)=0$ – palio Jun 2 '12 at 17:11
No, that's not (quite) correct. You need two (linear) equations to write down a line in the form $f(x, y, z) = 0$ The kernel of a linear map $\mathbb{R}^3\rightarrow\mathbb{R}^2$ of rank $2$ is one dimensional, just translate it (I'm too lazy right now to recall how to LaTeX a matrix).

format: plain


text:
# PDE11

 $u_{{xx}}=a^{{-2}}u_{t}\,$ $u(0,t)=10\,$ $u(10,t)=30\,$ $u(x,0)=0\,$ $00\,$

Write the solution as a sum of two functions, one independent of time.

$u(x,t)=s(x)+v(x,t)\,$

The new boundary conditions are:

$s(0)=10,\,\,\,s(10)=30\,$

$v(0,t)=0,\,\,\,v(10,t)=0\,$

$u(x,0)=s(x)+v(x,0)=0,\,\,\,v(x,0)=-s(x)\,$

Substituting the sum of functions into the DE gives a simple ODE and a PDE:

$s''=0\,$

$s(x)=Ax+B\,$

$s(0)=B=10\,$

$s(10)=10A+10=30,\,\,\,A=2\,$

• $s(x)=2x+10\,$

And the PDE

$v_{{xx}}=a^{{-2}}v_{t}\,$

Seperate variables to get two ODEs. Set them equal to a constant that will simplify calculations later:

$v(x,t)=X(x)T(t)\,$

$X''T=a^{{-2}}XT'\,$

${X'' \over X}=a^{{-2}}{T' \over T}=-\lambda ^{2}\,$

$X''+\lambda ^{2}X=0\,$

$X(x)=c_{1}\cos(\lambda x)+c_{2}\sin(\lambda x)\,$

$X(0)=c_{1}=0\,$

$X(10)=c_{2}\sin(\lambda 10)=0\,$

$\lambda _{n}={n\pi \over 10},\,\,\,n=1,2,3,...\,$

And

$T'+a^{2}\lambda ^{2}T=0\,$

$T(t)=c_{3}e^{{-a^{2}\lambda ^{2}t}}\,$

• $v(x,t)=\sum _{{n=1}}^{\infty }A_{n}\sin(\lambda _{n}x)e^{{-a^{2}\lambda ^{2}t}}\,$

$v(x,0)=\sum _{{n=1}}^{\infty }A_{n}\sin({n\pi x \over 10})=-2x-10\,$

This is a Fourier sine series for the function on the right. The coefficients are then given by:

$A_{n}={-1 \over 5}\int _{0}^{{10}}(2x+10)\sin({n\pi x \over 10})\,dx,\,\,\,n=1,2,3,...\,$

$={20(3(-1)^{n}-1) \over n\pi }\,$

The solution is

• $u(x,t)=2x+10+{20 \over \pi }\sum _{{n=1}}^{\infty }{3(-1)^{n}-1 \over n}\sin({n\pi x \over 10})e^{{-a^{2}{n^{2}\pi ^{2} \over 10^{2}}t}}\,$


text:
lilypond-devel
[Top][All Lists]

## Re: Is it possible to have a single \override for same context/Grob for

 From: Carl Sorensen Subject: Re: Is it possible to have a single \override for same context/Grob for different property values? Date: Mon, 24 Jan 2011 17:16:37 -0700

```

On 1/24/11 5:15 PM, "Neil Puttock" <address@hidden> wrote:

> On 25 January 2011 00:10, Carl Sorensen <address@hidden> wrote:
>
>> If you use pairs, instead of lists, for the property overrides then the
>> override list is a standard  alist:
>>
>> \overrideGrobProps #'Accidental #`((color . ,red) (font-size . 4)
>> (extra-offset . (1.5 . 3)))
>
> Yup, but then you have to type those annoying full stops (or periods,
> if you prefer. :)
>

This is true, but it's also consistent with the standard idiom of LilyPond
(another GLISS discussion, I suppose....)

Thanks,

Carl

```


text:
`ISRN Applied MathematicsVolume 2012 (2012), Article ID 124936, 10 pagesdoi:10.5402/2012/124936`
Research Article

## On Behavior of Solution of Degenerated Hyperbolic Equation

1Institute of Mathematics and Mechanics, Academy of Sciences, Baku, Azerbaijan
2The Sheki Institute of Teachers, Sheki Az5, Azerbaijan

Received 10 August 2012; Accepted 13 September 2012

Academic Editors: A. Bellouquid and F. Ding

Copyright © 2012 Tahir Gadjiev et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

The purpose of this paper is to learn some features of hyperbolic type of nonlinear equations. It is shown that the solution of the equation approaches to the endlessness in the inside of some initial conditions and time of the special marks. The local existence of the equation’s solution has been proved and the problem of unlimited increasing on the solution of nonlinear hyperbolic equations type during the finite time is investigated.

#### 1. Introduction

In this paper the unbounded increasing solution of the nonlinear hyperbolic-type equation for the finite times is considered. These type equations describe the processes of electron and ionic heat conductivity in plasma, diffusion of neutrons, and -particles, and so forth. Investigation of unbounded solution or regime of peaking solutions occurs in theory of nonlinear equations where one of the essential ideas is the representation called eigenfunction of nonlinear dissipative surroundings. It is well known that even a simple nonlinearity, subject to critical of exponent, the solution of nonlinear hyperbolic type equation for the finite time may increase unboundedly, that is, there is a number such that In [1] the existence of unbounded solution for finite time with a simple nonlinearity has been proved. [2] has shown that any nonnegative solution subject to critical exponent is unbounded increasing for the finite time. Similar results were obtained in [3] and corresponding theorems are called Fujita-Hayakawa’s theorems. More detailed reviews can be found in [46].

The paper is organizing as follows. In Section 2 we present some definitions and auxiliary results. In Section 3, we give the main results for type nonlinear equation, where blow up solutions are obtained and the local solution exists. This results generalize the corresponding result [7].

#### 2. Some Definitions and Auxiliary Results

Let us consider the equation in bounded domain with nonsmooth boundary. Let's denote , . . The functions are continuous with respect to uniformly in at any . Besides, the function is measurable by all arguments and does not decrease with respect to . Let's assume the fulfillment of the following Dirichlet boundary condition: with the initial conditions in some domain , where and are smooth functions. Let's assume that is measurable, nonnegative function satisfying the conditions and for any and some fixed where . From condition (2.4) it follows that and that , that is, Condition (2.6) is called Makhenkhoupt condition (see [8]). Besides, as in [8] we will assume that , that is,

Let's introduce the weight Sobolev space for any , where with the finite norm .

As the generalized solution of problem (2.1)–(2.3) in we will define the function such that where .

We will study the conditions at some generalized solution from at some .

At and at linear main part there are many works, such as [1, 2] on investigation properties of solutions having (2.10). We will show that if and are adequately big, then it holds (2.10). For small and if then , it doesn't depend on . Let's construct the sufficient condition on , at which any solution of problem (2.1)–(2.3) at , has “blowup” (without limitation of smallness on and ).

Let's formulate some auxiliary results from [9, 10], and let's determine the harmonic operator

Lemma 2.1 (see [9]). There exists the positive eigenvalue of spectral problem for the operator to which corresponds the positive in eigenfunction.

Lemma 2.2 (see [10]). Let on and for any nonnegative with . Then on whole domain .

#### 3. Main Results

Let be an eigenfunction of spectral problem for the operator corresponding to . Denote . Let's assume the fulfillment of the conditions:

Theorem 3.1. Let at . There exists such that if, or condition (3.2) is fulfilled. Then where .

Proof. Let's assume the opposite. Then is a solution of (2.1) in and condition (2.2) is fulfilled on . By the virtue of Lemma 2.2 in . Substitute in (2.9) , where in is an eigenfunction of spectral problem for the operator , corresponding to eigenvalue . Such eigenvalue exists by virtue of Lemma 2.1. As a result, we will obtain
Let’s formulate some transformations. After some simple manipulation we get Using condition (3.1), tending to zero at all , we obtain Here denoting we have Let's estimate first integral on the right-hand side of (3.8). Using the Cauchy inequality with we get Hence then So, from (3.8) we have By virtue of Holder inequality we obtain Thus, we have Let's multiply (3.14) by , then Hence Let's integrate by from 0 to , then from (3.16) we have or Denote If the condition (3.2) is fulfilled, that is, and if , then . Hence we have

Remark 3.2. This theorem generalizes the corresponding result [7].
That is why, (2.1) is not a solution in , satisfying the boundary condition (2.2), if is not smaller.
Now we will show that at small initial functions the solution of problem (2.1)–(2.3) exists in small in .

Theorem 3.3. Let one assume that . There exists such that if , then the solution of problem (2.1)–(2.3) exists in small in and does not depend on .

Proof. Let , where . Let in be an eigenfunction, corresponding to positive eigenvalue of the boundary-value problem Let's consider the function . We have if is sufficiently small. Inequality (3.23) is understood in weak sense (see [11]). From (3.23) and Lemma 2.2, it follows that . Let us determine the class of functions consisting from continuous in equal to zero at and such that is a subset at Banach space of continuous functions in with the norm .
Let Let's determine the operator on putting , where is a solution of linearizing problem. By virtue of obtained estimation above, maps to . This follows from the obtained estimation and theorem on the solution of the hyperbolic problems in at the small [11].
From Lerey-Shaudeer theorem, it follows that the operator has a fixed point . This shows the existence of solution in the small. The theorem is proved.

Note that the sufficient condition at which any nonnegative solution of problem (2.1)–(2.3) has “blow-up” is where .

Theorem 3.4. Let at , and be positive eigenvalue of problem (3.16) in which corresponds to the positive in eigenfunction. If , where , is a solution of problem (2.1)–(2.3) then it holds (3.25).

Proof. Analogously, as it is constructed in inequality (3.14), we will obtain
Let . From (3.25) it follows that . Hence at . Thus it tends to and . Consequently, also tends to infinity.

Corollary 3.5. Let at . Then there is no positive in solution of the problem (2.1)–(2.3).
Let one consider the following equation: in bounded domain , and let , then one will have eigenfunction of the problem in on , and the corresponding eigenvalue.

Theorem 3.6. Let at at . There exists such that if or condition (3.2) is fulfilled, then where .

