Solutions Manual
Advanced Calculus A Geometric View
By
James J. Callahan
( All Chapters Included - 100% Verified Solutions )
1
,Solutions: Chapter 1
Starting Points
1.1. The integrals are “improper,”, because there is an Therefore,
endpoint at ±∞. However, we know from the text that Z ∞ b
arctan x dx (arctan x)2 (π /2)2 π 2
Z b b = lim = = .
dx 0 1 + x2 b→∞ 2 0 2 8
= arctan x ;
a 1 + x2 a
1.5. a. The push-forward substitution u = ln w yields
basic facts about improper integrals (cf. Chap. 9 of the Z Z
text) then allow us to write dw du
= .
w(ln w) p up
Z ∞ b
dx
= lim arctan x = π /2; Hence
0 1 + x2 b→∞ 0
Z 1 1 Z ln(ln w), p = 1,
dx dw
= lim arctan x = π /4 − (−π /2) = 3π /4. = 1
−∞ 1 + x
2 a→−∞
a w(ln w) p , otherwise.
(1 − p)(lnw) p−1
1.2. We use the push-forward substitution u = 1 + x2, so 1.5. b. For the value of I to be finite, the expression just
that du = 2x dx and then given in part (a) must be finite as w → ∞. This rules out
Z Z 1 the first case (wherep = 1) and requires p − 1 > 0 in the
√ p
x dx 2 du second case. Thus, I is finite precisely when p > 1, and
= = 12 ln u = ln u = ln 1 + x2.
1 + x2 u then
b
1 −1
1.3. We use the pullback substitution x = R sin θ , so that I = = lim = .
b→∞ (1 − p)(lnw) p−1 (1 − p)(ln 2) p−1
dx = R cos θ d θ , R2 − x2 = R2 cos2 θ and θ = ±π /2 when 2
x = ±R. The result is
1.6. Assume first that the function x = ϕ (s) and its
Z Rp Z π /2
derivative are continuous on a ≤ s ≤ b. Then ϕ will have
R2 − x2 dx = R cos θ · R cos θ d θ an inverse on that interval if ϕ ′ (s) 6= 0 on a < s < b. To
−R −π /2
Z π /2 find the inverse we must solve the equation x = ϕ (s) for
1 + cos2θ
= R2 dθ the variable s.
−π /2 2
π /2 1.6. a. Here ϕ ′ = −1/s2, so ϕ is invertible on each of the
2 θ sin 2θ π R2 rays s > 0 and s < 0. The inverse is s = 1/x on each ray.
=R + = .
2 4 −π /2 2
1.6. b. Here ϕ ′ = 1 + 3s2 > 0 on the entire s-axis. The
inverse is the unique real root of s3 + s − x = 0 that is
1.4. Let u = arctan x (a push-forward substitution); then provided by Cardano’s formula:
du = dx/(1 + x2) so v v
u s u s
u 2 u
Z
arctan x dx
Z
u 2
(arctan x)2 t3 x x 1 t
3 x x2 1
= u du = = . s = + + + − + .
1 + x2 2 2 2 4 27 2 4 27
1
DVI file created at 14:27, 26 April 2011
2
,2 SOLUTIONS: CHAPTER 1. STARTING POINTS
1.6. c. Here ϕ ′ = (1 − s2 )/(1 + s2 )2 , so ϕ has an inverse 1.6. e. First of all, ϕ is defined only when −1 < s < 1. In
on each of the three domains s ≤ −1, −1 ≤ s ≤ 1, and that case,
1 ≤ s. The graph of ϕ , shown below, makes it clear that ϕ ϕ′ =
1
>0
has an inverse on each of the three sets. (1 − s2)3/2
so ϕ is invertible for −1 < s < 1. To get the formula for
0.4
the inverse, note that
0.2
s2
-20 -10 10 20 x2 = or x2 = s2 + s2 x2 .
-0.2 1 − s2
√
Solving for s gives s = ±x/ 1 + x2. Because x = ϕ (s)
-0.4
has the same sign as s, we must choose the plus sign in
To find a formula for the inverse, we need to solve the the formula above, so the inverse of ϕ is
equation x = s/(1 + s2) for s. We have
x
2 2
s= √ .
x + xs = s or xs − s + x = 0. 1 + x2
The roots of this quadratic equation are 1.6. f. Here ϕ (s) = ms + b is invertible everywhere pro-
√ vided ϕ ′ = m 6= 0. In that case, the inverse is
1 ± 1 − 4x2
s= , |x| ≤ 1/2.
