SECTION 3.4 129
k= ln(T0−Ta)/t −ln(T−Ta)/t = 0.1 ln(T0−Ta)−0.1 ln(T−Ta) = 0.0626
18. (a) a= 2.5, σa= 0.1, b= 0.05, σb= 0.01, w= 1.2, σw= 0.1, v=a+bw = 2.56
∂v
(b) σv=s∂v
∂a2
σ2
a+∂v
∂b 2
σ2
b+∂v
∂w 2
σ2
w,∂v
∂a = 1, ∂v
∂b =w= 1.2∂v
∂w =b= 0.05
19. (a) No, they both involve the quantities hand r.
Page 129
130 CHAPTER 3
20. X= 5.0, σX= 0.2, Y= 10.0, σY= 0.5
(b) U= 2Y/√X= 8.94, ln U= ln 2 −0.5 ln X+ ln Y
(c) U=X2+Y2= 125, ln U= ln(X2+Y2)
Page 130
SECTION 3.4 131
21. µ= 1.49, τ= 35.2, στ= 0.1, h= 12.0, σh= 0.3, V=τh/µ =τh/1.49 = 283.49,
ln V= ln τ+ ln h−ln 1.49
22. P1= 15.3, σP1= 0.2, P2= 25.8, σP2= 0.1, P3=√P1P2= 19.87, ln P3= 0.5 ln P1+ 0.5 ln P2
23. p= 4.3, σp= 0.1, q= 2.1, σq= 0.2, f=pq/(p+q) = 1.41, ln f= ln p+ ln q−ln(p+q)
Page 131
132 CHAPTER 3
24. (a) P= 224.51, σP= 0.04, V= 11.237, σV= 0.002, T=P V/8.31 = 303.59, ln T= ln P+ ln V−ln 8.31
(b) P= 224.51, σP= 0.04, T= 289.33, σT= 0.02, V= 8.31T/P = 10.71, ln V= ln 8.31 + ln T−ln P
(c) V= 11.203, σV= 0.002, T= 289.33, σT= 0.02, P= 8.31T/V = 214.62, ln V= ln 8.31 + ln T−ln V
25. θ1= 0.216, σθ1= 0.003, θ2= 0.456, σθ2= 0.005, n=sin θ1
sin θ2
= 0.487, ln n= ln(sin θ1)−ln(sin θ2)
Page 132
SECTION 3.4 133
26. (a) l= 10.0, σl/l = 0.005, d= 10.4, σd/d = 0.005, R=kl/d2= 0.0925k, ln R= ln k+ ln l−2 ln d
(b) σln R=rσl
+ 4 σd
27. F= 750, σF= 1, R= 0.65, σR= 0.09, L0= 23.7, σL0= 0.2, L1= 27.7, σL1= 0.2,
28. T= 50, k= 0.032, T0= 73.1, σT0= 0.1, Ta= 37.5, σTa= 0.2,
134 CHAPTER 3
∂t
∂Ta
=31.25
50 −Ta−31.25
T0−Ta
= 1.62219
29. r= 0.8, σr= 0.1, h= 1.9, σh= 0.1
(a) S= 2πr(h+ 2r) = 17.59, ln S= ln(2π) + ln r+ ln(h+ 2r)
(b) V=πr2(h+ 4r/3) = 5.965, ln V= ln(π) + 2 ln r+ ln(h+ 4r/3)
(c) R=c2πr(h+ 2r)
πr2(h+ 4r/3) =c2h+ 4r
rh + 4r2/3= 2.95c, ln R= ln c+ ln(2h+ 4r)−ln(rh + 4r2/3)
Page 134
30. P3=P0.5
1P0.5
2. The relative uncertainties are σP1
P1
= 0.05 and σP2
P2
= 0.02.
31. R=kld−2. The relative uncertainties are σk
Supplementary Exercises for Chapter 3
1. (a) X= 25.0, σX= 1, Y= 5.0, σY= 0.3, Z= 3.5, σZ= 0.2. Let U=XY +Z.
