Блог Анкара

In This Chapter

^ Fueling your knowledge of hydrocarbons

^ Naming alkanes, alkenes, and alkynes

^ Saturating and unsaturating hydrocarbons

I\ Ny study of organic chemistry begins with the study of Hydrocarbons. Hydrocarbons Ґ \ are some of the simplest and most important organic compounds. Organic compounds Are based on carbon skeletons. Hydrocarbon skeletons can be modified — you can dress them up with chemically interesting atoms like oxygen, nitrogen, halogens, phosphorus, silicon, or sulfur. This cast of atomic characters may seem like a rather small subset of the more than 100 elements in the periodic table. It’s true: Organic compounds typically use only a very small number of the naturally occurring elements. Yet these molecules include the most biologically important compounds in existence. As an introductory chemistry student, you won’t be expected to know more than the basic structure of organic molecules and how to name them. So relax and get organic.

Single File Now: Linking Carbons into Continuous Alkanes

The simplest of the hydrocarbons fall into the category of Alkanes. Alkanes are chains of carbon molecules connected by single covalent bonds. Chapter 5 describes how single covalent bonds result when atoms share pairs of valence electrons. Carbon molecules have four valence electrons. So, carbon atoms are eager to donate their four valence electrons to cova-lent bonds so they can receive four donated electrons in turn, filling their valence shell. In other words, carbon really likes to form four bonds. In alkanes, each of these is a single bond with a different partner.

As the name Hydrocarbon Suggests, these partners may be hydrogen or carbon. The simplest of the alkanes, called Continuous- Or Straight-chain alkanes, Consist of one straight chain of carbon atoms linked with single bonds. Hydrogen atoms fill all the remaining bonds. Other types of alkanes include closed circles and branched chains, but we begin with straight-chain alkanes because they make clear the basic strategy for naming hydrocarbons. From the standpoint of naming, the hydrogen atoms in a hydrocarbon are more or less "filler atoms." Alkanes’ names are based on the largest number of consecutively bonded carbon atoms. So, the name of a hydrocarbon tells you about that molecule’s structure.

To name a straight-chain alkane, simply match the appropriate chemical prefix with the suffix -ane. The prefixes relate to the number of carbons in the continuous chain and are listed in Table 21-1.

The naming method in Table 21-1 tells you how many carbons are in the chain. Because you know that each carbon has four bonds and because you are fiendishly clever, you can deduce the number of hydrogen atoms in the molecule as well. Consider the carbon structure of pentane, for example, shown in Figure 21-1.

Figure 21-1:

Pentane’s carbon skeleton.

-C-C-C-C-C—

Only four carbon-carbon bonds are required to produce the five-carbon chain of pentane. This leaves many bonds open — two for each interior carbon and three for each of the terminal carbons. These open bonds are satisfied by carbon-hydrogen bonds, thereby forming a hydrocarbon, as shown in Figure 21-2.

Figure 21-2:

Pentane’s hydrocarbon

Structure.

H HHH H

C CCC C H HHHHH

H

If you add up the hydrogen atoms in Figure 21-2, you get 12. So, pentane contains 5 carbon atoms and 12 hydrogen atoms.

As the organic molecules you study get more and more complicated, it will become more and more important to draw the molecular structure to visualize the molecule. In the case of straight-chain alkanes, the simplest of all organic molecules, you can remember a convenient formula for calculating the number of hydrogen atoms in the alkane without actually drawing the chain:

Number of hydrogen atoms = (2 X Number of carbon atoms) + 2

You can refer to the same molecule in a number of different ways. For example, you can refer to pentane by its name (ahem. . . Pentane), By its molecular formula, C5H12, or by the complete structure in Figure 21-2. Clearly, these different names include different levels of structural detail. A Condensed structural formula Is another naming method, one that straddles the divide between a molecular formula and a complete structure. For pentane, the condensed structural formula is CH3CH2CH2CH2CH3. This kind of formula assumes that you understand how straight-chain alkanes are put together. Here’s the lowdown:

Carbons on the end of a chain, for example, are only bonded to one other carbon, so they have three additional bonds that are filled by hydrogen and are labeled as CH3 in a condensed formula.

Interior carbons are bonded to two neighboring carbons and have only two hydrogen bonds, so they’re labeled CH2.

Your chemistry teacher will require you to draw structures of alkanes, given their names, and will require you to name alkanes, given their structure. If your teacher fails to make such requests, ask to see his credentials. You may be dealing with an impostor.

Q.

What is the name of the following structure, and what is its molecular formula?

HH HH

C – C – C – H

HH HH

Are four. Table 21-1 helpfully points out that four-carbon chains earn the prefix But-. What’s more, this molecule is an alkane (because it contains only single bonds), so it receives the suffix -ane. So, what you’ve got is butane. With four carbon atoms in a straight chain, ten hydrogen atoms are required to satisfy all the carbon bonds, so the molecular formula of butane is C4H10.

H

C

Butane; C4H10. First, count the number of carbons in the continuous chain. There

1. What is the name of the following structure, and what is its molecular formula?

HHH

HHH

Solve It

2. Draw the structure of straight-chain octane.

Solve It

Going Out on a Limb: Making Branched Alkanes by Substitution

Not all alkanes are straight-chain alkanes. That would be too easy. Many alkanes are so-called Branched alkanes. Branched alkanes differ from continuous-chain alkanes in that carbon chains substitute for a few hydrogen atoms along the chain. Atoms or other groups (like carbon chains) that substitute for hydrogen in an alkane are called Substituents.

Naming branched alkanes is slightly more complicated, but you need only to follow a simple set of steps to arrive at a proper (and often lengthy) name.

1. Count the longest continuous chain of carbons.

Tricky chemistry teachers often draw branched alkanes with the longest chain snaking through a few branches instead of obviously lined up in a row. Consider the two carbon structures shown in Figure 21-3. The two are actually the same structure, drawn differently! Yikes. In either case, the longest continuous chain in this structure has eight carbons.

H

C

Cc

C1 C2 C3 C4 Cb C

B

C7

7

C8

8

C

Figure 21-3: c

One carbon structure drawn two different ways.

2 3 4 b b

Cc

2. Number the carbons in the chain Starting with the end that’s closest to a branch.

You can always check to be sure you’ve done this step correctly by numbering the carbon chain from the opposite end as well. The correct numbering sequence is the one in which the substituent branches extend from the lowest-numbered carbons. For example, as it’s drawn and numbered in Figure 21-3, the alkane has substituent groups branching off of its third, fourth, and fifth carbons. If the carbon chain had been numbered backwards, these would be the fourth, fifth, and sixth carbons in the chain. Because the first set of numbers is lower, the chain is numbered properly. The longest chain in a branched alkane is called the Parent chain.

Count the number of carbons in each branch.

These groups are called Alkyl groups And are named by adding the suffix -yl To the appropriate alkane prefix (Table 21-1 awaits your visit). The three most common alkyl groups are the methyl (one carbon), ethyl (two carbons), and propyl (three carbons) groups. Figure 21-3 has two methyl groups, one ethyl group, and no propyl groups.

Be careful when you find yourself dealing with alkyl groups made up of more than just a few carbons. A tricky drawing may cause you to misnumber the parent chain!

C

C

7

C

C

Attach the number of the carbon from which each substituent branches to the front of the alkyl group name.

For example, if a group of two carbons is attached to the third carbon in a chain, like it is in Figure 21-3, the group is called 3-ethyl.