2x x−b
s= .
m
This formula defines all three inverse functions. first of
all, with the minus sign, the graph is:
1.6. g. Here ϕ ′ = sinh s; because ϕ ′ is nonzero on each of
1.0 the rays s < 0 and s > 0, ϕ will have an inverse on each
0.5
of these domains. To find the formulas, we just adapt the
procedure for sinh s, above. We have
-0.4 -0.2 0.2 0.4
-0.5 2x = es + e−s or (es )2 − 2xes + 1 = 0.
-1.0
This time the roots give us the two inverse functions
Clearly, this is the inverse for −1 ≤ s ≤ 1. Note that there ( √
is an indeterminate form when x → 0, but l’Hopital’s rule − ln x + x2 − 1 < 0,
s= √
shows that s → 0. With the plus sign, the formula defines ln x + x2 − 1 > 0.
the inverse on each of the other two domains:
−1/2 ≤ x < 0 ⇒ s ≤ −1; 0 < x ≤ 1/2 ⇒ 1 ≤ s. 1.6. h. Here ϕ ′ = 1 − 3s2 is nonzero on three separate do-
mains:
1.6. d. Here ϕ ′ = cosh s > 0 for all s, so ϕ is invertible −1 −1 1 1
s< √ , √ <s< √ , √ < s.
for all s. To get the formula for the inverse, we first write 3 3 3 3
2x = es − e−s or 2xes = e2s − 1. In each case the inverse is a root of the cubic equation
s3 − s + x = 0 as provided by Cardano’s formulas.
This is the quadratic equation (es )2 − 2xes − 1 = 0, and 1.6. i. Here ϕ ′ = sech2 s ≥ 1, so ϕ is invertible for all s.
has the roots To find the inverse, we have
√
2
2x ± 4x + 4 p
es = = x ± x2 + 1. es − e−s e2s − 1
2 x = = or e2s (x − 1) = −x − 1.
es + e−s e2s + 1
The minus sign is untenable, because es > 0, so we obtain Therefore,
the formula for the inverse as r
p 1 1+x 1+x
s = ln x + x2 + 1 . s = ln = ln .
2 1−x 1−x
DVI file created at 14:27, 26 April 2011
3
, 3
√
1.6. j. Here ϕ ′ = −2/(1 + s)2 < 0 for all s 6= −1. Thus when ∆s ≈ 0. In this question, ϕ (s) = s and s0 = 100,
ϕ has an inverse on each ray s < −1, −1 < s. To get the so the approximation is given by
formula, we note
√ ∆s
1−x 100 + ∆s ≈ 10 + , ∆s ≈ 0.
x + xs = 1 − s or s = 20
1+x
Therefore we have
on each ray. √
102 ≈ 10.1, (∆s = +2),
1.7. a. Let θ = arcsin s, so s = sin θ . Therefore √
p 99.4 ≈ 9.97, (∆s = −0.6).
p
cos θ = 1 − sin θ = 1 − s = f (s),
2 2
sin θ s 1.9. c. Calculator values to nine decimal places are
tan θ = =√ = g(s). √ √
cos θ 1−s 2
102 = 10.099 504 938, 99.4 = 9.969 954 864.
1.7. b. Applying the chain rule to cos(arcsin s), we get Therefore the microscope estimates are too large by about
0.000 5 and 0.000 004 5, respectively.
1 −s
f ′ (s) = − sin(arcsin s) · √ =√ . 1.9. d. The microscope values √ are given by the tangent
1−s2 1 − s2
line to the graph of x = ϕ (s) = s at s = 100. The graph
√
Direct computation of the derivative of f (s) = 1 − s2 itself is concave down there (because ϕ ′′ (100) < 0), so it
gives the same result. lies below its tangent line at that point.
Applying the chain rule to tan(arcsin s), we get 1.10. a. Here ϕ ′ (s) = −1/s2 , so ϕ ′ (2) = −1/4 and the
1 microscope equation is ∆x ≈ −∆s/4. Therefore the ap-
g′ (s) = sec2 (arcsin s) · √ proximation is given by
1 − s2
1 1 1
= ·√ ≈ 0.5 − 0.25∆s, ∆s ≈ 0.
cos2 (arcsin s) 1 − s2 2 + ∆s
1 1 1 This gives us the estimates
= 2
·√ = .