∂X 2
∂Y 2
∂Z 2
(b) X= 25.0, σX= 1, Y= 5.0, σY= 0.3, Z= 3.5, σZ= 0.2. Let U=Z/(X+Y).
∂X 2
∂Y 2
∂Z 2
Page 135
136 CHAPTER 3
(c) X= 25.0, σX= 1, Y= 5.0, σY= 0.3, Z= 3.5, σZ= 0.2. Let U=pX(ln Y+Z).
∂X 2
∂Y 2
∂Z 2
(d) X= 25.0, σX= 1, Y= 5.0, σY= 0.3, Z= 3.5, σZ= 0.2. Let U=XeZ2
−2Y.
∂X 2
∂Y 2
∂Z 2
2. The relative uncertainty in Xis σln X= 0.05, the relative uncertainty in Yis σln Y= 0.10, and the
relative uncertainty in Zis σln Z= 0.15.
(a) Let U=XY /Z, so ln U= ln X+ ln Y−ln Z.
(b) Let U=√XY 2Z3, so ln U= 0.5 ln X+ ln Y+ 1.5 ln Z.
Page 136
SUPPLEMENTARY EXERCISES FOR CHAPTER 3 137
(c) Let U=XY
Z1/3
, so ln U= (1/3) ln X+ (1/3) ln Y−(1/3) ln Z.
3. (a) Let X,Y, and Zrepresent the measured lengths of the components, and let T=X+Y+Zbe the
(b) Let σbe the required uncertainty in Xand Y. Then σT=√σ2+σ2+σ2= 0.05. Solving for σyields
σ= 0.029 mm.
4. t1= 20, t2= 40, m1= 17.7, σm1= 1.7, m2= 35.9, σm2= 5.8,
r= (m2−m1)/(t2−t1) = 0.05m2−0.05m1= 0.91
5. (a) γ= 9800, η= 0.85, ση= 0.02, Q= 60, σQ= 1, H= 3.71, σH= 0.10
Page 137
138 CHAPTER 3
(b) The relative uncertainty is σP
P=0.073 ×106
1.854 ×106= 0.039 = 3.9%.
6. (a) p= 0.360, σp= 0.048, q= 0.250, σq= 0.043, PAB =pq = 0.090
(b) The relative uncertainty is σPAB
PAB
=0.020
0.090 = 0.22 = 22%.
PAB
∂p 2
∂q 2
Page 138
SUPPLEMENTARY EXERCISES FOR CHAPTER 3 139
(c) σPAB =s∂PAB
∂p 2
σ2
p+∂PAB
∂q 2
σ2
q,∂PAB
∂p =q= 0.25 ∂PAB
∂q =p= 0.36
7. (a) H= 4, c= 0.2390, ∆T= 2.75, σ∆T= 0.02, m= 0.40, σm= 0.01
(b) The relative uncertainty is σC
C=0.17
6.57 = 0.026 = 2.6%.
C=σln C=s∂ln C
∂∆T2
∂m 2
(c) σC=s∂C
∂∆T2
σ2
∆T+∂C
∂m2
σ2
m,∂C
∂∆T=0.956
m= 2.39 ∂C
∂m =−0.956(∆T)
m2=−16.4312
8. (a) The estimate of the resistance is the average of the 16 measurements, which is X= 52.37. The
Page 139
9. (a) Let Xbe the measured northerly component of velocity and let Ybe the measured easterly component
of velocity. The velocity of the earth’s crust is estimated with V=√X2+Y2.
Now X= 22.10, σX= 0.34 and Y= 14.3, σY= 0.32, so V=√22.102+ 14.32= 26.32, and
∂X 2
∂Y 2
V= 26.32 ±0.33 mm/year
(b) Let T be the estimated number of years it will take for the earth’s crust to move 100 mm.
