text:
## I Don’t Have Any Good Ideas For Finding Cube Roots By Trigonometry

So I did a bit of thinking. There’s a prosthaphaeretic rule that lets you calculate square roots using nothing more than trigonometric functions. Is there one that lets you calculate cube roots?

And I don’t know. I don’t see where there is one. I may be overlooking an approach, though. Let me outline what I’ve thought out.

First is square roots. It’s possible to find the square root of a number between 0 and 1 using arc-cosine and cosine functions. This is done by using a trigonometric identity called the double-angle formula. This formula, normally, you use if you know the cosine of a particular angle named θ and want the cosine of double that angle:

$\cos\left(2\theta\right) = 2 \cos^2\left(\theta\right) - 1$

If we suppose the number whose square we want is $\cos^2\left(\theta\right)$ then we can find $\cos\left(\theta\right)$. The calculation on the right-hand side of this is easy; double your number and subtract one. Then to the lookup table; find the angle whose cosine is that number. That angle is two times θ. So divide that angle in two. Cosine of that is, well, $\cos\left(\theta\right)$ and most people would agree that’s a square root of $\cos^2\left(\theta\right)$ without any further work.

Why can’t I do the same thing with a triple-angle formula? … Well, here’s my choices among the normal trig functions:

$\cos\left(3\theta\right) = 4 \cos^3\left(\theta\right) - 3\cos\left(\theta\right)$

$\sin\left(3\theta\right) = 3 \sin\left(\theta\right) - 4\sin^3\left(\theta\right)$

$\tan\left(3\theta\right) = \frac{3 \tan\left(\theta\right) - \tan^3\left(\theta\right)}{1 - 3 \tan^2\left(\theta\right)}$

Yes, I see you in the corner, hopping up and down and asking about the cosecant. It’s not any better. Trust me.

So you see the problem here. The number whose cube root I want has to be the $\cos^3\left(\theta\right)$. Or the cube of the sine of theta, or the cube of the tangent of theta. Whatever. The trouble is I don’t see a way to calculate cosine (sine, tangent) of 3θ, or 3 times the cosine (etc) of θ. Nor to get some other simple expression out of that. I can get mixtures of the cosine of 3θ plus the cosine of θ, sure. But that doesn’t help me figure out what θ is.

Can it be worked out? Oh, sure, yes. There’s absolutely approximation schemes that would let me find a value of θ which makes true, say,

$4 \cos^3\left(\theta\right) - 3 \cos\left(\theta\right) = 0.5$

But: is there a way takes less work than some ordinary method of calculating a cube root? Even if you allow some work to be done by someone else ahead of time, such as by computing a table of trig functions? … If there is, I don’t see it. So there’s another point in favor of logarithms. Finding a cube root using a logarithm table is no harder than finding a square root, or any other root.

If you’re using trig tables, you can find a square root, or a fourth root, or an eighth root. Cube roots, if I’m not missing something, are beyond us. So are, I imagine, fifth roots and sixth roots and seventh roots and so on. I could protest that I have never in my life cared what the seventh root of a thing is, but it would sound like a declaration of sour grapes. Too bad.

If I have missed something, it’s probably obvious. Please go ahead and tell me what it is.

## How To Calculate A Square Root By A Method You Will Never Actually Use

Sunday’s comics post got me thinking about ways to calculate square roots besides using the square root function on a calculator. I wondered if I could find my own little approach. Maybe something that isn’t iterative. Iterative methods are great in that they tend to forgive numerical errors. All numerical calculations carry errors with them. But they can involve a lot of calculation and, in principle, never finish. You just give up when you think the answer is good enough. A non-iterative method carries the promise that things will, someday, end.

And I found one! It’s a neat little way to find the square root of a number between 0 and 1. Call the number ‘S’, as in square. I’ll give you the square root from it. Here’s how.

First, take S. Multiply S by two. Then subtract 1 from this.

Next. Find the angle — I shall call it 2A — whose cosine is this number 2S – 1.

You have 2A? Great. Divide that in two, so that you get the angle A.

Now take the cosine of A. This will be the (positive) square root of S. (You can find the negative square root by taking minus this.)

Let me show it in action. Let’s say you want the square root of 0.25. So let S = 0.25. And then 2S – 1 is two times 0.25 (which is 0.50) minus 1. That’s -0.50. What angle has cosine of -0.50? Well, that’s an angle of 2 π / 3 radians. Mathematicians think in radians. People think in degrees. And you can do that too. This is 120 degrees. Divide this by two. That’s an angle of π / 3 radians, or 60 degrees. The cosine of π / 3 is 0.5. And, indeed, 0.5 is the square root of 0.25.

I hear you protesting already: what if we want the square root of something larger than 1? Like, how is this any good in finding the square root of 81? Well, if we add a little step before and after this work, we’re in good shape. Here’s what.

So we start with some number larger than 1. Say, 81. Fine. Divide it by 100. If it’s still larger than 100, divide it again, and again, until you get a number smaller than 1. Keep track of how many times you did this. In this case, 81 just has to be divided by 100 the one time. That gives us 0.81, a number which is smaller than 1.

Twice 0.81 minus 1 is equal to 0.62. The angle which has 0.81 as cosine is roughly 0.90205. Half this angle is about 0.45103. And the cosine of 0.45103 is 0.9. This is looking good, but obviously 0.9 is no square root of 81.

Ah, but? We divided 81 by 100 to get it smaller than 1. So we balance that by multiplying 0.9 by 10 to get it back larger than 1. If we had divided by 100 twice to start with, we’d multiply by 10 twice to finish. If we had divided by 100 six times to start with, we’d multiply by 10 six times to finish. Yes, 10 is the square root of 100. You see what’s going on here.

(And if you want the square root of a tiny number, something smaller than 0.01, it’s not a bad idea to multiply it by 100, maybe several times over. Then calculate the square root, and divide the result by 10 a matching number of times. It’s hard to calculate with very big or with very small numbers. If you must calculate, do it on very medium numbers. This is one of those little things you learn in numerical mathematics.)

So maybe now you’re convinced this works. You may not be convinced of why this works. What I’m using here is a trigonometric identity, one of the angle-doubling formulas. Its heart is this identity. It’s familiar to students whose Intro to Trigonometry class is making them finally, irrecoverably hate mathematics:

$\cos\left(2\theta\right) = 2 \cos^2\left(\theta\right) - 1$

Here, I let ‘S’ be the squared number, $\cos^2\left(\theta\right)$. So then anything I do to find $\cos\left(\theta\right)$ gets me the square root. The algebra here is straightforward. Since ‘S’ is that cosine-squared thing, all I have to do is double it, subtract one, and then find what angle 2θ has that number as cosine. Then the cosine of θ has to be the square root.

