SM chapter 1 eorema (límite en coordenadas polares) Si 𝑓:𝑈 ⊂ ℝ2 ⟶ ℝ una función de dos variables, 𝑥⃗𝑜 = (0,0) un punto de acumulación del 𝐷𝑜𝑚(� PDF

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.8.1. Definición. Sea 𝑓:𝑈 ⊂ ℝ𝑛 ⟶ ℝ un función definida en una vecindad de
𝑥⃗𝑜(𝑉(𝑥𝑜, 𝜌)). Diremos que la función 𝑓 es continua en 𝑥⃗𝑜,si para cada 𝜀 > 0, existe
otro número 𝜌 > 0⁄𝑠𝑖 ‖𝑥 − 𝑥𝑜‖ < 𝜌 entonces |𝑓(𝑥) − 𝑓(𝑥𝑜
)| < �...

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1 Physics and Measurement ANSWERS TO QUESTIONS

CHAPTER OUTLINE 1.1 1.2 1.3 1.4 1.5 1.6

Standards of Length, Mass, and Time Matter and Model-Building Dimensional Analysis Conversion of Units Estimates and Order-ofMagnitude Calculations Significant Figures

* An asterisk indicates an item new to this edition. Q1.1

Density varies with temperature and pressure. It would be necessary to measure both mass and volume very accurately in order to use the density of water as a standard.

Q1.2

(a) 0.3 millimeters (b) 50 microseconds (c) 7.2 kilograms

*Q1.3

In the base unit we have (a) 0.032 kg (b) 0.015 kg (c) 0.270 kg (d) 0.041 kg (e) 0.27 kg. Then the ranking is c=e>d>a>b

Q1.4

No: A dimensionally correct equation need not be true. Example: 1 chimpanzee = 2 chimpanzee is dimensionally correct. Yes: If an equation is not dimensionally correct, it cannot be correct.

*Q1.5

The answer is yes for (a), (c), and (f ). You cannot add or subtract a number of apples and a number of jokes. The answer is no for (b), (d), and (e). Consider the gauge of a sausage, 4 kgⲐ2 m, or the volume of a cube, (2 m)3. Thus we have (a) yes (b) no (c) yes (d) no (e) no (f ) yes

*Q1.6

41 € ≈ 41 € (1 LⲐ1.3 €)(1 qtⲐ1 L)(1 galⲐ4 qt) ≈ (10Ⲑ1.3) gal ≈ 8 gallons, answer (c)

*Q1.7

The meterstick measurement, (a), and (b) can all be 4.31 cm. The meterstick measurement and (c) can both be 4.24 cm. Only (d) does not overlap. Thus (a) (b) and (c) all agree with the meterstick measurement.

*Q1.8

0.02(1.365) = 0.03. The result is (1.37 ± 0.03) × 107 kg. So (d) 3 digits are significant.

SOLUTIONS TO PROBLEMS Section 1.1 P1.1

Standards of Length, Mass, and Time

4 3 4 Modeling the Earth as a sphere, we find its volume as π r 3 = π ( 6.37 × 10 6 m) = 1.08 × 10 21 m 3. 3 3 m 5.98 × 10 24 kg Its density is then ρ = = 5 .52 × 10 3 kg m 3 . This value is intermediate 21 3 = V 1.08 × 10 m between the tabulated densities of aluminum and iron. Typical rocks have densities around 2 000 to 3 000 kgⲐm3. The average density of the Earth is significantly higher, so higher-density material

2

P1.2

Chapter 1 2 With V = ( base area ) ( height ) V = (π r ) h and ρ =

ρ=

m , we have V

⎛10 9 mm 3 ⎞ 1 kg m = π r 2h π (19.5 mm )2 (39.0 mm ) ⎜⎝ 1 m3 ⎟⎠

ρ = 2.15 × 104 kg m3 . P1.3

and ρ gold =

*P1.4

m for both. Then ρiron = 9.35 kg V V ⎛ 19.3 × 10 3 kg/m 3 ⎞ mgold = 23.0 kg . and mgold = 9.35 kg ⎜ = ⎝ 7.86 × 10 3 kg/m3 ⎟⎠ 9.35 kg

Let V represent the volume of the model, the same in ρ = m gold . Next, ρgold V ρiron