5. Check for repeated alkyl groups.

If multiple groups with the same number of carbons branch off the parent chain, don’t repeat the name. Rather, include multiple numbers, separated by commas, before the alkyl group name. Also, specify the number of instances of the alkyl group by using the prefixes Di-, tri-, tetra-, And so on. For example, if one-carbon groups (in other words, methyl groups) branch off carbons four and five of the parent chain, the two methyl groups appear together as "4,5-dimethyl."

6. Place the names of the substituent groups in front of the name of the parent chain In alphabetical order.

Prefixes like Di-, tri-, And Tetra – Don’t figure into the alphabetizing. So, the proper name of the organic molecule in Figure 21-3 is 3-ethyl-4,5-dimethyloctane.

Note that hyphens are used to connect all the naming elements except for the last connection to the parent chain (. . . dimethyl-octane is wrong).

Q.

Name the branched alkane shown in the A. Following structure:

/

CH.,

H, C-HC

\

CH

/

CH HC

/

H3C

3

\

CH

CH

CH,- C-CH,

CH

4-ethyl-2,6,6-trimethyloctane. First, notice how some of the bonds seem to zig-zag. This is a feature of many hydrocarbon structures. The longest continuous chain of carbon atoms in this alkane is eight carbons long. So, the parent chain is octane. Four alkyl groups branch off the parent chain: An ethyl group branches off the fourth carbon, two methyl groups branch off the sixth carbon, and another methyl group branches off the second carbon. Attaching appropriate prefixes and alphabetizing yields the name 4-ethyl-2,6,6-trimethyloctane.

3. Name the branched alkane shown in the following structure:

CH

H, C

3 \

H, C

\

CH

H3C

3

CH

CH

CH

CH

CH

H3C

3

CH CH

CH

CH

Solve It

4. Name the branched alkane shown in the following structure:

CH

H3C

3

CH

CH

Solve It

5. Draw the alkane 3,4-diethyl-5-propyldecane.

Solve It

6. Draw the alkane 3-propyl-2,2,4,4-tetra-methylheptane.

Solve It

Getting Unsaturated: Alkenes and Alkynes

JjjtJW^ Carbons can do more than enagage in four single bonds. There’s more to organic molecules 41/ M \, than substituent-for-hydrogen swaps. When carbons in an organic compound fill their

MM ) valence shells entirely with single bonds, we say the compound is Saturated. But many hydrocarbons contain carbons that bond to each other more than once, creating double or triple covalent bonds. We say these hydrocarbons are Unsaturated Because they have fewer than the maximum possible number of hydrogens or substituents. For every additional carbon-carbon bond formed in a molecule, two fewer covalent bonds to hydrogen are formed.

When neighboring carbons share four valence electrons to form a double bond, the resulting hydrocarbon is called an Alkene. Alkenes are characterized by these chemically interesting double bonds, which are more reactive than single carbon-carbon bonds (see Chapter 5 for a review of sigma and pi bonding). Double bonds also change the shape of a hydrocarbon, because the Sp2 Hybridized valence orbitals assume a trigonal planar geometry, as shown by the carbons of ethene in Figure 21-4. Saturated carbon is sp3 hybridized and has tetrahedral geometry (again, see Chapter 5 to review hybridization).

Figure 21-4:

Ethene, a

C

Alkene.

Two-carbon

Naming alkenes is slightly more complicated than naming alkanes. In addition to the number of carbons in the main chain and any branching substituents, you must also note the location of the double bonds in an alkene and incorporate that information into the name. Nevertheless, the essential naming strategy for alkenes is quite similar to that for alkanes in the previous section:

1. Locate the longest carbon chain, and number it, Starting at the end closest to the double bond.

In other words, double bonds trump substituents when it comes to numbering the parent chain. Build the name of the parent chain by using the same prefixes as used for alkanes (refer to Table 21-1), but match the prefix with the suffix -ene. A three-carbon chain with a double bond, for example, is called "propene."

2. Number and name substituents that branch off the alkene In the same way that you do for alkanes.

List the number of the substituted carbon, followed by the name of the substituent. Separate the substituent number and name with a hyphen.

3. Identify the lowest numbered carbon that participates in the double bond, and put that number Between the substituent names and the parent chain name (sandwiched by hyphens), but after all the substituent names.

For example, if the second and third carbons of a five-carbon alkene engage in a double bond, then the molecule is called 2-pentene, not 3-pentene. If that same molecule has a methyl substituent at the third carbon, then the molecule is called 3-methyl-2-pentene.

Alternately, and especially when there are substituents present, the position of an unsaturation is indicated between the prefix and suffix of the parent chain name. So, 3-methyl-2-pentene may also be written 3-methylpent-2-ene.

Alkynes Are hydrocarbons in which neighboring carbons share six electrons to engage in triple covalent bonds. The naming strategy for alkynes is the same as that for alkenes, except that the alkyne parent chain is named by matching the prefix with the suffix -yne.

The trick to drawing hydrocarbons is to start at the end of the name and work backwards. The prefix preceding the -ane, – ene, Or -yne Ending always tells you how many carbons are in the longest chain, so begin by drawing that parent chain. From there, work through the substituent groups, tacking them on as you go. Finally, add hydrogens into the structure wherever there are empty bonds, and voila! A portrait of a hydrocarbon.

KPLf

Q. Name the following alkene and the following alkyne:

H. C

3 *

.CH,

CH

HC

CH

CH

CH

C

The alkene is 2-methyl-2-butene (or 2-methylbut-2-ene); the alkyne is 1-butyne. The figure on the left shows a four-carbon alkene with the double bond between the second and third carbon atoms. Numbering the chain from either side yields the same numbers with respect to the double bond. However, numbering from right to left gives a lower number to the methyl substituent, so the compound is 2-methyl-2-butene (or 2-methylbut-2-ene). The figure on the right shows a four-carbon alkyne, with the triple bond located between the first and second carbon atoms. There are no substituents. The compound is therefore 1-butyne.

7. Name the following unsaturated hydrocarbon:

3 -

3

CH CH

H3C

3

CH2 CH3

CH C CH

H2C

2

CH

Solve It

8. Draw the compound 5-ethyl-5-methyl-3-octyne (or 5-ethyl-5-methyloct-3-yne).

Solve It

9. Draw the compound 4,4-dimethyl-2-pen-tene (or 4,4-dimethylpent-2-ene).

Solve It

Rounding ‘em Up: Circular Carbon Chains

The compounds we cover earlier in this chapter are linear or branched. However, hydrocarbons can be circular, or Cyclical. Among the cyclical carbons, there are two important categories, the Cyclic aliphatic Hydrocarbons and the Aromatic Hydrocarbons.

Chemists sometimes divide hydrocarbons into aliphatic and aromatic categories to highlight important differences in structure and reactivity. Without going into more technical detail than is useful here, aliphatic molecules and aromatic molecules have significantly different electronic configurations (which electrons go into which orbitals). As a result, the two types of hydrocarbons typically undergo different kinds of reactions. In particular, they tend to undergo different kinds of substitution reactions, ones in which some atom or group substitutes for hydrogen.

Wrapping your head around cyclic aliphatic hydrocarbons

Cyclic aliphatic hydrocarbons are like the hydrocarbons that we explain earlier in this chapter, except that they form a closed ring. The rules for naming these compounds build on the rules we provide earlier in this chapter. For example, a cyclical six-carbon alkane includes the name Hexane, But is preceded by the prefix Cyclo-, Making the final name Cyclohexane.