Oh, yeah, all right. There’s an extra little objection. In what world is it easier to take an arc-cosine (to figure out what 2θ is) and then later to take a cosine? … And the answer is, well, any world where you’ve already got a table printed out of cosines of angles and don’t have a calculator on hand. This would be a common condition through to about 1975. And not all that ridiculous through to about 1990.

This is an example of a prosthaphaeretic rule. These are calculation tools. They’re used to convert multiplication or division problems into addition and subtraction. The idea is exactly like that of logarithms and exponents. Using trig functions predates logarithms. People knew about sines and cosines long before they knew about logarithms and exponentials. But the impulse is the same. And you might, if you squint, see in my little method here an echo of what you’d do more easily with a logarithm table. If you had a log table, you’d calculate $\exp\left(\frac{1}{2}\log\left(S\right)\right)$ instead. But if you don’t have a log table, and only have a table of cosines, you can calculate $\cos\left(\frac{1}{2}\arccos\left(2 S - 1 \right)\right)$ at least.

Is this easier than normal methods of finding square roots? … If you have a table of cosines, yes. Definitely. You have to scale the number into range (divide by 100 some) do an easy multiplication (S times 2), an easy subtraction (minus 1), a table lookup (arccosine), an easy division (divide by 2), another table lookup (cosine), and scale the number up again (multiply by 10 some). That’s all. Seven steps, and two of them are reading. Two of the rest are multiplying or dividing by 10’s. Using logarithm tables has it beat, yes, at five steps (two that are scaling, two that are reading, one that’s dividing by 2). But if you can’t find your table of logarithms, and do have a table of cosines, you’re set.

This may not be practical, since who has a table of cosines anymore? Who hasn’t also got a calculator that does square roots faster? But it delighted me to work this scheme out. Give me a while and maybe I’ll think about cube roots.

## Reading the Comics, July 5, 2013

I’m surprised to discover it’s been over a month since I had a roster of mathematics-themed comic strips to share, but that’s how things happen to happen. It’s also been a month with repeated references to “finding square roots”, I suppose because that sounds like a really math-y thing to do. It’s certainly computationally challenging; the task of finding such is even a (very minor) moment in Isaac Asimov’s magnificent short story about arithmetic, “The Feeling Of Power”. I remember reading the procedure for finding them when I was a kid, and finding that with considerable effort, I was able to, though I’d probably refuse to do more than give a rough estimate of such a root nowadays.

Bill Watterson’s Calvin and Hobbes (June 4, rerun) is another entry in the long string of jokes about “why bother studying mathematics”, but Watterson’s craft lifts it above average. Admire that fourth panel: that’s every resistant student in one pose.

## I Know The Square Root Of Five, Too

Now I want to do a little more complicated problem of showing two numbers are equal because the difference between them is so tiny. It struck me that if I wanted to do that, I’d have to do some setup to even start. What I really meant to do was to show that some number was equal to the square root of five. I picked the square root of five because I had it burned into my memory from a children’s book that knowing the first few digits of an irrational number would be sufficient to immobilize the mind-controlled population of an all-powerful computer dictator, and I’ve kept it in mind just in case ever since. I’m also glad to know on double-checking that I remembered the first couple digits of the square root of five well (2.236). I’m shakier on the square root of seven (2.something) so if it’s a more advanced computer we’re up against I’m in trouble.

Still, most square roots would do. It’s a neat little property of the whole numbers that the square roots of them are either whole numbers themselves — the square root of 4 is 2, the square root of 169 is 13, the square root of 4,153,444 is not worth thinking about — or else they’re irrational numbers, going on without ending and without repetition. Most people who’d read a mathematics blog on purpose have heard about how the irrationality of the square root of 2 was proven in ancient days, and maybe heard the story of how the Pythagoreans murdered the person who let slip the horrifying secret that there were irrational numbers and they represented real things that might be of interest, and a few are even aware we don’t really know with certainty that the story’s actually true. (At this point, I suspect it’s too strong a claim to say we know anything about the Pythagoreans for certain, but I haven’t looked closely. Maybe matters are not quite that dismal.) Whether true or not the legend of the Pythagoreans turning to murder is a fine way to get an algebra class’s attention. I just fear that what the students take away from it is, “if you learn any of this math stuff a cabal of mathematicians will murder you” and they stay oblivious for reasonable self-protection.

But anyone who’s understood a proof that the square root of two is irrational is perfectly able to show that the square root of three is irrational as well, or the square root of five, or any other such desired number. The proof that way runs just about the same route, but takes longer to get there.

Similarly, if you have a rational number that comes to an end, such as 0.49, then the square root either is a rational number that comes to an end, in this case 0.7; or else it never comes to an and and never repeats. That’s easy to prove, if you have that idea about the square roots of whole numbers. The square root of 4.9, for example, is not a rational number, although I can’t promise anything for its ability to halt world-spanning computers.


text:
## Richard Bardgett

Print publication date: 2005

Print ISBN-13: 9780198525035

Published to Oxford Scholarship Online: April 2010

DOI: 10.1093/acprof:oso/9780198525035.001.0001

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# The soil environment

Chapter:
(p.1) 1 The soil environment
Source:
The Biology of Soil
Publisher:
Oxford University Press
DOI:10.1093/acprof:oso/9780198525035.003.0001

# Abstract and Keywords

Soil forms a thin mantle over the Earth's surface and acts as the interface between the atmosphere and lithosphere, the outermost shell of the Earth. It is a multiphase system, consisting of mineral material, plant roots, water and gases, organic matter at various stages of decay, and a variety of live organisms. The first step towards understanding what controls the abundance and activities of these organisms, and also the factors that lead to spatial and temporal variability in soil biological communities, is to gain an understanding of the physical and chemical nature of the soil matrix in which they live. This chapter provides background on the factors responsible for regulating soil formation, and hence the variety of soils in the landscape. It also discusses the key properties of the soil environment that most influence soil biota, leading to variability in soil biological communities across different spatial and temporal scales.

# 1.1 Introduction

Soil forms a thin mantle over the Earth’s surface and acts as the interface between the atmosphere and lithosphere, the outermost shell of the Earth. It is a multiphase system, consisting of mineral material, plant roots, water and gases, and organic matter at various stages of decay. The soil also provides a medium in which an astounding variety of organisms live. These organisms not only use the soil as a habitat and a source of energy, but also contribute to its formation, strongly influencing the soil’s physical and chemical properties and the nature of the vegetation that grows on it. Indeed, along with vegetation, the soil biota is one of five interactive soil-forming factors: parent material, climate, biota, relief, and time (Jenny 1941). The first step towards understanding what controls the abundance and activities of these organisms, and also the factors that lead to spatial and temporal variability in soil biological communities, is to gain an understanding of the physical and chemical nature of the soil matrix in which they live. This chapter provides background on the factors responsible for regulating soil formation, and hence the variety of soils in the landscape. It also discusses the key properties of the soil environment that most influence soil biota, leading to variability in soil biological communities across different spatial and temporal scales.

