[PDF] Log-Log Plots 11 mars 2004 In each

View - Download Log-Log Plots

Previous PDF

Next PDF

Basic MathematicsLog-Log PlotsR Horan & M LavelleThe aim of this package is to provide a short self assessment programme for students who wish to acquire an understanding of log-log

plots.Copyrightc?2003rhoran@plymouth.ac.uk,mlavelle@plymouth.ac.ukLast Revision Date: March 11, 2004 Version 1.0

Table of Contents1.Introduction2.Straight Lines from Curves3.Fitting Data4.Final QuizSolutions to ExercisesSolutions to Quizzes

Section 1: Introduction 31. IntroductionMany quantities in science can be described by equations of the form,y=Axn.It is, though, not easy to distinguish between graphs of

different power laws. Consider the data below:??xy0••••••••••••••It is not easy to see that theredpoints lie on a quadratic (y=Ax2)

and that thebluedata are on a quartic (y=Ax4). It is, however, clear that theblackpoints lie on a straight line! Results from the packages onLogarithmsandStraight Linesenable us to recast the power curves as straight lines and so extract bothnandA. Section 1: Introduction 4Example 1Consider the equationy=xn.This is a power curve, but if we take the logarithm of each side we obtain: log(y) = log(xn) =nlog(x)since log(xn) =nlog(x)IfY= log(y) andX= log(x) thenY=nX. This shows the linear relationship. PlottingYagainstX, i.e., log(y) against log(x), leads

←1→Herenis the slope of the line. Thus:from a log-log plot, we can directly read off the power,n.

←2→2-4- 2 |4|••abcd(a)a(b)b(c)c(d)dNote that the scales on the two axes are not the same.

Section 2: Straight Lines from Curves 62. Straight Lines from CurvesExample 2Consider the more general equationy=Axn.Again we

take the logarithm of each side: log(y) = log(Axn)

= log(A) + log(xn)since log(pq) = log(p) + log(q)?log(y) =nlog(x) +log(A)since log(xn) =nlog(x)The function log(y) is a linear function of log(x) and its graph is a

Section 2: Straight Lines from Curves 7QuizReferring to the lines,a,b,canddbelow, which of the following

c

d(a)Ifbcorresponds toy=x3, thendwould describey=x-3.(b)Linesaandccorrespond to curves with the same powern.(c)In the power law yieldingcthe coefficientAis negative.(d)Ifbis fromy=x3, then inathe powernsatisfies: 0< n <3.

Section 2: Straight Lines from Curves 8Exercise 1.Produce log-log plots for each of the following power

curves. In each case give the gradient and the intercept on the log(y)

axis. (Click on thegreenletters for the solutions).(a)y=x13(b)y= 10x5(c)y= 10x-2(d)y=13x-3QuizHow doeschanging the baseof the logarithm used (e.g., using

ln(x) instead of log10(x)), change a log-log plot?(a)The log-log plot is unchanged.(b)Only the gradient changes.(c)Only the intercept changes.(d)Both the gradient and the

intercept change.Notethat in an equation of the formy= 5+3x2,taking logs directly does not help. This is because there is no rule to simplify log(5+3x2).

Instead we have to subtract the constant from each side.We then get:y-5 = 3x2,which leads to the straight line equation:log(y-5) =

2log(x) + log(3).

Section 3: Fitting Data 93. Fitting DataSuppose we want to see if some experimental data fits a power law of

the form,y=Axn. We take logs of both sides and plot the points on a graph of log(y) against log(x). If they lie on a straight line (within experimental accuracy) then we conclude thatyandxare related by a power law and the parametersAandncan be deduced from the graph. If the points do not lie on a straight line, thenxandyare not

related by an equation of this form.Example 3Consider the following data:x23070100150y4.2416.425.130.036.7To see if it obeys,y=Axn, we take logarithms of both sides. Here

we use logarithms to the base 10. This gives the new table:log10(x)0.301.481.8522.18log10(y)0.631.211.401.481.56This is plotted on the next page.

the originalxandyvalues are related by a power lawy=Axn. To find the values ofAandn,we first continue the line to the log10(y) axis which it intercepts at the blue dot:log10(A) = 0.48.This means

thatA= 100.48= 3.0 (to 1d.p.).The gradient of the line is estimated using two of the pointsn=log(y2)-log(y1)log(x2)-log(x1)=1.56-0.632.18-0.3= 0.5(to 1d.p.)

So the original data lies on the curve:y= 3x12

Section 3: Fitting Data 11Exercise 2.In the exercises below click on thegreenletters for the solutions.(a)Rewritethe following expression in such a way that it gives the equation of a straight line y=⎷4x+ 2(b)What is the difference between two power laws if, when they are

plotted as a log-log graph, the gradients are the same, but thelog(y) intercepts differ by log(3)?(c)Produce a log-log plot for the following data, show it obeys a

power law andextract the law from the data.x515305095y109036010003610

Section 4: Final Quiz 124. Final QuizBegin QuizChoose the solutions from the options given.1.Theinterceptandsloperespectively of the log-log plot ofy=12x2(a)12&log(2)(b)-log(2)&2(c)log(2)&2(d)log(1/2)&log(2)2.If the log(y) axis intercept of the log-log plot ofy=Axnis

negative, which of the following statements is true.(a)n <0(b)A= 1(c)0< A <1(d)n=-A3.The data below obeys a power law,y=Axn. Obtain the equation

and select the correct statement.x515305095y109036010003610(a)n= 3(b)A=32(c)n= 4(d)A=12End Quiz

Solutions to Exercises 13Solutions to ExercisesExercise 1(a)Fory=x13, we get on taking logs:log(y) =13log(x).