A single substituent or unsaturation on a cyclic aliphatic hydrocarbon doesn’t require a number. So, a single bromine-for-hydrogen substitution on cyclohexane yields a compound with the name bromocyclohexane. Likewise, a lone double bond unsaturation on cyclohexane yields a compound with the name cyclohexene.

Multiple substitutions or unsaturations require numbering. In these cases, the same rules apply for deciding the rank of substituents. Triple bonds outrank double bonds. Double bonds outrank other substituents. So, number the carbons in the way that respects these rankings and produces the lowest overall numbers. A cyclohexane molecule with two methyl substituents on neighboring carbons, for example, is called 1,2-dimethylcyclohexane.

Sniffing out aromatic hydrocarbons

Aromatic hydrocarbons have special properties because of their electronic structure. Aromatics are Both cyclic and conjugated. Conjugation results from an alternation of double or triple bonds with single bonds. Noncyclic hydrocarbons can be conjugated, too, but they can’t be aromatic. Aromatic molecules have clouds of Delocalized pi electrons, Electrons that move freely through a set of overlapping P Orbitals. The model aromatic compound is benzene, the resonance structures for which are depicted in Chapter 5. Because of its cyclical, conjugated bonding, pi electrons delocalize evenly into rings above and below the plane of the flat benzene molecule. Aromatic compounds are very stable compared to their aliphatic counterparts.

Numbering substituents on aromatics follows the same basic pattern as followed for cyclic aliphatic compounds. A single substituent requires no numbering, as in bromobenzene. Multiple substituents are numbered by rank, with the highest-ranked substituent placed on carbon number one, and proceeding in a way that results in the lowest overall numbers. A benzene ring with chlorine and methyl substituents situated two carbons away from one another, for example, would be called 1-chloro-3-methylbenzene.

Name the following cyclic hydrocarbon:

3-methylcyclohexene. The structure is a six-carbon cyclic alkene with a methyl substituent. Numbering starts with the highest priority group, which is the double bond. Number so that the carbons of the double bond receive numbers 1 and 2, and the carbon to which the methyl group is attached gets the lowest possible number. This means numbering counterclockwise as shown in the figure. The name of the compound is 3-methylcyclohexene. You don’t need to specify the number of the carbon where the double bond appears; because it’s the highest-ranking feature, the double bond is assumed to start at carbon number one.

A.

10. Name the following cyclic hydrocarbon:

H, C.

,ChL

HC =

= CH

Solve It

11. Draw the compound 4-methylcyclopentene.

Solve It

C

12. Draw the compound 4-butyl-3-ethylcyclopentyne.

Solve It

Answers to Questions on Carbon Chains

Do you feel like you’re on the straight and narrow path to understanding carbon chains, or do you feel like you’ve been running around in circles? Take a deep breath, relax, and check your answers to the practice problems presented in this chapter.

D The structure is propane; its molecular formula is C3H8. The figure shows a three-carbon chain with only single bonds. Therefore, it is propane. Its molecular formula is C3H8.

CM The Oct – Prefix here tells you that this alkane is eight carbons long. Draw eight linked carbons and fill in the empty bonds with hydrogen. Your structure should look like this one:

H

H

H

H

H

H

H

H

HC

H3C

CH

CH

CH

CH

CH

CH

CH

CM 4-ethyl-3-methyl-5-propylnonane. The structure shown is a nine-carbon alkane, so its parent chain name is nonane. Begin numbering from the end that gives the substituent groups the lowest numbers. A one-carbon group extends from the third carbon (3-methyl), a two-carbon group extends from the fourth carbon (4-ethyl), and a three-carbon group extends from the fifth carbon (5-propyl). Alphabetize these substituents and tack on the nonane ending, and you have 4-ethyl-3-methyl-5-propylnonane.

MM 2,2-dimethylpropane. Note that this is the proper Systematic Name, and that other Common Names may also be used. The common name for this structure is neopentane.

CM The name 3,4-diethyl-5-propyldecane tells you first and foremost that this compound is a ten-carbon alkane. The substituent names indicate two ethyl groups extending from the third and fourth carbons and a propyl group extending from the fifth carbon in the chain. So your artwork should look like this figure:

H, CX

CH

H3C

/CH2^ /CH\ /CH2^ YCH2K /CH3

CH CH

CH2 CH2

22

CH2 CH2

22

H3C H2C

, 2 .

C

C

C

C

C

C

C

H

H

H

H

H

H

H

H

H

Or

CH,

MM The Heptane Ending tells you that you’re dealing with a seven-carbon alkane here, so begin by drawing a seven-carbon parent chain. Next, attach the substituents to the main chain by decoding their names. 3-propyl Tells you that a three-carbon substituent group extends from the third carbon in the chain, and 2,2,4,4-tetramethyl Tells you that four single-carbon substituent groups are attached to the chain, two on the second carbon and two on the fourth. Your drawing should look like this:

H3C

3

CH3 H3C

3 3 .

H3C

3

CH,

• CH3

CH

CH

CH

CH

H, C

CH3

3

EM 5,5-diethyl-2-methyl-3-heptene (or 5,5-diethyl-2-methylhept-3-ene). Note first that this structure contains a double bond, meaning it’s an alkene and will end with the suffix -ene. It has seven carbons in the parent chain, so its base name is Heptene. In general, you begin numbering the chain at the end closest to the double bond because it has the highest priority. However, in this case, the double bond is between carbons three and four (so it’s 3-heptene) no matter which end you begin numbering from. The substituents serve as the tiebreaker here to help you decide whether to number from the left or the right. Choose the end that yields the lowest substituent numbers and then name the substituents. You should get 5,5-diethyl-2-methyl-3-heptene (or 5,5-diethyl-2-methylhept-3-ene).

MM The ending 3-octyne Tells you that this compound is an eight-carbon alkyne with the triple bond between the third and fourth carbons. The substituent names 5-ethyl And 5-methyl Tell you that the chain includes both a two-carbon (ethyl) and a one-carbon (methyl) branch off the fifth carbon. If your drawing looks like the following figure, you’ve done well.

CH

CH

CH

H3C

3

CH, CH3

CH

H3C

3

C

C

C

MM The ending 3-pentene Tells you that this compound is a five-carbon alkene with the double bond

Between the second and third carbons. The substituent name 4,4-dimethyl Tells you that two single-carbon substituent groups are present, both extending from the fourth carbon in the chain. Your artwork should look like the following figure:

ChL

H, C -

3

CH -

CH

C

CH3

3

Jj 3,3-dimethylcyclopropene. This is a five-carbon cyclic alkene with two methyl substituents. Number the ring so the two carbons sharing the double bond receive the numbers 1 and 2, and the methyl substituents receive the lowest number. Doing so causes the two methyl groups to extend from the carbon numbered 3. The name is therefore 3,3-dimethylcyclopropene.

|f| The ending Cyclopentene Indicates that this is a five-carbon cyclic alkene. Draw a five-carbon ring containing a single double bond and number around the ring so the two carbons sharing the double bond receive the numbers 1 and 2. Then attach a one-carbon substituent group to the carbon numbered 4. Here’s what your drawing should look like:

CH

H3C

3

HC

CH

CH

CH

U The name 4-butyl-3-ethylcyclopentyne reveals that the compound is a five-carbon cyclic alkyne, so begin by drawing a five-carbon ring with one triple bond. Number around the ring so the two carbons sharing the triple bond receive the numbers 1 and 2, and then attach a four-carbon (butyl) substituent to the fourth carbon and a two-carbon (ethyl) substituent to the third carbon. Your drawing should look like this one:

H3C

CH

H2CN

2 ^

H3C

3

CH

CH

CH

HC

CH

C

C

The 5th Wave_By Rich Tennant

I think I’m having side ejects £rom my nevr prescription medication. I’m ieelino, nauseous and disoriented all day.’

