CHAPTER 16 EXPONENTIAL FUNCTIONS
CHAPTER 16 EXPONENTIAL FUNCTIONS EXERCISE 64 Page 136 1. Evaluate the following, correct to 4 significant figures: (a) e–1.8 (b) e– 0.78 (c) e10
Using a calculator: (a) e–1.8 = 0.1653, correct to 4 significant figures (b) e– 0.78 = 0.4584, correct to 4 significant figures (c) e10 = 22 030, correct to 4 significant figures
2. Evaluate the following, correct to 5 significant figures: (a) e1.629
(b) e–2.7483
(c) 0.62e4.178
Using a calculator: a) e1.629 = 5.0988, correct to 5 significant figures (b) e–2.7483 = 0.064037, correct to 5 significant figures (c) 0.62e4.178 = 40.446, correct to 5 significant figures
3. Evaluate correct to 5 decimal places: (a)
1 3.4629 e 7
(b) 8.52e–1.2651
(c)
5e 2.6921 3e1.1171
Using a calculator: (a)
1 3.4629 e = 4.55848, correct to 5 decimal places 7
(b) 8.52e–1.2651 = 2.40444, correct to 5 decimal places (c)
5e 2.6921 = 8.05124, correct to 5 decimal places 3e1.1171
4. Evaluate correct to 5 decimal places: (a)
5.6823 e −2.1347
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(b)
e 2.1127 − e −2.1127 2
(c)
4 (e −1.7295 − 1) e3.6817
© 2014, John Bird
Using a calculator: (a)
5.6823 = 48.04106, correct to 5 decimal places e −2.1347
(b)
e 2.1127 − e −2.1127 = 4.07482, correct to 5 decimal places 2
(c )
4 ( e −1.7295 − 1) e3.6817
= –0.08286, correct to 5 decimal places
5. The length of a bar l at a temperature θ is given by l = l 0 eαθ, where l 0 and α are constants. Evaluate l, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and α = 1.771 × 10−4 Using a calculator, = l l0= eα θ (2.587) e(321.7×1.771×10−4 ) = 2.739, correct to 4 significant figures
6. When a chain of length 2L is suspended from two points, 2D metres apart, on the same L + L2 + k 2 horizontal level: D = k ln k
D
. Evaluate D when k = 75 m and L = 180 m.
L + L2 + k 2 180 + 1802 + 752 180 + 195 = k ln = 75 ln = 75 ln 75 {ln 5} = 120.7 m k 75 75
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© 2014, John Bird
EXERCISE 65 Page 138 1. Evaluate 5.6e–1, correct to 4 decimal places, using the power series for ex.
e x =+ 1 x+
x 2 x3 x 4 + + + ... 2! 3! 4!
When x = –1, e −1 = 1 + (−1) +
(−1) 2 (−1)3 (−1) 4 (−1)5 (−1)6 + + + + + ... 2! 3! 4! 5! 6!
= 1 – 1 + 0.5 – 0.166666 + 0.041666 – 0.0083333 + 0.001388 – 0.000198 + 0.0000248 + … = 0.36788… Hence, 5.6e–1 = (5.6)(0.36788) = 2.0601, correct to 4 decimal places 2. Use the power series for ex to determine, correct to 4 significant figures, (a) e2 (b) e– 0.3 and check your result by using a calculator.
(a)
e x =+ 1 x+
When x = 2, e 2 = 1 + 2 +
x 2 x3 x 4 + + + ... 2! 3! 4! 2 2 23 2 4 25 2 6 + + + + + ... 2! 3! 4! 5! 6!
= 1 + 2 + 2 + 1.33333 + 0.66666 + 0.26666 + 0.08888 + 0.02540 + 0.00635 + 0.00141 + 0.00028 + 0.00005 + … = 7.389, correct to 4 significant figures, which may be checked with a calculator (b) When x = –0.3, e −0.3 = 1 − 0.3 +
(−0.3) 2 (−0.3)3 (−0.3) 4 (−0.3)5 + + + + ... 2! 3! 4! 5!
= 1 – 0.3 + 0.04500 – 0.00450 + 0.00034 – 0.00002 + … = 0.7408, correct to 4 significant figures 3. Expand (1 – 2x)e2x as far as the term in x4
(2 x) 2 (2 x)3 (2 x) 4 + + + ... (1 – 2x) e 2 x = (1 – 2x) 1 + 2 x + 2! 3! 4!
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4 2 = (1 – 2x) 1 + 2 x + 2 x 2 + x3 + x 4 3 3 = 1 + 2 x + 2 x2 +
4 3 2 4 8 x + x − 2 x − 4 x 2 − 4 x 3 − x 4 − ... 3 3 3
8 = 1 − 2 x 2 − x3 − 2 x 4 3 4. Expand (2e x2 )(x1/2) to six terms.
e x =+ 1 x+
1/2
(2x ) e
x2
x 2 x3 x 4 + + + ... 2! 3! 4!
2 ( x 2 ) 2 ( x 2 )3 ( x 2 ) 4 ( x 2 ) 2 = (2x ) 1 + x + + + + ... 2! 3! 4! 5!
1/2
x 4 x 6 x8 x10 + + ... = (2x1/2) 1 + x 2 + + + 2 6 24 120 1 2
= 2x + 2x
2+
1 2
4+
2x + 2
1 2
+
2x
6 +
6
1 2
8+
1
10
1
2x 2 2x 2 ... + + 24 120
1 5 9 1 13 1 17 1 21 = 2x 2 + 2x 2 + x 2 + x 2 + x 2 + x 2 ... 3 12 60
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EXERCISE 66 Page 139
1. Plot a graph of y = 3e 0.2 x over the range x = –3 to x = 3. Hence determine the value of y when x = 1.4 and the value of x when y = 4.5
x
–3
–2
–1
0
1
2
3
y = 3 e0.2 x 1.65 2.01 2.46 3.00 3.66 4.48 5.47
From the graph, when x = 1.4, y = 3.95 and when y = 4.5, x = 2.05
2. Plot a graph of y =
1 −1.5 x e over a range x = –1.5 to x = 1.5 and hence determine the value of y 2
when x = –0.8 and the value of x when y = 3.5
x y=
– 1.5 1 −1.5 x e 2
4.74
–1
– 0.5
0
0.5
1
1.5
2.24
1.06
0.50
0.24
0.11
0.05
A graph of y against x is shown below. 253
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From the graph, when x = –0.8, y = 1.65 and when y = 3.5, x = –1.30
3. In a chemical reaction the amount of starting material C cm 3 left after t minutes is given by: C = 40e −0.006t . Plot a graph of C against t and determine (a) the concentration C after 1 hour, and (b) the time taken for the concentration to decrease by half.