# 1.2 Soil formation

In order to understand the properties of soils that influence the biota that dwell therein, we must first consider some of the factors that lead to variations in soils and soil properties within the landscape. One of the most fascinating features of the terrestrial world is the tremendous variety in its landforms, reflecting a diversity of geological processes that have occurred (p.2) over millions of years; more recent as factors in the variation are biological processes and the influences of man. Similarly, within any landscape there is an incredible range of soils, resulting from almost infinite variation in soil-forming factors. These are highly interactive, in that they all play a part in the development of any particular soil. Combinations of these factors lead to the development of unique soil types, with a relatively predictable series of horizons (layers) that constitute the soil profile (Fig. 1.1). Of greatest interest to the soil ecologist are those horizons that are at, or close to, the soil surface; this is where most microbes and animals live and where most root growth and nutrient recycling occur. These horizons are referred to as the surface organic (O) horizon, which develops when decomposing organic matter accumulates on the soil surface, or the uppermost A horizon, which is composed largely of mineral material but also intermixed with organic matter derived from above. Soil ecologists are also concerned with the plant litter lying directly on the soil surface, deposited during the previous annual cycle of plant growth. This layer, referred to as the L layer, is often overlooked or even discarded in soil sampling regimes, but it is perhaps the most biologically active and functionally important zone of the soil profile.

Fig. 1.1 Schematic representation of a soil profile showing major surface and subsurface horizons.

(p.3)

Fig. 1.2 Podzol (Spodosol in US terminology) soil with deep O horizon (mor) and characteristic bleached Ea horizon above the red, depositional Bs or spodic horizon. (Image by Otto Ehrmann.)

Between this horizon and the A horizon are found layers of organic matter at intermediate stages of decomposition: the F layer, composed of partly decomposed litter from earlier years, and the H layer, made up of well decomposed litter, often mixed with mineral material from below.

While soil profiles vary greatly across landscapes, they can be classified into groups on the basis of their soil properties and soil-forming characteristics, each group having a unique set of ecosystem properties. For example, on free-draining, sandy parent material, in cold and wet climates, usually beneath coniferous forests, podzolic soils develop (Fig. 1.2). These soils, formed by podzolization (Box 1.1), have a deep, acidic surface O horizon, referred to as mor humus. They are subject to heavy leaching and are characterized by low rates of decomposition and plant nutrient availability, and hence low plant productivity. Typically the microbial biomass of these mor soils is dominated by fungi (rather than bacteria) and the fauna are characterized by high numbers of microarthropods (mites and Collembola), and an absence of earthworms. In contrast, on calcium-rich, clayey parent material, typically beneath grasslands and deciduous forests, brown earth soils are often found (Fig. 1.3). These soils have a mull humus composition that is often mildly acidic, owing to leaching of base cations (e.g. calcium) down the soil profile. The mull horizon is characterized by (p.4) (p.5) (p.6) intimate mixing of the surface organic and mineral-rich A horizon as a result of the high abundance and activity of soil biota, especially earthworms, leading to high rates of decomposition, nutrient availability, and plant growth. A total of 10 major soil groupings, termed soil orders, have been distinguished by the US Soil Taxonomy (Brady and Weil 1999) (Table 1.1), and 8 major soil groups are recognized by the Soil Survey of England and Wales (Avery 1980); each of these groupings has a unique set of ecosystem characteristics. Further details on these soils and their classification can be found in general soil science textbooks (White 1997; Brady and Weil 1999).

Fig. 1.3 A typical brown earth soil with mull horizon, under grassland. Note the lack of horizon differentiation caused by intense biological activity and the mixing of organic matter with mineral material from the A horizon (Image by Otto Ehrmann.).

Table 1.1 Soil taxonomy orders

Order

Brief description

Entisols

Recently formed azonal soils with no diagnostic horizons

Vertisols

Soils with swell-shrink clays and high base status

Inceptisols

Slightly developed soils without contrasting horizons

Aridosols

Soils of arid regions

Mollisols

Soils with mull humus

Spodosols

Podzolic soils with iron and humus B horizons

Alfisols

Soils with a clay B horizon and >35% base saturation

Ultisols

Soils with a clay B horizon and <35% base saturation

Oxisols

Sesquioxide-rich, highly weathered soils

Histosols

Organic hydromorphic soils (peats)

# 1.3 Soil-forming factors

As noted, within most landscapes there is a tremendous variety of soil types varying in physical and chemical make-up. The soil-forming factors are central to understanding the variability in soils at the landscape level and at the level of the individual soil profile. Being the central forces responsible for creating variety in soil conditions, and hence variations in the habitat of the soil biota, these factors require further consideration. The biota themselves, along with vegetation, constitute one of the main soil-forming factors; both can act as important determinants of soil formation and profile development. This section summarizes some of the important aspects of the main soil-forming factors. It is important to stress, however, that while soil-forming factors are considered individually, they operate interactively in nature, usually with a hierarchy of importance, with one or two of them being pre-eminent in soil development at a particular location.

## (p.7) 1.3.1 Parent material

Geological processes acting over millions of years determine the variations and distribution of parent materials from which soils develop. Soils are formed from the weathering of either consolidated rock in situ or from unconsolidated deposits—derived from erosion of consolidated rock—that have been transported by water, ice, wind, or gravity. The mineralogical composition of these deposits varies tremendously. For example, the mineralogy of igneous rocks, formed by solidification of molten magma in, or on, the Earth’s crust, ranges from base-rich basalts (basic lava) with high amounts of calcium (Ca) and magnesium (Mg) to acidic rhyolites (acid lava) which contain high amounts of silica (Si) and low amounts of Ca and Mg. Rocks of intermediate base status, such as andesites, also commonly occur. Parent material also determines grain size, which determines soil texture (relative proportions of sand, silt, and clay), which in turn affects many soil properties, such as the ability of the soil to retain cations (its cation exchange capacity), the moisture retaining capacity, and soil profile drainage. Such variation in the mineralogy of rocks, therefore, strongly influences the type of soils that are formed and the character of the vegetation that they support (Fig. 1.4). Soils formed from weathering of basic lava, for example, tend to be rich in minerals such as Ca, Mg, and potassium (K) and fine textured (clayey), and have a high ability to retain cations of importance to plant nutrition (e.g. $NH 4 +$, $Ca 2 +$). These soils are typically fertile brown earths with biologically active mull humus. In contrast, soils that are formed from acidic lava, such as granites and rhyolites, are low in Ca and Mg, coarse textured (sandy), and hence freely (p.8) drained, with low cation retention capacity. The soils that typically develop here are therefore strongly leached, nutrient-poor, acidic podzols with mor humus.