In this part. . .

Am Ppendix A contains three worksheets that are useful ¥ 9 When you’re choosing a Part D plan. The first helps you create your master list of the prescription drugs you take now, plus their dosages and how often you take them. You can use the second and third worksheets to help compare drug plans and Medicare health plans point by point.

Appendix B is the place to go for contact information for many sources of help. Here you find phone numbers and Web site addresses for the key government agencies (Medicare and Social Security), every State Health Insurance Assistance Program, and many consumer organizations. This appendix also suggests ways of getting updates on Medicare and Part D so you can stay informed on changes that come down the track.

Appendix C is a guide on how to get prescription drugs Safely If you purchase them by mail order from Canada, where they generally cost less. It explains how to avoid scams and counterfeit meds, how to assess pharmacies for safe and ethical service, and how to find prescreened licensed pharmacies you can trust.

Appendix A

In This Chapter

^ Working with surface areas and volumes of solid figures ^ Considering practical applications with prisms ^ Doing what you Can With a cylinder

7hree-dimensional figures are a part of our life, so they are measured, weighed, and stacked. A prism is a solid figure with a top and bottom that are exactly the same. Pyramids and cones come to a single point or vertex.

In this chapter, you see how problems are posed and answered by considering the various solid figures and their properties. The surface area of a solid figure is a measure of the number of squares covering the outside, and the volume of a solid figure is a measure of the number of cubes you can fit inside. The respective formulas for surface area and volume are a big help, but just as important is some common sense and being able to picture the situation and the solid figure being used.

The Pictures Speak Volumes

The volume of a prism is found by multiplying the area of the base of the prism by its height. The top and bottom or two bases of a prism are always congruent (exactly the same). The sides are always rectangles. The prisms being considered in this chapter are technically Right rectangular prisms, Meaning that the sides are perpendicular to the bases — the sides don’t slant like the Leaning Tower of Pisa — and the bases are parallel to one another.

Boxing up rectangular prisms

A rectangular prism has rectangles for its bases as well as its sides. Because all the surfaces of a rectangular prism are rectangles, you can actually turn it so that any pair of parallel sides are the bases. Pick your favorites.

The volume of a right rectangular prism is equal to the area of the base times the height. Because the base is also a rectangle, the volume formula is more conveniently written V = Iwh, Where L Is the length of the base rectangle, W Is the width of the base rectangle, and H Is the height or the distance between the two bases.

The Problem: What is the height of a rectangular prism that has a square base 6 inches on a side and a volume of 288 cubic inches?

Use the formula for the volume of a prism replacing the V With the 288 cubic inches, the L With 6, and the W With 6. Then solve for H.

V = Iwh 288 = 6 X 6 X H 288 = 36h

288 . 36

H, h=8

The height is 8 inches.

The Problem: A rectangular prism has a length that’s twice the width and a height that’s twice the length. If its volume is eight times the height, then what is the volume?

^VLA/y You use the formula for the volume, even though you don’t really have a numerical value to put in for V. The length, height, and volume can all be expressed in terms of the width. First, if The length is twice the width You write that L = 2w. Because the height is Twice the length, You write H = 2l = 2(2w) = 4w. And the volume is Eight times the height, So V = 8h = 8(4w) = 32w.

Now, using the formula for the volume, V = Lwh, Filling in the equivalences and solving for W,

V = Lwh 32w = (2w)( W)(4w)

32W= 8W3

4W= w3 0 = W3 – 4w

0 = W _w2- 4) = W(W – 2)(W + 2)

The solutions for W In the equation are 0, 2, and -2. You discard the 0 and -2, because, even though they’re solutions of the equation, they don’t answer the question. Letting W = 2, you get that the volume is 32(2) = 64 cubic units. The width of the prism is 2 units, the length is twice that or 4 units, and the height is 4 times the width, or 8 units.

Venturing out with pyramids

A pyramid is a solid figure that has a Polygon (a figure with segments for its sides) for a base and sides that are triangles — all meeting in one point called

The Vertex.

The volume of a pyramid is found by multiplying the area of the base of the pyramid by its height and taking one-third of that product. The height of a pyramid is the perpendicular distance from the center of the base up to the vertex. Figure 20-1 shows a square pyramid and a hexagonal pyramid. Traditionally, the name of the pyramid is determined by the shape of the base.

Figure 20-1:

Pyramids are found in Egypt and jewelry settings.

H \ \/»’/'h V \

The Problem: The volume of a pyramid with a square base is 216 cubic inches. If the sides of the square base are 9 inches, then what is the pyramid’s height?

Use the formula for the volume of a pyramid, V = 3 Bh And replace the VWith

216. The area of the base, B, Is the area of a square. The area of a square is found by squaring the length of a side. If the sides of the square base are 9 inches, then the area of the base is 81 square inches. Replace the B In the formula with 81 and solve for H.

216 = ^ (81) H 216 = 27h

O 1 C

~276 = H

H = 8

The pyramid is 8 inches high.

The Problem: A hexagonal-based pyramid has sides (of the base) measuring 8 inches. The length along any edge of the pyramid from a base corner to the vertex is 10 inches. What is the pyramid’s volume?

You need to find the pyramid’s height. Think of the height of the pyramid as being one side of a right triangle whose hypotenuse is the 10-inch edge. (See Figure 20-2 for a sketch of the pyramid and embedded right triangle.)

Figure 20-2:

The right triangle’s hypotenuse is an edge of the pyramid.

10

8

The hexagonal base is made up of equilateral triangles whose sides are all 8 inches, so the distance from the outer corner to the center of the hexagon is 8 inches. You have a right triangle whose hypotenuse is 10 inches with one leg of 8 inches. Using the Pythagorean theorem, you get that the measure of the other leg is 6 inches. That other leg is the height of the pyramid.

Now, armed with the height of the pyramid, 6 inches, you need to find the area of the base. Each of the six equilateral triangles making up the hexagonal base has sides of 8 inches. Use Heron’s formula to find the area of one of the triangles and multiply that area by 6. (If you need a refresher on using Heron’s formula to find the area of a triangle, refer to Chapter 10.)

A = /s (s – a)(s – b)(s – c) = /12 (12 – 8)(12 – 8)(12 – = /12 (4)3 = /12 (64) = /3 • 22 • 82 = 16/3

Six times this area is 96/3. Now use the formula for the volume of a pyramid, and insert the area of the base and the height.

V = 3 (96/3 J 6 = 192/3

The volume is about 333 cubic inches.

Fire-breathing dragon

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Dropping eaves with trapezoidal prisms

A trapezoid is a four-sided polygon with one side parallel to another side. A container with slanted sides so that the top or opening is larger than the bottom might have a trapezoidal cross-section. In Figure 20-3, you see a trapezoidal prism with the two parallel sides on the top and bottom. (Refer to Chapter 10 if you need a refresher on trapezoids and finding their area.)