Time t (min)
0
20
40
60
80
100 120
C = 40e −0.006t (cm 3 ) 40 35.3 31.5 27.9 24.8 22.0 19.5 A graph of C against t is shown below. From the graph, (a) When t = 60 min, C = 28 cm3 (b) When C has decreased by half, i.e. to 20 cm 3 , time, t = 116 min
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© 2014, John Bird
4. The rate at which a body cools is given by θ = 250e −0.05t , where the excess of temperature of a body above its surroundings at time t minutes is θ°C. Plot a graph showing the natural decay curve for the first hour of cooling. Hence determine (a) the temperature after 25 minutes and (b) the time when the temperature is 195°C.
t
0
10
20 30 40 50 60
θ = 250 e −0.05t 250 152 92 56 34 21 12
From the graph, (a) after t = 25 minutes, temperature θ = 70 °C (b) when the temperature is 195 °C , time t = 5 minutes
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EXERCISE 67 Page 142
1. Evaluate correct to 5 significant figures: (a)
1 ln 5.2932 3
(b)
ln 82.473 4.829
5.62 ln 321.62 e1.2942
(c)
Using a calculator: (a)
1 ln 5.2932 = 0.55547, correct to 5 significant figures 3
(b)
ln 82.473 = 0.91374, correct to 5 significant figures 4.829
(c)
5.62 ln 321.62 = 8.8941, correct to 5 significant figures e1.2942
2. Evaluate correct to 5 significant figures: (a)
1.786 ln e1.76 lg101.41
(b)
5e −0.1629 2 ln 0.00165
(c)
ln 4.8629 − ln 2.4711 5.173
Using a calculator,
(a)
1.786 ln e1.76 (1.786)(1.76) = = 2.2293, correct to 5 significant figures lg101.41 (1.41)
(b)
5e −0.1629 = –0.33154, correct to 5 significant figures 2 ln 0.00165
(c)
ln 4.8629 − ln 2.4711 = 0.13087, correct to 5 significant figures 5.173
3. Solve, correct to 4 significant figures: 1.5 = 4e2t
Rearranging 1.5 = 4e2t gives:
1.5 = e2t 4
1.5 2t Taking Napierian logarithms of both sides gives: ln = ln(e ) 4 Since log eeα = α, then
1.5 ln = 2t 4 257
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Hence,
t=
1 1.5 1 ln = (– 0.980829) = –0.4904, correct to 4 significant figures 2 4 2
4. Solve correct to 4 significant figures: 7.83 = 2.91e–1.7x
If 7.83 = 2.91e −1.7 x
i.e.
7.83 2.91
e −1.7 =
then
7.83 –1.7x = ln 2.91
and
x= −
7.83 ln e −1.7 x = ln 2.91
and
1 7.83 ln = –0.5822, correct to 4 significant figures 1.7 2.91
−
t
5. Solve correct to 4 significant figures: 16 = 24(1 – e 2 )
t − If= 16 24 1 − e 2
then
t 16 − = 1− e 2 24 −
t
16 24
from which,
e 2 = 1−
and
t 16 − = ln 1 − 2 24
16 t = −2 ln 1 − = 2.197, correct to 4 significant figures 24
and
x 6. Solve correct to 4 significant figures: 5.17 = ln 4.64 x x 5.17 From the definition of a logarithm, since 5.17 = ln then e = 4.64 4.64
Rearranging gives:
x = 4.64 e5.17
i.e.
x = 816.2, correct to 4 significant figures
1.59 7. Solve correct to 4 significant figures: 3.72 ln = 2.43 x
1.59 If 3.72 ln = 2.43 x
then
1.59 2.43 ln = x 3.72 258
© 2014, John Bird
2.43 1.59 = e 3.72 x
from which,
and
1.59
x=
e
2.43 3.72
= 1.59 e
2.43 − 3.72
= 0.8274, correct to 4 significant figures
8. Solve correct to 4 significant figures: ln x = 2.40
If
ln x = 2.40
then
x = e 2.4 = 11.02, correct to 4 significant figures
9. Solve correct to 4 significant figures: 24 + e 2 x = 45 Since 24 + e 2 x = 45 then e 2 x = 45 − 24 = 21 Then i.e. from which,
ln e 2 x = ln 21 2x = ln 21 x=
1 ln 21 = 1.522 2
10. Solve correct to 4 significant figures: = 5 e x+1 − 7 Since = 5 e x+1 − 7 then e x+1 = 12 Then
ln e x+1 = ln12
i.e.
x + 1 = ln 12
and
x = (ln 12) –1 = 2.485 –1 = 1.485
−x 5 8 1 − e 2 11. Solve correct to 4 significant figures: =
−x Rearranging= 5 8 1 − e 2 gives:
and
x 5 − =1– e 2 8
e
−
x 2
=1–
5 3 = 8 8 x
Taking the reciprocal of both sides gives:
e2 = 259
8 3
© 2014, John Bird
x 8 Taking Napierian logarithms of both sides gives: ln e 2 = ln 3
x 8 = ln 2 3
i.e.
8 x = 2 ln = 1.962, correct to 4 significant figures 3
from which,
12. Solve, correct to 4 significant figures: ln(x + 3) – ln x = ln(x – 1).
x+3 Since ln(x + 3) – ln x = ln(x – 1) then ln = ln ( x − 1) x x+3 = x
from which,
( x − 1)
i.e.
x + 3 = x(x – 1)
i.e.
x + 3 = x2 − x
Rearranging gives: Factorizing gives: from which,
x2 − 2x − 3 = 0 (x – 3)(x + 1) = 0 x = 3 or x = –1
In the given equation x = –1 is not a possible solution, hence x = 3 13. Solve, correct to 4 significant figures: ln ( x − 1) − ln 3= ln ( x − 1) 2
Since ln ( x − 1) − ln 3= ln ( x − 1) 2
from which,
then
( x − 1)2 ln ( x − 1) ln = 3
( x − 1) 2 = x −1 3
i.e.
x2 − 2x + 1 = x −1 3
and
x 2 − 2 x + 1 = 3x − 3
i.e.
x2 − 5x + 4 = 0
Factorizing gives: from which,
(x – 4)(x – 1) = 0 x = 4 or x = 1 260
© 2014, John Bird
The solution x = 1 is not possible in the given equation, hence, x = 4
14. Solve, correct to 4 significant figures: ln(x + 3) + 2 = 12 – ln(x – 2)
Rearranging ln(x + 3) + 2 = 12 – ln(x – 2) gives: ln(x + 3) + ln(x – 2) = 12 – 2 = 10 i.e.
ln[(x + 3)(x – 2)] = 10
i.e.
ln[ x 2 − 2 x + 3 x − 6 ] = 10
i.e.
ln[ x 2 + x − 6] = 10
x 2 + x − 6 = e10 and x 2 + x − 6 – e10 = 0 i.e.
from which,
x 2 + x − 22 032.5 = 0 x=
−1 ± 12 − 4(1)(−22032.5) −1 ± 296.87 = 147.9 or –1.001 = 2(1) 2
Neglecting the negative answer, x = 147.9 15. Solve correct to 4 significant figures: e( x +1) = 3e( 2 x −5) )
Since e( x +1) = 3e( 2 x −5) ) then
ln ( e( x +1) ) = ln ( 3e( 2 x −5) )
i.e.