Proof. Let's assume a contrary. Then is a solution of (3.27) in and condition (2.2) is fulfilled on . Substituting in corresponding integral identity    to get where . Tending to zero, we'll obtain where . Let’s consider first the integral on the right-hand side in (3.30). Let in and . Denote the surface of degeneration of (3.27), that is, the boundary of the solution . Since on and from the Green's formula, we get where is a derivative on direction of external norms to . Since is on and by virtue of continuity of flow at . Therefore, the last two integrals in (3.31) are equal to zero. Then we will obtain Using we will get Using the results of paper [12] we will obtain the required result.

#### References

1. S. Kaplan, “On the growth of solutions of quasi-linear parabolic equations,” Communications on Pure and Applied Mathematics, vol. 16, pp. 305–330, 1963.
2. H. Fujita, “On the blowing up of solutions of the Cauchy problem for ${u}_{t}=\mathrm{\Delta }u+{u}^{1+\alpha }$,” Journal of the Faculty of Science, vol. 13, pp. 109–124, 1966.
3. K. Hayakawa, “On nonexistence of global solutions of some semilinear parabolic differential equations,” Proceedings of the Japan Academy, vol. 49, pp. 503–505, 1973.
4. V. A. Galaktionov and H. A. Levine, “A general approach to critical Fujita exponents in nonlinear parabolic problems,” Nonlinear Analysis: Theory, Methods & Applications, vol. 34, no. 7, pp. 1005–1027, 1998.
5. K. Deng and H. A. Levine, “The role of critical exponents in blow-up theorems: the sequel,” Journal of Mathematical Analysis and Applications, vol. 243, no. 1, pp. 85–126, 2000.
6. A. A. Samarskii, V. A. Galaktinonov, S. P. Kurdyumov, and A. P. Mikahailov, Blow-Up in Quasi Linear Parabolic Equations, Nauka, Moscow, Russia, 1987.
7. Y. Ebihara, S. Kawashima, and H. A. Levine, “On solutions to ${u}_{tt}-{|x|}^{\alpha }\mathrm{\Delta }u=f\left(u\right)$,” Funkcialaj Ekvacioj, vol. 38, no. 3, pp. 539–544, 1995.
8. S. Chanillo and R. L. Wheeden, “Weighted Poincaré and Sobolev inequalities and estimates for weighted Peano maximal functions,” American Journal of Mathematics, vol. 107, no. 5, pp. 1191–1226, 1985.
9. P. Tolksdorf, “On quasilinear boundary value problems in domains with corners,” Nonlinear Analysis, vol. 5, no. 7, pp. 721–735, 1981.
10. D. Gilbarg and N. Trudinger, Elliptic Differential Equation With Partial Derivatives of the Second Order, Nauka, Moscow, Russia, 1989.
11. O. A. Ladyzhenskaya, V. A. Solonnikov, and N. N. Uralceva, Lineynie i Kvazilineynie Uravneniya Parabolicheskogo Tipa, Nauka, Moscow, Russia, 1967.
12. T. S. Hajiev and R. A. Rasulov, “Explosive solution of the nonlinear equation of a parabolic type,” Applied Mathematics Letters, vol. 24, no. 10, pp. 1676–1679, 2011.


text:
# Find the Z-transform of the anticausal sequence x(n) = {4,2,3,-1,-2,1}. (1 as the reference variable)

+1 vote
Find the Z-transform of the anticausal sequence x(n) = {4,2,3,-1,-2,1}. (1 as the reference variable)

(a) 4z^5 + 2z^4 + 3z^3 – z^2 – 2z + 1

(b) 4z^-5 + 2z^-4 + 3z^-3 -z^-2 – 2z^-1 + 1

(c) -4z^5 – 2z^4 – 3z^3 + z^2 + 2z – 1

(d) -4z^-5 – 2z^-4 – 3z^-3 + z^-2 + 2z^-1 – 1

I got this question during a job interview.

This interesting question is from Properties of Z-Transforms in section Z-Transform and Digital Filtering of Signals and Systems

by (42.1k points)
Right choice is (a) 4z^5 + 2z^4 + 3z^3 – z^2 – 2z + 1

To explain I would say: Given sequence values are :

x(-5)=4, x(-4)=2, x(-3)=3, x(-2)=-1, x(-1)=-2, x(0)=1

We know that

$X(Z) = \sum\limits_{n=-∞}^{∞} x(n) z^{-n}$

X(Z) = x(-5) z^5 + x(-4) z^4 + x(-3) z^3 + x(-2) z^2 + x(-1)z + x(0)

X(Z) = 4z^5 + 2z^4 + 3z^3 – z^2 – 2z + 1.

+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote
+1 vote


text:
# Analytical approximation of indefinite integral on a given interval to a given precision

I'm looking for an analytical approximation of $\int_a^b e^{-x^2}\mathrm{erf}(x+c) dx$ that would be accurate to precision $\varepsilon$ for $a,b,c$ within a certain range. How do I ask Mathematica for that?

I tried to Integrate the above expression, but no analytical solutions were found.

• "within a certain range" - can you please specify those ranges for completeness? – J. M.'s technical difficulties Aug 1 '16 at 23:44
• @J.M.: $-5<a,b,c<5$ would be good. – Michael Aug 1 '16 at 23:49
• Possible duplicate of Numerical underflow for a scaled error function – Artes Aug 2 '16 at 0:06
• What is your definition of analytical approximation? InterpolationFunction is an analytical approximation in the sense that it fits arrays of numbers to splines or Hermite polynomials, but I suspect that it is not what you are seeking. – bbgodfrey Aug 2 '16 at 1:12
• I don't agree with the proposed duplicate. OP's function has qualitatively different properties from the function in the other thread. Note that in this case both factors tend to a limit with increasing $b\gg c$, while the function in the other thread has an $\exp(x^2)$ factor whose blow-up is only mitigated by the complementary error function. – J. M.'s technical difficulties Aug 2 '16 at 4:14

Maybe so:

Expand with Series a function Erf[x+c] e.g for order 2.

n = 2;
func = Normal@Series[Erf[x + c], {x, 0, n}]

$-\frac{2 c e^{-c^2} x^2}{\sqrt{\pi }}+\frac{2 e^{-c^2} x}{\sqrt{\pi }}+\text{erf}(c)$

sol = Simplify@Integrate[Exp[-x^2]*func, {x, a, b}]

$\frac{-c e^{-c^2} \left(2 e^{-a^2} a-\sqrt{\pi } \text{erf}(a)-2 b e^{-b^2}+\sqrt{\pi } \text{erf}(b)\right)+2 e^{-c^2} \left(e^{-a^2}-e^{-b^2}\right)+\pi \text{erf}(c) (\text{erf}(b)-\text{erf}(a))}{2 \sqrt{\pi }}$

Check numerics:

 a = -4;
b = 4;
c = 4;
sol2 = NIntegrate[Exp[-x^2]*Erf[x + c], {x, a, b}]
(*1.77234*)

sol // N
(*1.77245*)

From Documentation Center Precision[x] is- Log[10, dx/x].

  -Log[10, Abs[(sol - sol2)/sol2]]
(*4.20019*)

I have 4 significant digit.

From OP qestion:

Can Mathematica predict for the given function and interval how many terms it needs in the Series for the accuracy ϵ on that interval?

Yes I use NonlinearModelFit.Generate data for order n from 1 to 20;

 data = Table[{n,
func = Normal@Series[Erf[x + c], {x, 0, n}];
sol = Simplify@Integrate[Exp[-x^2]*func, {x, a, b}];
a = -4;(*Interval (a,b)*)
b = 4;
c = 4;
sol2 = NIntegrate[Exp[-x^2]*Erf[x + c], {x, a, b}];
sol1 = (sol /. c -> 4 /. a -> -4 /. b -> 4) // N;
-Log[10, Abs[(sol1 - sol2)/sol2]]}, {n, 1, 20}]

(*{{1, 4.19844}, {2, 4.20019}, {3, 4.20019}, {4, 4.21306}, {5,
4.21306}, {6, 4.27069}, {7, 4.27069}, {8, 4.46324}, {9,
4.46324}, {10, 5.23787}, {11, 5.23787}, {12, 4.86359}, {13,
4.86359}, {14, 5.124}, {15, 5.124}, {16, 5.10868}, {17,
5.10868}, {18, 5.37338}, {19, 5.37338}, {20, 5.20063}}*)

nlm = NonlinearModelFit[data, a1 + a2*Exp[a3*n + a4], {a1, a2, a3, a4}, n]
Normal@nlm

Model:

$1.15801 e^{0.00463001 n+2.53736}-10.621$

 Show[Plot[Normal@nlm, {n, 0, 20}], ListPlot[data],
AxesOrigin -> {0, 0}, PlotRange -> {Automatic, {0, 5.5}},
AxesLabel -> {n, "significant digits"}]

for n=20 I have:

  nlm[20]
(*5.44438*)

5 significant digit.

for n=200 I have:

    nlm[200]
(*26.3475*)

26 significant digit.

You may change a model in NonlinearModelFit.Maybe my model is not the best.

• Thanks, Mariusz! One more question: can Mathematica predict for the given function and interval how many terms it needs in the Series for the accuracy $\epsilon$ on that interval? Or it's on me to determine n? – Michael Aug 2 '16 at 17:12
• @Michael.I edited my answer,see above. – Mariusz Iwaniuk Aug 2 '16 at 18:19

One can construct a Chebyshev series approximation to the integrand for an interval, such as -5 <= x <=5 mentioned in the comments, and integrate it to get a series expansion for the integral. It is well known that Chebyshev series representation have numerical advantages over power series. I saw a comment about a NPU-supported method, but I don't know what that is, this being a Mathematica site. Most numerical systems have access to an FFT, in one way or another, I think, and, of course, to trigonometric functions. That is all that is really needed.

Some auxiliary functions used here (code below):

• chebSeries[f, {a, b}, n, precision] computes the Chebyshev expansion of order n for f[x] over a <= x <= b.
• iCheb[c, {a, b}, k] integrates the Chebyshev series c, plus k.
• f = chebFunc[c,{a,b}], computes the function represented by the Chebyshev coefficients c = {c0, c1,...} over the interval {a, b}.

The first step is to compute the coefficients of the integrand as a function of x for a given value of c. The are saved (memoized) for the sake of speed, but it is not essential. An antiderivative of the integrand is computed with iCheb[coeff[c], {-5, 5}]. Depending on c, one needs a series of order 70 to 90+ to get a theoretical error of less than machine precision, so computing one of order 2^7 = 128 is sufficient for all c. (See Boyd, Chebyshev and Fourier Spectral Methods, Dover, New York, 2001, ch. 2., for a discussion of the convergence theory of Chebyshev series.)