Figure 20-3:

The

Trapezoidal prism lies on its side.

The volume of a trapezoidal prism is equal to the area of its base times its height, V = Bh. Replacing the B, For the area of the base, with the formula for the area of a trapezoid, the formula for volume becomes

V = ^ H1_b 1 + b2) $ hp. The subscripts T And P On the height measures are there to

Differentiate between the height of the trapezoid (the distance between the two parallel bases) and the height of the prism (the distance between the two trapezoidal bases.

The Problem: A watering trough for cattle is a trapezoidal prism. The trapezoid has a bottom base of 2 feet and an opening, at the top, of 3 feet. The trough is 2 feet deep. If the trough contains 60 cubic feet of water, how long is it?

You’re looking for the height of the prism — the distance between the two trapezoidal bases on either end of the trough. Find the area of the trapezoidal base (end) and insert it into the formula for the volume. Replace the V In the formula with 60 and solve for the height of the prism.

The area of the trapezoid is A = 2(2)(2 + 3) = 1 (5) = 5 square feet. Replacing

The V And B In the volume formula with 60 and 5, respectively, you get that 60 = 5H. Dividing each side of the equation by 5, your answer is that the height of the prism (in this case, the length of the trough) is 12 feet.

Most homes have rain gutters attached at the edge of the roof to catch water coming down the roof surface; the water is then directed to a downspout. The gutters are often trapezoidal prisms where one side of the trapezoid is perpendicular to the two parallel sides so that it forms a flat attachment to the side of the house.

The Problem: An aluminum gutter in the shape of a trapezoidal prism is to go across the 32-foot front portion of a house. If the two parallel sides of the trapezoid are 6 inches and 9 inches, then how deep must the trapezoid be so that the gutter can hold 10 cubic feet of water?

The first chore is to change all the units so they’re the same — right now they’re in inches and feet. You could change everything to inches, but a cubic foot is 12 x 12 x 12 cubic inches, and that makes the numbers a bit big to work with. Instead, you can change the 6 and 9 inches to feet and work with relatively nice fractions. The area of the trapezoidal end of the gutter is

5 j where the inches are changed to fractions

Of feet and the formula is simplified as much as possible. You’re looking for the depth of the gutter, which is the height of the trapezoid in this case. Now use the formula for the volume of the prism, replacing the V With 10, the area of the base, B, With the result of the area of the trapezoid, and the height of the prism, H, With 32 feet. Then solve for H.

1 h (1 + 3

2 h \ 2 + 4

1 H (5

2 H \ 4

V = B X h„ 10 = H (5 j X 32

10 = H

5

X 32

10 = H (20) -2°y = H

The height (depth) of the gutter is half a foot, or 6 inches.

Figure 20-4:

The prism is formed by folded cardboard and triangular ends.

Mailing triangular prisms

You want to mail an umbrella to a friend. An umbrella is long and has a handle. You don’t want to use a flat, rectangular prism, because the long package may get bent and damage the umbrella. A long tube or cylinder may work, but an even better idea is to use a triangular prism that has strength from the three sides and flatness to help pack up the umbrella and keep it from flopping about inside.

The surface area of a triangular prism is equal to the sum of the areas of the two bases (the triangles) plus the area of each of the three rectangular sides. Figure 20-4 shows you a triangular prism and, also, what the ends look like if they’re extended for tabs so they can be glued to the sides of the prism.

The Problem: You want to mail an umbrella that’s 3 feet long with a handle that’s 5 inches wide. You choose a mailing container that’s a triangular prism made of cardboard. The ends of the prism are equilateral triangles 6 inches on a side, and the length (height) of the prism is 2 inches longer than the umbrella. How much cardboard is needed for the container, if the sides of the triangles are extended by 1 inch for glue tabs?

Break this problem into three different area computations: the areas of the sides of the prisms, the areas of the two triangles that make up the bases, and then the areas of the glue tabs.

Each of the sides of the prism is a rectangle that’s 3 feet, 2 inches long and 6 inches wide. The prism has three sides. Changing the length measure to inches, the three rectangles are 38 inches by 6 inches or 228 square inches in area. Three of them makes 3(228) = 684 square inches.

The triangular bases are equilateral triangles measuring 6 inches on a side. Using Heron’s formula (refer back to Chapter 10 for more on this formula), the area of one triangle is

A = /s (s - A)(s - B)(s - C) = /9 (9 – 6)(9 – 6)(9 – 6) = /9 (3)3 = /9(27) = /32 • 32 • 3 = 9 /3

So two of them gives you 2 (9 /3 J = 18/3 square inches which is about 31 square inches.

The six tabs are each 6 inches by 1 inch or 6 square inches. The six tabs add a total of 36 square inches to the area. So the total area of the entire triangular prism is 684 + 31 + 36 = 751 square inches.

Folding Up the Sides for an Open Box

A classic calculus problem involves taking a rectangular piece of paper or cardboard or metal, cutting equal squares from the four corners, and folding up the sides to construct an open box. Figure 20-5 shows you how this is done.

Figure 20-5:

Fold up the sides and glue the edges together.

*.VLA*

The Problem: A 9-by-12-inch piece of cardboard is to be made into an open box by cutting slits into square-shaped corners, folding up the sides, and gluing the tabs to the ends. What is the volume of the open box if the corner squares are 2 inches on a side?

The volume of the right rectangular prism that’s formed is found with the formula V = Iwh. The height, H, Of the prism is 2 inches — the size of the squares at the corners. The length of the prism is 12 inches minus the two squares in the corners; 12 – 4 = 8 inches for the length. The width of the prism is

9 inches minus the two corners or 9 – 4 = 5 inches. The volume of the prism is V = 8(5)(2) = 80 cubic inches.

In the calculus problem involving the open box, you get to determine the size of squares in the corners that gives you the largest possible volume. If the squares are small, then the box isn’t very deep, but the area of the base is big. If the squares are large, then the box is deep, but the area of the base is small. Calculus allows you to balance the depth and the base area to find the best dimensions. In the next problem, you get to do somewhat the same process with a table of the possible values.

The Problem: A sandbox is to be formed from a 9 foot square piece of metal by cutting equal squares from the corners and folding up the sides. The edges are then welded together to form the box. What size squares should be cut from the corners to form a box that will hold the greatest amount of sand possible (have the greatest volume)?

First, determine the dimensions of the box in terms of the length in feet of the sides of the squares, X. The height is, of course, X. The length and width are both 9 minus the two squares (one on either end) or 9 – 2x feet. The volume of the sandbox is V = Iwh = (9 – 2x)(9 – 2x)x. In Table 20-1, you see measures for X In half-foot increments and the resulting volumes.

As you see, after 1.5 feet for the height, the volume is decreasing as the squares get bigger. In half-foot increments, the greatest possible volume is obtained when the sandbox is 1.5 feet deep and 6 feet on a side. Using calculus to solve this problem and find the greatest volume you get the same answer. (Of course, I picked a very convenient size box to demonstrate this.)

Following Postal Regulations

111

Postal services have regulations regarding the weight and size and contents of packages that you take to be mailed. You may remember taking a package in and watching the clerk wrap a chain around the middle of the box and then holding the remainder of the chain along the height. Regulations dictate that the height plus the Girth Of the box can’t exceed 108 inches. The Girth Is that measure around the two shorter dimensions.

The height plus the girth of a rectangular prism is equal to H + 2L + 2w.