1) ln ( e( x += ) ln 3 + ln ( e( 2 x−5) ) by the first law of logarithms
and
(x + 1) = ln 3 + (2x – 5)
Rearranging gives:
5 + 1 – ln 3 = 2x – x
i.e.
x = 6 – ln 3 = 4.901
16. Solve correct to 4 significant figures: ln ( x + 1) = 1.5 − ln ( x − 2 ) + ln ( x + 1) 2
Since
ln ( x + 1) = 1.5 − ln ( x − 2 ) + ln ( x + 1)
then
ln ( x + 1) + ln ( x − 2 ) − ln ( x + 1) = 1.5
2
2
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© 2014, John Bird
and
( x + 1) 2 ( x − 2) ) ln = 1.5 ( x + 1)
i.e.
ln ( ( x + 1)( x − 2) ) = 1.5
i.e.
ln ( x 2 − 2 x + x − 2 ) = 1.5
and
( x 2 − 2 x + x − 2 ) =e1.5
i.e.
x 2 − x − 2 − e1.5 = 0
i.e.
x 2 − x − 6.48169 = 0 2 − − 1 ± ( −1) − 4 (1)( −6.48169 ) 1 ± 26.92676 x= = 2 (1) 2
then
=
1 + 26.92676 1 − 26.92676 or 2 2
= 3.095 or –2.095, correct to 3 decimal places The solution x = –2.095 is not possible in the given equation, hence x = 3.095
17. Transpose: b = ln t – a ln D to make t the subject.
Since b = ln t – a ln D then ln t = b + a ln D b + a ln D = t e= eb= e a ln D eb eln Da
and
18. If
Since
i.e. t = eb D a
since eln x = x
P R1 = 10 log10 find the value of R1 when = P 160, = Q 8 and R2 = 5 Q R2 P R1 = 10 log10 Q R2
then
P R1 = log10 10Q R2 P
1010Q =
and
R1 R2 P
160
R1 = R2 ×1010Q = 5 ×10 80 = 5 ×102 = 5 × 100 = 500
from which,
W
19. If U 2 = U1 e PV make W the subject of the formula.
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Since U 2 = U1 e
W PV
W W U2 U2 then = e PV and ln = ln e PV U1 U1
U2 W ln = U1 PV
i.e.
U2 W = PV ln U1
from which,
m1 20. The velocity v2 of a rocket is given by v2= v1 + C ln where v1 is the initial rocket m2 velocity, C is the velocity of the jet exhaust gases, m1 is the mass of the rocket before the jet engine is fired, and m2 is the mass of the rocket after the jet engine is switched off. Calculate the velocity of the rocket given v1 = 600 m/s, C = 3500 m/s, m1 = 8.50 ×104 kg and
m2 = 7.60 ×104 kg. m1 8.50 ×104 v1 + C ln 600 + 3500 ln Velocity of rocket v2 = = m2 7.60 ×104 = 992 m/s
21. The work done in an isothermal expansion of a gas from pressure p1 to p2 is given by: p1 w = w0 ln p2 If the initial pressure p1 = 7.0 kPa, calculate the final pressure p2 if w = 3 w0
If w = 3 w0 then
p1 3 w0 = w0 ln p2
i.e.
p1 3 = ln p2
and
= e3
from which,
final pressure,= p2
p1 7000 = p2 p2 7000 = 7000 e −3 = 348.5 Pa e3
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EXERCISE 68 Page 145
1. The temperature, T°C, of a cooling object varies with time, t minutes, according to the equation:
T = 150 e − 0.04 t . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
(a) When t = 0, T = 150 e − 0.04 t = 150 e0 = 150°C (b) When t = 10, T == 150 e − 0.04 t 150 = e −0.04(10) 150 e −0.4 = 100.5°C
2. The pressure p pascals at height h metres above ground level is given by: p = p 0 e– h/C, where p 0 is the pressure at ground level and C is a constant. Find pressure p when p 0 = 1.012 ×105 Pa, height h = 1420 m and C = 71 500
Pressure, p = p0 e
−
h C
= (1.012 ×105 ) e
−
1420 71500
= (1.012 ×105 )(0.980336) = 9.921×104 Pa or 99.21 kPa
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by: v = 200 e
−
Rt L
,
where R = 150 Ω and L = 12.5 ×10– 3 H. Determine (a) the voltage when t = 160 × 10–6 s, and (b) the time for the voltage to reach 85 V.
(a) Voltage v = 200 e
− Rt L
= 200 e
(b) When v = 85 V, 85 = 200 e
and
Thus,
−
−
−
(150 )(160×10−6 ) 12.5×10−3
150 t 12.5×10−3
= 200 e −1.92 = 29.32 volts
from which,
150 t 85 − = e 12.5×10−3 200
150 t 85 = ln 3 − 12.5 ×10 200 time t = −
12.5 ×10−3 85 −6 ln = 71.31×10 s 150 200
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4. The length l metres of a metal bar at temperature t°C is given by l = l 0 eαt, where l 0 and α are constants. Determine (a) the value of l when l 0 = 1.894, α = 2.038 × 10 −4 and t = 250°C, and (b) the value of l 0 when l = 2.416, t = 310°C and α = 1.682 × 10−4 2.038×10−4 )( 250 ) (a) Length, l = l0 eα t 1.894 e(= = 1.894 = e0.05096 (1.894)(1.05227) = 1.993 m
l )( 310 ) 2.416 e − 0.052142 (b) Since l = l0 eα t , then l0 = = l = e −α t 2.416 e −(1.682×10−4= α t e
= (2.416)(0.949194) = 2.293 m 5. The temperature θ 2 °C of an electrical conductor at time t seconds is given by: θ2 = θ1 (1 – e–t/T), where θ 1 is the initial temperature and T seconds is a constant. Determine (a) θ 2 when θ 1 = 159.9°C, t = 30 s and T = 80 s, and
(b) the time t for θ 2 to fall to half the value of θ 1 if T remains at 80 s
30 − (a) θ 2 = θ 1(1 – e–t/T) = 159.9 1 − e 80 = 159.9 (1 − e − 0.375 )= 159.9 (1 − 0.687289...) = 50°C t θ2 − = 1− e T θ1
(b) Since θ 2 = θ1 (1 – e–t/T) then from which, Then
and from which,
e
−
t T
= 1−
θ2 θ1
θ2 − Tt ln e = ln 1 − θ1 t θ2 − = ln 1 − T θ1 1 θ2 time, t = –T ln 1 − = –80 ln 1 − = –80 ln 0.5 = 55.45 s 2 θ1
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6. A belt is in contact with a pulley for a sector of θ = 1.12 radians and the coefficient of friction between these two surfaces is µ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side is given by T 0 = 22.7 newtons, given that these quantities are related by the law T = T0 eµθ. Tension T = T0 e µθ = 22.7 e( 0.26)(1.12) = 22.7 e0.2912 = 30.37 N
7. The instantaneous current i at time t is given by: i = 10e–t/CR when a capacitor is being charged. The capacitance C is 7 × 10− 6 farads and the resistance R is 0.3 × 106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and (b) the time for the instantaneous current to fall to 5 amperes. Sketch a curve of current against time from t = 0 to t = 6 seconds.