ClearAll[coeffs];
coeffs[c_?NumericQ] := coeffs[c] =
Module[{
cc,       (* Chebyshev coefficients *)
pg = 16,  (* Precision goal:  *)
sum,      (*   sum of tail = error bound *)
max,      (*   max coefficient to measure rel. error *)
len},     (*   how many terms of the tail to drop *)
cc = chebSeries[Function[x, Exp[-x^2] Erf[x + c]], {-5, 5}, 128];
max = Max@Abs@cc;
sum = 0;
(* trim tail of series of unneeded coefficients (smaller than desired precision) *)
len = LengthWhile[Reverse[cc], (sum += Abs[#1]) < 10^-pg max &];
Drop[cc, -len]
];

Next we can define the user's sought-after function in terms of the antiderivative:

func[a_, b_, c_] :=
With[{antiderivative = chebFunc[
N@iCheb[coeffs[SetPrecision[c, Infinity]], {-5, 5}, 0],
{-5, 5}]},
antiderivative[b] - antiderivative[a]
];

The following computes the relative and absolute error of func on a hundred random inputs, using a high-precision NIntegrate[] to compute the "true" value.

cmp[a_, b_, c_] := {(#1 - #2)/#2, #1 - #2} &[
func[a, b, c],
NIntegrate[Exp[-x^2] Erf[x + SetPrecision[c, Infinity]], {x, a, b},
WorkingPrecision -> 40]
]

ListLinePlot[
Transpose@RealExponent[cmp @@@ RandomReal[{-5, 5}, {100, 3}]],
PlotRange -> {-20, 0}, PlotLabel -> "Error",
PlotLegends -> {"Rel.", "Abs."}]

The yellow line shows the absolute error is limited to a few ulps. The theoretical bound on the error does not take into account rounding error (in both the coefficients and the evaluation of the series). The lines that drop off the bottom are the result of an error of zero.

Auxiliary functions

(* Chebyshev extreme points *)
chebExtrema::usage =
"chebExtrema[n,precision] returns the Chebyshev extreme points of order n";
chebExtrema[n_, prec_: MachinePrecision] :=
N[Cos[Range[0, n]/n Pi], prec];

(* Chebyshev series approximation to f *)
Clear[chebSeries];
chebSeries::usage =
"chebSeries[f,{a,b},n,precision] computes the Chebyshev expansion \
of order n for f[x] over a <= x <= b.";
chebSeries[f_, {a_, b_}, n_, prec_: MachinePrecision] :=
Module[{x, y, cc},
x = Rescale[chebExtrema[n, prec], {-1, 1}, {a, b}];
y = f /@ x; (* function values at Chebyshev points *)
cc = Sqrt[2/n] FourierDCT[y, 1]; (* get coeffs from values *)
cc[[{1, -1}]] /= 2;  (* adjust first & last coeffs *)
cc
];

(* Integrate a Chebyshev series -- cf. Clenshaw-Norton, Comp.J., 1963, p89, eq. (12) *)
Clear[iCheb];
iCheb::usage = "iCheb[c, {a, b}, k] integrates the Chebyshev series c, plus k";
iCheb[c0_, {a_, b_}, k_: 0] := Module[{c, i, i0},
c[1] = 2 First[c0];
c[n_] /; 1 < n <= Length[c0] := c0[[n]];
c[_] := 0;
i = 1/2 (b - a) Table[(c[n - 1] - c[n + 1])/(2 (n - 1)), {n, 2, Length[c0] + 1}];
i0 = i[[2 ;; All ;; 2]];
Prepend[i, k - Sum[(-1)^n*i0[[n]], {n, Length[i0]}]]]

(* chebFunc[c,...] computes the function represented by a Chebyshev series *)
chebFunc::usage =
"f = chebFunc[c,{a,b}], c = {c0,c1,...} Chebyshev coefficients, over the interval {a,b}
y = chebFunc[c,{a,b}][x] evaluates the function";
chebFunc[c_, dom_][x_] := chebFunc[c, dom, x];
chebFunc[c_?VectorQ, {a_, b_}, x_] :=
Cos[Range[0, Length[c] - 1] ArcCos[(2 x - (a + b))/(b - a)]].c;

Update: Comparison of Chebyshev and power series

Perhaps it would be worth illustrating the difference between power series and Chebyshev series approximations for those who are not familiar with it. (One should become familiar with it, for Chebyshev expansions are to functions what decimal expansions are to numbers.)

Key differences:

• Symbolic series expansion of the function Erf[x + c] grows extremely fast and takes a much longer time to evaluate than the DCT-I used to compute the Chebyshev coefficients. Attempting a degree 40 expansion hanged the computer and I had to kill the kernel.

• Aside from not being able to compute the series expansion to arbitrary order, it is probably impossible to get convergence for a fixed precision due to rounding error. At machine precision, you cannot even get 2 digits throughout the interval {c, -5, 5}, for a = -4, b = 4 up to order 25. OTOH, the Chebyshev series has an exponential order of convergence and can nearly achieve machine precision with machine precision coefficients.

• It is fairly easy to figure out when you have enough terms of a Chebyshev series $\sum a_j T_j$, because the error is bounded by the tail $\sum |a_j|$ and the $a_n \rightarrow 0$ roughly geometrically on average.

• If you don't have fast trigonometric functions, then instead of the code in chebFunc above, you can use Clenshaw's algorithm (see chebeval) to evaluate the series.

Here's another implementation of Mariusz's power series idea. I speed up the integration with a "power rule" int[{n}] for $$\int \exp\left(-x^2\right) x^n \; dx\,.$$ Of course it turned out that Series was the bigger bottleneck.

ClearAll[sol, int];
int[{0}] = Integrate[Exp[-x^2], x, Assumptions -> x ∈ Reals];
int[{n_}] = Integrate[Exp[-t^2] t^n, {t, 0, x},
Assumptions -> n > 0 && n ∈ Integers && t ∈ Reals];
$seriesCoefficientPart = 3; sol[n_] := sol[n] = Total@MapIndexed[ First@Differences[#1 int[#2 - 1] /. {{x -> a}, {x -> b}}] &, Series[Erf[x + c], {x, 0, n}][[$seriesCoefficientPart]]
];

(* Times *)
First@*AbsoluteTiming@*sol /@ {2, 10, 20, 22, 23, 24, 25}
(*  {0.013267, 0.071065, 2.01908, 7.6296, 23.4752, 33.1553, 48.3542}  *)

(* Sizes *)
LeafCount /@ sol /@ {2, 10, 20, 22, 23, 24, 25}
(*  {163, 693, 2413, 2765, 1100459, 1779935, 2879267}  *)

Chebyshev speed for c = 4, per evaluation of c:

First@AbsoluteTiming@func[a, b, 4, #] & /@ {2, 10, 20, 25}
(*  {0.002047, 0.000184, 0.000248, 0.000276}  *)

To illustrate the issue with power series, the graphics below shows the error in approximating Erf[x + c] by its Taylor series (times Exp[-x^2] for c = -4, -2, 0, 2, 4 and various orders. It does pretty good for Abs[x] < 1 as the order increases, but it gets worse for Abs[x] > 4.

GraphicsRow[
Table[
Plot[
Evaluate@Table[
Exp[-x^2] (Erf[x + c] - Normal@Series[Erf[x + c], {x, 0, n}]) //
RealExponent,
{c, -4, 4, 2}],
{x, -5, 5},
PlotRange -> {-18, 0}, Frame -> True, Axes -> False,
PlotLabel -> Row[{"Order ", n}], AspectRatio -> 1,
FrameLabel -> {"x", "Log error"}], {n, {2, 10, 20, 25}}],
PlotLabel -> "Error in approximating integrand by power series"]

The two plots below compare the absolute error of approximating by power series and by Chebyshev series. The convergence of the Chebyshev series is remarkable by comparison.

Plot[Evaluate@Table[
sol[n] - exact[a, b, c] /. {a -> -4, b -> 4} // RealExponent,
{n, {2, 10, 20, 25}}],
{c, -5, 5},
PlotRange -> {-18, 0}, Frame -> True, Axes -> False,
GridLines -> {None, -Range[2, 16, 2]},
PlotLabel -> "Log error for power series of order n",
FrameLabel -> {"c", "Log error"},
PlotLegends -> {2, 10, 20, 25}]

Plot[Evaluate@Table[
func[a, b, c, n] - exact[a, b, c] /. {a -> -4, b -> 4} // RealExponent,
{n, {2, 10, 20, 30, 40, 50, 60}}],
{c, -5, 5},
PlotRange -> {-18, 0}, Frame -> True, Axes -> False,
GridLines -> {None, -Range[2, 16, 2]},
PlotLabel -> "Log error for Chebyshev series of order n",
FrameLabel -> {"c", "Log error"},
PlotLegends -> {2, 10, 20, 30, 40, 50, 60}]

Determining the order of the approximation:

trim[cc_, eps_] := Module[{sum, max, len},
max = Max@Abs@cc;
sum = 0;
len = LengthWhile[Reverse[cc], (sum += Abs[#1]) < eps max &];
Drop[cc, -len]
]

Manipulate[
With[{cc = iCheb[chebSeries[Exp[-#^2] Erf[# + c] &, -5, 5, 128, 32], {-5, 5}]},
With[{order = Length@trim[cc, 10^-accuracy]},
ListPlot[
RealExponent@cc,
PlotLabel -> Column[{
Row[{"Chebyshev coefficients a[n] for ",
HoldForm["c"] -> Chop[N[c, {2, 1.5}], 0.05], ", "}],
Row[{"For accuracy ", SetPrecision[10^-accuracy, 2], " use order ", order}]
}, Alignment -> Center],
Frame -> True, FrameLabel -> {"n", "exponent of a[n]"},
GridLines -> {{order}, {-accuracy}}, PlotRange -> {-31, 1}]
]],
{{c, 4}, -5, 5, 1/10}, {{accuracy, 16.}, 2, 28}]

Addendum: Failed ideas - Maybe someone can make them work...