Refer to Figure 20-6 for a picture of how the height and girth are measured.

Figure 20-6:

The girth is the distance around the middle.

W

Finding the right size

With the limitations on the total of the height plus the girth, you have to be careful what size package you prepare to send. Everything has to fit into a carton that passes muster. It takes a balance between a package being wide enough and long enough but not too much of either.

The Problem: How tall a box will the postal service accept if it has a base that’s 16 inches by 18 inches and if the height plus the girth can’t exceed 108 inches?

The measures of 16 and 18 inches represent the width and length of the base. Put these measures into the formula H + 2L + 2W And set the expression equal to 108. Then solve for H.

H + 2 (16) + 2 (18) = 108 H + 32 + 36 = 108 H + 68 = 108 H = 40

A height of 40 inches is a fairly long box — that’s over 3 feet!

If you’re sending something that will fit into any size box and are restricted only by the total volume, then you have more flexibility with the postal rules. For example, if you’re sending Ping-Pong balls or nails, the shape of the box doesn’t matter very much.

The Problem: You have to send wrapped candies that take up 1,000 cubic inches of space. Because of the weight of the candy, the rule is that the height plus the girth can’t exceed 50 inches. If the base of the box is a square, then what are the dimensions of the carton you’ll be using?

^VLA* Write two equations: one about the volume and the other about the height

And girth. Find the common solution or solutions to the system of equations. The volume of this box is found with V = Wwh, Because the base is a square. Replacing the V With 1,000, the equation becomes 1,000 = W2h. Write the equation involving the height and girth with 50 = H + 2w + 2w or 50 = H + 4w. Solve the second equation for H, And you get H = 50 – 4w. Substitute this equivalence for H Into the volume formula and solve.

1,000 = W2 h 1,000 = W2(50 – 4w) 1,000 = 50w2 – 4w3 4w3 – 50w2 + 1,000 = 0 2w3 – 25w2 + 500 = 0

The cubic equation can be factored. You get to use some more powerful algebra in the form of the rational root theorem and synthetic division to find the factors. (These topics are covered thoroughly in Algebra II For Dummies.)

2W3 – 25w2 + 500 = 0 (w – 10)(2w2 – 5w – 50) = 0

The first factor gives you the answer that W = 10. The second factor doesn’t simplify any farther, so you need the quadratic formula to find the solutions.

5 +J25 – 4 (2)(-50) W =–2 (2)-

5 + 7425

=-4-. 6.4 or -3.9

Discard the negative number and just consider the solution that W = 6.4. It appears that two different sets of dimensions satisfy the requirements that the box contain 1,000 cubic inches of space and that the height plus the girth doesn’t exceed 50 inches.

If W = 10, then H + 4w = 50 yields H + 40 = 50 and H = 10. The box is a perfect cube that’s 10 by 10 by 10.

If W = 6.4, then H + 4w = 50 yields H + 25.6 = 50 and H = 24.4. The box is roughly 6.4 by 6.4 by 24.4. The choice is which works better for the supplier and the customer.

Maximizing the possible Volume

If the mailing regulations dictate that you can’t send a rectangular box whose height plus girth exceeds 108 inches, then how can you make the most of that 108 inches? What’s the greatest volume you can create?

The Problem: What are the dimensions of a rectangular prism (box) with a square base that has the greatest possible volume if the height plus the girth can be at most 108 inches?

Write an expression for the volume using the relationship that H + 4w = 108. The base of the box is a square, so W Represents both the length and the width. Solving for H, You get H = 108 – 4w. Substitute that into the volume formula, and you have V = w2(108 – 4w). Create a table, trying out different values for W And seeing what volumes you obtain (see Table 20-2).

The volumes are starting to decrease, so you settle on the 18-inch square base and height of 36 inches to obtain the greatest possible volume. Are you wondering how I knew to jump to the higher numbers and skip up to where

The width is 18 inches? The answer is simple: calculus! I got impatient with putting in values and used calculus to tell me the exact answer — and then just looked like a hero by going to the right spot.

Making the Most of a 12-Ounce Can

Cylindrical shapes are found in grocery stores, at oil refineries, and attached to space shuttles. A cylinder has circles for its top and bottom and a rectangle wrapped around its sides. The circular and rectangular shapes lend themselves to the formulas needed to find the volume and surface area of a cylinder. The volume of a cylinder is found by multiplying the area of the circular base times the height of the cylinder (the distance between the circles). The surface area is found by adding the areas of the two circles to the area of the rectangle; the width of the rectangle is the same as the circle’s circumference.

The formula for the volume of a cylinder is V = Nr1h Where R Is the radius of the circular base and H Is the height between the bases or circles. To find the surface area of a cylinder, use SA = 2N? + 2nrh = 2nr(r + h).

Filling a cylindrical tank

The volume of a cylindrical tank may be measured in cubic feet or cubic yards, or you may get to work with liquid measures such as gallons or quarts. Equivalences are used to change from one unit to another.

Equivalences:

1 cubic foot ~ 7.481 gallons 1 gallon ~ 0.134 cubic foot 1 ounce ~ 1.805 cubic inches

The Problem: A cylindrical tank has a height that’s equal to its diameter and it contains 128,000n cubic feet of liquid. What is the height of the tank?

^VLA/V Use the formula for the volume of a cylinder, replacing the V With the number of cubic feet. The radius of a circle is half the diameter, so, if the height is equal to the diameter, then the height is equal to twice the radius. Replace the H In the volume formula with 2R And solve for R.

V = nr2 H 128,000n = Nr 2( 2r) 128,000n = 2nr3

128,000N 3 2n =R 64,000 = R3 40 = R

The radius of the tank is 40 feet, so the height is 80 feet.

The Problem: A cylindrical tank is to be filled with vanilla ice cream. The tank is 6 feet tall and has a radius of 4 feet. How many gallons of ice cream will it take to fill the tank?

First, find the volume in terms of cubic feet. Then use the equivalence for changing cubic feet to gallons to determine the amount of ice cream needed. The volume of the tank is V = n(4)2(6) = 96n~ 301.44 cubic feet. One cubic foot is equal to about 7.481 gallons, so multiply 301.44 by 7.481 to get 2,255.07 gallons of vanilla ice cream. Yum!

Economizing with the surface area

You see 12-ounce containers all over the place in the form of cans for soft drinks, fruit beverages, and more potent brews. The traditional 12-ounce container is about 4.5 inches tall and has a diameter of about 2.5 inches. The shapes and exact sizes vary a bit, but the size is pretty standard. When you go to the grocery store, you need to check the shelves and see how many different heights and diameters of cans all contain 12 ounces. Looks can be deceiving. In Figure 20-7, you see two cylinders drawn to the approximate size of the traditional 12-ounce container and the most Efficient 12-ounce container. What do you think?

Figure 20-7:

The cans hold the same amount of fluid.

Fizzle

The Problem: A 12-ounce beverage can has a diameter of 2.5 inches and a height of 4.5 inches. The optimum size for a 12-ounce container (one that uses less material to contain the same amount of fluid) is one that has a diameter of 3 inches and a height of 3 inches. How much more material is used to produce the traditional beverage can?

Use the formula for surface area to determine the amount of material needed for each size can.

Diameter 2.5 = radius 1.25, height 4.5: SA = 2Nr(r + h) = 2n(1.25)(1.25 + 4.5) ~ 45.14

Diameter 3 = radius 1.5, height 3: SA = 2Nr(r + h) = 2n(1.5)(1.5 + 3) ~ 42.39

The difference in the amount of material is almost 3 square inches. Multiply that by millions of cans, and it’s a significant amount of material.