(a) Current, i = 10 e (b) When i = 5,
−
2.5 7×10−6 × 0.3×106
5 = 10 e
−
= 3.04 A
t 2.1
from which,
t 5 ln = − 2.1 10
Thus, Time t
0
Current i 10
1
and 2
3
t 5 − = e 2.1 10
time, t = −(2.1) ln 0.5 = 1.46 s 4
5
6
6.21 3.86 2.40 1.49 0.92 0.57
A graph of current against time is shown below.
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8. The amount of product x (in mol/cm3) found in a chemical reaction starting with 2.5 mol/cm3 of reactant is given by: x = 2.5(1 – e– 4t) where t is the time, in minutes, to form product x. Plot a graph at 30-second intervals up to 2.5 minutes and determine x after 1 minute. Amount of product, x = 2.5(1 – e– 4t) Time t (min) 0 x (mol/cm3)
0.5 1.0
1.5
2.0
2.5
0 2.16 2.45 2.49 2.50 2.50
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The amount of product after 1 minute = 2.45 mol/cm3 9. The current i flowing in a capacitor at time t is given by: i = 12.5(1 – e–t/CR) where resistance R is 30 kΩ and the capacitance C is 20 µF. Determine (a) the current flowing after 0.5 seconds, and (b) the time for the current to reach 10 amperes.
t 0.5 − − (a) Current, = i 12.5 1 − e CR = 12.5 1 − e 20×10−6 ×30×103 = 7.07 A t t 10 − − (b) When i = 10 A, = from which, 10 12.5 1 − e 0.6 = 1 − e 0.6 12.5
Thus,
i.e.
e
−
t 0.6
= 1−
10 12.5
and
−
t 10 = ln 1 − 0.6 12.5
10 time, t = −0.6 ln 1 − = 0.966 s 12.5
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10. The amount A after n years of a sum invested P is given by the compound interest law: A = Per n/100 when the per unit interest rate r is added continuously. Determine, correct to the nearest pound, the amount after eight years for a sum of £1500 invested if the interest rate is 6% per annum.
rn
6×8
Amount after eight years, A= = P e100 (1500) = e 100 1500 e0.48 = (1500)(1.6160744) = £2424 correct to the nearest pound
11. The percentage concentration C of the starting material in a chemical reaction varies with time t according to the equation C = 100 e − 0.004t . Determine the concentration when (a) t = 0, (b) t = 100 s, (c) t = 1000 s. Percentage concentration, C = 100 e − 0.004t (a) When t = 0, C = 100 e0 = 100% (b) When t = 100, C = 100 e −0.004(100) = C = 100 e − 0.4 = 67.03% (c) When t = 1000, C = 100 e −0.004(1000) = C = 100 e − 4 = 1.83% 12. The current i flowing through a diode at room temperature is given= by: i iS ( e 40V − 1) amperes. Calculate the current flowing in a silicon diode when the reverse saturation current iS = 50 nA and the forward voltage V = 0.27 V Current in diode, i = iS ( e 40V − 1) = 50 ×10−9 ( e 40(0.27) − 1) = 50 ×10−9 ( e10.8 − 1) = 2.45 ×10−3 A = 2.45 mA
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t − 13. A formula for chemical decomposition is given by: C = A 1 − e 10 where t is the time in
seconds. Calculate the time, in milliseconds, for a compound to decompose to a value of C = 0.12 given A = 8.5
t − C = A 1 − e 10
hence
t − 0.12 = 8.5 1 − e 10
t 0.12 − = 1 − e 10 8.5
i.e.
from which,
e
−
t 10
= 1−
and
t − 0.12 10 ln e= ln 1 − 8.5
Hence,
0.12 time, t = −10 ln 1 − = 0.142 s = 142 ms 8.5
−
i.e.
0.12 8.5
t 0.12 = ln 1 − 10 8.5
14. The mass m of pollutant in a water reservoir decreases according to the law m = m0 e − 0.1t where t is the time in days and m0 is the initial mass. Calculate the percentage decrease in the mass after 60 days, correct to 3 decimal places. − 0.1t 0 Initially, i.e. at time t = 0, = m m0 e= m= m0 0e
−6 0.1(60) When t = 60, = m m0= e − 0.1t m0 e −= m= 0.00248m0 0e
Percentage decrease =
1 − 0.00248 ×100% = 99.752% 1
15. A metal bar is cooled with water. Its temperature, in °C, is given by: θ = 15 + 1300 e − 0.2 t where t is the time in minutes. Calculate how long it will take for the temperature, θ, to decrease to 36°C, correct to the nearest second. Temperature, θ = 15 + 1300 e − 0.2 t When θ = 36°,
36 = 15 + 1300 e − 0.2 t
and 36 – 15 = 1300 e − 0.2 t 270
© 2014, John Bird
Hence,
i.e.