The Chebyshev series of Exp[-x^2] can be computed over {-x0, x0} exactly in terms of modified Bessel functions BesselI[]. Therefore, so can the series for Erf[x]. I was seduced into trying to come up with a way to compute the OP's function in this way but the +c in Erf[x + c] was too ornery.

One thing that would be needed is a way to write ChebyshevT[n, x + c] as a Chebyshev series in ChebyshevT[n, x]. The coefficients would be polynomials in c (with integer coefficients), which themselves could be represented as Chebyshev expansions. This can be done, in fact, but it's a bit cumbersome and slow. Further the Chebyshev coefficients for n = 64 get bigger than 2^100, and I worried about numerical stability. For the moment, I have given up without testing it. The way above seems superior, in simplicity, as well as (probably) in speed and numerics.

• You forgot to include the definition for chebExtrema[]; otherwise, I'm glad you caught the Chebyshev bug... :) – J. M.'s technical difficulties Aug 3 '16 at 19:40
• @J.M. Thanks, x2. :) – Michael E2 Aug 3 '16 at 21:57
• @MichaelE2 Why is this voted 0.00001 times it deserves? ;) I don't get it. This is amazing. Thanks a lot! – MathX Nov 16 '16 at 20:07


text:
# Open SourceMolecular Dynamics

### Sidebar

#### For Developers

This is an old revision of the document!

In this exercise, you will be asked to calculate the adsorption energy of CO molecule on the graphene surface, in an attempt to reproduce a part of the experiments presented in this paper.

## Lattice constant optimization

As you have seen in earlier exercises, the actual energy – and therefore also the stress tensor – depends on many parameters, like the selected functional. This means that geometrical parameters like the lattice constant may also vary and therefore need to be optimized first when building a new geometry. While this could be done using CP2K's CELL_OPT run type, optimizing both the lattice/cell constants and the geometry simultaneously, we are going to do it manually here, especially since we can assume that only the lattice constant will actually change.

What we are using to determine the center volume (the volume for which the energy is minimal) is the Birch–Murnaghan equation of state (to be precise: the BM equation integrated over pressure), which links the energy and the volume using the minimal energy $E_0$, the center volume $V_0$, the bulk modulus $B_0$ and its derivative $B_1$:

\begin{align*} E(V) = E_0 + \frac{9 V_0 B_0}{16} \Bigg\{ \left[ \left(\frac{V_0}{V}\right)^{2/3} - 1 \right]^3 B_1 \; + \left[ \left(\frac{V_0}{V}\right)^{2/3} - 1 \right]^2 \left[ 6 - 4 \left(\frac{V_0}{V}\right)^{2/3} \right] \Bigg\} \end{align*}

Use the following input file as a starting point, and an adapted version of the script you documented in a previous exercise to generate a number of input files for different lattice constants and run the respective calculation. A good interval for the fraction of the lattice constant is $0.90-1.10$ with a step size of $0.025$.

Extract the energies and fit $E_0$, $V_0$, $B_0$, $B_1$ using the Birch–Murnaghan EOS and using the new $V0$ determine the lattice constant.

graphene.inp
&GLOBAL
PROJECT graphene
RUN_TYPE ENERGY
PRINT_LEVEL MEDIUM
&END GLOBAL

&FORCE_EVAL
METHOD Quickstep
&DFT
BASIS_SET_FILE_NAME  BASIS_MOLOPT
POTENTIAL_FILE_NAME  POTENTIAL

&POISSON
PERIODIC XYZ
&END POISSON
&SCF
SCF_GUESS ATOMIC
EPS_SCF 1.0E-6
MAX_SCF 300

# The following settings help with convergence:
CHOLESKY INVERSE
&SMEAR ON
METHOD FERMI_DIRAC
ELECTRONIC_TEMPERATURE [K] 300
&END SMEAR
&DIAGONALIZATION
ALGORITHM STANDARD
&END DIAGONALIZATION
&MIXING
METHOD BROYDEN_MIXING
ALPHA 0.2
BETA 1.5
NBROYDEN 8
&END MIXING
&END SCF
&XC
&XC_FUNCTIONAL PBE
&END XC_FUNCTIONAL
&END XC
&PRINT
&PDOS
# print all projected DOS available:
NLUMO -1
# split the density by quantum number:
COMPONENTS
&END
&END
&END DFT

&SUBSYS
&CELL
# create a hexagonal unit cell:
ABC 2.4612 2.4612 15.0
ALPHA_BETA_GAMMA 90. 90. 60.
SYMMETRY HEXAGONAL
PERIODIC XYZ
&END CELL
&COORD
SCALED
C  1./3.  1./3.  0.
C  2./3.  2./3.  0.
&END
&KIND C
ELEMENT C
BASIS_SET DZVP-MOLOPT-GTH
POTENTIAL GTH-PBE
&END KIND
&END SUBSYS

&END FORCE_EVAL
The following commands may be useful.

Doing calculations on the command line using the bc tool:

bc -l <<< "5.6 * 12.3"

# you can also use variables and capture the output again in a variable:
x=1.025
a=$(bc -l <<< "$x * 2.4612")

Replacing numbers (or any text) inside a file and write the changed file to a new file:

a=3.54
sed -e "s|2.4612|$a|g" graphene.inp > "graphene_V-${x}.inp"
Be careful when fitting values for the Birch-Murnaghan EOS: the volume is usually the volume per atom (and the total volume of the cell you can also get from the CP2K output).

Adsorb one CO molecule on a graphene 6X6X1 supercell at the top(T), bridge(B) and center(C) sites and optimize the geometry. See the paper for the definitions as well as the initial distances.

You need change the RUN_TYPE to GEO_OPT and also specify the (absolute) coordinates by yourself.

You can get a 6x6x1 unit cell with absolute coordinates by using MULTIPLE_UNIT_CELL for the original input file like shown in previous examples, run it with CP2K and get the calculated absolute coordinates from the CP2K output (you can interrupt the actual calculation since the coordinates are printed before the actual calculation starts):
[...]
MODULE QUICKSTEP:  ATOMIC COORDINATES IN angstrom

Atom  Kind  Element       X           Y           Z          Z(eff)       Mass

1     1 C    6    1.267080    0.731549    0.000000      4.00      12.0107
2     1 C    6    2.534160    1.463098    0.000000      4.00      12.0107
3     1 C    6    3.801240    0.731549    0.000000      4.00      12.0107
4     1 C    6    5.068320    1.463098    0.000000      4.00      12.0107
5     1 C    6    6.335400    0.731549    0.000000      4.00      12.0107
[...]

The adsorption energy is given by:$E_{ad} = E_{CO+graphene} - E_{CO} - E_{graphene}$

This means that you also have to make an auxiliary geometry optimization calculation using CO in vacuum, using the same settings as for the other calculations except for the periodicity. Use a large enough cell (~ 15 Å) and CENTER_COORDINATES for this.

Find the most stable adsorption site and study the coverage effect such like 1/2 and 1. What do you observe when increasing the coverage?


text:
How do I make a Bezier curve with control points in pgfplots

I would like to create a Bezier curve with control points. I'm having no success finding what I want in the pgfplots manual. I was hoping someone could help me.

What I would like to do is create a curve by joining piecewise pieces defined using Bezier curves and control points. This is to create graphs for calculus students where I give them the graph of a function and they have to create a graph of the derivative. So, I would like to be able to precisely control where inflection points, extrema, etc....

Here's an example, but I have no clue what's going on or why the points have been connected in the order that they've been connected in.

\documentclass[border=6pt]{standalone}
\usepackage{pgfplots}
\usepgfplotslibrary{patchplots}
\begin{document}

\begin{tikzpicture}
\begin{axis}[nodes near coords={(\coordindex)},
title={\texttt{patch type=cubic spline}}]
mark=*,
patch,
patch type=cubic spline]
coordinates {
(2,2)        [0]
(0,2)        [0]
(2,0)        [0]
(0,0)        [1]
};
\end{axis}
\end{tikzpicture}

\end{document}

My understanding of what I read was that the tags [0] would help define control points. I expected this curve to look something like an S shape starting at (2,2) and ending at (0,0) where (0,2) and (2,0) were control points.

I thought I could accomplish this with something like:

\documentclass[border=6pt]{standalone}
\usepackage{pgfplots}
\usepgfplotslibrary{patchplots}
\begin{document}

\begin{tikzpicture}
\begin{axis}
[nodes near coords={(\coordindex)},
title={\texttt{patch type=cubic spline}}
]

\draw (axis cs:2,2) .. controls (axis cs:0,2) and (axis cs:2,0) .. (axis cs:0,0);

\end{axis}

\end{tikzpicture}

\end{document}

But pgfplots seems to ignore my path for its bounding box.

The feature patch type=cubic spline expects interpolation points, i.e. points which are on the curve. It chooses the unique cubic spline which passes through (= interpolates) the four points.

\draw ... controls <A> and <B> .. is the TikZ instruction for bezier drawing operations and does what you expect it to do.

In pgfplots, only coordinates inside of an \addplot ...; contribute to axis limits.

It sounds as if you two choices:

1. to use an interpolatory basis together with pgfplots or
2. to rely on a pure tikz solution with \draw .. controls .. (which can be concatenated, by the way).

Solution 2. can be drawn inside of a pgfplots axis; in this case, axis limits need to be given using xmin=-1,xmax=3,ymin=-1,ymax=3 or something like that:

\documentclass[border=6pt]{standalone}
\usepackage{pgfplots}
\usepgfplotslibrary{patchplots}

% 1.11 does not need "axis cs:", i.e. (axis cs:2,2) is equivalent to (2,2)
\pgfplotsset{compat=1.11}
\begin{document}

\begin{tikzpicture}
\begin{axis}
[
title={\texttt{patch type=cubic spline}},
xmin=-1,xmax=3,ymin=-1,ymax=3,
]

\draw (2,2) .. controls (0,2) and (2,0) .. (0,0);

\node at (2,2) {$0$};
\node at (0,2) {$1$};
\node at (2,0) {$2$};
\node at (0,0) {$3$};

\end{axis}
\end{tikzpicture}

\end{document}

• I particularly appreciate the comment about (axis cs:2,2) vs (2,2) – A.Ellett Sep 29 '14 at 21:07


text:
• ### Announcements

#### Archived

This topic is now archived and is closed to further replies.