Fig. 1.4 Schematic classification of igneous rocks and their resulting soils.

## 1.3.2 Climate

Historically, climate has been considered pre-eminent in soil formation, owing largely to the striking associations that exist, on continental scales, among regional climate, vegetation type, and associated soils. Indeed, these broad climatic associations led to the development in Russia of one of the first soil classification systems—the zonal concept of soils (Dokuchaev 1879). This system identified so-called zonal soils—those that are influenced over time more by regional climate than by any other soil-forming factor. While climate may play a crucial role in soil development on continental scales, for example, across Russia and Australia, it is arguably not as important in areas such as the subtropics and tropics where land surfaces are much older and more eroded, or in younger landscapes such as Britain where most soils are developed on recent, glacial deposits. Here, the factors of topography and parent material are of greater importance.

The effects of climate on soil development are largely due to temperature and precipitation, which vary considerably across climatic zones. These factors strongly govern the rates of chemical reactions and the growth and activities of biota in soil, which in turn affect the soil-forming processes of mineral weathering and decomposition of organic material. The effects of temperature on soil biological activity are well known; it is generally accepted that there is an approximate doubling of microbial activity and enzyme-catalysed reaction rates in soil for each 10 °C rise in temperature, up to around 30–35 °C. Above this temperature, however, most enzyme-catalysed reactions decline markedly, as proteins and membranes become denatured. Some microbes can live at extreme temperatures; for example, cold-tolerant fungi occur in polar soils and remain physiologically active down to −7 °C (Robinson and Wookey 1997). These cold-tolerant microbes are called pychrophiles, whereas microbes that live in extremely high temperatures are called thermophiles.

The effects of temperature and precipitation on soil formation are especially marked at high altitudes and latitudes. For example, soil organic matter content is often found to increase with increasing elevation, commonly reaching a peak in montane forests (Körner 1999). (At higher altitudes, above the treeline, the organic matter content of soils declines and reaches almost zero in unvegetated substrates in the upper alpine zone.) This increase in soil organic matter content is largely due to declines in temperature and high precipitation, which reduce microbial activity and rates of decomposition. Similarly, in high-latitude regions of Europe, vast peatlands have developed in areas where the combined effects of high (p.9) rainfall and low temperature, and minimal evapotranspiration, have led to anaerobic conditions (waterlogging) and the retardation of decomposition of organic matter. The consequence of this has been the accumulation of great masses of peat (blanket peats), especially in topographically uniform areas where drainage is reduced (Fig. 1.5). Dramatic changes in the physiology and productivity of dominant plants also occur along altitudinal and latitudinal gradients (Díaz et al. 1998), altering the nutritional quality (e.g. N content) of the leaf litter that is produced annually. As will be discussed in Chapter 4, such changes in organic matter quality resulting from shifts in plant community composition can have profound effects on the decomposability of organic matter, and hence the accumulation of organic matter on the soil surface.

Fig. 1.5 Blanket peat at Moor House National Nature Reserve in northern England. Here, deep peats have developed at high altitudes where high rainfall and low evapotranspiration combine to cause excessive soil wetness that retards decomposition. These peatlands are of special significance because they represent a significant (ca. 30%) store of global terrestrial C. Indeed, the majority of the UK’s terrestrial C is stored in the peat soils of northern Britain (Image by Richard Bardgett.).

## 1.3.3 Topography

Variations in topography influence soil development, largely through effects on soil drainage and erosion. Soil drainage is primarily affected by the position of a soil on a slope; soils at or near the top of a slope tend to be freely (p.10) drained with a water table at some depth, whereas those at or near the bottom of the valley tend to be poorly drained with a water table close to the soil surface. These differences in drainage strongly influence soil development, leading to the development of a hydrological sequence (Fig. 1.6): welldrained soils on hilltops have deep, orange-brown subsurface horizons, indicative of oxidation processes (iron in ferric state); as drainage deteriorates towards the valley bottom, the soil profile becomes increasingly anaerobic and blue-grey in colour, indicative of a dominance of reduction processes (p.11) (iron in ferrous state) and the process of gleying (Box 1.1). In extremely wet valley bottom soils, deep O horizons develop on the soil surface as a result of retardation of decomposition processes under anaerobic conditions.

Fig. 1.6 (a) Section of a slope and valley bottom showing a hydrological soil sequence, and (b) changes in soil profile morphology. (Redrawn with permission from Blackwell Science; White 1997)

Slope characteristics also greatly influence soil erosion processes, which in turn affect soil formation. In general, soils on ridges and steeper parts of slopes are shallower than those on lower slopes and valley bottoms, owing to the movement of soil particles down-slope by wash and soil creep. Because erosion preferentially moves finer particles down-slope, the soils of lower slopes and valley bottoms also differ in their mineralogical composition, being more fine textured. In humid regions, soils of lower slopes and valley bottoms also tend to be enriched in base cations and salts, owing to seepage of solutes from higher slopes and hilltops. Soil movement down-slope also leads to the formation of distinct morphological features on slopes, such as terraces. These features are especially common in alpine regions where steep slopes and freeze–thaw motion lead to instability of the surface soil and consequent down-slope creeping. This process is called solifluction. What happens here is that soils become saturated with water and freeze, and then melt; the expansion associated with freezing makes the surface soil very unstable when it thaws. This leads to downward movement of soil even on the gentlest slopes, especially if the subsurface soil is frozen.

## 1.3.4 Time

The age of the soil is a major factor underlying variations in soils and ecosystem properties. Soils become increasingly weathered over time, and, consequently, soil profiles generally become more differentiated, with more abundant and thicker horizons. This weathering process involves progressive leaching downwards of elements and minerals in percolating water. In particular, during the process of podzolization, progressive downward movement of iron (Fe) and aluminium (Al) leads to the development of a spodic (Box 1.1) subsurface horizon that is enriched with these minerals, and an upper Ea horizon, whence these elements have been removed, that is bleached in appearance. Over time, clay minerals also become leached down-profile, a process called lessivage (Box 1.1), leading to the development of subsurface argillic horizons. Other important changes in soils that occur with age are increases in soil organic matter and nitrogen (N) content (Crocker and Major 1955) and, over very long timescales (hundreds of thousands or millions of years), a progressive reduction in the availability of soil phosphorus (P) owing to its loss from the system and fixation in mineral forms that are not available to plants (Walker and Syers 1976).

