**Unformatted text preview: **Math 118: Honours Calculus II
Winter, 2014
Assignment 6
March 23 due April 3
1. Determine which of the following improper integrals converge and which diverge.
In the cases where they converge, except for question (e), compute their values.
(a) ∞
0 dx
.
a2 + x2 If a = 0, the integral diverges (due to the contributions near x = 0). For a = 0 we
ﬁnd
T
dx
T
π
1
lim
= lim arctan =
.
2 + x2
T →∞ 0 a
T →∞ a
a
2 |a| (b)
1− √ 0
δ lim δ→1− 0 dx
π
√
= lim arcsin δ = .
2
1 − x2 δ→1− (c) √
1+ x
√
dx.
x ∞
1 Since and √
∞
1 dx/ x dx
.
1 − x2 √
1
1+ x
√
>√
x
x
diverges, we know from the Comparison Theorem that
√
∞
1+ x
√
dx
x
1 also diverges. (d)
1 log x dx.
0+
1 lim t→0+ t 1 · log x dx = lim t→0+ x log x − x
dx
x 1
t = lim [x log x − x]1
t
t→0+ = lim (−1 − t log t + t) = −1 + lim
t→0+ t→0+ Hence the integral converges to the value −1. 1 1/t
= −1.
−1/t2 (e)
sin2 x
dx.
1 + x3/2 ∞
0 Since 1
sin2 x
≤ 3/2
3/2
1+x
x 0≤
and ∞ 1
x3/2 1 dx converges, we know by the Comparison Theorem that
∞
1 sin2 x
dx
1 + x3/2 converges. Hence
∞
0 sin2 x
dx =
1 + x3/2 1
0 sin2 x
dx +
1 + x3/2 ∞
1 sin2 x
dx
1 + x3/2 also converges. (f)
2
−2 1
dx.
(x − 1)4/5 We need to split the integral into two pieces:
2 t t + lim 5(x − 1)1/5
−2
t→1+
t→1−
t→1+ t
t→1− −2
√
5
= [0 − 5(−3)1/5 ] + [5 − 0] = 5 + 5 3.
√
Thus this improper integral is convergent and has the value 5 + 5 5 3.
(x − 1)−4/5 dx + lim lim (x − 1)−4/5 dx = lim 5(x − 1)1/5 2
t (g)
2
−2 1
dx.
(x − 1)7/3 We need to split the integral into two pieces:
t
t→1− 2 (x−1)−7/3 dx+ lim lim −2 t→1+ (x−1)−7/3 dx = lim t→1− t 3
− (x − 1)−4/3
4 t + lim
−2 t→1+ 3
− (x − 1)−4/3
4 Since these individual limits do not exist, the integral is divergent. 2. Let p and q be positive real numbers such that 1/p + 1/q = 1.
(a) Prove that the function f (x) = xp
p + x−q
q ≥ 1 on (0, ∞). We note that f ′ (x) = xp−1 − x−q−1 = 0 when xp+q = xpq = 1; in other words, when
x = 1. Since this is the only critical point and lim f (x) = lim f (x) = ∞, we see
x→0+ x→∞ that f has a global minimum at x = 1, where f (1) = 1/p + 1/q = 1. 2 2 .
t (b) Apply part (a) to a suitable choice for x to prove Young’s inequality: for all
real numbers A ≥ 0 and B ≥ 0,
AB ≤ Ap B q
+
.
p
q If either A or B is zero the result is trivial. Otherwise, on letting x = A/(AB)1/p , we
see that 1/x = (AB)1/p /A = (AB)1−1/q /A = B/(AB)1/q , so that Young’s inequality
follows immediately from part (a). (c) Given two positive integrable functions f and g on [a, b], show that the
choices
g(x)
f (x)
and
B=
A=
1/p
1/q
b p
b q
f (x) dx
g (x) dx
a
a
leads to H¨lder’s inequality:
o
b
a 1/p b fg ≤ f g a a fg ≤ F 1/p 1/q G b fp
a F p + b gq
a G =F q q . a
b p
a f (x) dx On integrating part (b) we see, denoting F =
b 1/q b p 1/p 1/q G 1 1
+
p q b q
a g (x) dx, and 1/p b f = that a 1/q b p g q . a 3. A piece of wire is bent into a quarter unit circle centered about the origin with
endpoints at (1, 0) and (0, 1).