ρ = m / V and V = ( 4 / 3)π r 3 = ( 4 / 3)π (d / 2) 3 = π d 3 / 6 where d is the diameter. Then ρ = 6 m / π d = 3

17

3

6 (1 .67 × 10 −27 kg ) = 2 .3 × 1017 kg/m3 π (2 .4 × 10 −15 m ) 3 3

3

2.3 ×10 kg/m /(11.3 ×10 kg/m ) = it is 20 × 1012 times the density of lead . P1.5

4 3 4 For either sphere the volume is V = π r3 and the mass is m = ρV = ρ π r . We divide 3 3 this equation for the larger sphere by the same equation for the smaller: mᐉ ρ 4π ᐉr3 3 ᐉr3 = = = 5. ms ρ 4π rs3 3 rs3 Then rᐉ = rs 3 5 = 4 .50 cm(1 .71 ) = 7 .69 cm .

Section 1.2 P1.6

From the figure, we may see that the spacing between diagonal planes is half the distance between diagonally adjacent atoms on a flat plane. This diagonal distance may be obtained from the Pythagorean theorem, Ldiag = L2 + L2 . Thus, since the atoms are separated by a distance 1 2 2 L = 0.200 nm, the diagonal planes are separated by L + L = 0. 141 nm . 2

Section 1.3 P1.7

P1.8

Matter and Model-Building

Dimensional Analysis

(a)

This is incorrect since the units of [ ax ] are m s , while the units of [ v] are m s .

(b)

−1 This is correct since the units of [ y] are m, and cos ( kx ) is dimensionless if [ k ] is in m .

2

(a)

Circumference has dimensions of L.

(b)

Volume has dimensions of L3 .

(c)

Area has dimensions of L2 . Expression (i) has dimension L ( L2 )

1/ 2

2

= L2 , so this must be area (c).

Expression (ii) has dimension L so it is (a)

Physics and Measurement

P1.9

3

Inserting the proper units for everything except G, ⎡ kg m ⎤ G [kg ] . ⎢⎣ s2 ⎥⎦ = [ m]2 2

2 Multiply both sides by [ m ] and divide by [ kg ] ; the units of G are 2

Section 1.4 P1.10

m3 . kg ⋅ s2

Conversion of Units

Apply the following conversion factors: 9 1 in = 2. 54 cm, 1 d = 86 400 s, 100 cm = 1 m, and 10 nm = 1 m −2 9 ⎛ 1 in day⎞ ( 2.54 cm in )( 10 m cm) ( 10 nm m ) = 9.19 nm s . ⎠ ⎝ 32 86 400 s day This means the proteins are assembled at a rate of many layers of atoms each second!

P1.11

Conceptualize: We must calculate the area and convert units. Since a meter is about 3 feet, we should expect the area to be about A ≈ 30 m 50 m = 1500 m 2 .

(

)(

)

Categorize: We model the lot as a perfect rectangle to use Area = Length × Width. Use the conversion:1 m = 3.281 ft. 1m ⎞ 1m ⎞ = 1 390 m 2 = 1 .39 × 10 3 m 2 . Analyze: A = LW = (100 ft) ⎛ ( 150 ft ) ⎛ ⎝ 3. 281 ft ⎠ ⎝ 3.281 ft ⎠ Finalize: Our calculated result agrees reasonably well with our initial estimate and has the proper units of m 2 . Unit conversion is a common technique that is applied to many problems. P1.12

(a)

V = (40.0 m ) (20.0 m ) (12.0 m ) = 9. 60 × 10 3 m3 V = 9.60 × 103 m3 (3.28 ft 1 m ) = 3. 39 × 105 ft3 3

(b)

The mass of the air is m = ρair V = (1. 20 kg m 3) ( 9.60 × 10 3 m 3 ) = 1.15 × 10 4 kg. The student must look up weight in the index to find Fg = mg = (1.15 × 10 4 kg ) (9.80 m s 2 ) = 1.13 × 10 5 N . Converting to pounds, Fg = (1 .13 ×10 5 N ) (1 lb 4.45 N ) = 2 .54 × 10 4 lb .

*P1.13

The area of the four walls is (3.6 + 3.8 + 3.6 + 3.8)m (2.5 m) = 37 m2. Each sheet in the book has area (0.21 m) (0.28 m) = 0.059 m2. The number of sheets required for wallpaper is 37 m2Ⲑ0.059 m2 = 629 sheets = 629 sheets(2 pagesⲐ1 sheet) = 1260 pages. The pages from volume one are inadequate, but the full version has enough pages.