from which,
36 − 15 = e − 0.2t 1300
and
36 − 15 − 0.2 t ln = ln e 1300
36 − 15 – 0.2t = ln 1300 time, t =
1 36 − 15 ln = 20.63 minutes − 0.2 1300
0.63 min = 0.63 × 60 s = 37.8 s = 38 s correct to the nearest second. Hence, time for the temperature to decrease to 36°C = 20 min 38 s
271
© 2014, John Bird
Using a calculator: (a) e–1.8 = 0.1653, correct to 4 significant figures (b) e– 0.78 = 0.4584, correct to 4 significant figures (c) e10 = 22 030, correct to 4 significant figures
2. Evaluate the following, correct to 5 significant figures: (a) e1.629
(b) e–2.7483
(c) 0.62e4.178
Using a calculator: a) e1.629 = 5.0988, correct to 5 significant figures (b) e–2.7483 = 0.064037, correct to 5 significant figures (c) 0.62e4.178 = 40.446, correct to 5 significant figures
3. Evaluate correct to 5 decimal places: (a)
1 3.4629 e 7
(b) 8.52e–1.2651
(c)
5e 2.6921 3e1.1171
Using a calculator: (a)
1 3.4629 e = 4.55848, correct to 5 decimal places 7
(b) 8.52e–1.2651 = 2.40444, correct to 5 decimal places (c)
5e 2.6921 = 8.05124, correct to 5 decimal places 3e1.1171
4. Evaluate correct to 5 decimal places: (a)
5.6823 e −2.1347
249
(b)
e 2.1127 − e −2.1127 2
(c)
4 (e −1.7295 − 1) e3.6817
© 2014, John Bird
Using a calculator: (a)
5.6823 = 48.04106, correct to 5 decimal places e −2.1347
(b)
e 2.1127 − e −2.1127 = 4.07482, correct to 5 decimal places 2
(c )
4 ( e −1.7295 − 1) e3.6817
= –0.08286, correct to 5 decimal places
5. The length of a bar l at a temperature θ is given by l = l 0 eαθ, where l 0 and α are constants. Evaluate l, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and α = 1.771 × 10−4 Using a calculator, = l l0= eα θ (2.587) e(321.7×1.771×10−4 ) = 2.739, correct to 4 significant figures
6. When a chain of length 2L is suspended from two points, 2D metres apart, on the same L + L2 + k 2 horizontal level: D = k ln k
D
. Evaluate D when k = 75 m and L = 180 m.
L + L2 + k 2 180 + 1802 + 752 180 + 195 = k ln = 75 ln = 75 ln 75 {ln 5} = 120.7 m k 75 75
250
© 2014, John Bird
EXERCISE 65 Page 138 1. Evaluate 5.6e–1, correct to 4 decimal places, using the power series for ex.
e x =+ 1 x+
x 2 x3 x 4 + + + ... 2! 3! 4!
When x = –1, e −1 = 1 + (−1) +
(−1) 2 (−1)3 (−1) 4 (−1)5 (−1)6 + + + + + ... 2! 3! 4! 5! 6!
= 1 – 1 + 0.5 – 0.166666 + 0.041666 – 0.0083333 + 0.001388 – 0.000198 + 0.0000248 + … = 0.36788… Hence, 5.6e–1 = (5.6)(0.36788) = 2.0601, correct to 4 decimal places 2. Use the power series for ex to determine, correct to 4 significant figures, (a) e2 (b) e– 0.3 and check your result by using a calculator.
(a)
e x =+ 1 x+
When x = 2, e 2 = 1 + 2 +
x 2 x3 x 4 + + + ... 2! 3! 4! 2 2 23 2 4 25 2 6 + + + + + ... 2! 3! 4! 5! 6!
= 1 + 2 + 2 + 1.33333 + 0.66666 + 0.26666 + 0.08888 + 0.02540 + 0.00635 + 0.00141 + 0.00028 + 0.00005 + … = 7.389, correct to 4 significant figures, which may be checked with a calculator (b) When x = –0.3, e −0.3 = 1 − 0.3 +
(−0.3) 2 (−0.3)3 (−0.3) 4 (−0.3)5 + + + + ... 2! 3! 4! 5!
= 1 – 0.3 + 0.04500 – 0.00450 + 0.00034 – 0.00002 + … = 0.7408, correct to 4 significant figures 3. Expand (1 – 2x)e2x as far as the term in x4
(2 x) 2 (2 x)3 (2 x) 4 + + + ... (1 – 2x) e 2 x = (1 – 2x) 1 + 2 x + 2! 3! 4!
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© 2014, John Bird
4 2 = (1 – 2x) 1 + 2 x + 2 x 2 + x3 + x 4 3 3 = 1 + 2 x + 2 x2 +
4 3 2 4 8 x + x − 2 x − 4 x 2 − 4 x 3 − x 4 − ... 3 3 3
8 = 1 − 2 x 2 − x3 − 2 x 4 3 4. Expand (2e x2 )(x1/2) to six terms.
e x =+ 1 x+
1/2
(2x ) e
x2
x 2 x3 x 4 + + + ... 2! 3! 4!
2 ( x 2 ) 2 ( x 2 )3 ( x 2 ) 4 ( x 2 ) 2 = (2x ) 1 + x + + + + ... 2! 3! 4! 5!
1/2
x 4 x 6 x8 x10 + + ... = (2x1/2) 1 + x 2 + + + 2 6 24 120 1 2
= 2x + 2x
2+
1 2
4+
2x + 2
1 2
+
2x
6 +
6
1 2
8+
1
10
1
2x 2 2x 2 ... + + 24 120
1 5 9 1 13 1 17 1 21 = 2x 2 + 2x 2 + x 2 + x 2 + x 2 + x 2 ... 3 12 60
252
© 2014, John Bird
EXERCISE 66 Page 139
1. Plot a graph of y = 3e 0.2 x over the range x = –3 to x = 3. Hence determine the value of y when x = 1.4 and the value of x when y = 4.5
x
–3
–2
–1
0
1
2
3
y = 3 e0.2 x 1.65 2.01 2.46 3.00 3.66 4.48 5.47
From the graph, when x = 1.4, y = 3.95 and when y = 4.5, x = 2.05
2. Plot a graph of y =
1 −1.5 x e over a range x = –1.5 to x = 1.5 and hence determine the value of y 2
when x = –0.8 and the value of x when y = 3.5
x y=
– 1.5 1 −1.5 x e 2
4.74
–1
– 0.5
0
0.5
1
1.5
2.24
1.06
0.50
0.24
0.11
0.05
A graph of y against x is shown below. 253
© 2014, John Bird
From the graph, when x = –0.8, y = 1.65 and when y = 3.5, x = –1.30
3. In a chemical reaction the amount of starting material C cm 3 left after t minutes is given by: C = 40e −0.006t . Plot a graph of C against t and determine (a) the concentration C after 1 hour, and (b) the time taken for the concentration to decrease by half.