# Vector along a plane

## Recommended Posts

How do I rotate a vector so that it is parallel to a plane? I can do this using sin/cos but is there a way to do it in directX? The info I have is 1) The Plane as a D3dPlane 2) The normal of the plane as a d3dvector 3) The vector i want to rotate as a d3dvector I can use the the D3dxmatrixReflect to reflect the vector, but this is more for bounce, and I''m trying to get acceleration along a slope. Learning to fly is easy, but as a tortoise, the landings are really rough.

##### Share on other sites
The fastest way I know of is to algebraically simplify the following:

A = input_vector;N = plane_normal_vector;P = Projection(A, N); // A onto NB = A-P; // the remainder, which is aligned to the planeB = B * Length(A) / Length(B); // fixes the length// output is B.

I'm about 70% sure that there is a way to do it without the Length() calculations that would be pretty slow. You may be able to remove that part of the equation completely if you simplify the equations algebraically.

(edit)
You're using this for your ball/landscape thing, right? If so, this might not be the best solution for that situation.

[edited by - Nypyren on March 7, 2003 4:09:39 AM]

##### Share on other sites
Why not for the ball thing (and thank''s for remembering).

Learning to fly is easy, but as a tortoise, the landings are really rough.

• ### Forum Statistics

• Total Topics
628285
• Total Posts
2981835

• 10
• 10
• 10
• 11
• 17


text:
# Identify the formula of the compound iron (III) phosphide.

###### Question:

Identify the formula of the compound iron (III) phosphide.

### Where does rev. dimmesdale hold his vigil why is it important?

Where does rev. dimmesdale hold his vigil why is it important?...

### Which reason best explains why “The Beginnings of the Maasai” can be classified as a myth? It has fantastic features that help to explain the reason for the Maasai’s traditions. It has supernatural features that help to explain the volcanic eruption in the Maasai’s home. It has supernatural features that help to explain the conflict with the Maasai’s cattle. It has fantastic features that help to explain the purpose of the Maasai’s weather.

Which reason best explains why “The Beginnings of the Maasai” can be classified as a myth? It has fantastic features that help to explain the reason for the Maasai’s traditions. It has supernatural features that help to explain the volcanic eruption in the Maasai’s home. It has supernatural ...

### Spontaneous depolarization of the nodal tissue in the sa node occurs at ____________ action potentials per minute:

Spontaneous depolarization of the nodal tissue in the sa node occurs at ____________ action potentials per minute:...

### Algebra 2 help - DO NOT GUESS! EXPLAIN YOUR ANSWER PLEASE :) How far does the barnacle travel in one revolution of the water wheel?Just the last question in the picture ^^ I got the other two! This question is very confusing and i forget how to do it so some help would be great :,)WILL REPORT ANY RANDOM GUESSES**

Algebra 2 help - DO NOT GUESS! EXPLAIN YOUR ANSWER PLEASE :) How far does the barnacle travel in one revolution of the water wheel?Just the last question in the picture ^^ I got the other two! This question is very confusing and i forget how to do it so some help would be great :,)WILL REPORT ANY RA...

### Is it appropriate to elect a variety of English language as the only correct one to teach?

Is it appropriate to elect a variety of English language as the only correct one to teach?...

### When reading electronics texts it is important to

When reading electronics texts it is important to...

### Which of the following was the most significant outcome of the Seven Years' War? American Indians won protection for some ancestral lands. France as an imperial power disappeared from North America. Colonists rallied behind Great Britain in supporting the war effort. Britain and Spain transferred some territorial claims between them.

Which of the following was the most significant outcome of the Seven Years' War? American Indians won protection for some ancestral lands. France as an imperial power disappeared from North America. Colonists rallied behind Great Britain in supporting the war effort. Britain and Spain transferred so...

### (1.1)3 + 8.6 ÷ 2.15 – 0.086 = Plisss answer the question

(1.1)3 + 8.6 ÷ 2.15 – 0.086 = Plisss answer the question...

### Les just bought a megaphone of root beer. As he drinks from the giant cup, he eventually becomes full. One of his friend's comments, "If you don't stop drinking that stuff, you will get sick." Les replies, "Hey, I bought it, and I am not going to waste one drop of it." Les's behavior could best be described by which of the following mental biases?

Les just bought a megaphone of root beer. As he drinks from the giant cup, he eventually becomes full. One of his friend's comments, "If you don't stop drinking that stuff, you will get sick." Les replies, "Hey, I bought it, and I am not going to waste one drop of it." Les's behavior could best be d...

### What is the distance between the points (4, 7) and (-8, -9) in the coordinate plane

What is the distance between the points (4, 7) and (-8, -9) in the coordinate plane...

### . Which country did iraq invade in 1990, leading to the outbreak of the persian gulf war?

. Which country did iraq invade in 1990, leading to the outbreak of the persian gulf war?...

### Factors affecting military capabilities of Nations include

factors affecting military capabilities of Nations include...

### Cabeza de Vaca __________________. a. wandered for 8 years, seeing the continent from east to west b. was part of Coronado's expeditions c. was a missionary in Florida and never made it west d. found the Seven Cities of Gold

Cabeza de Vaca __________________. a. wandered for 8 years, seeing the continent from east to west b. was part of Coronado's expeditions c. was a missionary in Florida and never made it west d. found the Seven Cities of Gold...

### THE SUM OF A NUMBER X AND 4 ALL DIVIDED BY 3

THE SUM OF A NUMBER X AND 4 ALL DIVIDED BY 3...

### Economic, population, and battle maps are all examples of __________.

Economic, population, and battle maps are all examples of __________....

### E._____________________

E._____________________...


text:
Problem #91

 91 The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$? $[asy]size(140); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("A",A,W); label("B",B,W); label("C",C,NE); label("D",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]$ This problem is copyrighted by the American Mathematics Competitions.
Note: you aren't logged in. If you log in, we'll keep a record of which problems you've solved.

Instructions for entering answers:

• Reduce fractions to lowest terms and enter in the form 7/9.
• Numbers involving pi should be written as 7pi or 7pi/3 as appropriate.
• Square roots should be written as sqrt(3), 5sqrt(5), sqrt(3)/2, or 7sqrt(2)/3 as appropriate.
• Exponents should be entered in the form 10^10.
• If the problem is multiple choice, enter the appropriate (capital) letter.
• Enter points with parentheses, like so: (4,5)
• Complex numbers should be entered in rectangular form unless otherwise specified, like so: 3+4i. If there is no real component, enter only the imaginary component (i.e. 2i, NOT 0+2i).

For questions or comments, please email markan@eudelic.com.

Find out how your skills stack up!

Try our new, free contest math practice test. All new, never-seen-before problems.

AMC/AIME classes

I offer online AMC/AIME classes periodically. Join the mailing list to be informed next time they're offered.

Private coaching is also available.


text:
New Industry Products

# Zetex Intros SOD323 Packaged Varactor Diodes - ZMV830 & ZMV900 Series

May 02, 1999 by Jeff Shepard

Zetex (Commack, NY) introduced its new SOD323 packaged Varactor Diodes, which are claimed to offer the same high-performance operation as their larger SOT23 packaged alternatives. Zetex guarantees a minimum quality factor Q at test conditions of 50MHz and a low Vr of 3V or 4V. The Hyperabrupt ZMV830 Series of SOD323 Tuners offer a nominal capacitance range between 8.2pF and 68pF for a Vr of 2V and a frequency of 1MHz. The ZMV900 series of High Ratio Hyperabrupt Varactor Diodes are claimed to demonstrate a much faster change in capacitance for a given change in bias voltage. The mid-range ZMV933 device provides capacitance of 42pF at 1Vr, falling to 12pF as bias rises to 4V. Breakdown voltage for the product range is 12V. Pricing for the new devices ranges from $0.46 to$0.54 each, in quantities of 5,000.


text:
Physical Meaning of Vanishing of the Norm of the Weyl Tensor

The BBMB Black Hole is a solution of the action:

$$S = \int d^{4}x\sqrt{-g}\left[\cfrac{1}{16\pi G}R - \cfrac{1}{2}\nabla^{\mu}\phi \nabla_{\mu}\phi - \cfrac{1}{12}R\phi^{2}\right]$$

where $$\phi$$ is a scalar field.

Solving this action, the metric and the scalar field finally read (setting $$8\pi G=1$$):

$$ds^{2} = - \left(1-\cfrac{m}{r}\right)^2dt^{2} + \left(1-\cfrac{m}{r}\right)^{-2}dr^{2} + r^{2}\big(d\theta^{2} + \sin^{2}\theta d\varphi^{2} \big)$$

$$\phi (r) = \pm \sqrt{6}\cfrac{m}{r-m}$$

Computing the Weyl invariant we find:

$$C_{abcd}C^{abcd} = \frac{48 m^2 (m-r)^2}{r^8}$$

The metric is the extermal Reissner Nordstrom one and the scalar field diverges at the black hole horizon: $$r_H=m$$. Is there any physical meaning behind the vanishing of the Weyl invariant at the point of divergence of the scalar field?

Since the extermal Reissner - Nordstrom spacetime is the same with the BBMB black hole the Weyl invariant will be the same. In that case how do we interpret the vanishing of the Weyl invariant at the black hole horizon?

Let us leave scalar field aside for now and discuss just the geometry, which is the same as that of an extreme Reissner–Nordström (ERN) solution.

The ERN metric is type D of Petrov classification, so vanishing of Weyl invariant implies that the Weyl tensor is zero at the horizon (and not just null). Under suitable choice of NP tetrad there is only one Weyl scalar of NP formalism: $$\Psi_2 =\frac{m(m-r)}{r^4},$$ which describes “Coulomb-like” gravitational field of an isolated charge. To clarify the physical meaning of this structure of Weyl tensor one could consider the stresses induced within some extended body (“gravitational compass”) by tidal accelerations of its different parts as was discussed by Szekeres here, see case (c) in this image:

So the vanishing of Weyl tensor means that the stresses from tidal accelerations on such a device would vanish when it approaches the black hole horizon.

Note, that for a non-extreme RN solutions there is still a hypersurface where Weyl tensor vanishes located at the minimum of $$g_{00}$$ metric component, $$r=Q^2/M$$ in Schwarzschild coordinates. For non-extreme RN black hole this hypersurface is spacelike and located between inner and outer horizons, while for over-extreme RN solutions it is timelike and serves as a boundary between regions where gravitational attraction toward the central singularity at larger $$r$$ changes to gravitational repulsion at smaller $$r$$.

