Higher Engineering Mathematics, Sixth Edition

34 Higher Engineering Mathematics

11. 5= 8

(

1 −e

−x

2

)

[1.962]

12. ln(x+ 3 )−lnx=ln(x− 1 ) [3]


13. ln(x− 1 )^2 −ln3=ln(x− 1 ) [4]


14. ln(x+ 3 )+ 2 = 12 −ln(x− 2 ) [147.9]


15. e(x+^1 )=3e(^2 x−^5 ) [4.901]


16. ln(x+ 1 )^2 = 1. 5 −ln(x− 2 )
    +ln(x+ 1 ) [3.095]


17. Transpose:b=lnt−alnDto maketthe
    subject.
    [t=eb+alnD=ebealnD=ebelnD
    a

i.e.t=ebDa]

18. If

P

Q

=10log 10

(

R 1

R 2

)

find the value ofR 1

whenP= 160 ,Q=8andR 2 =5. [500]

19. IfU 2 =U 1 e

(W

PV

)

makeWthe subject of the

formula.

[

W=PVln

(

U 2

U 1

)]

20. 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