10. (a) Let M1represent the estimated molar mass of the unknown gas, and let M2represent the molar mass
of carbon dioxide. Then M2= 44, and R= 1.66, σR= 0.03.
dM1
M1= 121.2±4.4 g/mol
(b) The relative uncertainty is σM1
M1
= 4.4/121.2 = 0.036 = 3.6%.
11. Let X1and X2represent the estimated thicknesses of the outer layers, and let Y1, Y2, Y3represent the
estimated thicknesses of the inner layers. Then X1=X2= 0.50, Y1=Y2=Y3= 6.25, σXi= 0.02 for
12. Let Xbe the estimated concentration in the first experiment and let Ybe the estimated concentration
in the second experiment. Then X= 43.94, σX= 1.18, and Y= 48.66, σY= 1.76.
13. (a) The relative uncertainty in λis σλ/λ =σln λ. The given relative uncertainties are σln V= 0.0001,
σln I= 0.0001, σln A= 0.001, σln l= 0.01, σln a= 0.01. ln λ= ln V+ ln I+ ln A−ln π−ln l−ln a.
14. Consider the cable to be made of four strands of each of three types of wire. Let X1=X2=X3=X4
Page 141
142 CHAPTER 3
(a) Let Ddenote the strength of the cable estimated by the ductile wire method.
15. (a) Yes, since the strengths of the wires are all estimated to be the same, the sum of the strengths is
the same as the strength of one of them multiplied by the number of wires. Therefore the estimated
16. A= 80, σA= 5, B= 90, σB= 3. The relative increase is R= (B−A)/A =B/A −1.
17. (a) r= 3.00, σr= 0.03, v= 4.0, σv= 0.2. Q=πr2v= 113.1.
Page 142
SUPPLEMENTARY EXERCISES FOR CHAPTER 3 143
(b) r= 4.00, σr= 0.04, v= 2.0, σv= 0.1. Q=πr2v= 100.5.
∂Q
(c) The relative uncertainty is σQ
Q=σln Q.
18. (a) C0= 0.2, t= 300, C= 0.174, σC= 0.005
k= (1/t)(ln C0−ln C) = −0.005365 −(1/300) ln C= 0.000464
Page 143
144 CHAPTER 3
(d) ln(t1/2) = ln(ln 2) −ln k. Therefore σln t1/2=σln k=σk
k= 0.21 = 21% from part (b).
19. (a) C0= 0.03, t= 40, C= 0.0023, σC= 0.0002. k= (1/t)(1/C −1/C0) = (1/40)(1/C)−5/6 = 10.04
(b) C0= 0.03, t= 50, C= 0.0018, σC= 0.0002. k= (1/t)(1/C −1/C0) = (1/50)(1/C)−2/3 = 10.4
(c) Let k= (b
k1+b
k2)/2 = 0.5b
k1+ 0.5b
k2. From parts (a) and (b), σb
b
k1
b
k2
20. (a) s= 1, v= 3.2, σv= 0.1, a1=v2/2s=v2/2 = 5.12
Page 144
(b) s= 1, t= 0.63, σt= 0.01, a2= 2s/t2= 2/t2= 5.04
(c) The weighted average with the smallest uncertainty is cbesta1+ (1 −cbest )a2, where cbest =σ2
a2
.
21. (a) Let sbe the measured side of the square. Then s= 181.2, σs= 0.1.
The estimated area is S=s2= 32,833.
22. (a) r= 3.0, σr= 0.1, A=πr2= 28.27433, dA
dr = 2πr = 18.85
146 CHAPTER 3
dA
23. (a) P1= 8.1, σP1= 0.1, P2= 15.4, σP2= 0.2, P=√P1P2= 11.16871
∂P3
∂P1
=P2
2√P1P2
= 0.689426
SECTION 4.1 147
Chapter 4
Section 4.1
1. (a) X∼Bernoulli(p), where p= 0.4. µX=p= 0.4, σ2
X=p(1 −p) = 0.24.
2. (a) pX= 0.2
3. (a) pX= 0.05
Page 147