Mathematical Methods for Physics and Engineering : A Comprehensive Guide

2.4 HINTS AND ANSWERS

2.45 IfJris the integral
∫∞

0

xrexp(−x^2 )dx

show that

(a) J 2 r+1=(r!)/2,

(b)J 2 r=2−r(2r−1)(2r−3)···(5)(3)(1)J 0.

2.46 Find positive constantsa,bsuch thatax≤sinx≤bxfor 0≤x≤π/2. Use
this inequality to find (to two significant figures) upper and lower bounds for the
integral

I=

∫π/ 2

0

(1+sinx)^1 /^2 dx.

Use the substitutiont=tan(x/2) to evaluateIexactly.
2.47 By noting that for 0≤η≤1,η^1 /^2 ≥η^3 /^4 ≥η, prove that

2

3

≤

1

a^5 /^2

∫a

0

(a^2 −x^2 )^3 /^4 dx≤

π

4

.

2.48 Show that the total length of the astroidx^2 /^3 +y^2 /^3 =a^2 /^3 ,whichcanbe
parameterised asx=acos^3 θ,y=asin^3 θ,is6a.
2.49 By noting that sinhx<^12 ex<coshx,andthat1+z^2 <(1 +z)^2 forz>0, show
that, forx>0, the lengthLof the curvey=^12 exmeasured from the origin
satisfies the inequalities sinhx<L<x+sinhx.
2.50 The equation of a cardioid in plane polar coordinates is

ρ=a(1−sinφ).

Sketch the curve and find (i) its area, (ii) its total length, (iii) the surface area of

the solid formed by rotating the cardioid about its axis of symmetry and (iv) the

volume of the same solid.

2.4 Hints and answers

2.1 (a) 3; (b) 2x+1,2,0;(c)cosx.
2.3 Use: the product rule in (a), (b), (d) and (e)[ 3 factors ]; the chain rule in (c), (f)
and (g); logarithmic differentiation in (g) and (h).
(a) (x^2 +2x)expx;(b)2(cos^2 x−sin^2 x)=2cos2x;
(c) 2 cos 2x;(d)sinax+axcosax;
(e) (aexpax)[(sinax+cosax)tan−^1 ax+(sinax)(1 +a^2 x^2 )−^1 ];
(f) [a(xa−x−a)]/[x(xa+x−a)]; (g) [(ax−a−x)lna]/(ax+a−x); (h) (1 + lnx)xx.
2.5 (a)−6(2x+3)−^4 ;(b)2sec^2 xtanx;(c)−9cosech^33 xcoth 3x;
(d)−x−^1 (lnx)−^2 ;(e)−(a^2 −x^2 )−^1 /^2 [sin−^1 (x/a)]−^2.
2.7 Calculatedy/dtanddx/dtand divide one by the other. (t+2)^2 /[2(t+1)^2 ].
Alternatively, eliminatetand finddy/dxby implicit differentiation.
2.9 −sinxin both cases.
2.11 The required conditions are 8n−4=0and4n^2 − 8 n+ 3 = 0; both are satisfied
byn=^12.
2.13 The stationary points are the zeros of 12x^3 +12x^2 − 24 x. The lowest stationary
value is−26 atx=−2; other stationary values are 6 atx= 0 and 1 atx=1.
2.15 Use logarithmic differentiation. Setdy/dx= 0, obtaining 2x^2 +2xlna+1=0.
2.17 See figure 2.14.

2.19

dy

dx

=−

(y

x

) 1 / 3

;

d^2 y

dx^2

=

a^2 /^3

3 x^4 /^3 y^1 /^3

.