Hints and answers

2.4 HINTS AND ANSWERS
2.45
If Jr is the integral
∞
xr exp(−x2 ) dx
0
show that
(a) J2r+1 = (r!)/2,
(b) J2r = 2−r (2r − 1)(2r − 3) · · · (5)(3)(1) J0 .
2.46
Find positive constants a, b such that ax ≤ sin x ≤ bx for 0 ≤ x ≤ π/2. Use
this inequality to find (to two significant figures) upper and lower bounds for the
integral
π/2
I=
(1 + sin x)1/2 dx.
0
2.47
2.48
2.49
2.50
Use the substitution t = tan(x/2) to evaluate I exactly.
By noting that for 0 ≤ η ≤ 1, η 1/2 ≥ η 3/4 ≥ η, prove that
a
1
2
π
(a2 − x2 )3/4 dx ≤ .
≤ 5/2
3
a
4
0
Show that the total length of the astroid x2/3 + y 2/3 = a2/3 , which can be
parameterised as x = a cos3 θ, y = a sin3 θ, is 6a.
By noting that sinh x < 12 ex < cosh x, and that 1 + z 2 < (1 + z)2 for z > 0, show
that, for x > 0, the length L of the curve y = 12 ex measured from the origin
satisfies the inequalities sinh x < L < x + sinh x.
The equation of a cardioid in plane polar coordinates is
ρ = a(1 − sin φ).
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
2.3
2.5
2.7
2.9
2.11
2.13
2.15
2.17
2.19
(a) 3; (b) 2x + 1, 2, 0; (c) cos x.
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) (x2 + 2x) exp x; (b) 2(cos2 x − sin2 x) = 2 cos 2x;
(c) 2 cos 2x; (d) sin ax + ax cos ax;
(e) (a exp ax)[(sin ax + cos ax) tan−1 ax + (sin ax)(1 + a2 x2 )−1 ];
(f) [a(xa − x−a )]/[x(xa + x−a )]; (g) [(ax − a−x ) ln a]/(ax + a−x ); (h) (1 + ln x)xx .
(a) −6(2x + 3)−4 ; (b) 2 sec2 x tan x; (c) −9 cosech3 3x coth 3x;
(d) −x−1 (ln x)−2 ; (e) −(a2 − x2 )−1/2 [sin−1 (x/a)]−2 .
Calculate dy/dt and dx/dt and divide one by the other. (t + 2)2 /[2(t + 1)2 ].
Alternatively, eliminate t and find dy/dx by implicit differentiation.
− sin x in both cases.
The required conditions are 8n − 4 = 0 and 4n2 − 8n + 3 = 0; both are satisfied
by n = 12 .
The stationary points are the zeros of 12x3 + 12x2 − 24x. The lowest stationary
value is −26 at x = −2; other stationary values are 6 at x = 0 and 1 at x = 1.
Use logarithmic differentiation. Set dy/dx = 0, obtaining 2x2 + 2x ln a + 1 = 0.
See figure 2.14.
y 1/3
dy
d2 y
a2/3
;
= 4/3 1/3 .
=−
dx
x
dx2
3x y
81
PRELIMINARY CALCULUS
y
2a
x
πa
2πa
Figure 2.14 The solution to exercise 2.17.
2.21
2.23
2.25
2.27
2.29
2.31
2.33
2.35
2.37
2.39
2.41
2.43
2.45
2.47
2.49
(a) 2(2 − 9 cos2 x) sin x; (b) (2x−3 − 3x−1 ) sin x − (3x−2 + ln x) cos x; (c) 8(4x3 +
30x2 + 62x + 38) exp 2x.
(a) f(1) = 0 whilst f (1) = 0, and so f(x) must be negative in some region with
x = 1 as an endpoint.
(b) f (x) = tan2 x > 0 and f(0) = 0; g (x) = (− cos x)(tan x − x)/x2 , which is
never positive in the range.
The false result arises because tan nx is not differentiable at x = π/(2n), which
lies in the range 0 < x < π/n, and so the conditions for applying Rolle’s theorem
are not satisfied.
The relationship is x dy/dx = (2 − x)y.
By implicit differentiation, y (x) = (3x2 − 12)/(8 − 3y 2 ), giving y (±2) = 0. Since
y(2) = 4 and y(−2) = 0, the curve touches the x-axis at the point (−2, 0).
(a) Express in partial fractions; J = 13 ln[(x − 1)4 /(x + 2)] + c.
(b) Divide the numerator by the denominator and express the remainder in
partial fractions; J = x2 /4 + 4 ln(x + 2) − 3 ln(x + 3) + c.
(c) After division of the numerator by the denominator, the remainder can be
expressed as 2(x + 3)−1 − 5(x + 3)−2 ; J = 3x + 2 ln(x + 3) + 5(x + 3)−1 + c.
(d) Set x4 = u; J = (4a4 )−1 tan−1 (x4 /a4 ) + c.
Writing b2 − 4ac as ∆2 > 0, or 4ac − b2 as ∆ 2 > 0:
(i) ∆−1 ln[(2ax + b − ∆)/(2ax + b + ∆)] + k;
(ii) 2∆ −1 tan−1 [(2ax + b)/∆ ] + k;
(iii) −2(2ax + b)−1 + k.
f (x) = (1 + sin x)/ cos2 x = f(x) sec x; J = ln(f(x)) + c = ln(sec x + tan x) + c.
Note that dx = 2(b − a) cos θ sin θ dθ.
(a) π; (b) π(b − a)2 /8; (c) π(b − a)/2.
(a) (2 − y 2 ) cos y + 2y sin y − 2; (b) [(y 2 ln y)/2] + [(1 − y 2 )/4];
(c) y sin−1 y + (1 − y 2 )1/2 − 1;
−1
(d) ln(a2 + 1) − (1/y) ln(a2 + y 2 ) +√(2/a)[tan−1 (y/a)
√ − tan (1/a)].
Γ(n + 1) = nΓ(n); (a) (i) n!, (ii) 15 π/8; (b) −2 π.
By integrating twice, recover a multiple of In .
J2r+1 = rJ2r−1 and 2J2r = (2r − 1)J2r−2 .
Set η = 1 − (x/a)2 throughout, and x = a sin θ in one of the bounds.
1/2
x
dx.
L = 0 1 + 14 exp 2x
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