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Calculus Homework
Assignment 5
1. Find volume of the largest right circular cone that can be inscribed in a sphere of radius 4.
[like §4.5 #12]
Sol. Let x denote the base radius of the right circular cone inscribed in the sphere with radius 4, and y be the perpendicular
distance form the base of the right circular cone to the center
of the sphere. As shown in following graph
4
y
4
x
p
We immediately have that x = 16 −p
y 2 . Since the volume of
the cone is V = 31 πr2 h, where r = x = 16 − y 2 and h = 4 + y,
thus we can express the volume of the cone as a function of y,
that is,
π
V (y) = (16 − y 2 )(4 + y)
3
π
0
So V (y) = 3 [(−2y)(4 + y) + (16 − y 2 )] = − π3 (4 + y)(−4 + 3y),
this implies V 0 (y) = 0 ⇔ y = −4 or y = 43 . But by definition
of y, y = −4 is impossible, so y = 43 is the only critical point
of V (y). Since V 00 (y) = − 2π
(4 + 3y) ⇒ V 00 ( 43 ) − 16π
< 0, and
3
3
4
−4 ≤ y ≤ 4 ⇒ V (−4) = V (4) = 0, so as y = 3 , we have the
.
maximal volume of the cone as V ( 43 ) = 2048π
81
2. Find the limits.
sin(a + 2h) − sin a
h→0
√ h
x
b. lim+ √
x→0
sin 3x
[like §4.6 #20,28]
Sol.
a. Since sin(a + 2h) − sin a = sin a − sin a = 0 as h = 0,
sin(a + 2h) − sin a
so
is an indeterminate form as h = 0.
h
Apply L’Hôpital’s Rule, we get
a. lim
sin(a + 2h) − sin a
2 cos(a + 2h)
= lim
= 2 cos a
h→0
h→0
h
1
lim
b.
√
x
lim+ √
=
x→0
sin 3x
r
lim+
x→0
x
1
1
= lim+ q
= lim+ √ q
x→0
sin 3x x→0
sin 3x
3 sin3x3x
x
1
1
= √ q
3
lim+
x→0
sin 3x
3x
1
1
= √ ·1= √
3
3
3. Evaluate
Z 1
2
a.
2 − cot x dx
2
Z
3 csc θ
b.
dθ
csc θ − sin θ
[like §4.8 #52,54]
Sol.
a.
Z Z
Z
Z
Z
1
1
1
2
2
2 − cot x dx =
(csc2 x − 1)dx
2dx −
cot xdx = 2 dx −
2
2
2
Z
Z
Z
1
1
= 2 dx −
csc2 xdx +
dx
2
2
Z
Z
5
1
5x + cot x
=
dx −
csc2 xdx =
+C
2
2
2
b.
Z
Z
Z
3
3 csc θ
3
sin θ
dθ =
dθ =
dθ
1
csc θ − sin θ
− sin θ
1 − sin2 θ
sin θ
Z
Z
dθ
= 3
= 3 sec2 θdθ = 3 tan θ + C
cos2 θ
4. Solve the initial value problem
y (4) = − cos x + 8 sin 2x;
y 000 (0) = 0, y 00 (0) = y 0 (0) = 1, y(0) = 3
[like §4.8 #86]
Sol. Since y (4) (x) = − cos x + 8 sin 2x, so
Z
000
y (x) = (− cos x + 8 sin 2x)dx = − sin x − 4 cos 2x + C1
Since y 000 (0) = 0, so C1 = 4 ⇒ y 000 (x) = − sin x − 4 cos 2x + 4.
Thus
Z
00
y (x) = (− sin x − 4 cos 2x + 4)dx = cos x − 2 sin 2x + 4x + C2
Since y 00 (0) = 1, so C2 = 0 ⇒ y 00 (x) = cos x − 2 sin 2x + 4x.
Thus
Z
0
y (x) = (cos x − 2 sin 2x + 4x)dx = sin x + cos 2x + 2x2 + C3
Since y 0 (0) = 1, so C3 = 0 ⇒ y 0 (x) = sin x + cos 2x + 2x2 . Thus
Z
1
2
y(x) = (sin x+cos 2x+2x2 )dx = − cos x+ sin 2x+ x3 +C4
2
3
Since y(0) = 3, so C4 = 4. Therefore, the solution of the initial
value problem is of the form
1
2
y(x) = − cos x + sin 2x + x3 + 4
2
3
5. Find the curve y = f (x) in the xy-plane that passes
√ through
the point (9, 4) and whose slope at each point is 4 x.
