Hints and answers

5.14 HINTS AND ANSWERS
5.33
If
1
I(α) =
0
xα − 1
dx,
ln x
α > −1,
what is the value of I(0)? Show that
d α
x = xα ln x,
dα
and deduce that
5.34
d
1
I(α) =
.
dα
α+1
Hence prove that I(α) = ln(1 + α).
Find the derivative, with respect to x, of the integral
3x
exp xt dt.
I(x) =
x
5.35
The function G(t, ξ) is defined for 0 ≤ t ≤ π by
#
− cos t sin ξ
for ξ ≤ t,
G(t, ξ) =
− sin t cos ξ
for ξ > t.
Show that the function x(t) defined by
π
x(t) =
G(t, ξ)f(ξ) dξ
0
satisfies the equation
d2 x
+ x = f(t),
dt2
where f(t) can be any arbitrary (continuous) function. Show further that x(0) =
[dx/dt]t=π = 0, again for any f(t), but that the value of x(π) does depend upon
the form of f(t).
[ The function G(t, ξ) is an example of a Green’s function, an important
concept in the solution of differential equations and one studied extensively in
later chapters. ]
5.14 Hints and answers
5.1
5.3
5.5
5.7
5.9
5.11
5.13
5.15
5.17
(a) (i) 2xy, x2 ; (ii) 2x, 2y; (iii) y −1 cos(x/y), (−x/y 2 ) cos(x/y);
(iv) −y/(x2 + y 2 ), x/(x2 + y 2 ); (v) x/r, y/r, z/r.
(b) (i) 2y, 0, 2x; (ii) 2, 2, 0; (v) (y 2 + z 2 )r−3 , (x2 + z 2 )r−3 , −xyr−3 .
(c) Both second derivatives are equal to (y 2 − x2 )(x2 + y 2 )−2 .
2x = −2y − x. For g, both sides of equation (5.9) equal y −2 .
∂2 z/∂x2 = 2xz(z 2 + x)−3 , ∂2 z/∂x∂y = (z 2 − x)(z 2 + x)−3 , ∂2 z/∂y 2 = −2z(z 2 + x)−3 .
(0, 0), (a/4, −a) and (16a, −8a). Only the saddle point at (0, 0).
The transformed equation is 2(x2 + y 2 )∂f/∂v = 0; hence f does not depend on v.
Maxima, equal to 1/8, at ±(1/2, −1/2), minima, equal to −1/8, at ±(1/2, 1/2),
saddle points, equalling 0, at (0, 0), (0, ±1), (±1, 0).
Maxima equal to a2 e−1 at (±a, 0), minima equal to −2a2 e−1 at (0, ±a), saddle
point equalling 0 at (0, 0).
Minimum at (0, 0); saddle points at (±1, ±1). To help with sketching the contours,
determine the behaviour of g(x) = f(x, x).
The Lagrange multiplier method gives z = y = x/2, for a maximal area of 4.
185
PARTIAL DIFFERENTIATION
5.19
5.21
5.23
5.25
5.27
5.29
5.31
5.33
5.35
The cost always includes 2αh, which can therefore be ignored in the optimisation.
With Lagrange multiplier λ, sin θ = λw/(4β) and β sec θ − 12 λw tan θ = λh, leading
to the stated results.
√
√
√
The envelope of the lines x/a + y/(c − a) − 1 = 0, as a is varied, is x + y = c.
2
Area = c /6.
(a) Using α = cot θ, where θ is the initial angle a jet makes with the vertical, the
equation is f(z, ρ, α) = z−ρα+[gρ2 (1+α2 )/(2v02 )], and setting ∂f/∂α = 0 gives
α = v02 /(gρ). The water bell has a parabolic profile z = v02 /(2g) − gρ2 /(2v02 ).
(b) Setting z = 0 gives the minimum diameter as 2v02 /g.
Show that (∂G/∂P )T = V and (∂G/∂T )P = −S. From each result, obtain an
expression for ∂2 G/∂T ∂P and equate these, giving (∂V /∂T )P = −(∂S/∂P )T .
Find expressions for (∂S/∂V )T and (∂S/∂T )V , and equate ∂2 S/∂V ∂T with
−1
∂2 S/∂T ∂V . U(V
∞, T ) = cT − aV .
dI/dy = −Im[ 0 exp(−xy + ix) dx] = −1/(1 + y 2 ). Integrate dI/dy from 0 to ∞.
I(∞) = 0 and I(0) = J.
Integrate the RHS of the equation by parts, before differentiating with respect
to y. Repeated application of the method establishes the result for all orders of
derivative.
I(0) = 0; use Leibnitz’
t rule.
π
Write x(t) = − cos t 0 sin ξ f(ξ) dξ − sin t t cos ξ f(ξ) dξ and differentiate each
term as a product to obtain dx/dt. Obtain d2 x/dt2 in a similar way. Note that
integrals
πthat have equal lower and upper limits have value zero. The value of
x(π) is 0 sin ξ f(ξ) dξ.
186