The relationship between time and soil development is best illustrated by examining soil chronosequences, which are places where, for various reasons, a sequence of differently aged, but otherwise similar, geologic substrates exists. Glacier Bay, on the coast of southeast Alaska, is one of the (p.12) best known places for research on soil and ecosystem development because of the continuous retreat of the glaciers since 1794. Furthermore, records of the retreat, over some 100 km, have been maintained since this time, so the age of the glacial moraines is known. This has resulted in a site chronology over a period of 200 years, and from this, patterns of soil development can be tracked along with the succession in vegetation from the initial pioneer plant communities on recent moraine through to the climax spruce forest on the oldest moraines (Box 1.2). This chronosequence represents stages in the development of a podzol; as time progresses, the soil profile becomes progressively deeper and differentiated, the oldest soils being acidic in nature and having a thick organic surface horizon, a thin bleached horizon, and subsurface spodic horizon (Crocker and Major 1955). As organic matter steadily accumulates in the surface organic horizon, the amount of N in soil also increases; the organic carbon (C) and N content of underlying mineral soil also builds up (Fig. 1.7). Of particular significance for organic matter and N accumulation is the early stage of vigorous alder growth on terrain that has been ice-free for some 75 years. Here, N accumulates at a rate of about 4.9 g N m−2 yr−1, reaching values of some 250 g N m−2 within around 50 years of soil development (Crocker and Major 1955). Although there is no single explanation for the soil development sequence at Glacier Bay, it appears that establishment of plants and intensive leaching have played key roles (Matthews 1992).

Studies at Glacier Bay demonstrate progressive soil development over relatively short timescales towards climax, while other sites can be used to demonstrate soil change over hundreds of thousands, or even millions of years. The Hawaiian island archipelago, for example, presents a chronology of soil development over some 4.5 million years; Kilauea volcano on the Island of Hawaii is active, while Kauai, the oldest site, on the northwest end of the high islands is estimated to be 4.5 million years old (Fig. 1.8). A range of intermediate aged sites is also present, and all sites are derived from volcanic lava of similar mineralogy and have vegetation that is dominated by the same tree, Metrosideros polymorpha (Fig. 1.9). A key feature of this chronology is that it demonstrates a reduction in P availability in the oldest sites, which causes a dramatic decline in plant productivity (Crews et al. 1995). This fall in P availability follows the theoretical model of soil development that was proposed by Walker and Syers (1976): as soils age, P becomes surrounded, or occluded, by Fe and Al hydrous oxides, rendering the P largely unavailable to plants and also the soil biota (Fig. 1.10). This process of occlusion is especially likely to occur in very old soils because prolonged weathering of minerals leads to the formation of Fe and Al oxides that have a strong affinity for P. This P limitation to vegetation is further exacerbated in old soils because low soil fertility sets in motion a feedback whereby reductions in biological activity in soil reduce decomposition of plant litter, further intensifying nutrient limitation (p.13) (p.14) (p.15) (p.16) (Crews et al. 1995). Another excellent example of P limitation in very old sites is the large sand dune systems on the subtropical coast of Queensland, Australia. These dune systems provide a chronology of soil development, without interruption from glaciation, dating back some 400,000 years (Thompson 1981). Here, progressive weathering and leaching in the free-draining sand over this period has led to the development of so-called giant podzols (Thompson 1981) with soil profiles of some 20 m depth, which are characterized by extreme P limitation and stunted tree growth (Fig. 1.11). A detailed account of this soil development sequence can be found in Thompson (1992).

Fig. 1.7 Total N content of soils recently uncovered by glacial retreat at Glacier Bay, Alaska. Plant succession is shown along the top. (Data from Crocker and Major 1955) (Reprinted with permission of Pearson Education; Kerbs 2001, using data from Crocker and Major 1955)

Fig. 1.8 The Hawaiian Island archipelago presents a chronology of soil development of some 4.5 million years: Kilauea volcano on the Island of Hawaii is active, while Kauai, the oldest site, on the northwest end of the high islands is estimated to be 4.5 million years old. (Redrawn from Crews et al. 1995).

Fig. 1.9 A Metrosideros polymorpha (‘ohi’a lehua) forest on the Hawaiian island of Molokai with low stature and biomass. This ecosystem developed in the absence of catastrophic disturbances during the past 1.4 million years. Depletion of P at this site has led to a decline in forest productivity. (Image by Richard Bardgett.)

Fig. 1.10 Generalized effects of long-term weathering and soil development on the distribution and availability of P in soil (Adapted from Walker and Syers 1976).

## 1.3.5 Human influences

An increasing proportion of the Earth’s surface is under some form of management by man, and human impacts on ecosystems are significant and growing (Vitousek et al. 1997). The most obvious effects of humans on soil development result from the widespread destruction of natural vegetation for agricultural use, for example, in the tropics where some 8% of rainforest is thought to be cut down per decade (Watson 1999). Human activities also have dramatic effects on soil development by directly modifying the chemical and physical nature of the soil environment by fertilization, irrigation, and drainage, and by ploughing for cultivation. Humans can also have important indirect effects on soil development. For example, increasing concentrations of CO2 in the atmosphere have an effect on vegetation, (p.17) which in turn affects soil processes and soil biota. The introduction of invasive species into natural ecosystems can also have profound effects on soil properties and biota, for example, by changing the quality of litter inputs into soil and/or soil nutrient availability. Effects of human activities on soil biota and soil properties will be discussed in detail in Chapter 6.

Fig. 1.11 Stunted tree-growth resulting from P limitation in old-growth (400,000-year-old) eucalyptus forest on the subtropical coast of Queensland, Australia. (Image by Richard Bardgett.)

# 1.4 Soil properties

Variation in soil-forming factors determine the physical and chemical nature of soils, which in turn influences greatly the nature of the soil biota and hence ecosystem properties of decomposition and nutrient cycling. Variation in soil properties, especially the physical matrix of the soil, also greatly influences the movement of water and associated materials both (p.18) within and between ecosystems. This section examines some of the key soil properties that most strongly influence the soil biota and their activities, and the nature of ecosystems.

## 1.4.1 Soil texture and structure

The term soil texture refers to the relative proportions of various-sized particles—sand (0.05–2.0 mm), silt (0.002–0.05 mm), and clay (<0.002 mm)—within the soil matrix (Fig. 1.12). It is primarily dependent on the parent material from which the soil is formed and the rate at which it is weathered, as discussed above. Soil texture is of importance largely because it determines the ability of the soil to retain water and nutrients: clay minerals have a higher surface area to volume ratio than sand and silt, and hence soils with a high clay content are better able to hold water by adsorption and to retain cations on their charged surfaces. The ability of a soil to retain cations (e.g. Ca2+, Mg2+, $NH 4 +$) is referred to as its cation exchange capacity, which reflects the capacity of clay minerals to hold cations on negatively charged surfaces. This retention of cations on clay minerals represents a major short-term store of nutrients for plant and microbial uptake.