4

P1.14

Chapter 1

(a)

Seven minutes is 420 seconds, so the rate is r=

(b)

30 .0 gal = 7. 14 × 10 −2 gal s . 420 s

Converting gallons first to liters, then to m 3, ⎛ 3.786 L ⎞ ⎛ 10−3 m 3⎞ 2 r = (7.14 × 10 − gal s) ⎜ ⎝ 1 gal ⎟⎠ ⎜⎝ 1 L ⎟⎠ r = 2. 70 × 10− 4 m3 s .

(c)

At that rate, to fill a 1-m3 tank would take ⎛ 1 m3 ⎞⎛ 1 h ⎞ = 1 .03 h . t=⎜ ⎝ 2.70 × 10 −4 m 3 s ⎠⎟ ⎜⎝ 3 600 ⎠⎟

P1.15

From Table 14.1, the density of lead is 1. 13 × 10 4 kg m 3 , so we should expect our calculated value to be close to this number. This density value tells us that lead is about 11 times denser than water, which agrees with our experience that lead sinks. Density is defined as mass per volume, in ρ =

m . We must convert to SI units in the calculation. V

3

ρ=

23.94 g ⎛ 1 kg ⎞ ⎛ 100 cm⎞ 23 .94 g ⎛ 1 kg ⎞ ⎛ 1 000 000 cm 3 ⎞ 3 4 = ⎟⎠ = 1 .14 ×10 kg m 2.10 cm3 ⎜⎝1000 g ⎟⎠ ⎝⎜ 1 m ⎟⎠ 2.10 cm3 ⎜⎝ 1000 g ⎟⎠ ⎜⎝ 1 m3

At one step in the calculation, we note that one million cubic centimeters make one cubic meter. Our result is indeed close to the expected value. Since the last reported significant digit is not certain, the difference in the two values is probably due to measurement uncertainty and should not be a concern. One important common-sense check on density values is that objects which sink in water must have a density greater than 1 g cm3, and objects that float must be less dense than water. P1.16

The weight flow rate is 1 200

P1.17

(a) (b)

P1.18

ton ⎛ 2 000 lb ⎞ ⎛ 1 h ⎞ ⎛ 1 min ⎞ = 667 lb s . h ⎝ ton ⎠ ⎝ 60 min ⎠ ⎝ 60 s ⎠

⎛ 8 × 1012 $ ⎞ ⎛ 1 h ⎞ ⎛ 1 day ⎞ ⎛ 1 yr ⎞ ⎜⎝ 1 000 $ s ⎟⎠ ⎜⎝ 3 600 s ⎟⎠ ⎝ 24 h ⎠ ⎜⎝ 365 days⎟⎠ = 250 years The circumference of the Earth at the equator is 2π ( 6. 378 × 103 m) = 4. 01× 107 m. The length of one dollar bill is 0.155 m so that the length of 8 trillion bills is 1. 24× 1012 m. Thus, the 8 trillion dollars would encircle the Earth 1.24 × 1012 m = 3 .09 × 10 4 times . 4.01 × 1 07 m

⎡ (13.0 acres) (43 560 ft2 acre)⎦⎤ 1 Bh = ⎣ ( 481 ft) 3 3 = 9.08 × 10 7 ft 3 ,

V=

h

B

or −2 m 3⎞ 3 ⎛ 2.83 × 10 7 V = (9.08 × 10 ft ) ⎜ 3 ⎟⎠ ⎝ 1 ft

= 2.57 × 10 6 m 3

FIG. P1.18

Physics and Measurement

P1.19

Fg = ( 2 .50 tons block )( 2 .00 ×10 6 blocks) ( 2 000 lb ton) = 1. 00 × 1010 lbs

P1.20

(a)

5

⎛d ⎞ 300 ft ⎛ ⎞ = 6.79 × 10 −3 ft , or d nucleus, scale = d nucleus, real⎜ atom, scale ⎟ = ( 2 .40 × 10 −15 m) ⎝ d 1.06 ×10 −10 m ⎠ ⎝ atom, real ⎠ dnucleus, scale = ( 6. 79 × 10− 3 ft) ( 304. 8 mm 1 ft) = 2.07 mm 3