1−s 1−s 2 (1 − s2 )3/2
√ 1
≈ 0.4925, (∆s = 0.03),
Direct computation of the derivative of g(s) = s/ 1 − s2 2.03
gives the same result. 1
≈ 0.505, (∆s = −0.02).
1.8. From the previous exercise we know that if we take 1.98
x = arcsin s, then
1.10. b. Calculator values are
p 3 ds
cos3 x = 1 − s2 = (1 − s2)3/2 , dx = √ . 1
= 0.492 610 837,
1
= 0.050 505 050.
1 − s2 2.03 1.98
Consequently, This time the microscope estimates are too small, by about
Z Z 3 −0.000 1 and −0.000 05, respectively.
s3 sin x
cos3 x dx = (1 − s2 ) ds = s − = sin x − ,
1.10. c. This time the tangent line (which gives the micro-
3 3
scope estimates) lies below the graph, because the graph
using s = sin x. is concave up: ϕ ′′ (2) > 0.
√
1.9. a. We have ϕ ′ (s) = 1/(2 s), so ϕ ′ (100) = 1/20 1.11. Let ϕ (s) = √s, and let s0 = 1. The approximation
and the microscope equation is we use (see the solution to Exercise 1.9) is
∆s ϕ (s0 + ∆s) ≈ ϕ (s0 ) + ϕ ′ (s0 )∆s, ∆s ≈ 0.
∆x ≈ ϕ ′ (100)∆s = .
20
Here ∆s = 2h, ϕ (s0 ) = 1, and ϕ ′ (s0 ) = 1/2, so our equa-
1.9. b. The microscope equation is used in the setting tion becomes
√
ϕ (s0 + ∆s) = ϕ (s0 ) + ∆x ≈ ϕ (s0 ) + ϕ ′ (s0 )∆s 1 + 2h ≈ 1 + h, h ≈ 0.
DVI file created at 14:27, 26 April 2011
4
Advanced Calculus A Geometric View
By
James J. Callahan
( All Chapters Included - 100% Verified Solutions )
1
,Solutions: Chapter 1
Starting Points
1.1. The integrals are “improper,”, because there is an Therefore,
endpoint at ±∞. However, we know from the text that Z ∞ b
arctan x dx (arctan x)2 (π /2)2 π 2
Z b b = lim = = .
dx 0 1 + x2 b→∞ 2 0 2 8
= arctan x ;
a 1 + x2 a
1.5. a. The push-forward substitution u = ln w yields
basic facts about improper integrals (cf. Chap. 9 of the Z Z
text) then allow us to write dw du
= .
w(ln w) p up
Z ∞ b
dx
= lim arctan x = π /2; Hence
0 1 + x2 b→∞ 0
Z 1 1 Z ln(ln w), p = 1,
dx dw
= lim arctan x = π /4 − (−π /2) = 3π /4. = 1
−∞ 1 + x
2 a→−∞
a w(ln w) p , otherwise.
(1 − p)(lnw) p−1
1.2. We use the push-forward substitution u = 1 + x2, so 1.5. b. For the value of I to be finite, the expression just
that du = 2x dx and then given in part (a) must be finite as w → ∞. This rules out
Z Z 1 the first case (wherep = 1) and requires p − 1 > 0 in the
√ p
x dx 2 du second case. Thus, I is finite precisely when p > 1, and
= = 12 ln u = ln u = ln 1 + x2.
1 + x2 u then
b
1 −1
1.3. We use the pullback substitution x = R sin θ , so that I = = lim = .
b→∞ (1 − p)(lnw) p−1 (1 − p)(ln 2) p−1
dx = R cos θ d θ , R2 − x2 = R2 cos2 θ and θ = ±π /2 when 2
x = ±R. The result is
1.6. Assume first that the function x = ϕ (s) and its
Z Rp Z π /2
derivative are continuous on a ≤ s ≤ b. Then ϕ will have
R2 − x2 dx = R cos θ · R cos θ d θ an inverse on that interval if ϕ ′ (s) 6= 0 on a < s < b. To
−R −π /2
Z π /2 find the inverse we must solve the equation x = ϕ (s) for
1 + cos2θ
= R2 dθ the variable s.