Fig. 1.12 Composition of the textural classes of soils based on percentages of sand, silt, and clay. For example, a soil with 60% sand, 10% silt, and 30% clay is a sandy clay loam.

(p.19) Soil structure reflects the binding of the above various-sized mineral particles into larger aggregates or a ped. The actual formation of stable aggregates requires the action of physical, chemical, and biological factors: (1) freeze–thaw and soil shrinkage and swelling help to mold the soil into aggregates; (2) the mechanical impact of rain and ploughing reorganizes soil materials; (3) the activities of burrowing animals, such as earthworms, lead to the mixing of mineral and organic materials and the formation of stable organo-mineral complexes; (4) the faeces of soil animals can act as nuclei for aggregate formation; (5) fine roots and microbes produce a range of polysaccharide glues which bind soil particles together, and fungal hyphae literally hold together mineral particles and organic matter. Together, these factors combine to produce stable aggregates in soil. As will be discussed in Chapter 3, large soil animals, such as earthworms and termites, can also substantially affect soil structure by creating macropores and channels as a consequence of their feeding and burrowing activities; this in turn increases infiltration rates, and hence drainage of water through soil.

Soil aggregation and structure is of concern to the soil ecologist, not only because the activity of the soil biota strongly affects it, but also because the structure of soil determines the physical nature of the living space. Aggregation determines the pore distribution of soil, which affects both the distribution of water in soil (specifically the degree to which pores are filled with water) and the extent to which biota are able to enter and occupy pore space, which is controlled by pore neck diameter and the size of the organism. For example, nematodes are approximately 30 μm in diameter, so their migration in soil is restricted by pores of diameter <30 μm and is optimal in soils with particle sizes in the region of 150–250 μm. Fungi and bacteria are much smaller in size, so they can occupy much smaller pores (1–6 μm) than can nematodes. Because the primary food source of many nematodes is microbes, these smaller pores can therefore provide a refuge for microbes from their predators, thereby affecting the nature and intensity of trophic interactions in soil.

In sum, good soil structure is recognized as a key attribute of fertile and biologically active soil, because it increases the flow of water and gases through soil, reducing the possibility of the development of anaerobic conditions, which would be detrimental to soil biota and their activities, and harmful to plant growth. Good soil structure promotes free movement of biota, thus increasing opportunities for trophic interactions; it allows roots to proliferate and enables aerobic microbial processes to dominate.

## 1.4.2 Soil organic matter

The organic matter content of soil varies tremendously in terms of both its chemical composition and its quantity. As already discussed, this depends (p.20) on a variety of interacting factors such as vegetation type, climate, parent material, soil drainage, and the activity of soil biota: a particular combination of these factors generally leads to the formation of either mull or mor humus forms. Soil organic matter is of importance because it promotes soil structural stability (see above), thereby preventing soil erosion. It is also extremely effective in retaining water within the soil matrix. Soil organic matter is of particular importance for biota because it is their primary source of nutrients and C. Through a range of activities, soil biota decompose organic matter, converting it into simple organic molecules that they can assimilate and use for energy and growth. Most components of soil organic matter occur as large molecules (e.g. cellulose, amino sugars, proteins, nucleic acid, lignin, etc.), which, over time, are depolymerized by extracellular microbial enzymes to yield simpler units which can then be assimilated by the microbes for energy and C. For example, the enzyme cellulase breaks down cellulose—the principal component of plant tissue—into glucose units that are readily assimilated by microbes and used for energy and C. Microbial enzymes also produce soluble organic nutrients (e.g. amino acids), which are either absorbed by microbes to meet their nutrient requirement (e.g. for C and N) or, in some circumstances, taken up directly by plants and mycorrhizal fungi (Chapter 3). Soluble organic nutrients that are absorbed by microbes are either retained to meet their C and N needs, or, when they require only C to support their energy needs, the excess N is secreted into the soil environment as inorganic N (ammonium), which is then available for plant uptake. This process is called mineralization; it is of key importance at the ecosystem scale because it determines the availability of inorganic nutrients for plant uptake, and hence plant productivity (Chapter 3). Soil fauna also contribute to the decomposition process and nutrient mineralization both directly, by mixing and fragmenting organic matter into smaller units that are more accessible to microbial attack, and indirectly, by feeding on microbes, affecting their growth and activity, and excreting nutrients that are in excess of the animals’ requirements into the soil environment (Chapter 3).

The rate at which organic inputs to soil are decomposed depends primarily on their quality, which is dependent on the type of compounds that are present within them. Rapidly decomposing materials, such as the litter of deciduous trees and animal faeces, generally contain high amounts of labile substances, such as amino acids and sugars, and low concentrations of recalcitrant compounds such as lignin. In contrast, the litter of coniferous trees decomposes slowly, being rich in large, complex structural compounds such as lignin and defence compounds such as polyphenols; this material is also unpalatable to soil fauna, further slowing down its decomposition. The importance of variation in the rate of decomposition at the ecosystem scale relates to the production of CO2 from heterotrophic microbial activity and its evolution into the atmosphere, and to the (p.21) conversion of organic nutrient forms to simple inorganic nutrients (e.g. ammonium and phosphate) that are available for plant uptake, a strong determinant of plant productivity. These factors will be discussed in detail in later chapters.

## 1.4.3 Soil water

It is water that renders the soil environment habitable. Together with dissolved nutrients it makes up the soil solution, an important medium for supplying nutrients and water to growing plants. It also provides a medium for many soil biota to live and move around in. Nematodes and protozoa, for example, live in water films and in free water, along with the bacteria that they feed upon. Larger fauna, such as microarthropods (mites and Collembola), live in open pore spaces but are very sensitive to desiccation, migrating out of soil when it becomes dry.

There are certain water-holding characteristics of soils that determine the amount of water that is available to plants and soil biota. Under normal conditions, soil pores will contain air as well as water, and most water is held in pore spaces as films or as water absorbed onto soil particles. Under these conditions the soil is said to be unsaturated. After a period of heavy rain or irrigation, however, pore spaces in soil become filled with water and the soil becomes saturated. After a period of drainage, when the amount of water held in the soil is in equilibrium with gravitational suction (5–10 kPa), field capacity is reached and no more water drains from the soil. The pores that are drained at field capacity are the macropores, which are those that are created by the burrowing activities of animals and by plant roots. As plant roots continue to absorb water from the soil, the amount of water held in films is reduced, and the remaining water adheres tightly to soil particles. At this stage, water moves through thin films to plant roots in response to a gradient of water potential; if water is not replenished and plants continue to take it up, a situation is reached when plants can no longer remove water that is strongly adhered to particle surfaces. This is the permanent wilting point, and the difference between this measure and field capacity is referred to as the water-holding capacity of the soil. In general, the water-holding capacity is greater in soils that contain large amounts of clay and/or organic matter, because these components have high surface areas that readily retain water. Also, clay soils have more small pores that readily retain water under gravitational suction than do sandy soils, which have larger pores that are more easily drained.

