34 Higher Engineering Mathematics
- 5= 8
(
1 −e
−x
2
)
[1.962]
- ln(x+ 3 )−lnx=ln(x− 1 ) [3]
- ln(x− 1 )^2 −ln3=ln(x− 1 ) [4]
- ln(x+ 3 )+ 2 = 12 −ln(x− 2 ) [147.9]
- e(x+^1 )=3e(^2 x−^5 ) [4.901]
- ln(x+ 1 )^2 = 1. 5 −ln(x− 2 )
+ln(x+ 1 ) [3.095] - Transpose:b=lnt−alnDto maketthe
subject.
[t=eb+alnD=ebealnD=ebelnD
a
i.e.t=ebDa]
- If
P
Q
=10log 10
(
R 1
R 2
)
find the value ofR 1
whenP= 160 ,Q=8andR 2 =5. [500]
- IfU 2 =U 1 e
(W
PV
)
makeWthe subject of the
formula.
[
W=PVln
(
U 2
U 1
)]
- The work done in an isothermal expansion of
a gas from pressurep 1 top 2 is given by:
w=w 0 ln
(
p 1
p 2
)
If the initial pressurep 1 = 7 .0 kPa, calculate
the final pressurep 2 ifw= 3 w 0.
[p 2 = 348 .5Pa]
4.5 Laws of growth and decay
The laws of exponential growth and decay are of the
formy=Ae−kxandy=A( 1 −e−kx), whereAandkare
constants. When plotted, the form of each of these equa-
tions is as shown in Fig. 4.5. The laws occur frequently
in engineering and science and examples of quantities
related by a natural law include.
(i) Linear expansion l=l 0 eαθ
(ii) Change in electrical resistance
with temperature Rθ=R 0 eαθ
(iii) Tension in belts T 1 =T 0 eμθ
(iv) Newton’s law of cooling θ=θ 0 e−kt
(a)
y
x
A
0
y 5 Ae^2 kx
(b)
y
x
A
0
y 5 A(1 2 e^2 kx)
Figure 4.5
(v) Biological growth y=y 0 ekt
(vi) Discharge of a capacitor q=Qe−t/CR
(vii) Atmospheric pressure p=p 0 e−h/c
(viii) Radioactive decay N=N 0 e−λt
(ix) Decay of current in an
inductive circuit i=Ie−Rt/L
(x) Growth of current in a
capacitive circuit i=I( 1 −e−t/CR)
Problem 17. The resistanceRof an electrical
conductor at temperatureθ◦Cisgivenby
R=R 0 eαθ,whereαis a constant and
R 0 = 5 × 103 ohms. Determine the value ofα,
correct to 4 significant figures, when
R= 6 × 103 ohms andθ= 1500 ◦C. Also, find the
temperature, correct to the nearest degree, when the
resistanceRis 5. 4 × 103 ohms.
TransposingR=R 0 eαθgives
R
R 0
=eαθ.
Taking Napierian logarithms of both sides gives:
ln
R
R 0
=lneαθ=αθ