Time t (min)
0
20
40
60
80
100 120
C = 40e −0.006t (cm 3 ) 40 35.3 31.5 27.9 24.8 22.0 19.5 A graph of C against t is shown below. From the graph, (a) When t = 60 min, C = 28 cm3 (b) When C has decreased by half, i.e. to 20 cm 3 , time, t = 116 min
254
© 2014, John Bird
4. The rate at which a body cools is given by θ = 250e −0.05t , where the excess of temperature of a body above its surroundings at time t minutes is θ°C. Plot a graph showing the natural decay curve for the first hour of cooling. Hence determine (a) the temperature after 25 minutes and (b) the time when the temperature is 195°C.
t
0
10
20 30 40 50 60
θ = 250 e −0.05t 250 152 92 56 34 21 12
From the graph, (a) after t = 25 minutes, temperature θ = 70 °C (b) when the temperature is 195 °C , time t = 5 minutes
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© 2014, John Bird
256
© 2014, John Bird
EXERCISE 67 Page 142
1. Evaluate correct to 5 significant figures: (a)
1 ln 5.2932 3
(b)
ln 82.473 4.829
5.62 ln 321.62 e1.2942
(c)
Using a calculator: (a)
1 ln 5.2932 = 0.55547, correct to 5 significant figures 3
(b)
ln 82.473 = 0.91374, correct to 5 significant figures 4.829
(c)
5.62 ln 321.62 = 8.8941, correct to 5 significant figures e1.2942
2. Evaluate correct to 5 significant figures: (a)
1.786 ln e1.76 lg101.41
(b)
5e −0.1629 2 ln 0.00165
(c)
ln 4.8629 − ln 2.4711 5.173
Using a calculator,
(a)
1.786 ln e1.76 (1.786)(1.76) = = 2.2293, correct to 5 significant figures lg101.41 (1.41)
(b)
5e −0.1629 = –0.33154, correct to 5 significant figures 2 ln 0.00165
(c)
ln 4.8629 − ln 2.4711 = 0.13087, correct to 5 significant figures 5.173
3. Solve, correct to 4 significant figures: 1.5 = 4e2t
Rearranging 1.5 = 4e2t gives:
1.5 = e2t 4
1.5 2t Taking Napierian logarithms of both sides gives: ln = ln(e ) 4 Since log eeα = α, then
1.5 ln = 2t 4 257
© 2014, John Bird
Hence,
t=
1 1.5 1 ln = (– 0.980829) = –0.4904, correct to 4 significant figures 2 4 2
4. Solve correct to 4 significant figures: 7.83 = 2.91e–1.7x
If 7.83 = 2.91e −1.7 x
i.e.
7.83 2.91
e −1.7 =
then
7.83 –1.7x = ln 2.91
and
x= −
7.83 ln e −1.7 x = ln 2.91
and
1 7.83 ln = –0.5822, correct to 4 significant figures 1.7 2.91
−
t
5. Solve correct to 4 significant figures: 16 = 24(1 – e 2 )
t − If= 16 24 1 − e 2
then
t 16 − = 1− e 2 24 −
t
16 24
from which,
e 2 = 1−
and
t 16 − = ln 1 − 2 24
16 t = −2 ln 1 − = 2.197, correct to 4 significant figures 24
and
x 6. Solve correct to 4 significant figures: 5.17 = ln 4.64 x x 5.17 From the definition of a logarithm, since 5.17 = ln then e = 4.64 4.64
Rearranging gives:
x = 4.64 e5.17
i.e.
x = 816.2, correct to 4 significant figures
1.59 7. Solve correct to 4 significant figures: 3.72 ln = 2.43 x
1.59 If 3.72 ln = 2.43 x
then
1.59 2.43 ln = x 3.72 258
© 2014, John Bird
2.43 1.59 = e 3.72 x
from which,
and
1.59
x=
e
2.43 3.72
= 1.59 e
2.43 − 3.72
= 0.8274, correct to 4 significant figures
8. Solve correct to 4 significant figures: ln x = 2.40
If
ln x = 2.40
then
x = e 2.4 = 11.02, correct to 4 significant figures
9. Solve correct to 4 significant figures: 24 + e 2 x = 45 Since 24 + e 2 x = 45 then e 2 x = 45 − 24 = 21 Then i.e. from which,
ln e 2 x = ln 21 2x = ln 21 x=
1 ln 21 = 1.522 2
10. Solve correct to 4 significant figures: = 5 e x+1 − 7 Since = 5 e x+1 − 7 then e x+1 = 12 Then
ln e x+1 = ln12
i.e.
x + 1 = ln 12
and
x = (ln 12) –1 = 2.485 –1 = 1.485
−x 5 8 1 − e 2 11. Solve correct to 4 significant figures: =
−x Rearranging= 5 8 1 − e 2 gives:
and
x 5 − =1– e 2 8
e
−
x 2
=1–
5 3 = 8 8 x
Taking the reciprocal of both sides gives:
e2 = 259
8 3
© 2014, John Bird
x 8 Taking Napierian logarithms of both sides gives: ln e 2 = ln 3
x 8 = ln 2 3
i.e.
8 x = 2 ln = 1.962, correct to 4 significant figures 3
from which,
12. Solve, correct to 4 significant figures: ln(x + 3) – ln x = ln(x – 1).
x+3 Since ln(x + 3) – ln x = ln(x – 1) then ln = ln ( x − 1) x x+3 = x
from which,
( x − 1)
i.e.
x + 3 = x(x – 1)
i.e.
x + 3 = x2 − x
Rearranging gives: Factorizing gives: from which,
x2 − 2x − 3 = 0 (x – 3)(x + 1) = 0 x = 3 or x = –1
In the given equation x = –1 is not a possible solution, hence x = 3 13. Solve, correct to 4 significant figures: ln ( x − 1) − ln 3= ln ( x − 1) 2
Since ln ( x − 1) − ln 3= ln ( x − 1) 2
from which,
then
( x − 1)2 ln ( x − 1) ln = 3
( x − 1) 2 = x −1 3
i.e.
x2 − 2x + 1 = x −1 3
and
x 2 − 2 x + 1 = 3x − 3
i.e.
x2 − 5x + 4 = 0
Factorizing gives: from which,
(x – 4)(x – 1) = 0 x = 4 or x = 1 260
© 2014, John Bird
The solution x = 1 is not possible in the given equation, hence, x = 4
14. Solve, correct to 4 significant figures: ln(x + 3) + 2 = 12 – ln(x – 2)
Rearranging ln(x + 3) + 2 = 12 – ln(x – 2) gives: ln(x + 3) + ln(x – 2) = 12 – 2 = 10 i.e.
ln[(x + 3)(x – 2)] = 10
i.e.
ln[ x 2 − 2 x + 3 x − 6 ] = 10
i.e.
ln[ x 2 + x − 6] = 10
x 2 + x − 6 = e10 and x 2 + x − 6 – e10 = 0 i.e.
from which,
x 2 + x − 22 032.5 = 0 x=
−1 ± 12 − 4(1)(−22032.5) −1 ± 296.87 = 147.9 or –1.001 = 2(1) 2
Neglecting the negative answer, x = 147.9 15. Solve correct to 4 significant figures: e( x +1) = 3e( 2 x −5) )
Since e( x +1) = 3e( 2 x −5) ) then
ln ( e( x +1) ) = ln ( 3e( 2 x −5) )
i.e.