## 1.4.4 Soil pH

A large proportion of the Earth’s soils are acidic, especially in the tropics, where ecosystems persist at soil pH values of 4 or less (pH here is a measure (p.22) of the concentration of H+ ions in soil water). Many northern ecosystems also have very acidic soils: the pH values of the soils of Boreal forests and heathlands are often of 4 or less. In many parts of the world, soil acidity is further exacerbated by the use of inorganic fertilizers and acid rain (Kennedy 1992). Soils become acidic if base cations (e.g. Ca2+, Mg2+, K+) are leached from the soil profile, to be replaced by H+and Al3+ ions on cation exchange sites. Acidity in soils can come from various sources: (1) carbonic acid, which is formed by the dissolution of CO2 in water, dissociates to yield H+ ions; (2) microbial oxidation of ammonium ions $(NH 4 + )$ to nitrate $(NO 3 - )$, the former being derived from mineralization and fertilizer inputs, also yields H+ ions; (3) atmospheric pollution (acid rain) and natural sources of acids, including volcanic eruptions and thunderstorms that yield sulphur dioxide and oxides of N, respectively, produce sulphuric and nitric acids that acidify soils (it has been proposed that the widespread occurrence of acidic soils in tropical and subtropical regions is, in part, a result of high thunderstorm activity in these regions. Long-term weathering and leaching of cations also contributes significantly to acidity in these old soils); and (4) decomposition of organic matter that has high concentrations of phenolic and carboxyl groups liberates H+ ions.

Soil pH is of concern to the soil ecologist because it controls nutrient availability and it directly impacts on soil biota. Acidic soils, for example, are characteristically high in soluble aluminium (Al3+), which can be toxic to plants and microbes. This is because aluminium occurs largely as insoluble forms, for example, as part of clay minerals whose structure becomes unstable at low pHs (4–5), releasing aluminium ions into soil solution (Kennedy 1992). The P availability in soil is typically low under acidic conditions, owing to the formation of iron and aluminium phosphates. These phosphates dissolve to release P into soil solution as pH rises, making it available for plant uptake; the availability of P is typically greatest between pH 6 and 7 (Chapter 3).

The effects of pH on soil organisms are well documented, and approximate tolerances of major groups of soil organisms are known. Most microbes grow within the pH range 4–9, although acidophiles are known to survive at pHs as low as 1, for example, sulphate-oxidizing bacteria (species of the genus Thiobacillus) which occur in hot springs and mine wastes. Growth in extremely alkaline conditions is restricted to a few fungi and bacteria, but such conditions tend to occur only in soda lakes and deserts, for example, those in Egypt which have pH values between 9 and 11. Soil animals are also very sensitive to acidic conditions in soil. For example, earthworms occur in very low numbers, with few species, in most acidic soils, and they become progressively more abundant as soil pH increases to neutrality (Edwards and Bohlen 1996). Acid soils tend to be dominated by enchytraeid worms (potworms), reaching densities as (p.23) high as 100 × 103 m−2 in acid peat soils (Cole et al. 2002a). Under such conditions, these enchytraeid worms replace earthworms as the functionally dominant soil animal.

# 1.5 Conclusions

Within most landscapes, there is a tremendous variety of soil types, reflecting spatial variations in the operation of a range of interacting soil-forming factors and pedogenic process, and the influences of man. The need for the soil ecologist to recognize this variation, at both spatial and temporal scales, cannot be stressed enough because it means that the roles of biota, relative to other factors, in controlling ecosystem processes (e.g. nutrient cycling, hydrological fluxes, and plant productivity) and soil formation itself will be context dependent. In other words, the roles of biota and the factors that regulate their community structure and activities will vary from soil to soil, depending on the dominant physical and chemical characteristics of that soil. In view of this, it is essential that studies on the soil biota are accompanied by detailed characterization of the physical and chemical nature of their habitat, and an appreciation of the soil-forming factors that led to the development of that soil in the first place.

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# Math Help - sphere cone problem

1. ## sphere cone problem

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.

2. ## Volume of sphere and come

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.
The volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.
Let s denote the slanted height, r the radius of the base circle and h the height of the cone. Then the curved surface of a cone is a sector of a circle with radius s. The area of the curved surface, the area of the complete circle, the circumference of the base circle = arc of the sector and the circumference of the complete circle are directly proportional:

$\dfrac{a_{cs}}{\pi s^2} = \dfrac{2\pi r}{2\pi s}~\implies~a_{cs} = \dfrac{2\pi r}{2\pi s} \cdot \pi s^2 = \boxed{\pi r s}$

Hello GAdamsThe volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

Hmm...not sure.

5. ## Surface area of cone

earboth has given you the formula for the curved surface area of a cone. You now need to work out the value of $s$, which you can do using Pythagoras' Theorem. (It's the easiest right-angled triangle of all!)

Then plug the values of $r$ and $s$ into this formula, and you're there.

earboth has given you the formula for the curved surface area of a cone. You now need to work out the value of $s$, which you can do using Pythagoras' Theorem. (It's the easiest right-angled triangle of all!)

Then plug the values of $r$ and $s$ into this formula, and you're there.

s = root 7

Is teh answer 3 pi root7 ?

... The radius of the base of a cone is also 3 cm.
...
[SIZE=3]...

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

...
Hmm...not sure.
Since r = 3 cm and h = 4 cm the slanted height s is:

$s^2 = r^2 + h^2~\implies~ s = 5\ cm$

Therefore the curved surface area is:

$a_{cs} = \pi\cdot 3\ cm \cdot 5\ cm = 15 \pi\ cm^2$

8. Originally Posted by earboth
Since r = 3 cm and h = 4 cm the slanted height s is:

$s^2 = r^2 + h^2~\implies~ s = 5\ cm$

Therefore the curved surface area is:

$a_{cs} = \pi\cdot 3\ cm \cdot 5\ cm = 15 \pi\ cm^2$
Ah yes, I made a basic error. I have it now. Thanks!

9. ## Dont understand

Hello GAdamsThe volume of a sphere is $\frac{4}{3}\pi r^3$, and the volume of a cone is $\frac{1}{3}\pi r^2h$. So:

$\frac{4}{3}\pi r^3 = 3\times \frac{1}{3}\pi r^2h = \pi r^2h$

$\Rightarrow h = \frac{4}{3}r = 4$ cm.

Can you finish it now?

Hi i dont understand why you hve multiplied the volume of the cone by 3

10. ## Sphere and cone

Hello osmosis786
Originally Posted by osmosis786
Hi i dont understand why you hve multiplied the volume of the cone by 3

I think you're going to kick yourself!

