Embed Size (px)
Citation preview
© 2014, John Bird
249
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) 17
e3.4629 (b) 8.52e–1.2651 (c) 2.6921
1.1171
5e3e
Using a calculator:
(a) 17
e3.4629 = 4.55848, correct to 5 decimal places
(b) 8.52e–1.2651 = 2.40444, correct to 5 decimal places
(c) 2.6921
1.1171
5e3e
= 8.05124, correct to 5 decimal places
4. Evaluate correct to 5 decimal places: (a) 2.1347
5.6823e−
(b) 2.1127 2.1127e e
2
−− (c) 1.7295
3.6817
4 (e 1)e
− −
© 2014, John Bird
250
Using a calculator:
2.1347
5.6823( )e
a−
= 48.04106, correct to 5 decimal places
2.1127 2.1127e e( )2
b−− = 4.07482, correct to 5 decimal places
( )1.7295
3.6817
4 e 1( )
ec
− − = –0.08286, correct to 5 decimal places
5. The length of a bar l at a temperature θ is given by l = l0eαθ, where l0 and α are constants. Evaluate l, correct to 4 significant figures, where l0 = 2.587, θ = 321.7 and α = 1.771 × 410− Using a calculator, ( )4321.7 1.771 10
0 e (2.587)el l αθ −× ×= = = 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
horizontal level: 2 2
ln L L kD kk
+ + = . Evaluate D when k = 75 m and L = 180 m.
2 2 2 2180 180 75 180 195ln 75 ln 75 ln 75 ln 5
75 75L L kD k
k + + + + + = = = =
= 120.7 m
© 2014, John Bird
251
EXERCISE 65 Page 138
1. Evaluate 5.6e–1, correct to 4 decimal places, using the power series for ex.
2 3 4e 1 ...
2! 3! 4!x x x xx= + + + + +
When x = –1, 1e− = 2 3 4 5 6( 1) ( 1) ( 1) ( 1) ( 1)1 ( 1) ...
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) 2 3 4
e 1 ...2! 3! 4!
x x x xx= + + + + +
When x = 2, 2e = 2 3 4 5 62 2 2 2 21 2 ...
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, 0.3e− = 2 3 4 5( 0.3) ( 0.3) ( 0.3) ( 0.3)1 0.3 ...
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
(1 – 2x) 2e x = (1 – 2x)2 3 4(2 ) (2 ) (2 )1 2 ...
2! 3! 4!x x xx + + + + +
© 2014, John Bird
252
= (1 – 2x) 2 3 44 21 2 23 3
x x x x + + + +
= 2 3 4 2 3 44 2 81 2 2 2 4 4 ...3 3 3
x x x x x x x x+ + + + − − − − −
= 2 3 481 2 23
x x x− − −
4. Expand (2e 2x )(x1/2) to six terms.
2 3 4e 1 ...
2! 3! 4!x x x xx= + + + + +
(2x1/2) 2ex = (2x1/2) ( )222 2 2 3 2 42 ( ) ( ) ( )1 ...
2! 3! 4! 5!xx x xx
+ + + + +
= (2x1/2)4 6 8 10
21 ...2 6 24 120x x x xx + + + + + +
= 1 1 1 14 6 8 101 1 2 2 2 22
2 22 2 2 22 2 ...
2 6 24 120x x x xx x
+ + +++ + + + +
= 1 5 9 13 17 212 2 2 2 2 2
1 1 12 2 ...3 12 60
x x x x x x+ + + + +
© 2014, John Bird
253
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 0.2e 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 = 12
e 1.5 x− over a range x = –1.5 to x = 1.5 and hence determine the value of y
when x = –0.8 and the value of x when y = 3.5
x – 1.5 – 1 – 0.5 0 0.5 1 1.5
y = 12
e 1.5 x− 4.74 2.24 1.06 0.50 0.24 0.11 0.05
A graph of y against x is shown below.