(b)

3

3 ⎞ ⎛ d ⎛ 1.06 × 10 −10 m⎞ 4 π ratom / 3 ⎛ ratom ⎞ V atom = =⎜ = ⎜ atom ⎟ = ⎜ 3 ⎟ ⎝ 2.40 × 10 −15 m ⎟⎠ Vnucleus 4π rnucleus / 3 ⎝ rnucleus ⎠ ⎝ dnucleus ⎠

= 8 .62 × 10

13

3

times as large

V 3 .78 ×10 −3 m3 = = 1. 51 × 10 −4 m (or 151 µ m) 25. 0 m 2 A

P1.21

V = At so t =

P1.22

(a)

2 ⎛ (6 .37 × 10 6 m )(100 cm m )⎞ ⎛ rEarth ⎞ AEarth 4π rEarth = = = ⎜ ⎟ = 13.4 2 ⎜r 1. 74 × 108 cm A Moon 4π rMoon ⎝ Moon ⎟⎠ ⎠ ⎝

(b)

3 ⎛ (6.37 × 106 m ) (100 cm m )⎞ / 3 ⎛ rEarth ⎞ V Earth 4π rEarth =⎜ = =⎜ ⎟ = 49.1 3 ⎟ 1 .74 ×10 8 cm VMoon 4 π rMoon / 3 ⎝ rMoon ⎠ ⎝ ⎠

2

2

3

3

P1.23

To balance, mFe = mAl or ρ FeVFe = ρ AlVAl ⎛ 4 ρFe ⎜ 4 ⎟⎞ π rFe 3 = ρAl ⎛ ⎞ π rAl 3 ⎝ 3⎠ ⎝3⎠ ⎛ρ ⎞ rAl = rFe ⎜ Fe ⎟ ⎝ρ ⎠ Al

P1.24

1 /3

7 .86 ⎞ = (2 .00 cm ) ⎛ ⎝ 2. 70⎠

1 /3

= 2. 86 cm .

The mass of each sphere is mAl = ρAl VAl =

4 π ρAl rAl 3

m Fe = ρFeV Fe =

4 π ρFe rFe 3 . 3

3

and

Setting these masses equal, 4 π ρ Al rAl 3 4π ρFe rFe 3 and r = r 3 ρ Fe . = Al Fe ρ Al 3 3 The resulting expression shows that the radius of the aluminum sphere is directly proportional to the radius of the balancing iron sphere. The sphere of lower density has larger radius. The ρ Fe fraction is the factor of change between the densities, a number greater than 1. Its cube root ρ Al is a number much closer to 1. The relatively small change in radius implies a change in volume sufficient to compensate for the change in density.

6

Chapter 1

Section 1.5 P1.25

Estimates and Order-of-Magnitude Calculations

Model the room as a rectangular solid with dimensions 4 m by 4 m by 3 m, and each ping-pong ball as a sphere of diameter 0.038 m. The volume of the room is 4 × 4 × 3 = 48 m3, while the volume of one ball is 4 π ⎛ 0.038 m ⎞ = 2.87 ×10 −5 m 3. ⎠ 2 3 ⎝ 48 Therefore, one can fit about ~ 106 ping-pong balls in the room. 2.87 × 10−5 3

As an aside, the actual number is smaller than this because there will be a lot of space in the room that cannot be covered by balls. In fact, even in the best arrangement, the so-called “best packing fraction” is 1π 2 = 0. 74 so that at least 26% of the space will be empty. Therefore, the 6 above estimate reduces to 1.67 × 10 6 × 0 .740 ~ 10 6 . P1.26

A reasonable guess for the diameter of a tire might be 2.5 ft, with a circumference of about 8 ft. Thus, the tire would make 50 000 mi 5 280 ft mi 1 rev 8 ft = 3 × 10 7 rev ~ 107 rev .

P1.27

Assume the tub measures 1.3 m by 0.5 m by 0.3 m. One-half of its volume is then

(

)(

)(

)

V = ( 0.5 ) (1. 3 m ) (0. 5 m ) (0. 3 m ) = 0.10 m 3 . The mass of this volume of water is 2 m water = ρ waterV = ( 1 000 kg m3 )( 0.10 m3 ) = 100 kg ~ 10 kg .