Vanishing of Weyl tensor on the horizon also means that the near-horizon geometry is conformally flat. This near horizon limit of ERN geometry is Bertotti–Robinson spacetime $$AdS_2 \times S^2$$.

Weyl tensor also vanishes for large distances $$r\to \infty$$ and there is in fact a discrete conformal isometry that exchanges extremal black hole horizons (future and past, $$\mathcal{H}^+$$ and $$\mathcal{H}^-$$) and future and past null infinities, $$\mathcal{I}^+$$ and $$\mathcal{I}^-$$ discovered by Couch and Torrence in 1984. With ERN metric written in “isotropic” coordinates: $$ds^2=−\left(1 +\frac{Q}{r}\right)^{−2}dt^2+\left(1 +\frac{Q}{r}\right)^2(dr^2+r^2dΩ_2^2),$$ where horizon is located at $$r=0$$, this conformal isometry is just an inversion: $$\hat{r}=\frac{Q^2}{r},\qquad d\hat{s}^2=\frac{Q^2}{r^2}ds^2,$$ The new metric is the same as ERN but with a new radial coordinate $$\hat{r}$$. In recent works such as this this symmetry has been used to relate conserved quantities at horizon and at null infinity.

Now, let us discuss the scalar field of the BBMB solution. To me the intuitive reason why scalar field must diverge at the horizon is simple: the spatial distance to the horizon as measured by an integral $$\int \sqrt{|g_{rr}|} dr$$ along $$t=\mathrm{const}$$ hypersurface is infinite. And since $$\phi$$ is dependent only on $$r$$ and its SE tensor remains bounded on the horizon the field variation must also be infinite as it reaches horizon. Note, that scalar field singularity at the horizon is rather mild from the point of view of both neutral particles and particles carrying scalar charge as has been demonstrated by Bekenstein. Also, the existence of enhanced $$AdS_2$$ near-horizon symmetry, which maps time independent scalar field onto time dependent but still having a static SE tensor, suggests to me that there must also be a dynamical version of BBMB black hole with time dependent scalar field that sources precisely the same geometry and has a nicer behavior across horizon. There is probably a connection with “Aretakis charges” describing polynomial growth of scalar field perturbations near ERN horizon. While Aretakis' original work concerns test fields, there must also be a nonlinear version of the results. One could also look for a version of Couch–Torrence inversion symmetry relating both metric (such as asymptotic values of NP scalars) and scalar field quantities at horizon and at spatial infinity.

All this suggests that there must be some interesting physics happening with conformal scalar field near the horizon of BBMB black hole.

• Thank you for the answer! Unfortunately i do not have the background to understand everything in your answer, but i think it will come handy in the near future!!! Thank you! Mar 16, 2021 at 10:51
• @ApolloRa The Riemann tensor consists of he Ricci tensor describing how spacetime volumes increase or decrease and the Weyl tensor describing how they stretch. So the norm of the Weyl tensor is a measure of how the tidal forces scale. This corresponds to the statements of A.V.S. referring to tidal accelerations and conformally flat geometry. The vanishing norm of the Weyl tensor implies that the scalar field cancels the tidal forces at the horizon. Any corrections are welcome. Mar 22, 2021 at 14:34


text:
Over the last few days I've converted some Subversion repositories used for Debian packaging to Git. None of these Subversion repositories contained upstream sources because of Subversion's storage inefficiency. With Git I wanted to change that and also track upstream sources in my repositories. To move a Git repository from a debian-only to debian+upstream layout I found zack's recipe very helpful.

The actual conversion to Git was done with git-svn. I cloned the Subversion repositories with git-svn's --prefix=svn-import/ and --stdlayout options so that the trunk and all branches and tags were imported as remote branches with svn-import/ prepended. Then I created local branches of the previous Subversion branches and removed the remote remnants, for example:

git checkout -b debian/backports/etch svn-import/branches/etch-backports
git branch -d -r svn-import/branches/etch-backports

The Subversion tags are imported as svn-import/tags/<version> where <version> is the Debian version number. I don't need branches for every version of my packages but I wanted to convert these tags to real Git tags but without losing the actual commit dates and messages. To achieve this I wrote this small script:

#!/bin/bash

for branch in git branch -r; do
if [ echo $branch | egrep "svn-import/tags/.+$" ]; then
version=basename $branch subject=git log -1 --pretty=format:"%s"$branch
export GIT_COMMITTER_DATE=git log -1 --pretty=format:"%ci" $branch echo "Tag$version [Y/n]?"
if [ -z $yesno ] || [$yesno = "Y" ]; then
git tag -s -f -m "$subject" "debian/$version" "$branch^" git branch -d -r$branch

For each remote branch that contains svn-import/tags/ it gets the version number, the commit message of the tag and exports GIT_COMMITTER_DATE (man git-tag) with the value of the date the version was tagged in Subversion and then asks you if it should tag the parent of the tag branch with the original commit date and message and remove the then unnecessary tag branch. The "safety measure" is there because people sometimes commit directly to Subversion tags (ugh!) and then "\$branch^" would not be the commit you want to tag. To decide which Subversion tags are safe to tag with this script, gitk --all can be of great help.


text:
GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 07 Aug 2020, 10:47

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# A boat traveled upstream 90 miles at an average speed of

Author Message
TAGS:

### Hide Tags

Intern
Joined: 06 Sep 2010
Posts: 16
Schools: HBS
WE 1: Management Consulting- 2 years
WE 2: Private Equity- 2 years
A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

09 Sep 2010, 16:29
18
1
143
00:00

Difficulty:

55% (hard)

Question Stats:

71% (03:13) correct 29% (03:24) wrong based on 1082 sessions

### HideShow timer Statistics

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1
Math Expert
Joined: 02 Sep 2009
Posts: 65854
Re: Difficult PS Problem- Help!  [#permalink]

### Show Tags

09 Sep 2010, 16:54
30
53
jjewkes wrote:

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

-2.5
-2.4
-2.3
-2.2
-2.1

Trip upstream took $$\frac{90}{v-3}$$ hours and trip downstream took $$\frac{90}{v+3}$$ hours. Also given that the difference in times was $$\frac{1}{2}$$ hours --> $$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$;

$$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$ --> $$\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}$$ --> $$\frac{90*6}{v^2-9}=\frac{1}{2}$$ --> $$v^2=90*6*2+9$$ --> $$v^2=9*(10*6*2+1)$$ --> $$v^2=9*121$$ --> $$v=3*11=33$$;

Trip downstream took $$\frac{90}{v+3}=\frac{90}{33+3}=2.5$$ hours.

_________________
VP
Joined: 02 Jul 2012
Posts: 1091
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: A boat traveled upstream a distance of 90miles at an average  [#permalink]

### Show Tags

19 Nov 2012, 00:10
14
Amateur wrote:
A boat traveled upstream a distance of 90miles at an average speed of (v-3) miles per hour and then traveled down stream at an average of (v+3) miles per hour. If upstream look 30 min more than downstream time, how much time did the boat travel downstream?
a) 2.5
b) 2.4
c) 2.3
d) 2.2
e) 2.1

how to solve these kind of questions in 2 min? any short way...
\frac{90}{(v-3)}=\frac{90}{(v+3)} +1/2
solving for v, value of \frac{90}{(v+3)} is tedious... and answer choices are in too close range to make an estimation.... how should I be tackling these?

The equation itself is simple enough..

$$\frac{90}{v-3}- \frac{90}{v+3} = \frac{1}{2}$$

$$\frac{540}{v^2-9} = \frac{1}{2}$$

$$v^2 - 9 = 1080$$

$$v^2 = 1089$$ $$= 3^2 * 11^2$$

v = 3*11 = 33

Kudos Please... If my post helped.
##### General Discussion
Joined: 31 Dec 1969
WE: Supply Chain Management (Energy and Utilities)
Re: Difficult PS Problem- Help!  [#permalink]

### Show Tags

11 Sep 2010, 20:33
1
2
90/(v-3)=90(v+3)+.5

90[v+3-v+3]/(v*v-9)=.5
90*6/(v*v-9)=.5
90*12+9=v*v
v*v=1089
v=33
v+3=36

Time required to travel downstream=90/36=2.5 hours
Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 311
Location: Milky way
Schools: ISB, Tepper - CMU, Chicago Booth, LSB

### Show Tags

30 Sep 2010, 21:43
1
1
Michmax3 wrote:
A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then traveled the same distance downstream at an average speed of (V+3) miles per hour. If the upstream trip took half an hour longer than the downstream trip, how many hours did it take the boat to travel downstream?

A)2.5
B)2.4
C)2.3
D)2.2
E)2.1

Given 90/(v-3) = 90/(v+3) + (1/2)
90[(1/(v-3)) - (1/(v+3))] = 1/2
90[6/(v^2 - 9)] = 1/2

90*6*2 = v^2 - 9 => v^2 = 1089 => v = 33.

Downstream time -- 90/(v+3) => 90/36 => 2.5 (A).
Retired Moderator
Joined: 02 Sep 2010
Posts: 701
Location: London

### Show Tags

30 Sep 2010, 21:45
3
Michmax3 wrote:
A boat traveled upstream a distance of 90 miles at an average speed of (V-3) miles per hour and then traveled the same distance downstream at an average speed of (V+3) miles per hour. If the upstream trip took half an hour longer than the downstream trip, how many hours did it take the boat to travel downstream?

A)2.5
B)2.4
C)2.3
D)2.2
E)2.1

Upstream time = downstream time + 0.5

90/(v-3) = 90/(v+3) + 1/2
90 (1/(v-3) - 1/(v+3))= 1/2
90 (6/(v^2-9)) = 1/2
v^2 = 1089
V = 33

Downstream time = 90/36 = 2.5

_________________
Intern
Joined: 05 Oct 2010
Posts: 43

### Show Tags

22 Nov 2010, 01:19
2
shrouded1 wrote:
Michmax3 wrote:
v^2 = 1089
V = 33

Downstream time = 90/36 = 2.5

Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?

many thanks
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India

### Show Tags

22 Nov 2010, 05:57
12
20
hafgola wrote:

Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?

many thanks

Let me take your question first:

It is a good idea to remember squares of numbers till 20. They come in handy sometimes.
To find the square root of bigger numbers (which are perfect squares), observe the following:
1. A square of a number only ends in 0/1/4/5/6/9.
-If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.)
-If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 1[highlight]9[/highlight], root 441 = 2[highlight]1[/highlight])
-If it ends in 4, the square root will end in 2/8
-If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer)
-If it ends in 6, the square root will end in 4/6.
-If it ends in 9, the square root will end in 3/7.
These are all derived by just observing the last digits when we square a number.. You don't need to learn this.

