Higher Engineering Mathematics, Sixth Edition

48 Higher Engineering Mathematics

(a) y=40 ch

x

40

,andwhenx=25,

y=40 ch

25

40

=40 ch 0. 625

= 40 ( 1. 2017536 ...)= 48. 07

(b) When y= 54. 30 , 54. 30 =40ch

x

40

, from which

ch

x

40

=

54. 30

40

= 1. 3575

Hence,

x

40

=cosh−^11. 3575 =± 0. 822219 ....

(see Fig. 5.2 for the reason as to why the answer

is±)from which,x= 40 (± 0. 822219 ....)=± 32. 89

Equations of the formachx+bshx=c,wherea,band

care constants may be solved either by:

(a) plotting graphs of y=achx+bshx and y=c

and noting the points of intersection, or more

accurately,

(b) by adopting the following procedure:

(i) Change shx to

(

ex−e−x

2

)

and chx to

(

ex+e−x

2

)

(ii) Rearrange the equation into the form

pex+qe−x+r=0, wherep,qandrare

constants.

(iii) Multiplyeach term by ex, which produces

an equation of the form p(ex)^2 +rex+

q=0(since(e−x)(ex)=e^0 = 1 )

(iv) Solvethequadraticequationp(ex)^2 +rex+

q=0forexby factorising or by using the

quadratic formula.

(v) Given ex=a constant (obtained by solv-

ing the equation in (iv)), take Napierian

logarithms of both sides to give

x=ln(constant)

This procedure is demonstrated in Problem 22.

Problem 22. Solve the equation

2 .6chx+ 5 .1shx= 8. 73 ,correct to 4 decimal

places.

Following the above procedure:

(i) 2.6chx+ 5 .1shx= 8. 73

i.e. 2. 6

(

ex+e−x

2

)

+ 5. 1

(

ex−e−x

2

)

= 8. 73

(ii) 1.3ex+ 1 .3e−x+ 2 .55ex− 2 .55e−x= 8. 73

i.e. 3.85ex− 1 .25e−x− 8. 73 = 0

(iii) 3. 85 (ex)^2 − 8 .73ex− 1. 25 = 0

(iv) ex

=

−(− 8. 73 )±

√

[(− 8. 73 )^2 − 4 ( 3. 85 )(− 1. 25 )]

2 ( 3. 85 )

=

8. 73 ±

√

95. 463

7. 70

=

8. 73 ± 9. 7705

7. 70

Hence ex= 2 .4027 or ex=− 0. 1351

(v) x=ln2.4027 orx=ln(−0.1351) which has no

real solution.

Hencex=0.8766, correct to 4 decimal places.

Now try the following exercise

Exercise 22 Further problems on

hyperbolic equations

In Problems 1 to 8, solve the given equations

correct to 4 decimal places.

1. (a) sinhx=1(b)shA=− 2. 43
   [(a) 0.8814 (b)− 1 .6209]


2. (a) coshB= 1 .87 (b) 2chx= 3
   [(a)± 1 .2384 (b)± 0 .9624]


3. (a) tanhy=− 0 .76 (b) 3thx= 2. 4
   [(a)− 0 .9962 (b) 1.0986]


4. (a) sechB= 0 .235 (b) sechZ= 0. 889
   [(a)± 2 .1272 (b)± 0 .4947]


5. (a) cosechθ= 1 .45 (b) 5 cosechx= 4. 35
   [(a) 0.6442 (b) 0.5401]


6. (a) cothx= 2 .54 (b) 2cothy=− 3. 64
   [(a) 0.4162 (b)− 0 .6176]


7. 3.5shx+ 2 .5chx=0[− 0 .8959]