[like §4.8 #87]
√
Sol. Since the slope of the tangent line at each point is 4 x, so
√
dy
= f 0 (x) = 4 x. Thus we have that
dx
Z
Z
√
1
8 3
f (x) = 4 xdx = 4 x 2 dx = x 2 + C
3
Since the curve passes through (9, 4), that is, f (9) = 4, so
C = −68. Hence the curve is defined by the equation
8 3
y = x 2 − 68
3
6. (a) Inscribe a regular n-sided polygon inside a circle of radius
2 and compute the area of one of the n congruent triangles
formed by drawing radii to the vertices of the polygon.
(b) Compute the limit of the area of the inscribed polygon as
n → ∞.
[like §5.1 #22]
Sol.
(a) Note that for each congruent isosceles triangle has central
angle with measure 2π
. Thus the area of each isosceles
n
triangle is
1
2π
2π
= 2 sin
AT (n) = · 22 sin
2
n
n
(b) The area of the inscribed regular n-sided polygon is AP (n) =
nAT (n) = 2n sin 2π
. Hence
n
sin 2π
2π
lim AP (n) = lim 2n sin
= 4π lim 2π n = 4π
n→∞
n→∞
n→∞
n
n
7. Graph f (x) = 2 sin x + 1 over the interval [−π, π]. Partition
the interval into four subintervals of equal length. Then add
to your sketch the rectangles associated with the Riemann sum
4
P
f (ck )∆xk , given that ck is the (a) righthand endpoint, (b)
k=1
midpoint of the kth subinterval. (Make a separate sketch for
each set of rectangles.)
[like §5.2 #32]
Sol.
(a)
3
2
1
−π
–3
–2
−π/ 2
–1
0
0
–1
1
π/2
2
3
π
It is clear that ∆xk = π2 for each k = 1, 2, 3, 4. And the
righthand endpoint ck for kth subintervals are
π
π
c1 = − , c2 = 0, c3 = , c4 = π
2
2
respectively. Thus the associate Riemann sum is
4
X
f (ck )∆xk =
k=1
π
(−1 + 1 + 3 + 1) = 2π
2
(b)
3
2
1
−π
–3
–2
−π/ 2
–1
0
1
π/2
2
3
π
–1
It is clear that ∆xk = π2 for each k = 1, 2, 3, 4. And the
righthand endpoint ck for kth subintervals are
π
π
3π
3π
, c2 = − , c3 = , c4 =
4
4
4
4
respectively. Thus the associate Riemann sum is
c1 = −
4
X
k=1
f (ck )∆xk =
√
√
√
√ π
(1 − 2) + (1 − 2) + (1 + 2) + (1 + 2)
2
= 2π
8. For the function f (x) = 2x + 3x2 , find a formula for the upper
sum obtained by dividing the interval [0, 1] into n equal subintervals. Then take a limit of these sums as n → ∞ to calculate
the area under the curve over [0, 1].
[like §5.2 #40]
Sol. Since f 0 (x) = 2 + 6x which is positive on [0, 1], so f is
increasing on [0, 1]. Thus we can use right endpoints to obtain
upper sums. Now ∆x = n1 , and xk = nk , k = 1, · · · , n. So upper
sum is
n
n n
k1 X
X
X
2k 3k 2 1
Rn =
=
+ 2
f (xk )∆x =
f
n n k=1 n
n
n
k=1
k=1
n
n
2 X
3 X 2
=
k+ 3
k
n2 k=1
n k=1
3 n(n + 1)(2n + 1)
2 n(n + 1)
+ 3·
·
2
n
2
n
6
2
4n + 5n + 1
=
2n2
Therefore the area under the curve over [0, 1] is
=
4 + n5 +
4n2 + 5n + 1
=
lim
n→∞
n→∞
2n2
2
4
=
= 2.
2
lim Rn =
n→∞
lim
1
n2