Pennies are now mostly zinc, but consider copper pennies filling 50% of the volume of the tub. The mass of copper required is m copper = ρ copperV = (8 920 kg m3 ) (0.10 m3 ) = 892 kg ~ 10 3 kg . *P1.28

The time required for the task is 1 bad yr ⎞ ⎛ 1 s ⎞ 1 h ⎞ ⎛ 1 working day ⎞ ⎛ 109 $ ⎜ ⎟ ⎛ = 58 yr ⎜ ⎝ ⎝ ⎠ ⎝ 300 working days ⎟⎠ ⎝ 1 $⎠ 3600 s ⎠ 16 h Since you are already around 20 years old, you would have a miserable life and likely die before accomplishing the task. You have better things to do. Say no.

P1.29

Assume: Total population = 107; one out of every 100 people has a piano; one tuner can serve about 1 000 pianos (about 4 per day for 250 weekdays, assuming each piano is tuned once per year). Therefore, ⎛ 1 tuner ⎞ ⎛ 1 piano ⎞ # tuners ~⎜ (107 people) = 100 tuners . ⎝ 1 000 pianos ⎟⎠ ⎜⎝ 10 0 people ⎟⎠

Section 1.6 P1.30

Significant Figures

METHOD ONE We treat the best value with its uncertainty as a binomial ( 21.3 ± 0.2) cm ( 9.8± 0.1) cm, A = [21.3 ( 9.8 ) ± 21.3 ( 0. 1) ± 0.2 ( 9.8 ) ± ( 0.2 ) ( 0. 1 )] cm 2 . The first term gives the best value of the area. The cross terms add together to give the uncertainty and the fourth term is negligible. A = 209 cm 2 ± 4 cm 2 . METHOD TWO

Physics and Measurement

P1.31 P1.32

(a)

3

(b)

4

3

(c)

(d)

7

2

r = (6.50 ± 0.20) cm = ( 6.50 ± 0.20 ) × 10 −2 m m = (1 .85 ±0 .02) kg m ρ= 4 ( 3 ) π r3 also, δ ρ = δ m + 3δ r . m r ρ In other words, the percentages of uncertainty are cumulative. Therefore,

δ ρ 0 .02 3 (0 .20 ) = + = 0 .103 , 6 .50 ρ 1 .85 ρ=

1.85

( )π ( 6. 5 × 10 4 3

−2

m)

3

3 3 = 1. 61× 10 kg m

and

ρ ± δ ρ = ( 1. 61± 0. 17) × 103 kg m 3 = (1. 6 ± 0. 2) × 103 kg m 3. P1.33

(a)

756.?? 37.2? 0.83 + 2.5? 796./ 5 / 3 = 797

P1.34

(b)

0.003 2 (2 s.f. ) ×356 .3 (4 s.f.) =1.140 16 = ( 2 s.f.)

(c)

5.620 (4 s.f.) × π ( >4 s.f.) = 17.656=( 4 s.f.)

1.1

17.66

We work to nine significant digits: ⎛365.242 199 d ⎞ ⎛ 24 h ⎞ ⎛ 60 min ⎞ ⎛ 60 s ⎞ 1 yr =1 yr ⎜ ⎟⎠ ⎝ 1 d ⎠ ⎝ 1 h ⎠ ⎝1 min ⎠ = 31 556 926. 0 s . ⎝ 1 yr

*P1.35

The tax amount is $1.36 − $1.25 = $0.11. The tax rate is $0.11Ⲑ$1.25 = 0.088 0 = 8.80%

*P1.36

(a)

We read from the graph a vertical separation of 0.3 spaces = 0.015 g .

(b)

Horizontally, 0.6 spaces = 30 cm2 .

(c)

Because the graph line goes through the origin, the same percentage describes the vertical and the horizontal scatter: 30 cm2Ⲑ380 cm2 = 8% .

(d)

Choose a grid point on the line far from the origin: slope = 0.31 gⲐ600 cm2 = 0.000 52 gⲐcm2 = (0.000 52 gⲐcm2)(10 000 cm2Ⲑ1 m2) = 5.2 g/m2 .

(e)

For any and all shapes cut from this copy paper, the mass of the cutout is proportional to its area. The proportionality constant is 5.2 g/m2 ± 8%, where the uncertainty is estimated.