One of the factors that must have determined the size and shape of the traditional 12-ounce container is the cost of materials. Before the days of pop-tops, you had to open the containers with can openers that puncture the top of the can. The material had to be a bit stronger and heavier — and was more expensive.

The Problem: What if the cost of the material on the sides of a 12-ounce container is 1<t per square inch and the cost of the material on the top and bottom is 2<t per square inch. How much more does it cost for the material in a cylinder that’s 3 inches high with a 1.5-inch radius than it does for a cylinder that’s 4.5 inches high with a 1.25-inch radius?

The side of a cylinder is a rectangle that’s as long as the circumference of the circle and as high as the height of the cylinder. Multiply the length times the height to get the area. The two bases of the cylinder are circles. Find the area of one and double it. Multiply the areas times their respective costs.

The cylinder with a radius of 1.5 inches has a circumference of 2nr = 2n(1.5) ~ 9.42 inches. Multiply that by the height, 3, and the area of the side is about 28.26 square inches. The circular bases have an area of Nr2 = n(1.5)2 ~ 7.07 square inches. The total cost for the material is 28.26(1) + 2(7.07)(2) = 56.54 cents.

The cylinder with a radius of 1.25 inches has a circumference of 2nr = 2n(1.25) ~ 7.85 inches. Multiply that by the height, 4.5, and the area of the side is about 35.33 square inches. The circular bases have an area of Nr2 = n(1.25)2 ~ 4.91 square inches. The total cost for the material is 35.33(1) + 2(4.91)(2) = 54.97 cents.

The difference in the cost is about 1.5<t. Multiply that by millions of cans. . . .

Piling It On with a Conical Sand Pile

6

A cone has a circle for a base. The sides slant upward to meet at a single point called the vertex. The volume of a cone is one-third the volume of a cylinder that has the same base and same height.

The volume of a circular cone is V = ^ Nr2 h, where R Is the radius of the cone

And H Is the height — the distance from the center of the circular base to the vertex or top of the cone.

The Problem: What is the height of a cone if its diameter is twice the height and the volume of the cone is 72n cubic inches?

^VLA* Use the formula for the volume of a cone, replacing the VWith 72n and the height, H, With r. Why replace the height measure with R? The diameter of a circle is twice the radius, so the diameter is equal to 2r. If the diameter is twice the height, then the diameter, 2r = 2h. The length of the radius is equal to the height.

The radius is 6 inches. Because the radius and height are the same, the height is also 6 inches.

Have you ever tried to pile sand? It doesn’t cooperate all that well. The grains of sand don’t stick together unless they’re wet. When sand is poured from a container, it tends to form a cone-shaped pile, spreading out farther than it is high.

The Problem: Sand is falling off a conveyer belt and forming a conical shape as the falling sand runs down the sides of the pile. If the height of the pile is always one-third the diameter, then by how much does the volume of the pile change when the pile grows from 10 feet tall to 12 feet tall?

^VLA* Find the volume of a pile of sand that’s 10 feet tall and compare it to the

Volume of a pile of sand that’s 12 feet all. The pile of sand that’s 10 feet tall has a diameter of 30 feet — which means a radius of 15 feet. The pile of sand that’s 12 feet tall has a diameter of 36 feet — or a radius of 18 feet.

10-foot pile: V = I n (15)2(10) . 2,355 cubic feet 3

12-foot pile: V = I n (18)2(12) . 4,069 cubic feet 3

The pile of sand has grown by over 1,700 cubic feet.

Part V

In This Chapter

^ Philosophising rationally ^ Taking responsibility for your feelings ^ Enhancing your psychological health Staying interested

I\ S we discuss many times in this book, the attitudes you hold about your-Ґ \ self, other people, and the world greatly affect your ability to respond successfully to negative life events. Even in the absence of unusual or difficult circumstances, your core philosophies influence your overall experience of life. People who hold rational philosophies are generally less prone to emotional disturbances, such as anxiety and depression, and are more readily able to solve problems.

This chapter offers ten rational philosophical standpoints that are good for your psychological health. Read them, re-read them, think them through, and test out acting upon them to see for yourself.

Assuming Emotional Responsibility: \lou Feel the Way \lou Think

Bad or unfortunate things, such as splitting up from a partner, being made redundant, or having a car accident, can happen to anyone. You may reasonably have negative feelings in response to such events. Experiencing extreme sadness or annoyance in the face of misfortune is wholly understandable.

In some instances, bad things occur through no fault of your own. In other cases, you may have some personal responsibility. We don’t suggest that you blame yourself for every bad thing that comes your way. However, try to assess a given situation and determine whether you have any Legitimate responsibility For its development and look for a resolution.

Even if you’re not personally responsible for a negative event, you can still take responsibility for your emotional and behavioural Responses To the event. People who deny their part in creating their own emotional problems in the face of negative events don’t recognise how their thoughts and actions can make a bad situation worse. They hand over their personal power to make things better by waiting passively for someone or something to step into the breach.

When you hold an attitude of personal responsibility for your feelings and actions, you’re more able to find creative solutions, and your belief in your ability to cope with adversity is heightened. You empower yourself by focusing on your ability to influence the way you feel even if you can’t control events.

On a cheerier note, when good things happen, you can also assess the extent to which they’re a result of your own efforts – and then give yourself credit where due. You can appreciate good fortune without sabotaging your positive feelings with worries that your luck may run out.

Thinking Flexibly

Making demands and commands – thinking in terms of ‘must’, ‘should’, and ‘have to’ – about yourself, the world around you, and other people has a fundamental problem: Such thinking limits your flexibility to adapt to reality. The human capacity to adapt creatively to what’s going on is one of the hallmarks of the species’ success. However, humans are fallible, and the world continues to be an imperfect place. Insisting ‘It shouldn’t be this way!’ can leave you irate, depressed, or anxious and much less able to focus on how to cope with and adapt to reality.

Although circumstances may well be Desirable, preferable, And even Better \i The situation were different, they don’t Have To be a particular way. Accepting reality and striving to improve it where wise and achievable can help you save your energy for creative thought and action. Refer to Chapter 2 for more on demands, and Chapter 12 for more on developing realistic attitudes towards yourself.

Valuing \lour Individuality

You can express your individuality in many ways, such as in your dress sense, musical tastes, political opinions, or choice of career. Yet perhaps you’re hesitant to express your individuality openly because you fear the reaction of others. People who develop the ability to value their idiosyncrasies and to express them Respectfully Tend to be well-adjusted and content. Accepting that you’re an individual and have the right to live your life, just as other people have the right to live theirs, is a pretty good recipe for happiness.

As social animals, humans like to feel part of a group or social structure, and tend to be happier when interacting meaningfully with other humans. However, the ability to go against group mentality when it’s at odds with your own personal views or values is a tremendous skill. You can be both socially integrated and true to your values by accepting yourself as an individual and by being a selective non-conformist. Check out Chapter 12 for more on accepting yourself.

Accepting That Life Can Be Unfair

Sometimes, life’s just plain unfair. Sometimes, people treat you unjustly and nothing gets done to put the balance right. Bad things happen to the nicest of people, and people who don’t seem to have done a deserving thing in their lives get a winning ticket. On top of being unfair, life’s unpredictable and uncertain a great deal of the time. And really, that’s just the way life is.