−π /2 2
π /2 1.6. a. Here ϕ ′ = −1/s2, so ϕ is invertible on each of the
2 θ sin 2θ π R2 rays s > 0 and s < 0. The inverse is s = 1/x on each ray.
=R + = .
2 4 −π /2 2
1.6. b. Here ϕ ′ = 1 + 3s2 > 0 on the entire s-axis. The
inverse is the unique real root of s3 + s − x = 0 that is
1.4. Let u = arctan x (a push-forward substitution); then provided by Cardano’s formula:
du = dx/(1 + x2) so v v
u s u s
u 2 u
Z
arctan x dx
Z
u 2
(arctan x)2 t3 x x 1 t
3 x x2 1
= u du = = . s = + + + − + .
1 + x2 2 2 2 4 27 2 4 27
1
DVI file created at 14:27, 26 April 2011
2
,2 SOLUTIONS: CHAPTER 1. STARTING POINTS
1.6. c. Here ϕ ′ = (1 − s2 )/(1 + s2 )2 , so ϕ has an inverse 1.6. e. First of all, ϕ is defined only when −1 < s < 1. In
on each of the three domains s ≤ −1, −1 ≤ s ≤ 1, and that case,
1 ≤ s. The graph of ϕ , shown below, makes it clear that ϕ ϕ′ =
1
>0
has an inverse on each of the three sets. (1 − s2)3/2
so ϕ is invertible for −1 < s < 1. To get the formula for
0.4
the inverse, note that
0.2
s2
-20 -10 10 20 x2 = or x2 = s2 + s2 x2 .
-0.2 1 − s2
√
Solving for s gives s = ±x/ 1 + x2. Because x = ϕ (s)
-0.4
has the same sign as s, we must choose the plus sign in
To find a formula for the inverse, we need to solve the the formula above, so the inverse of ϕ is
equation x = s/(1 + s2) for s. We have
x
2 2
s= √ .
x + xs = s or xs − s + x = 0. 1 + x2
The roots of this quadratic equation are 1.6. f. Here ϕ (s) = ms + b is invertible everywhere pro-
√ vided ϕ ′ = m 6= 0. In that case, the inverse is
1 ± 1 − 4x2
s= , |x| ≤ 1/2.
2x x−b
s= .
m
This formula defines all three inverse functions. first of
all, with the minus sign, the graph is:
1.6. g. Here ϕ ′ = sinh s; because ϕ ′ is nonzero on each of
1.0 the rays s < 0 and s > 0, ϕ will have an inverse on each
0.5
of these domains. To find the formulas, we just adapt the
procedure for sinh s, above. We have
-0.4 -0.2 0.2 0.4
-0.5 2x = es + e−s or (es )2 − 2xes + 1 = 0.
-1.0
This time the roots give us the two inverse functions
Clearly, this is the inverse for −1 ≤ s ≤ 1. Note that there ( √
is an indeterminate form when x → 0, but l’Hopital’s rule − ln x + x2 − 1 < 0,
s= √
shows that s → 0. With the plus sign, the formula defines ln x + x2 − 1 > 0.
the inverse on each of the other two domains:
−1/2 ≤ x < 0 ⇒ s ≤ −1; 0 < x ≤ 1/2 ⇒ 1 ≤ s. 1.6. h. Here ϕ ′ = 1 − 3s2 is nonzero on three separate do-
mains:
1.6. d. Here ϕ ′ = cosh s > 0 for all s, so ϕ is invertible −1 −1 1 1
s< √ , √ <s< √ , √ < s.
for all s. To get the formula for the inverse, we first write 3 3 3 3
2x = es − e−s or 2xes = e2s − 1. In each case the inverse is a root of the cubic equation
s3 − s + x = 0 as provided by Cardano’s formulas.
This is the quadratic equation (es )2 − 2xes − 1 = 0, and 1.6. i. Here ϕ ′ = sech2 s ≥ 1, so ϕ is invertible for all s.
has the roots To find the inverse, we have
√
2
2x ± 4x + 4 p
es = = x ± x2 + 1. es − e−s e2s − 1
2 x = = or e2s (x − 1) = −x − 1.
es + e−s e2s + 1
The minus sign is untenable, because es > 0, so we obtain Therefore,
the formula for the inverse as r
p 1 1+x 1+x
s = ln x + x2 + 1 . s = ln = ln .