The radius of a sphere is 3 cm. The radius of the base of a cone is also 3 cm.
The volume of the sphere is 3 times the volume of the cone.

Work out the curved surface area of the cone, giving your answer as a multiple of pi.

Thanks.


text:
# 2. You roll a single die number 1 to 6. What are the odds of rolling number? an ODD

###### Question:

2. You roll a single die number 1 to 6. What are the odds of rolling
number?
an ODD

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### What is the solutions to this system of equations? x+ 2y - z = 3 2x - y + 2z = 6 x- 3y + 3z = 4 ____________ A) (-6,7,14) B) (2,-3,5) C) infinite solutions D) no solutions

What is the solutions to this system of equations? x+ 2y - z = 3 2x - y + 2z = 6 x- 3y + 3z = 4 ____________ A) (-6,7,14) B) (2,-3,5) C) infinite solutions D) no solutions...

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### If 2a^2 +3^1a - 5a^2 = 9, then a - a^2 =(A)3(B) 6(C) 9(D) 12

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### What is the value of x? Which theorem will you use? Thank you

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### Which quantity is directly related to the temperature of a substance? A.) average kinetic energy of the particles in the substance B.) average size of the particles in the substance C.) composition of the substance D.) pH of the substance Even if you can't help, thanks for taking the time to look at my question and see if you can solve it. :)

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### Rachel’s mother tells her she can play at the arcade as long as the cost does not exceed $24. Each game costs$1.50.
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text:
# Catching a train

1. Apr 13, 2004

### jennypear

A train pulls away from a station with a constant acceleration of 0.21 m/s2. A passenger arrives at the track 6.0 s after the end of the train passed the very same point. What is the slowest constant speed at which she can run and catch the train?

Xtrain=Xo+Vot+1/2(a*t^2)
Xtrain=3.78m + 1.26t + .5*(.21m/s^2)t^2
Vtrain=Vo + at
Vtrain=1.26m/s + (.21m/s^2)t

Xperson=Vperson*t
Vperson=X/t

**at the time the person catches the train the
Vtrain=Vperson &
Xtrain=Xperson

Vtrain=Vperson
1.26m/s + (.21m/s^2)t=X/t
since the X's are the same substitute Xtrain for X on right side of equation.
1.26m/s+(.21m/s^2)t=[3.78m + (1.26m/s)t +(.105m/s^2)t^2]/t
(1.26m/s)t + (.21m/s^2)t^2=3.78m + (1.26m/s)t +(.105m/s^2)t^2
(.105m/s^2)t^2=3.78m
t^2=36s^2
t=6s

plug it in
X=Vot + 1/2*a*t^2
X=1.26m/s*(6s) + .105m/s^2(36s^2)
X=7.56m+3.78m=11.34m

**which in theory would make min velocity needed 11.34m/6s=1.89m/s

but this isnt correct
Can anyone see what I have done wrong w/this problem?

Thanks!

2. Apr 13, 2004

### jennypear

nevermind I got it. I forgot to add the initial 3.78m the train traveled before the person started running.

3. Apr 13, 2004

### Chen

Why did you decide that Xo of the train is 3.78m and its Vo is 1.26m/s? Instead of calculating how much the train has passed in 6 seconds (which is what I suspect you have done, and maybe also where your error lies), I would just give it a 6 seconds fore on the person by changing "t" in its equations with "t + 6".

So we have:

$$x_{train} = x_{person}$$
$$v_{train} = v_{person}$$

$$\frac{1}{2}a(t + 6s)^2 = vt$$
$$a(t + 6s) = v$$

Substituting v we get:

$$\frac{1}{2}a(t + 6s)^2 = a(t + 6s)t$$

And indeed t = 6s. Now just plug that into the second equation above:

$$v = a(t + 6s) = a(6s + 6s) = 2.52 m/s$$

4. Apr 13, 2004

### outandbeyond2004

I find your explanation bewildering. I am restating the problem as I understand it.

At time = 0 the end of the train is at x = 0 and v = 0. Six seconds later, xend = 0 + 0*t + (0.21)*t^2/2 = 3.78 meters
vend0 = left to you to calc.

At this point, xpass = 0, vpass = something, call it vpass

xpass = vpass*t

By the time the passenger catches the train,

xpass = xend
vpass = vend

So,
vpass*t = 3.78 + vend0*t + 0.21*t*t/2
vend = vpass = 0.21*t

I believe that's all you need -- solve for t then solve for vpass.


text:
# solve for t

• September 8th 2008, 01:05 PM
dm10
solve for t
50000=100*2^(t/3)

Solve for t.

I'm not sure how to solve for t when it's an exponent and a fraction.
• September 8th 2008, 01:24 PM
mr fantastic
Quote:

Originally Posted by dm10
50000=100*2^(t/3)

Solve for t.

I'm not sure how to solve for t when it's an exponent and a fraction.

$\Rightarrow 500 = 2^{t/3} \Rightarrow \log_{10} 500 = \log_{10} 2^{t/3} = \frac{t}{3} \log_{10} 2 \Rightarrow \frac{t}{3} = \frac{\log_{10} 500}{\log_{10} 2} \Rightarrow t = \, ....$
• September 8th 2008, 01:41 PM
dm10
thanks for the help.

I have one more if you don't mind.

100000=500*2^t/(1/2)
• September 8th 2008, 03:11 PM
mr fantastic
Quote:

Originally Posted by dm10
thanks for the help.

I have one more if you don't mind.

100000=500*2^t/(1/2)

Do you mean $100000 = (500) \times 2^{t/(1/2)}$?
• September 8th 2008, 03:24 PM
dm10
Yes that's right. It's alright though. I solved it by myself but I'd appreciate if you posted your answer so I can check my work.
• September 8th 2008, 03:34 PM
mr fantastic
Quote:

Originally Posted by mr fantastic
Do you mean $100000 = (500) \times 2^{t/(1/2)}$?

Quote:

Originally Posted by dm10
Yes that's right. It's alright though. I solved it by myself but I'd appreciate if you posted your answer so I can check my work.

$\Rightarrow 200 = 2^{t/(1/2)} = 2^{2t}$ since $\frac{t}{1/2} = 2t$.

Then $\log_{10} 200 = \log_{10} 2^{2t} = 2t \log_{10} 2 \Rightarrow 2t = \frac{\log_{10} 200 }{\log_{10} 2} \Rightarrow t = \frac{\log_{10} 200 }{2 \log_{10} 2}$.

This answer can also be written as $t = \frac{\log_{10} 200 }{\log_{10} 4}$.

The reason I use base 10 rather than base 2 is so that a scientific calculator can be used to get a decimal approximation of the answer if required.


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