What can you do? You can whine and moan and make yourself thoroughly miserable about the lamentable state of the world. Or you can accept things and get on with the business of living. No matter how much you insist that the world should be fair and you should be given certainty about how things are going to pan out, you ain’t going to get it.

Life’s unfair to pretty much Everyone From time to time – in which case, perhaps things aren’t as desperately unfair as you thought. If you can accept the cold hard reality of injustice and uncertainty, you’re far more likely to be able to bounce back when life slaps you in the face with a wet fish. You’re also likely to be less anxious about making decisions and taking risks. You can still strive to play fair yourself, but if you accept that unfairness exists you may be less outraged and less horrified if and when justice simply doesn’t prevail.

Understanding That Approval from Others Isn’t Necessary

Receiving approval from someone important to you is nice. Getting a bit of praise from a boss or a friend can feel good. But if you believe that you Need The approval of significant others or, indeed, everyone you meet, then you probably spend a lot of time feeling unhappy and unsure of yourself. Many people get depressed because they believe they’re only as good as the opinions others hold of them. These people can’t feel good about themselves unless they get positive feedback or reassurance from others.

Accept yourself, independent of overt approval from other people in your life. Having a Preference For being liked, appreciated, and approved of by others – but not believing that you Need Approval – means that your self-opinion can be stable and you can weather disapproval. You may still behave in ways that are more likely to generate approval than disapproval, but you can also assert yourself without fear. You can consider praise and compliments a bonus rather than something you must cling to and work over-hard to maintain.

If you hold the belief that you Need Rather than Desire Approval, you may pay emotionally for it somewhere along the line. You’re likely to feel anxious about whether approval’s forthcoming – and when you get approval you may worry about losing it. If you fail to get obvious approval or – horror of all horrors – someone criticises you, you’re likely to put yourself down and make yourself depressed. Refer to Chapter 9 for more on combating anxiety, and Chapter 10 for tackling depression.

You cannot please all the people all the time – and if that’s what you try to do, you’re almost certainly going to be overly passive. If you can take the view that disapproval isn’t the end of the world, not intolerable, and not an indication that you’re less than worthy, you can enjoy approval when you get it and still accept yourself when you don’t.

Realising LoVe’s besirable, Not Essential

Some people would rather be in any relationship – even an unsatisfying or abusive one – than in no relationship at all. This need may stem from a belief that they can’t cope with feelings of loneliness or get through life in general if

They’re alone. Other people consider themselves worthy or lovable only when they’re reassured by being in a relationship.

Romantic relationships Can Enhance your enjoyment of life, but they’re not essential for you to enjoy life. Holding this attitude can help you to feel good about yourself when you’re not part of a couple and may lead you to make more discerning partner choices in future since you will choose, rather than be compelled. Believing that your basic lovability is relatively constant, regardless of whether a significant other actively loves you, can help you to feel secure within a relationship.

People who strongly Prefer Having a partner and yet believe that they can survive a break up tend to experience little romantic jealousy. Jealousy can be a big obstacle to relationship satisfaction – jealous people tend to believe that they Must Keep their partner and end up focusing on signs of infidelity or waning interest rather than on the pleasure of the relationship. Jealousy’s turned many a relationship sour. A jealous partner can end up alienating the other person through constant reassurance-seeking or monitoring, leaving both members of the couple feeling that mutual trust doesn’t exist between them.

Preferring Instead of Demanding To have a relationship helps you to retain your independence and individuality. Then when you Are In a relationship, you’re less likely to fall into the trap of trying to be the perfect partner -which means you can continue to attend to your own interests while being able to negotiate compromises when appropriate.

Tolerating Short-Term discomfort

Healthy, robust, and successful people are often able to tolerate temporary discomfort in the pursuit of longer-term goals. They practise self-denial and delay gratification when doing so is in their long-term interests. These people are the ones who are able to eat healthily, exercise regularly, save money, study effectively, and so on.

You Can Experience intense pleasure in the present and the future, but often some degree of pain and effort Today Are necessary to win you greater pleasure Tomorrow. This will be true for many of the achievements you’ve already made in life. Putting up with temporary discomfort is also going to be crucial in Reducing Painful feelings of anxiety and depression. Refer to Chapters 9, 10, and 11 for more on overcoming these problems.

Enacting Enlightened Self-Interest

Enlightened self-interest Is about putting yourself first most of the time and one, two, or a small handful of selected others a very close second. Enlightened self-interest is about looking after your own needs and interests while also being mindful of the needs of your loved ones and other people living on the planet.

So why put yourself first? When you reach a certain age, you need to look after yourself because nobody else is going to do so for you. If you can keep yourself healthy and content, you are better able to turn your attention to caring for the people in your life that you love.

Many people make the mistake of always suppressing their own needs and end up tired, unhappy, or ill. People may think they’re doing the right thing by putting others first all the time, but in fact they’re left with very little to give.

Of course you Will Experience times when putting someone else’s needs before your own and making personal sacrifices is a good choice. For example, parents frequently put the welfare of their children before their own. But you must still make space for your own pursuits too.

If you’re starting to get concerned that ‘self-interest’ translates to ‘selfish beast’, stop! To clarify: Self-interest involves taking responsibility for looking after yourself because you understand that you’re worth taking care of. Self-interest means being able to care for others very deeply. When you’re self-interested, you’re able to meet your own needs and take a keen interest in the welfare of other people in the world around you. You can also determine when you’re going to put yourself Second For a period of time because someone else’s need is greater than your own – which is where the ‘enlightened’ part comes into play.

Selfishness is not – we stress, Not! – The same animal as self-interest. Ultimately, selfish people put their own wants and needs first, To the exclusion and detriment of other people. Selfishness is much less about taking responsibility for looking after yourself and much more about demanding that you get what you want, when you want, and to hell with everybody else. The two concepts are very different – so don’t be scared. Head to Chapter 16 for more on building a lifestyle that promotes taking care of yourself.

Pursuing Interests and Acting Consistently u/ith \lour Values

Loads of evidence indicates that people are happier and healthier if they pursue interests and hobbies. Have you let your life become dominated by work or chores at home, and do you spend your evenings sitting in front of the television as a means of recharging? If your answer to this question is ‘Yes!’, then you’re in extremely good, but not optimally healthy, company.

One of the arts of maximising your happiness is to pursue personally meaningful goals, such as furthering your education, participating in sports and exercise, developing skills, improving relationships, or acting in ways that contribute to the sort of world you’d like to live in, for example by doing some voluntary work. Try to structure your life to ensure that you have some time for personally meaningful pursuits. Check that the things you do in life reflect what you believe is important.

As far as we can tell, life isn’t a dress rehearsal. Will you really look back and regret missing a bit of TV because you dragged yourself out to spend time on a hobby, to exercise, to enjoy a night out with your friends, or to participate in some charity work?

Healthy and productive people tend to be prepared to tolerate a degree of risk and uncertainty. Demanding certainty and guarantees in an uncertain world is a sure-fire recipe for worry and inactivity. Safety comes at a cost – fewer rewards, less excitement, fewer new experiences.

The fact that you don’t know what the future holds is grounds for Calculated risks And Experiments, Not avoidance, reassurance-seeking, or safety precautions. You can make educated decisions and take calculated risks, but if you accept that 100 per cent certainty is exceptionally rare, you can reduce undue anxiety and worry.