(f )

This result should be expected if the paper has thickness and density that are uniform i hi h i l i h l i h ld i f h i

8

Chapter 1

*P1.37

15 players = 15 players (1 shift Ⲑ1.667 player) = 9 shifts

*P1.38

Let o represent the number of ordinary cars and s the number of trucks. We have o = s + 0.947s = 1.947s, and o = s + 18. We eliminate o by substitution: s + 18 = 1.947s 0.947s = 18 and s = 18Ⲑ0.947 = 19 .

*P1.39

Let s represent the number of sparrows and m the number of more interesting birds. We have sⲐm = 2.25 and s + m = 91. We eliminate m by substitution: m = sⲐ2.25 s + sⲐ2.25 = 91 1.444s = 91 s = 91Ⲑ1.444 = 63 .

*P1.40

For those who are not familiar with solving equations numerically, we provide a detailed solution. It goes beyond proving that the suggested answer works. 4 3 The equation 2 x − 3 x + 5 x − 70 = 0 is quartic, so we do not attempt to solve it with algebra. To find how many real solutions the equation has and to estimate them, we graph the expression:

x y = 2 x4 − 3x3 + 5 x − 70

−3

−2

−1

0

1

2

3

4

158

−24

−70

−70

−66

−52

26

270

We see that the equation y = 0 has two roots, one around x = −2.2 and the other near x = +2.7. To home in on the first of these solutions we compute in sequence: When x = −2.2, y = −2.20. The root must be between x = −2.2 and x = −3. When x = −2.3, y = 11.0 . The root is between x = −2.2 and x = −2.3. When x = −2.23, y = 1.58 . The root is between x = −2.20 and x = −2.23. When x = −2.22, y = 0.301. The root is between x = −2.20 and −2.22. When x = −2. 215, y = −0.331. The root is between x = −2.215 and −2.22. We could next try x = −2.218, but we already know to three-digit precision that the root is x = −2.22. *P1.41

y

x

FIG. P1.40

sinθ We require sin θ = −3 cos θ, or = −3 , or tan θ = −3. tan θ cosθ For tan− 1( − 3) = arc tan( − 3) , your calculator may return −71.6°, but this angle is not between 0° and 360° as the problem 0 requires. The tangent function is negative in the second quadrant (between 90° and 180°) and in the fourth quadrant (from 270° to 360°). The solutions to the equation are then 360° − 71.6° = 288° and 180° − 71.6° = 108° .

θ 360°

FIG. P1.41

Physics and Measurement

*P1.42

9

We draw the radius to the initial point and the radius to the final point. i The angle θ between these two radii has its sides perpendicular, right side 35.0° to right side and left side to left side, to the 35° angle between the original R f and final tangential directions of travel. A most useful theorem from N geometry then identifies these angles as equal: θ = 35°. The whole θ W E circumference of a 360° circle of the same radius is 2π R. By proportion, S 2 840 m . then π R = 360° 35° FIG. P1.42 360 ° 840 m 840 m 3 = = 1.38 × 10 m R= 0.611 2 π 35° We could equally well say that the measure of the angle in radians is 840 m ⎛ 2π radians⎞ = 0.611 rad = θ = 35° = 35° . ⎝ 360° ⎠ R Solving yields R = 1. 38 km.

*P1.43

Mass is proportional to cube of length: m = kᐉ3

mƒ Ⲑmi = (ᐉf Ⲑᐉi)3.

Length changes by 15.8%: ᐉf = ᐉi + 0.158 ᐉi = 1.158 ᐉi . Mass increase: mf = mi + 17.3 kg. mf 3 = 1.158 = 1.553 Eliminate by substitution: m f −17.3 kg mf = 1.553 mf − 26.9 kg *P1.44

26.9 kg = 0.553 mf

mf = 26.9 kg Ⲑ0.553 = 48. 6 kg .

We use substitution, as the most generally applicable method for solving simultaneous equations. We substitute p = 3q into each of the other two equations to eliminate p: ⎧3qr = qs ⎪ 1 2 1 2. ⎨1 2 ⎪⎩ 2 3qr + 2 qs = 2 qt 2 ⎧3 r = s 3 r2 + ( 3 r ) = t 2 . We solve for the These simplify to ⎨ 2 2 2 . We substitute to eliminate s: 12r2 = t 2 ⎩3 r + s = t combination t : r t2 = 12 . r2

*P1.45

t = either 3.46 or ...