Higher Engineering Mathematics, Sixth Edition
42 Higher Engineering Mathematics Hencesinhxand tanhxare both odd functions (see Section 5.1), as also are cosechx ( = 1 sinhx ) ...
Hyperbolic functions 43 Now try the following exercise Exercise 20 Further problemson evaluating hyperbolic functions In Problem ...
44 Higher Engineering Mathematics x 0 1 2 3 shx 0 1.18 3.63 10.02 chx 1 1.54 3.76 10.07 y=thx= shx chx 0 0.77 0.97 0.995 y=cothx ...
Hyperbolic functions 45 5.3 Hyperbolic identities For every trigonometric identity there is a corres- pondinghyperbolicidentity. ...
46 Higher Engineering Mathematics (c) Dividing each term in equation (1) by sh^2 x gives: ch^2 x sh^2 x − sh^2 x sh^2 x = 1 sh^2 ...
Hyperbolic functions 47 Now try the following exercise Exercise 21 Further problemson hyperbolic identities In Problems 1 to 4, ...
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) ...
Hyperbolic functions 49 2 shx+3chx=5[0.6389 or− 2 .2484] 4 thx− 1 =0[ 0 .2554] A chain hangs so that its shape is of the formy= ...
50 Higher Engineering Mathematics In the series expansion for shx,letx= 2 θ, then: sh2θ= 2 θ+ ( 2 θ)^3 3! + ( 2 θ)^5 5! +··· = 2 ...
Chapter 6 Arithmetic and geometric progressions 6.1 Arithmetic progressions When a sequence has a constant difference between su ...
52 Higher Engineering Mathematics Then’th term of an AP isa+(n− 1 )d The 6th term is: a+ 5 d= 17 (1) The 13th term is:a+ 12 d= 3 ...
Arithmetic and geometric progressions 53 Hence 35 × 2 7 = 2 a+ 7. 2 10 = 2 a+ 7. 2 Thus 2 a= 10 − 7. 2 = 2 .8, from which a= 2. ...
54 Higher Engineering Mathematics Problem 10. An oil company bores a hole 80m deep. Estimate the cost of boring if the cost is £ ...
Arithmetic and geometric progressions 55 Subtracting equation (1) from equation (2) gives Sn= a(rn−1) (r−1) which is valid whenr ...
56 Higher Engineering Mathematics The sum of 9 terms, S 9 = a( 1 −rn) ( 1 −r) = 72. 0 ( 1 − 0. 89 ) ( 1 − 0. 8 ) = 72. 0 ( 1 − 0 ...
Arithmetic and geometric progressions 57 (a) Let the GP bea,ar,ar^2 ,...,arn The first terma=£100 The common ratior= 1. 08 Hence ...
Chapter 7 The binomial series 7.1 Pascal’s triangle Abinomialexpressionisone whichcontains twoterms connected by a plus or minus ...
The binomial series 59 Hence (a+x)^7 =a^7 + 7 a^6 x+ 21 a^5 x^2 + 35 a^4 x^3 + 35 a^3 x^4 + 21 a^2 x^5 + 7 ax^6 +x^7 Problem 2. ...
60 Higher Engineering Mathematics Whena=2andn=7: ( 2 +x)^7 = 27 + 7 ( 2 )^6 x+ ( 7 )( 6 ) ( 2 )( 1 ) ( 2 )^5 x^2 + ( 7 )( 6 )( 5 ...
The binomial series 61 ( 1 +x)n= 1 +nx+ n(n− 1 ) 2! x^2 + n(n− 1 )(n− 2 ) 3! x^3 +··· ( 1. 002 )^9 =( 1 + 0. 002 )^9 Substitutin ...
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