© 2014, John Bird
254
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
© 2014, John Bird
255
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
0.05250e tθ −= 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
© 2014, John Bird
256
© 2014, John Bird
257
EXERCISE 67 Page 142
1. Evaluate correct to 5 significant figures:
(a) 13
ln 5.2932 (b) ln82.4734.829
(c) 1.2942
5.62ln 321.62e
Using a calculator:
(a) 13
ln 5.2932 = 0.55547, correct to 5 significant figures
(b) ln82.4734.829
= 0.91374, correct to 5 significant figures
(c) 1.2942
5.62ln 321.62e
= 8.8941, correct to 5 significant figures
2. Evaluate correct to 5 significant figures:
(a) 1.76
1.41
1.786ln elg10
(b) 0.16295e
2ln 0.00165
− (c) ln 4.8629 ln 2.4711
5.173−
Using a calculator,
1.76
1.41
1.786ln e(a)lg10
= (1.786)(1.76)(1.41)
= 2.2293, correct to 5 significant figures
0.16295e(b)2 ln 0.00165
− = –0.33154, correct to 5 significant figures
ln 4.8629 ln 2.4711(c)5.173− = 0.13087, correct to 5 significant figures
3. Solve, correct to 4 significant figures: 1.5 = 4e2t
Rearranging 1.5 = 4e2t gives: 1.54
= e2t
Taking Napierian logarithms of both sides gives: ln 1.54
= ln(e2t)
Since logeeα = α, then ln 1.54
= 2t
© 2014, John Bird
258
Hence, t = 12
ln 1.54
= 12
(– 0.980829) = –0.4904, correct to 4 significant figures
4. Solve correct to 4 significant figures: 7.83 = 2.91e–1.7x
If 1.77.83 2.91e x−= then 1.7 7.83e2.91
− = and 1.7 7.83ln e ln2.91
x− =
i.e. –1.7x = 7.83ln2.91
and x = 1 7.83ln1.7 2.91
−
= –0.5822, correct to 4 significant figures
5. Solve correct to 4 significant figures: 16 = 24(1 – 2et
− )
If 216 24 1 et
− = −
then 216 1 e24
t−= −
from which, 216e 124
t− = −
and 16ln 12 24t − = −
and t = 162ln 124
− −
= 2.197, correct to 4 significant figures
6. Solve correct to 4 significant figures: 5.17 = ln4.64
x
From the definition of a logarithm, since 5.17 = ln4.64
x
then e5.17 = 4.64
x
Rearranging gives: x = 5.174.64e
i.e. x = 816.2, correct to 4 significant figures
7. Solve correct to 4 significant figures: 3.72 ln 1.59x
= 2.43
If 1.593.72ln 2.43x
=
then 1.59 2.43ln3.72x
=
© 2014, John Bird
259
from which, 2.433.72
1.59 ex
=
and x = 2.433.72
2.433.72
1.59 1.59ee
−
= = 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 = 2.4e = 11.02, correct to 4 significant figures
9. Solve correct to 4 significant figures: 224 e 45x+ =
Since 224 e 45x+ = then 2e 45 24 21x = − =
Then 2ln e ln 21x =
i.e. 2x = ln 21
from which, x = 1 ln 212
= 1.522
10. Solve correct to 4 significant figures: 15 e 7x+= −
Since 15 e 7x+= − then 1e 12x+ =
Then 1ln e ln12x+ =
i.e. x + 1 = ln 12
and x = (ln 12) –1 = 2.485 –1 = 1.485
11. Solve correct to 4 significant figures: 25 8 1 ex− = −
Rearranging 25 8 1 ex− = −
gives: 5
8 = 1 – 2e
x−
and 2ex
− = 1 – 58
= 38
Taking the reciprocal of both sides gives: 2ex = 8
3
© 2014, John Bird
260
Taking Napierian logarithms of both sides gives: ln 2ex = ln 8
3
i.e. 2x = ln 8
3
from which, x = 2 ln 83
= 1.962, correct to 4 significant figures
12. Solve, correct to 4 significant figures: ln(x + 3) – ln x = ln(x – 1).
Since ln(x + 3) – ln x = ln(x – 1) then ( )3ln ln 1x xx+ = −
from which, ( )3 1x xx+ = −
i.e. x + 3 = x(x – 1)
i.e. x + 3 = 2x x−
Rearranging gives: 2 2 3 0x x− − =
Factorizing gives: (x – 3)(x + 1) = 0
from which, 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: ( ) ( )2ln 1 ln 3 ln 1x x− − = −
Since ( ) ( )2ln 1 ln 3 ln 1x x− − = − then ( ) ( )21
ln ln 13
xx
− = −
from which, 2( 1) 1
3x x−
= −
i.e. 2 2 1 1
3x x x− +
= −
and 2 2 1 3 3x x x− + = −
i.e. 2 5 4 0x x− + =
Factorizing gives: (x – 4)(x – 1) = 0
from which, x = 4 or x = 1
© 2014, John Bird
261
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[ 2 2 3 6x x x− + − ] = 10 i.e. ln[ 2 6]x x+ − = 10 2 6x x+ − = 10e
and 2 6x x+ − – 10e = 0
i.e. 2 22 032.5 0x x+ − =
from which, x = 21 1 4(1)( 22032.5) 1 296.87
2(1) 2− ± − − − ±
= = 147.9 or –1.001
Neglecting the negative answer, x = 147.9
15. Solve correct to 4 significant figures: ( )2 5)( 1)e 3e xx −+ =
Since ( )2 5)( 1)e 3e xx −+ = then ( )( ) ( )( )1 2 5ln e ln 3ex x+ −=
i.e. ( )( ) ( )( )1 2 5ln e ln 3 ln ex x+ −= + 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: ( ) ( ) ( )2ln 1 1.5 ln 2 ln 1x x x+ = − − + +
Since ( ) ( ) ( )2ln 1 1.5 ln 2 ln 1x x x+ = − − + + then ( ) ( ) ( )2ln 1 ln 2 ln 1 1.5x x x+ + − − + =
© 2014, John Bird
262
and ( )2( 1) 2)ln 1.5
( 1)x x
x + −
= +
i.e. ( )ln ( 1)( 2) 1.5x x+ − = i.e. ( )2ln 2 2 1.5x x x− + − = and ( )2 1.52 2 ex x x− + − = i.e. 2 1.52 e 0x x− − − = i.e. 2 6.48169 0x x− − =
then x = ( ) ( )( )
( )
21 1 4 1 6.48169 1 26.926762 1 2
− − ± − − − ± =
= 1 26.926762
+ or 1 26.926762
−
= 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
and ln ln lne e e e e ab a D b a D b Dt += = = i.e. eb at D= since lne x x=
18. If 110
210 logP R
Q R =
find the value of 1R when 160, 8P Q= = and 2 5R =
Since 110
210 logP R
Q R =
then 110
2log
10P RQ R
=
and 110
210
PQ
RR
=
from which, 1R = 160
210 802 10 5 10 5 10PQR × = × = × = 5 × 100 = 500
19. If 2 1 eWPVU U
= make W the subject of the formula.