1) ln ( e( x += ) ln 3 + ln ( e( 2 x−5) ) by the first law of logarithms
and
(x + 1) = ln 3 + (2x – 5)
Rearranging gives:
5 + 1 – ln 3 = 2x – x
i.e.
x = 6 – ln 3 = 4.901
16. Solve correct to 4 significant figures: ln ( x + 1) = 1.5 − ln ( x − 2 ) + ln ( x + 1) 2
Since
ln ( x + 1) = 1.5 − ln ( x − 2 ) + ln ( x + 1)
then
ln ( x + 1) + ln ( x − 2 ) − ln ( x + 1) = 1.5
2
2
261
© 2014, John Bird
and
( x + 1) 2 ( x − 2) ) ln = 1.5 ( x + 1)
i.e.
ln ( ( x + 1)( x − 2) ) = 1.5
i.e.
ln ( x 2 − 2 x + x − 2 ) = 1.5
and
( x 2 − 2 x + x − 2 ) =e1.5
i.e.
x 2 − x − 2 − e1.5 = 0
i.e.
x 2 − x − 6.48169 = 0 2 − − 1 ± ( −1) − 4 (1)( −6.48169 ) 1 ± 26.92676 x= = 2 (1) 2
then
=
1 + 26.92676 1 − 26.92676 or 2 2
= 3.095 or –2.095, correct to 3 decimal places The solution x = –2.095 is not possible in the given equation, hence x = 3.095
17. Transpose: b = ln t – a ln D to make t the subject.
Since b = ln t – a ln D then ln t = b + a ln D b + a ln D = t e= eb= e a ln D eb eln Da
and
18. If
Since
i.e. t = eb D a
since eln x = x
P R1 = 10 log10 find the value of R1 when = P 160, = Q 8 and R2 = 5 Q R2 P R1 = 10 log10 Q R2
then
P R1 = log10 10Q R2 P
1010Q =
and
R1 R2 P
160
R1 = R2 ×1010Q = 5 ×10 80 = 5 ×102 = 5 × 100 = 500
from which,
W
19. If U 2 = U1 e PV make W the subject of the formula.
262
© 2014, John Bird
Since U 2 = U1 e
W PV
W W U2 U2 then = e PV and ln = ln e PV U1 U1
U2 W ln = U1 PV
i.e.
U2 W = PV ln U1
from which,
m1 20. The velocity v2 of a rocket is given by v2= v1 + C ln where v1 is the initial rocket m2 velocity, C is the velocity of the jet exhaust gases, m1 is the mass of the rocket before the jet engine is fired, and m2 is the mass of the rocket after the jet engine is switched off. Calculate the velocity of the rocket given v1 = 600 m/s, C = 3500 m/s, m1 = 8.50 ×104 kg and
m2 = 7.60 ×104 kg. m1 8.50 ×104 v1 + C ln 600 + 3500 ln Velocity of rocket v2 = = m2 7.60 ×104 = 992 m/s
21. The work done in an isothermal expansion of a gas from pressure p1 to p2 is given by: p1 w = w0 ln p2 If the initial pressure p1 = 7.0 kPa, calculate the final pressure p2 if w = 3 w0
If w = 3 w0 then
p1 3 w0 = w0 ln p2
i.e.
p1 3 = ln p2
and
= e3
from which,
final pressure,= p2
p1 7000 = p2 p2 7000 = 7000 e −3 = 348.5 Pa e3
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© 2014, John Bird
EXERCISE 68 Page 145
1. The temperature, T°C, of a cooling object varies with time, t minutes, according to the equation:
T = 150 e − 0.04 t . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
(a) When t = 0, T = 150 e − 0.04 t = 150 e0 = 150°C (b) When t = 10, T == 150 e − 0.04 t 150 = e −0.04(10) 150 e −0.4 = 100.5°C
2. The pressure p pascals at height h metres above ground level is given by: p = p 0 e– h/C, where p 0 is the pressure at ground level and C is a constant. Find pressure p when p 0 = 1.012 ×105 Pa, height h = 1420 m and C = 71 500
Pressure, p = p0 e
−
h C
= (1.012 ×105 ) e
−
1420 71500
= (1.012 ×105 )(0.980336) = 9.921×104 Pa or 99.21 kPa
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by: v = 200 e
−
Rt L
,
where R = 150 Ω and L = 12.5 ×10– 3 H. Determine (a) the voltage when t = 160 × 10–6 s, and (b) the time for the voltage to reach 85 V.
(a) Voltage v = 200 e
− Rt L
= 200 e
(b) When v = 85 V, 85 = 200 e
and
Thus,
−
−
−
(150 )(160×10−6 ) 12.5×10−3
150 t 12.5×10−3
= 200 e −1.92 = 29.32 volts
from which,
150 t 85 − = e 12.5×10−3 200
150 t 85 = ln 3 − 12.5 ×10 200 time t = −
12.5 ×10−3 85 −6 ln = 71.31×10 s 150 200
264
© 2014, John Bird
4. The length l metres of a metal bar at temperature t°C is given by l = l 0 eαt, where l 0 and α are constants. Determine (a) the value of l when l 0 = 1.894, α = 2.038 × 10 −4 and t = 250°C, and (b) the value of l 0 when l = 2.416, t = 310°C and α = 1.682 × 10−4 2.038×10−4 )( 250 ) (a) Length, l = l0 eα t 1.894 e(= = 1.894 = e0.05096 (1.894)(1.05227) = 1.993 m
l )( 310 ) 2.416 e − 0.052142 (b) Since l = l0 eα t , then l0 = = l = e −α t 2.416 e −(1.682×10−4= α t e
= (2.416)(0.949194) = 2.293 m 5. The temperature θ 2 °C of an electrical conductor at time t seconds is given by: θ2 = θ1 (1 – e–t/T), where θ 1 is the initial temperature and T seconds is a constant. Determine (a) θ 2 when θ 1 = 159.9°C, t = 30 s and T = 80 s, and
(b) the time t for θ 2 to fall to half the value of θ 1 if T remains at 80 s
30 − (a) θ 2 = θ 1(1 – e–t/T) = 159.9 1 − e 80 = 159.9 (1 − e − 0.375 )= 159.9 (1 − 0.687289...) = 50°C t θ2 − = 1− e T θ1
(b) Since θ 2 = θ1 (1 – e–t/T) then from which, Then
and from which,
e
−
t T
= 1−
θ2 θ1
θ2 − Tt ln e = ln 1 − θ1 t θ2 − = ln 1 − T θ1 1 θ2 time, t = –T ln 1 − = –80 ln 1 − = –80 ln 0.5 = 55.45 s 2 θ1
265
© 2014, John Bird
6. A belt is in contact with a pulley for a sector of θ = 1.12 radians and the coefficient of friction between these two surfaces is µ = 0.26. Determine the tension on the taut side of the belt, T newtons, when tension on the slack side is given by T 0 = 22.7 newtons, given that these quantities are related by the law T = T0 eµθ. Tension T = T0 e µθ = 22.7 e( 0.26)(1.12) = 22.7 e0.2912 = 30.37 N
7. The instantaneous current i at time t is given by: i = 10e–t/CR when a capacitor is being charged. The capacitance C is 7 × 10− 6 farads and the resistance R is 0.3 × 106 ohms. Determine: (a) the instantaneous current when t is 2.5 seconds, and (b) the time for the instantaneous current to fall to 5 amperes. Sketch a curve of current against time from t = 0 to t = 6 seconds.