(a) Compute the coordinates (x, y) of the centroid of the wire using integration. √
A quarter circle can be parametrized as y(x) = 1 − x2 for x ∈ [0, 1], for which
√
√
y ′ (x) = −x/ 1 − x2 and ds = 1/ 1 − x2 = 1/y. In terms of the length L = π/2 of
the wire, we ﬁnd
1 x = L−1 x ds = L−1
0 √ x
dx = L−1 −
2
1−x 1 − x2 1
0 = L−1 = 2
π and, as expected from the symmetry principle,
1 y = L−1 y ds = L−1 1 dx = L−1 =
0 2
.
π (b) Use Pappus’ ﬁrst theorem to check your answers in part (a).
We check the x centroid by computing the surface area 2πxL = 2π generated by
revolving about the y axis, which is indeed the surface area of a hemisphere. One can
also check the y centroid by revolving around the x axis. 3 (c) A quadratic Bezier approximation to a quarter unit circle is described by
the curve x(t) = 1 − t2 and y(t) = 2t − t2 , with t ∈ [0, 1]. Show that the arc
√ √
length of this curve is 1 + log(1 + 2)/ 2.
1 1 √
=2 2 1 t2
0
1
2 √
=4 2
√ √
1
− t + dt = 2 2
2 τ2 0
sinh−1 (1) 2 cosh u du = 2 = 1
t−
2 0 τ2 0 √ 1√
sinh−1 (1)
2
1+1+
2
2 1 1
2 √
1
+ dτ = 4 2
4 0 √ [−2t]2 + [2 − 2t]2 dt = 2 0 0 = 1 [x′ (t)]2 + [y ′ (t)]2 dt = L= 2 1
2 √
1
+ dt = 2 2
4 √
1
+ dτ = 2 2
4 1
u
2 sinh(2u) +
4
2 2t2 − 2t + 1 dt 0 1
−2 sinh−1 (1) 0
sinh−1 (1) τ2 + sinh2 u + 1 √ 1
= 2 sinh u
2
0
√
−1
log(1 + 2)
sinh (1)
√
√
=1+
,
=1+
2
2 where we have used the substitution t − 1
2 = τ followed by τ = 1
2 1
dτ
4
1
cosh u du
2 u
1 + sinh u +
2 sinh u. Note: The resulting arc length, 1.623, is about 3% greater than the arc length of a
true quarter unit circle. However, if one uses a cubic Bezier approximation, the arc
length is then accurate to 0.01%. 4. Calculate the surface area generated by revolving the curve y = 2x3/2 /3 from
x = 0 to x = 1 about the y axis.
The surface area is given by
1 1+ x 2π
0 dy
dx 2 1 dx = 2π √
x 1 + x dx. 0 Now use the substitution u = 1 + x to obtain
2 2π
1 2 5/2 2 3/2 2
u − u
5
3
1
1
10 √
3
1 5/2 1 3/2 1 1
12 √
5
2−
2−
= 4π 2 − 2 − +
+
= 4π
5
3
5 3
15
15
15 15
√
8
π 2+1 .
=
15 √
(u − 1) u du = 2π 2 (u3/2 − u1/2 ) du = 2π 5. Consider the curve γ described in polar coordinates as r(θ) = (θ − π)2 for
θ ∈ [0, 2π].
(a) Sketch γ in the x–y plane. 4 sinh−1 (1) 2 0 y x (b) Determine the portion of the area enclosed by γ that lies outside the circle
r(θ) = π 2 /4.