© 2014, John Bird
263
Since 2 1 eWPVU U
= then 2
1e
WPV
UU
= and 2
1ln ln e
WPV
UU
=
i.e. 2
1ln U W
U PV =
from which, 2
1ln UW PV
U =
20. The velocity 2v of a rocket is given by 12 1
2ln mv v C
m = +
where 1v is the initial rocket
velocity, C is the velocity of the jet exhaust gases, 1m is the mass of the rocket before the jet
engine is fired, and 2m is the mass of the rocket after the jet engine is switched off. Calculate
the velocity of the rocket given 1v = 600 m/s, C = 3500 m/s, 1m = 48.50 10× kg and
2m = 47.60 10× kg.
Velocity of rocket
412 1
42
8.50 10ln 600 3500ln7.60 10
mv v Cm
× = + = + ×
= 992 m/s 21. The work done in an isothermal expansion of a gas from pressure 1p to 2p is given by:
10
2ln pw w
p
=
If the initial pressure 1p = 7.0 kPa, calculate the final pressure 2p if 03w w=
If 03w w= then 03w = 10
2ln pw
p
i.e. 3 = 1
2ln p
p
and 13
2 2
7000e pp p
= =
from which, final pressure, 323
7000 7000ee
p −= = = 348.5 Pa
© 2014, John Bird
264
EXERCISE 68 Page 145
1. The temperature, T°C, of a cooling object varies with time, t minutes, according to the
equation: T = 0.04150e t− . Determine the temperature when (a) t = 0, (b) t = 10 minutes.
(a) When t = 0, T = 0.04 0150e 150et− = = 150°C
(b) When t = 10, T = 0.04 0.04(10) 0.4150e 150e 150et− − −= = = 100.5°C
2. The pressure p pascals at height h metres above ground level is given by: p = p0e– h/C, where p0 is
the pressure at ground level and C is a constant. Find pressure p when p0 = 1.012 ×105 Pa, height
h = 1420 m and C = 71 500
Pressure, p =1420
5 5715000 e (1.012 10 )e (1.012 10 )(0.980336)hCp
−− = × = × = 49.921 10× Pa or 99.21 kPa
3. The voltage drop, v volts, across an inductor L henrys at time t seconds is given by: v = 200eRtL
− ,
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 200eRtLv
−
= = ( )( )6
3
150 160 101.9212.5 10200e 200e
−
−
×− −× = = 29.32 volts
(b) When v = 85 V, 3150
12.5 1085 200et−
−×= from which, 3
15012.5 10
85 e200
t−
−×=
and 3
150 85ln12.5 10 200
t−
− = ×
Thus, time t = 312.5 10 85ln
150 200
−× −
= 671.31 10 s−×
© 2014, John Bird
265
4. The length l metres of a metal bar at temperature t°C is given by l = l0eαt, where l0 and α are
constants. Determine (a) the value of l when l0 = 1.894, α = 2.038 × 10 4− and t = 250°C, and
(b) the value of l0 when l = 2.416, t = 310°C and α = 1.682 × 410−
(a) Length, l = ( )( )42.038 10 250 0.050960 e 1.894e 1.894e (1.894)(1.05227)tl α −×= = = = 1.993 m
(b) Since l = 0 e tl α , then 0l = ( )( )41.682 10 310 0.052142e 2.416e 2.416ee
tt
l l αα
−− ×− −= = =
= (2.416)(0.949194) = 2.293 m
5. The temperature θ2°C of an electrical conductor at time t seconds is given by: θ2 = θ1(1 – e–t/T),
where θ1 is the initial temperature and T seconds is a constant.