2 1−x 1−x
DVI file created at 14:27, 26 April 2011
3
, 3
√
1.6. j. Here ϕ ′ = −2/(1 + s)2 < 0 for all s 6= −1. Thus when ∆s ≈ 0. In this question, ϕ (s) = s and s0 = 100,
ϕ has an inverse on each ray s < −1, −1 < s. To get the so the approximation is given by
formula, we note
√ ∆s
1−x 100 + ∆s ≈ 10 + , ∆s ≈ 0.
x + xs = 1 − s or s = 20
1+x
Therefore we have
on each ray. √
102 ≈ 10.1, (∆s = +2),
1.7. a. Let θ = arcsin s, so s = sin θ . Therefore √
p 99.4 ≈ 9.97, (∆s = −0.6).
p
cos θ = 1 − sin θ = 1 − s = f (s),
2 2
sin θ s 1.9. c. Calculator values to nine decimal places are
tan θ = =√ = g(s). √ √
cos θ 1−s 2
102 = 10.099 504 938, 99.4 = 9.969 954 864.
1.7. b. Applying the chain rule to cos(arcsin s), we get Therefore the microscope estimates are too large by about
0.000 5 and 0.000 004 5, respectively.
1 −s
f ′ (s) = − sin(arcsin s) · √ =√ . 1.9. d. The microscope values √ are given by the tangent
1−s2 1 − s2
line to the graph of x = ϕ (s) = s at s = 100. The graph
√
Direct computation of the derivative of f (s) = 1 − s2 itself is concave down there (because ϕ ′′ (100) < 0), so it
gives the same result. lies below its tangent line at that point.
Applying the chain rule to tan(arcsin s), we get 1.10. a. Here ϕ ′ (s) = −1/s2 , so ϕ ′ (2) = −1/4 and the
1 microscope equation is ∆x ≈ −∆s/4. Therefore the ap-
g′ (s) = sec2 (arcsin s) · √ proximation is given by
1 − s2
1 1 1
= ·√ ≈ 0.5 − 0.25∆s, ∆s ≈ 0.
cos2 (arcsin s) 1 − s2 2 + ∆s
1 1 1 This gives us the estimates
= 2
·√ = .
1−s 1−s 2 (1 − s2 )3/2
√ 1
≈ 0.4925, (∆s = 0.03),
Direct computation of the derivative of g(s) = s/ 1 − s2 2.03
gives the same result. 1
≈ 0.505, (∆s = −0.02).
1.8. From the previous exercise we know that if we take 1.98
x = arcsin s, then
1.10. b. Calculator values are
p 3 ds
cos3 x = 1 − s2 = (1 − s2)3/2 , dx = √ . 1
= 0.492 610 837,
1
= 0.050 505 050.
1 − s2 2.03 1.98
Consequently, This time the microscope estimates are too small, by about
Z Z 3 −0.000 1 and −0.000 05, respectively.
s3 sin x
cos3 x dx = (1 − s2 ) ds = s − = sin x − ,
1.10. c. This time the tangent line (which gives the micro-
3 3
scope estimates) lies below the graph, because the graph
using s = sin x. is concave up: ϕ ′′ (2) > 0.
√
1.9. a. We have ϕ ′ (s) = 1/(2 s), so ϕ ′ (100) = 1/20 1.11. Let ϕ (s) = √s, and let s0 = 1. The approximation
and the microscope equation is we use (see the solution to Exercise 1.9) is
∆s ϕ (s0 + ∆s) ≈ ϕ (s0 ) + ϕ ′ (s0 )∆s, ∆s ≈ 0.
∆x ≈ ϕ ′ (100)∆s = .
20
Here ∆s = 2h, ϕ (s0 ) = 1, and ϕ ′ (s0 ) = 1/2, so our equa-
1.9. b. The microscope equation is used in the setting tion becomes
√
ϕ (s0 + ∆s) = ϕ (s0 ) + ∆x ≈ ϕ (s0 ) + ϕ ′ (s0 )∆s 1 + 2h ≈ 1 + h, h ≈ 0.
DVI file created at 14:27, 26 April 2011
4