This describes a line that passes through the origin and has slope 13. Solutions to Exercises 14Exercise 1(b)Fory= 10x5, we get on taking logarithms of each side:log(y) = 5log(x) + log(10). This describes a line that passes ←1→Click on thegreensquare to return? ←1→Click on thegreensquare to return? Solutions to Exercises 16Exercise 1(d)Ify=13x-3, thenlog(y) =-3log(x) + log(13). This can also be written aslog(y) =-3log(x)-log(3). It is the equation of a line with slope-3 and intercept at-log(3). The line is sketched ←1→Click on thegreensquare to return? Solutions to Exercises 17Exercise 2(a)y=⎷4x+4can be re-expressed as follows. Subtract

4 from each side

y-4 =⎷4x y-4 = 2⎷x y-4 = 2x12 Taking logarithms of each side yieldslog(y-4)=12log(x)+ log(2) Thus plottinglog(y-4)againstlog(x)would givea straight line with slope

12and intercept log(2) on the log(y-4) axis.Click on thegreensquare to return?

Solutions to Exercises 18Exercise 2(b)Ify=Axnthen the log-log plot is the graph of the straight line log(y) =nlog(x) + log(A) So if the slope is the same the powernis the same in each case.

If the coefficientsA1andA2differ by

log(A1)-log(A2) = log(3) then log?A1A2?

= log(3)since log(p/q) = log(p)-log(q)so it follows that the coefficients are related byA1= 3A2.Click on thegreensquare to return?

Solutions to Exercises 19Exercise 2(c)To see if it obeys,y=Axn, we take logarithms to the

base 10 of both sides. The table and graph are below:log10(x)0.701.181.481.701.98log10(y)11.952.5633.56??log(x)log(y)0••••••3-2.0-1.0--1-|1|2The data points are fitted by a line that intercepts the log(y) axis atlog(A) =-0.40, soA= 10-0.40= 0.4.Thegradientcan be calcu-

lated fromn= (3-1)/(1.70-0.70)= 2. So the data lie ony= 0.4x2.

Click on thegreensquare to return?

Solutions to Quizzes 20Solutions to QuizzesSolution to Quiz:The curve isy=x2.Taking logs of both sides

gives:log(y)= log(x2) =2log(x), i.e., the log-log plot is astraight

line through the origin with gradient 2.Linebpasses though the origin and through the point (x= 2,y= 4).

From the package onStraight Lineswe know that the gradient, m, of a straight line passing through (log(x1), log(y1)) and (log(x2), log(y2)) is given by m=log(y2)-log(y1)log(x2)-log(x1) we see that thegradient of linebis given by m b=4-02-0= 2This is therefore the correct log-log plot.End Quiz c dIfccorresponds toy=Axn, thenlog(y) =nlog(x) + log(A).The intercept of linecon the log(y) axis is negative. This implies that log(A)<0, which means that0< A <1.It doesnotsignify thatA itself is negative. (Of course we also cannot take the logarithm of a negative number like this.) It may be checked that the other statements are correct.End Quiz

Solutions to Quizzes 22Solution to Quiz:The equation of a log-log plot is:log(y) =nlog(x) + log(A)If we change the base of the logarithm that is used, then the gradient

nis unchanged but the intercept, log(A), is altered. For example the log-log plot ofy= 3x4in terms oflogarithms to the

base 10is:log10(y) = 4log10(x) + log10(3)which has aninterceptat log10(3) =0.477(to 3d.p.) Using natural

logarithms the equation would becomeln(y) = 4ln(x) + ln(3)This has the same gradient, but theintercepton the ln(y) axis is now

at ln(3) =1.099(to 3d.p.) The onlyexceptionto this is ifA= 1, since logN(1) = 0 for allN.End Quizquotesdbs_dbs44.pdfusesText_44

[PDF] resoudre systeme avec ln

[PDF] toutes les écritures comptables pdf

[PDF] manhattan kaboul instruments

[PDF] premiere professionnelle apres seconde generale

[PDF] casio fx 92 college mode d'emploi

[PDF] griffith's improved vacuum apparatus for removing dust from carpets

[PDF] evolution de l'aspirateur au cours du temps

[PDF] frise chronologique de l'aspirateur

[PDF] aspirateur 1920

[PDF] evolution de l'aspirateur

[PDF] aspirateur 1950

[PDF] aspirateur 1905

[PDF] evolution du balai

[PDF] equation fonction

[PDF] équation fonctionnelle