Determine (a) θ2 when θ1 = 159.9°C, t = 30 s and T = 80 s,
and (b) the time t for θ2 to fall to half the value of θ1 if T remains at 80 s
(a) θ2 = θ1(1 – e–t/T) = ( ) ( )30
0.37580159.9 1 e 159.9 1 e 159.9 1 0.687289...− − − = − = −
= 50°C
(b) Since θ2 = θ1(1 – e–t/T) then 2
11 e
tT
θθ
−= −
from which, 2
1e 1
tT
θθ
− = −
Then 2
1ln e ln 1
tT
θθ
− = −
and 2
1ln 1t
Tθθ
− = −
from which, time, t = –T 2
1ln 1 θ
θ −
= –80 1ln 12
−
= –80 ln 0.5 = 55.45 s
© 2014, John Bird
266
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 T0 = 22.7 newtons, given that these
quantities are related by the law T = T0eµθ.
Tension 0 eT T µθ= = ( )( )0.26 1.12 0.291222.7e 22.7e= = 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 × 610− 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 = 6 62.5
7 10 0.3 1010e −−
× × × = 3.04 A
(b) When i = 5, 2.15 10et
−= from which, 2.15 e
10
t−=
Thus, 5ln10 2.1
t = −
and time, t = (2.1) ln 0.5− = 1.46 s
Time t 0 1 2 3 4 5 6
Current i 10 6.21 3.86 2.40 1.49 0.92 0.57 A graph of current against time is shown below.
© 2014, John Bird
267
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 0.5 1.0 1.5 2.0 2.5
x (mol/cm3) 0 2.16 2.45 2.49 2.50 2.50
© 2014, John Bird
268
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.
(a) Current, 12.5 1 et
CRi − = −
= 6 30.5
20 10 30 1012.5 1 e −−
× × × −
= 7.07 A
(b) When i = 10 A, 0.610 12.5 1 et
− = −
from which, 0.610 1 e
12.5
t−= −
Thus, 0.610e 1
12.5
t− = − and 10ln 1
0.6 12.5t − = −
i.e. time, t = 100.6 ln 112.5
− −
= 0.966 s
© 2014, John Bird
269
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.
Amount after eight years, A = 6 8
0.48100 100e (1500)e 1500er n
P×
= = = (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 0.004100e tC −= . Determine the concentration when (a) t = 0,
(b) t = 100 s, (c) t = 1000 s.
Percentage concentration, 0.004100e tC −= (a) When t = 0, 0100eC = = 100% (b) When t = 100, 0.004(100)100eC −= = 0.4100eC −= = 67.03% (c) When t = 1000, 0.004(1000)100eC −= = 4100eC −= = 1.83% 12. The current i flowing through a diode at room temperature is given by: ( )40e 1VSi i= − amperes.
Calculate the current flowing in a silicon diode when the reverse saturation current Si = 50 nA
and the forward voltage V = 0.27 V
Current in diode, ( ) ( ) ( )40 9 40(0.27) 9 10.8e 1 50 10 e 1 50 10 e 1VSi i − −= − = × − = × − = 32.45 10−× A = 2.45 mA
© 2014, John Bird
270
13. A formula for chemical decomposition is given by: C = 101 et
A − −
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
C = 101 et
A − −
hence 0.12 = 108.5 1 et
− −
i.e. 100.12 1 e8.5
t−= − from which, 10
0.12e 18.5
t− = −
and 100.12ln e ln 18.5
t− = −
i.e. 0.12ln 1
10 8.5t − = −
Hence, time, t = 0.1210ln 18.5
− −
= 0.142 s = 142 ms
14. The mass m of pollutant in a water reservoir decreases according to the law 0.10 e tm m −=
where t is the time in days and 0m is the initial mass. Calculate the percentage decrease in the
mass after 60 days, correct to 3 decimal places.
Initially, i.e. at time t = 0, 0.1 00 0 0e etm m m m−= = =
When t = 60, 0.1 0.1(60) 60 0 0 0e e e 0.00248tm m m m m− − −= = = =
Percentage decrease = 1 0.00248 100%1
−× = 99.752%
15. A metal bar is cooled with water. Its temperature, in °C, is given by: θ = 15 + 1300 0.2e 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 0.2e t− When θ = 36°, 36 = 15 + 1300 0.2e t− and 36 – 15 = 1300 0.2e t−
© 2014, John Bird
271
Hence, 0.236 15 e1300
t−−
= and 0.236 15ln ln e1300
t−− =
i.e. – 0.2t = 36 15ln1300−
from which, time, t = 1 36 15ln0.2 1300
− −
= 20.63 minutes
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