Since 1089 ends in a 9, its square root will end in 3/7.
Also $$30^2 = 900$$ so root of 1089 will be greater than 30.
But $$40^2 = 1600$$ so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check.

Also, an equation like $$\frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}$$, doesnt need to be solved.

99% chance is that v will be an integer.
The options give me values for $$\frac{90}{{v + 3}}$$

Let me say if it is 2.5, then 90/2.5 = 36 and v = 33
Then $$\frac{90}{{v - 3}}$$ = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition.

Had 2.5 not been the first option, you might say that it would take me a long time to check each option.
No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24
90/2.3. No ways 900 is divisible by 23
90/2.2. No ways 900 is divisible by 11
90/2.1. No ways 900 is divisible by 7
In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct.
_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator
Joined: 16 Nov 2010
Posts: 1162
Location: United States (IN)
Concentration: Strategy, Technology
Re: Difficult PS Problem- Help!  [#permalink]

### Show Tags

27 Mar 2011, 00:18
3
1
90/(v-3) = 90/(v+3) + 1/2

=> 90( V + 3 - v + 3)/(v-3)(v+3) = 1/2

=> 90 * 6 * 2 = v^2 - 9

=> v^2 = 1080 + 9

=> v^2 = 9 (120 + 1)

=> v^2 = 3^2 * 11^2

=> v = 3*11 = 33

So Downstream time = 90/36 = 30/12 = 2.5 hrs

_________________
Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

GMAT Club Premium Membership - big benefits and savings
Senior Manager
Joined: 01 Feb 2011
Posts: 490
Re: Difficult PS Problem- Help!  [#permalink]

### Show Tags

25 Aug 2011, 16:19
2
upstream v-3 t+1/2

downstream v+3 t

(v+3)t = (v-3)(t+1/2)

vt + 3t = vt +v/2 - 3t - 3/2

=> t = (v-3)/12 = 90/v+3 => v = 33

=> t = 30/12 = 2.5 hours

Manager
Joined: 27 Dec 2012
Posts: 126

### Show Tags

14 Mar 2013, 01:11
1
VeritasPrepKarishma wrote:
hafgola wrote:

Hi I struggled with this one, I am wondering
Is there any good shortcut in how to take the root of 1089 ? how do you do that without spending too much time on it ?

many thanks

Let me take your question first:

It is a good idea to remember squares of numbers till 20. They come in handy sometimes.
To find the square root of bigger numbers (which are perfect squares), observe the following:
1. A square of a number only ends in 0/1/4/5/6/9.
-If it ends in 0, it will have even number of 0s at the end and the square root will end in 0 (I know 1340 cannot be a perfect square.)
-If it ends in 1, the square root will end in 1/9 (e.g. root 361 = 19, root 441 = 21)
-If it ends in 4, the square root will end in 2/8
-If it ends in 5, there will be a 2 right before it and its square root will end in 5 (e.g. root 625 = 25, root 675 is not an integer)
-If it ends in 6, the square root will end in 4/6.
-If it ends in 9, the square root will end in 3/7.
These are all derived by just observing the last digits when we square a number.. You don't need to learn this.

Since 1089 ends in a 9, its square root will end in 3/7.
Also $$30^2 = 900$$ so root of 1089 will be greater than 30.
But $$40^2 = 1600$$ so root of 1089 will be much less than 40. The number that can be the square root of 1089 is 33 (if 1089 is a perfect square). You can square 33 to check.

Also, an equation like $$\frac{90}{{v -3}} - \frac{90}{{v + 3}} = \frac{1}{2}$$, doesnt need to be solved.

99% chance is that v will be an integer.
The options give me values for $$\frac{90}{{v + 3}}$$

Let me say if it is 2.5, then 90/2.5 = 36 and v = 33
Then $$\frac{90}{{v - 3}}$$ = 3. I get the difference of 1/2 an hour. 2.5 satisfies my condition.

Had 2.5 not been the first option, you might say that it would take me a long time to check each option.
No. 90/2.4 = 900/24 which is not an integer since 900 is not divisible by 8 so it will not be divisible by 24
90/2.3. No ways 900 is divisible by 23
90/2.2. No ways 900 is divisible by 11
90/2.1. No ways 900 is divisible by 7
In fact, when I look at the options, the first one I try is 2.5. With practice, you will gain the instinct.

Hi karishma,
I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill??
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India

### Show Tags

14 Mar 2013, 19:36
1
sauravleo123 wrote:
Hi karishma,
I have been able to solve most of the questions from work, speeed distane & mixtures. However i always solve them using the algebraic approach which eats up a lot of time. I have seen you solving questions by a logical approach that saves a considerable bit of time. I have tried solving that way but I just cant get going. Could you please advice me on how to develop this skill??

It needs practice. You need to believe that you can solve the question very quickly for you to be able to think up of logical approaches. In GMAT, you can solve almost every question within two minutes so if you get a big complicated equation, you can be sure that you are missing something. If the question looks really complicated, we know there has to be a trick (In fact, one tends to overlook the so called short-cuts when questions are simple and don't take much time anyway).
I would suggest you to read up and practice the logical solutions - once you understand the logic, try using it in other questions. Don't worry about the time factor. I discuss some logic based ideas regularly on my blog. Check it out:

http://www.veritasprep.com/blog/categor ... om/page/3/
_________________
Karishma
Veritas Prep GMAT Instructor

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

10 Feb 2014, 21:12
3
1
CfaMiami wrote:
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

Here is a post discussing how to solve such equations easily: http://www.veritasprep.com/blog/2013/03 ... culations/
_________________
Karishma
Veritas Prep GMAT Instructor

Retired Moderator
Joined: 17 Sep 2013
Posts: 315
Concentration: Strategy, General Management
GMAT 1: 730 Q51 V38
WE: Analyst (Consulting)
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

20 Mar 2014, 12:19
1
2.5 hrs..A set of best possible option spares u the horror of solving the equations..

90 miles is the distance traveled downstream..2.5 is the only only option that divides 90..I am pretty sure the speeds are never (25.12+3) & (25.12-3)..So we can safely assume an integer..2.5 & 3 are the set of speeds that aptly fit in.

took me just 10 secs..
Intern
Joined: 18 Jan 2016
Posts: 19
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

08 Feb 2016, 15:03
This is the hardest problem I have seen in my GMAT studies so far. Not because of the maths involved, but the time required. Even when I know the answer, I can't go through the calculations in under 5 mins (I just spent an hour trying over and over). Ouch.
GMAT Club Legend
Joined: 11 Sep 2015
Posts: 5000
GMAT 1: 770 Q49 V46
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

08 Feb 2016, 16:18
2
Quote:
A boat travelled upstream a distance of 90 miles at an average speed of (v-3) miles per hour and then travelled downstream at an average speed of (V+3) miles per hour. If the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?
A) 2.5
B) 2.4
C) 2.3
D) 2.2
E) 2.1

I like to begin with a "word equation."
We can write:
travel time upstream = travel time downstream + 1/2

Time = distance/rate
So, we can replace elements in our word equation to get:
90/(v-3) = 90/(v+3) + 1/2

Now solve for v (lots of work here)
.
.
.
v = 33

So, travel time downstream = 90/(v+3)
= 90/(33+3)
= 90/36
= 5/2
= 2 1/2 hours

Cheers,
Brent
_________________
If you enjoy my solutions, you'll love my GMAT prep course.

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 10797
Location: Pune, India
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

09 Feb 2016, 02:55
1
lamode wrote:
This is the hardest problem I have seen in my GMAT studies so far. Not because of the maths involved, but the time required. Even when I know the answer, I can't go through the calculations in under 5 mins (I just spent an hour trying over and over). Ouch.

Check out these posts lamode:

http://www.veritasprep.com/blog/2013/03 ... culations/
http://www.veritasprep.com/blog/2013/03 ... s-part-ii/

You will learn how to do such questions much faster.
_________________
Karishma
Veritas Prep GMAT Instructor

Manager
Joined: 28 Apr 2016
Posts: 81
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

25 Aug 2016, 05:04
I am a little confused as to how we arrived at the highlighted equation, from the previous equation. How did we get 90 * 6. Can someone please explain?

Bunuel wrote:
jjewkes wrote:

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

-2.5
-2.4
-2.3
-2.2
-2.1

Trip upstream took $$\frac{90}{v-3}$$ hours and trip downstream took $$\frac{90}{v+3}$$ hours. Also given that the difference in times was $$\frac{1}{2}$$ hours --> $$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$;

$$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$ --> $$\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}$$ --> $$\frac{90*6}{v^2-9}=\frac{1}{2}$$ --> $$v^2=90*6*2+9$$ --> $$v^2=9*(10*6*2+1)$$ --> $$v^2=9*121$$ --> $$v=3*11=33$$;

Trip downstream took $$\frac{90}{v+3}=\frac{90}{33+3}=2.5$$ hours.

Math Expert
Joined: 02 Sep 2009
Posts: 65854
Re: A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

25 Aug 2016, 07:15
1
ameyaprabhu wrote:
I am a little confused as to how we arrived at the highlighted equation, from the previous equation. How did we get 90 * 6. Can someone please explain?

Bunuel wrote:
jjewkes wrote:

A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

-2.5
-2.4
-2.3
-2.2
-2.1

Trip upstream took $$\frac{90}{v-3}$$ hours and trip downstream took $$\frac{90}{v+3}$$ hours. Also given that the difference in times was $$\frac{1}{2}$$ hours --> $$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$;

$$\frac{90}{v-3}-\frac{90}{v+3}=\frac{1}{2}$$ --> $$\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}$$ --> $$\frac{90*6}{v^2-9}=\frac{1}{2}$$ --> $$v^2=90*6*2+9$$ --> $$v^2=9*(10*6*2+1)$$ --> $$v^2=9*121$$ --> $$v=3*11=33$$;

Trip downstream took $$\frac{90}{v+3}=\frac{90}{33+3}=2.5$$ hours.

$$\frac{90(v+3)-90(v-3)}{v^2-9}=\frac{1}{2}$$;

$$\frac{90v+3*90-90v+3*90}{v^2-9}=\frac{1}{2}$$;

$$\frac{3*90+3*90}{v^2-9}=\frac{1}{2}$$;

$$\frac{90*6}{v^2-9}=\frac{1}{2}$$.