Since γ intersects the circle of radius π 2 /4 at 3π/2 and π/2 and the area of the
semicircle is π 5 /32, the desired area is
π/2 π5
32
0
π/2
π5
(θ − π)4 dθ −
=
32
0 1
A= 2
2 r 2 (θ) dθ − (θ − π)5
5
13 5
= π .
80 π/2 = (c) Use the fact that
to γ at θ = π/2. dy
dx = dy dx
/
dθ dθ 0 − π5
32 to compute the equation of the tangent line Since x(θ) = r(θ) cos θ and y(θ) = r(θ) sin θ we ﬁnd at θ = π/2 that
dy
2(θ − π) sin θ + (θ − π)2 cos θ
2(−π/2)
4
=
=
= .
2 sin θ
2
dx
2(θ − π) cos θ − (θ − π)
−(π/2)
π
So the equation is y = π2
4
x+ .
π
4 6. Let y = f (x) = x−3/2 − 1 and consider the curve γ = {(x, f (x)) : x ∈ [δ, 1]} in
the limit δ → 0+ . (a) Does the area bounded by the curve and the lines y = 0, x = 0, and x = 1
approach a ﬁnite value as δ → 0+ ? Why or why not? If it does, compute the
limiting area.
No. The corresponding improper integral is divergent:
1 lim δ→0+ δ x−3/2 − 1 dx = lim −2x−1/2 − x
δ→0+ 5 1
δ = ∞. (b) If the area in part (a) is fully rotated around the y axis, use the method of
cross-sections to determine whether or not the resulting volume approaches a
ﬁnite value as δ → 0+ . If it does, compute the limiting volume.
On slicing the y axis, we ﬁnd that the limiting volume is ﬁnite:
∞ π
0 ∞ 1
dy = π
(y + 1)4/3 ∞ (y + 1)−4/3 dy = π −3(y + 1)−1/3 0 = 3π. 0 (c) If the area in part (a) is fully rotated around the y axis, use the method of
cylindrical shells to determine whether or not the resulting volume approaches
a ﬁnite value as δ → 0+ . If it does, compute the limiting volume. We now slice the x axis to obtain shells of radius x, height f (x), and thickness dx.
The limiting volume is then
1 1 2π
0+ 1
(x−1/2 − x) dx = 2π 2x1/2 − x2
2
0+ x(x−3/2 − 1) dx = 2π 1 = 3π,
0+ in agreement with part (b). 7. The Euler-Mascheroni constant γ is deﬁned as the limiting diﬀerence between
the harmonic series and the logarithm:
n .
γ = lim n→∞ k=1 1
− log n .
k (a) Prove for every n ∈ N that
n k=1 n
k =1 1
log 1 +
k n log =
k=1
n+1 =
k=2 k+1
k 1
k = log(n + 1). n log 1 + n =
k=1 log(k + 1) − log(k)
k=1 n log(k) − k=1 log(k) = log(n + 1) − log(1) = log(n + 1). (b) Evaluate
lim log n→∞ = log lim n→∞ n
n+1 n
n+1 = log 1 = 0, noting that the natural logarithm is continuous. 6 (c) Use parts (a) and (b) to prove that
∞ 1
− log
k k=1 k+1
k = γ. In view of part (a) we ﬁnd that
∞ k =1 1
− log
k ∞ k+1
k n 1
1
− log 1 +
k
k =
k=1 n = lim n→∞ k=1 = lim n→∞ k=1 n
n+1 1
− log n + log
k 1
− log(n + 1)
k
= γ + 0 = γ. (d) Compute
∞ 1
1
.
−
k k+1 k=1 n = lim n→∞ k=1
n 1
−
k = lim n→∞ k=1 n k=1
n+1 1
−
k n 1
k+1 k=2 1
k = lim n→∞ = lim n→∞ k=1 1− 1
−
k 1
n+1 n+1
k=2 1
k = 1. (e) In the ﬁgure below, use parts (c) and (d) to determine the sum of the
brown+green shaded areas: 1 and the sum of the green areas alone: γ.
y y= (1,1) 1 2 ... k 7 1
x k+1 ... x ...

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