(a) Current, i = 10 e (b) When i = 5,
−
2.5 7×10−6 × 0.3×106
5 = 10 e
−
= 3.04 A
t 2.1
from which,
t 5 ln = − 2.1 10
Thus, Time t
0
Current i 10
1
and 2
3
t 5 − = e 2.1 10
time, t = −(2.1) ln 0.5 = 1.46 s 4
5
6
6.21 3.86 2.40 1.49 0.92 0.57
A graph of current against time is shown below.
266
© 2014, John Bird
8. The amount of product x (in mol/cm3) found in a chemical reaction starting with 2.5 mol/cm3 of reactant is given by: x = 2.5(1 – e– 4t) where t is the time, in minutes, to form product x. Plot a graph at 30-second intervals up to 2.5 minutes and determine x after 1 minute. Amount of product, x = 2.5(1 – e– 4t) Time t (min) 0 x (mol/cm3)
0.5 1.0
1.5
2.0
2.5
0 2.16 2.45 2.49 2.50 2.50
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The amount of product after 1 minute = 2.45 mol/cm3 9. The current i flowing in a capacitor at time t is given by: i = 12.5(1 – e–t/CR) where resistance R is 30 kΩ and the capacitance C is 20 µF. Determine (a) the current flowing after 0.5 seconds, and (b) the time for the current to reach 10 amperes.
t 0.5 − − (a) Current, = i 12.5 1 − e CR = 12.5 1 − e 20×10−6 ×30×103 = 7.07 A t t 10 − − (b) When i = 10 A, = from which, 10 12.5 1 − e 0.6 = 1 − e 0.6 12.5
Thus,
i.e.
e
−
t 0.6
= 1−
10 12.5
and
−
t 10 = ln 1 − 0.6 12.5
10 time, t = −0.6 ln 1 − = 0.966 s 12.5
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10. The amount A after n years of a sum invested P is given by the compound interest law: A = Per n/100 when the per unit interest rate r is added continuously. Determine, correct to the nearest pound, the amount after eight years for a sum of £1500 invested if the interest rate is 6% per annum.
rn
6×8
Amount after eight years, A= = P e100 (1500) = e 100 1500 e0.48 = (1500)(1.6160744) = £2424 correct to the nearest pound
11. The percentage concentration C of the starting material in a chemical reaction varies with time t according to the equation C = 100 e − 0.004t . Determine the concentration when (a) t = 0, (b) t = 100 s, (c) t = 1000 s. Percentage concentration, C = 100 e − 0.004t (a) When t = 0, C = 100 e0 = 100% (b) When t = 100, C = 100 e −0.004(100) = C = 100 e − 0.4 = 67.03% (c) When t = 1000, C = 100 e −0.004(1000) = C = 100 e − 4 = 1.83% 12. The current i flowing through a diode at room temperature is given= by: i iS ( e 40V − 1) amperes. Calculate the current flowing in a silicon diode when the reverse saturation current iS = 50 nA and the forward voltage V = 0.27 V Current in diode, i = iS ( e 40V − 1) = 50 ×10−9 ( e 40(0.27) − 1) = 50 ×10−9 ( e10.8 − 1) = 2.45 ×10−3 A = 2.45 mA
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© 2014, John Bird
t − 13. A formula for chemical decomposition is given by: C = A 1 − e 10 where t is the time in
seconds. Calculate the time, in milliseconds, for a compound to decompose to a value of C = 0.12 given A = 8.5
t − C = A 1 − e 10
hence
t − 0.12 = 8.5 1 − e 10
t 0.12 − = 1 − e 10 8.5
i.e.
from which,
e
−
t 10
= 1−
and
t − 0.12 10 ln e= ln 1 − 8.5
Hence,
0.12 time, t = −10 ln 1 − = 0.142 s = 142 ms 8.5
−
i.e.
0.12 8.5
t 0.12 = ln 1 − 10 8.5
14. The mass m of pollutant in a water reservoir decreases according to the law m = m0 e − 0.1t where t is the time in days and m0 is the initial mass. Calculate the percentage decrease in the mass after 60 days, correct to 3 decimal places. − 0.1t 0 Initially, i.e. at time t = 0, = m m0 e= m= m0 0e
−6 0.1(60) When t = 60, = m m0= e − 0.1t m0 e −= m= 0.00248m0 0e
Percentage decrease =
1 − 0.00248 ×100% = 99.752% 1
15. A metal bar is cooled with water. Its temperature, in °C, is given by: θ = 15 + 1300 e − 0.2 t where t is the time in minutes. Calculate how long it will take for the temperature, θ, to decrease to 36°C, correct to the nearest second. Temperature, θ = 15 + 1300 e − 0.2 t When θ = 36°,
36 = 15 + 1300 e − 0.2 t
and 36 – 15 = 1300 e − 0.2 t 270
© 2014, John Bird
Hence,
i.e.
from which,
36 − 15 = e − 0.2t 1300
and
36 − 15 − 0.2 t ln = ln e 1300
36 − 15 – 0.2t = ln 1300 time, t =
1 36 − 15 ln = 20.63 minutes − 0.2 1300
0.63 min = 0.63 × 60 s = 37.8 s = 38 s correct to the nearest second. Hence, time for the temperature to decrease to 36°C = 20 min 38 s
271
© 2014, John Bird