Hope it's clear.
_________________
Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 11419
Location: United States (CA)
A boat traveled upstream 90 miles at an average speed of  [#permalink]

### Show Tags

26 Aug 2016, 08:55
1
DSGB wrote:
A boat traveled upstream 90 miles at an average speed of (v-3) miles per hour and then traveled the same distance downstream at an average speed of (v+3) miles per hour. If the trip upstream took a half hour longer than the trip downstream, then how many hours did it take the boat to travel downstream?

A. 2.5
B. 2.4
C. 2.3
D. 2.2
E. 2.1

Since time = distance/rate, therefore the time going upstream = 90/(v – 3) and the time going downstream = 90/(v + 3). Since the time going upstream is ½ hour more than the time going downstream, we can set up an equation as follows:

90/(v – 3) = 90/(v + 3) + 1/2

Let’s multiply the equation by 2(v – 3)(v + 3) to get rid of the denominators:

2(90)(v + 3) = 2(90)(v – 3) + (v – 3)(v + 3)

180v + 540 = 180v – 540 + v^2 – 9

540 = v^2 – 549

v^2 = 1089

v = $$\sqrt{1089}$$

v = 33

Since the time going downstream = 90/(v + 3) and we’ve found that v = 33, so the time going downstream = 90/(33 + 3) = 90/36 = 5/2 = 2.5 hours.

_________________

# Scott Woodbury-Stewart | Founder and CEO | Scott@TargetTestPrep.com

250 REVIEWS

5-STAR RATED ONLINE GMAT QUANT SELF STUDY COURSE

NOW WITH GMAT VERBAL (BETA)

See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews

A boat traveled upstream 90 miles at an average speed of   [#permalink] 26 Aug 2016, 08:55

Go to page    1   2    Next  [ 27 posts ]


text:
# All Questions

3 views

### Changing font of to address on letter.cls

I am trying to change the font of different pieces of a letter.cls letter so my cover letter better matches my resume. Right now, I have been able to change the body font, but not the font of the ...
7 views

### Use array within a gather

I have the following code: \begin{gather} u(x,t) = 0 \ \mathrm{ si\ } x = 0, x = 1\\ \label{ex1bc} \\ u(x,0) = \left\{ \begin{array}{ll} 2x, & \mathrm{si\ } 0 \le x \le 1/2 \\ ...
14 views

### absolute value in numerator of fraction

I am having trouble writing a term of an equation which is fractional where only the numerator is inside the absolute value symbol. The syntax I used is: \frac{\abs{\lambda_f}}{8\pi^2} But I keep ...
4 views

### How can I use an algorithm2e function macro inside an algorithm2e caption?

Algorithm2e lets you define functions using \SetKwFunction{Fn}{Function} which you can then reference anywhere in the document (not just in an algorithm environment) using e.g. \Fn (which will output ...
12 views

### Spacing between custom command and subsequent text [duplicate]

I have defined a custom command for the name of my project; e.g. \newcommand{\projectname}{Gizmo}. When I used the command in my document, I've found that I need to place a ~ after the command in ...
9 views

### Alternative to a Minipage page breaks

I would like to have a replacement for my minipage design of this exercise environment. I recently learned that minipages are unbreakable across page boundaries. And as a novice to the TeX/LaTeX ...
15 views

### Section titles formatted as normal text

Can I get output like that of this \documentclass{article} \begin{document} \noindent Sections whose titles are set as normal text.---I would like some of those. \end{document} but such that the ...
14 views

41 views

### Error when using tikz \node, \subnode

In the code below, building off solutions found here, I am trying to use the \node and \subnode options to draw the arrow, but I cannot get the file to compile. Can you help me to fix the code to get ...
41 views

### Bring enumerate back to numbering default

I'm using a nested list, where the outer items should be with numerals and the inner ones with numbers. I'm trying this with the package enumerate: \begin{enumerate}[I] \item numeral I. This is ...
25 views

### “Extra \fi” error when defining mutually exclusive package options with conditionals

I have been trying to create a package, which loads the babel package and sets the language with an option. I've been getting strange errors when trying to define mutually exclusive package options, ...
27 views

### How can I insert vertical text spanning multiple rows into a table?

I would like to add a column to my table that looks like the right most column in the included picture. As of now, I am using tabularx to create the table \begin{tabularx}{\textwidth}{|X|c|} ...
39 views

### How to use lettrine in Cyrillic?

There are many topics about Latin initial capitals but is it possible to make something similar to that picture in Cyrillic?
28 views

### Make the Header space a little shorter

Basically I am trying to ensure that text on page two of any chapter respects the header. I sort of resolved this by including \headsep = 45pt (it ensures that the not-first chapter-pages respect the ...
18 views

### Three and four images with captions in a Beamer framer

I have often used two columns and two images side by side in Beamer frame. But I could not get three one large and second column contains two small images or four images with captions in a single ...
20 views

### IPE: An error occurred during the Pdflatex run

I have the same problem as here: Ipe latex, error with pdflatex command but I don't understand the answer :( Problem: When I run IPE, I get this error: "An error occurred during the Pdflatex run ...
25 views

### How to make ClassicThesis work?

I'm trying to write a test document in classicthesis, but I think I have a problem with the package. I wrote this \documentclass[a4paper,11pt]{scrbook} \usepackage{lipsum} ...
15 views

### How do I increase the white space between a section's title and the page header?

I would like something like this:
18 views

### Adding shell commands to hide LaTeX Aux files after compiling in TeXShop or TexStudio

I run LaTeX under OSX, and I would like to attempt to hide all of the LaTeX helper files from the finder after compiling. It strikes me that the simplest thing to do is not to move them or delete them ...
34 views

### Where do I send patches to epstopdf? I made a patch to force B&W PDFs

I made a small patch to epstopdf to force the PDF to B&W. I had a document with some color, and I needed to convert it to a B&W PDF to avoid triggering color separations. Is there a place to ...
17 views

### Put ';' after page number in index

I'm trying to have dictionary like index, where some terms with different variations in text are defined. it seems this is easier with indexing. I try to have subitems of one index item on same line ...
41 views

### Highlight matrix cell and have arrows pointing from cell with text [duplicate]

This code is building from the solutions posted here. I am trying the approach suggested from the duplicate solutions. I am having two issues: I am not sure how to get the arrow to point to the ...
16 views

### how to change font-size of a tikz tree

Is there a possibility to change the font size of the text right on the line of the following tree? I would like to reduce it, so that the tree does not appear such big. I hope someone may help me. ...
16 views

### colorbox highlighting: overfull hbox

For a current LaTeX-Document, I've done some highlighting using colorboxes around specific words or formulas. This works quite well, although, of course, the colorbox doesn't do line breaks when it is ...
20 views

### Arithmetic overflow error using bytefield package while attempting to constrain width

I'm attempting to draft a document that makes very heavy use of the bytefield package and running into the situation that it seems no good solution exists for fields that have to be 64 bits wide (even ...
9 views

### pgf-umlcd: underscore in class name failing

I have a simple class diagram drawn with pgf-umlcd, but using an underscore within the class name leads to an error message. \begin{tikzpicture} \begin{class}{foo\_bar}{0,0} % some stuff here ...
20 views

### Why do my lists of floats get collapsed?

I defined two types of new floats, but when I try to list each of them separately, they get collapsed into the same list. Here is an MWE: \documentclass{book} \usepackage{float} ...
19 views

### How put two glyphs to mp image and get kerning

Is possible to put two glyphs in any place in MetaPost and get kerning between it? for example using Lato-Regular.ttf 'ab' in 1.7cm,2cm and get information kerning 'a' and 'b' or put 'a' in ...
5 views

### Force natbib+apalike to display the first two authors for a specific item

I am using natbib with apalike bibliography style. Some of the entries in my bib file are papers characterized by first co-authorship. So far, when I \cite them, only the first author is reported ...
24 views

### Indentation issues with highlighting cell using tikzmarkin

I am trying to eliminate offset/indentation issues when using tikzmarkin to highlight regions of a table. \documentclass{beamer} \RequirePackage{booktabs,colortbl,caption,tabularx,chngcntr, tikz, ...
18 views

### Undefined citation

I'm doing my first bibliography and I'm getting errors for one hour now. The .tex file : \documentclass[francais,10pt,twosides,a4paper]{report} \usepackage[francais,polutonikogreek]{babel} ...
140 views

### Expandable macro with loops and advanced string functions?

This might be a very basic problem (as a follow-up to this question) but I cannot figure out how to do it: I need a macro \mycmd{sometext} that yields a string depending on the first character of its ...
39 views

### How do I create a template for thesis title page?

My university wants my thesis title page to be formatted in a certain way (see left image below) and I was able to do it in LaTeX (see right image below): I used the following LaTeX code to do the ...
100 views

### Straight Table Borders

I was just wondering whether there was a simple way to make the borders of a table straight? My code currently is: \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{setspace} ...
22 views

### How to set column width in tabularx?

I have a table, how to set width of 6th row, so the widths of columns be 25%|25%|25%|25% ? right now my code of it looks like this: \documentclass{article} \usepackage[utf8]{inputenc} ...


text:
# Time Limit

Well, apparently 15 minutes is not enough to complete the practical exam of measuring the acceleration caused by gravity by measuring the period of a pendulum.

Reasons:
– I forgot (read: didn’t think) to bring my calculator. Not that I have one anyway. The reason: I think the calculations are not going to be insane. I WAS WRONG. $g = \dfrac{4\pi^2 l}{T^2}$ is hard to compute. Especially since I’m also asked to compute the frequency $f = \dfrac{1}{T}$ although I very well won’t need it. (Eventually I changed so I compute $g = (2\pi f)^2 l$.)
– The stopwatch worked only 50% of the time. Imagine working alone, fixating your view to the swinging pendulum, just to find out that the stopwatch didn’t start. 30 seconds wasted.
– Delicate handwork. I took like 3 minutes to tie the nylon string that is weighed to the pivot. Double that, as I need to do the experiment twice with different nylon lengths.
– Stupid misguiding. When you’re told “10 minutes”, you think that there is 10 minutes left. Not too long after that (about 3 minutes), while I was still doing the experiment accurately (but hence slowly), I was told “1.5 minutes left”. Imagine my immediate panic.

Must play more time management games, not rhythm games.