1. No category

MA1104
Multivariable Calculus
AY 2005/2006 Sem 1
NATIONAL UNIVERSITY OF SINGAPORE
MATHEMATICS SOCIETY
PAST YEAR PAPER SOLUTIONS
with credits to Zheng Shaoxuan
MA1104 Multivariable Calculus
AY 2005/2006 Sem 1
Question 1
(a) Let P be the point of the given ellipsoid, closest to the given plane than any other point on the
ellipsoid. Furthermore, let f (x, y, z) = x2 + y 2 + 2x2 − 1 and g(x, y, z) = x + y + z − 100. Hence,
∇f = h2x, 2y, 4zi and ∇g = h1, 1, 1i.
By the given condition, the normal vector of the ellipsoid at the point P must be parallel to the
normal vector of the plane. Hence, ∇f = λ∇g and f (x, y, z) = 0. This yields the system of
equations:




2x
2y
4z


 2
2
x + y + 2z 2
=
=
=
=
λ;
λ;
λ;
1.
By substituting λ from the first three equations to the fourth,
1 2 1 2 1 2
λ + λ + λ = 1
4
4
8
r
λ = ±
8
.
5
Hence, by substituting λ into the above three equations, the point P has coordinates
r r r !
r
r !
r
2
2
1
2
2
1
,
,
,−
,−
or
−
5
5
10
5
5
10
A point on the given plane is h100, 0, 0i. Hence, a vector pointing from the plane to P is
*r
* r
r r +
r
r +
2
2
1
2
2
1
− 100,
,
or
−
− 100, −
,−
5
5
10
5
5
10
The distance of the former vector to the plane is given by
Dq
q q E
2
2
1
5 − 100,
5,
10 · h1, 1, 1i √
12 + 1 2 + 1 2
q
q
q
1
100 − 25 − 25 − 10
√
=
.
3
The distance of the latter vector to the plane is given by
D q
q
q E
1
− 25 − 100, − 25 , − 10
·
h1,
1,
1i
√
2 + 12 + 12
1
q
q
q
2
2
1
5 +
5 +
10 + 100
√
=
.
3
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AY 2005/2006 Sem 1
Since the former distance value is smaller, that is the smallest distance to be considered between
the ellipsoid and the plane. Hence, after simplification, the distance between the ellipsoid and the
plane is
√
√
200 3 − 30
.
6
(b)
(i) They are the cone and the hyperbolic paraboloid.
(ii) Choosing the cone with equation of z 2 = x2 + y 2 (the reason for this choice is that it is
the easiest one to explain), all the lines that lie on the surface can be described as r(t) for
0 ≤ θ ≤ 2π, t ∈ R, where
r(t) = ht cos θ, t sin θ, ti.
This is because r(t) is a line for each particular value of θ since the x, y and z coordinates
varies linearly with the variable t. Also, by substituting x = t cos θ and y = t sin θ into the
equation of the cone, we find that, indeed, z 2 = t2 cos2 θ+t2 sin2 θ = t2 , and hence, z = t or −t
which coincides with the particular z value that the line holds.
Question 2
(a) We have,
sin(x4 + y 4 )
(x,y)→(0,0) sin(x2 + y 2 )
lim
sin(r4 (cos4 θ + sin4 θ))
r→0 sin(r 2 (cos2 θ + sin2 θ))
r2
r4 (cos4 θ + sin4 θ)
sin(r4 (cos4 θ + sin4 θ))
·
·
= lim
r→0
sin r2
r2
r4 (cos4 θ + sin4 θ)
sin(r4 (cos4 θ + sin4 θ))
r2
= lim
·
lim
· lim r2 (cos4 θ + sin4 θ)
r→0
r→0 sin r 2 r→0
r4 (cos4 θ + sin4 θ)
= 1·1·0
= lim
= 0.
(b)
(i) We have,
lim
f (x, y)
(x,y)→(0,0)
=
x2 y 2
(x,y)→(0,0) x2 + y 2
lim
r4 cos2 θ sin2 θ
r→0 r 2 cos2 θ + r 2 sin2 θ
= lim r2 cos2 θ sin2 θ
= lim
r→0
= 0
= f (0, 0).
Therefore, f is continuous at (0, 0).
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AY 2005/2006 Sem 1
(ii) By the definition of partial differentiation,
f (h, 0) − f (0, 0)
h
0−0
= lim
h→0
h
= 0.
fx (0, 0) =
lim
h→0
By the definition of partial differentiation,
(0, h) − f (0, 0)
h
0−0
= lim
h→0
h
= 0.
fy (0, 0) =
lim
h→0
(iii) Note that for all real h,
f (h + t, 0) − f (h, 0)
t→0
t
0−0
= lim
h→0
t
= 0.
fx (h, 0) = lim
f (0, h + t) − f (0, h)
t→0
t
0−0
= lim
h→0
t
= 0.
fy (0, h) = lim
f (t, h) − f (0, h)
t→0
t
fx (0, h) = lim
= lim
t2 h2
t2 +h2
t
th2
= lim 2
t→0 t + h2
0
=
0 + h2
= 0.
t→0
f (h, t) − f (h, 0)
t→0
t
fy (h, 0) = lim
= lim
t2 h2
t2 +h2
t
th2
= lim 2
t→0 t + h2
0
=
0 + h2
= 0.
t→0
Hence,
fx (h, 0) − fx (0, 0)
h→0
h
0−0
= lim
h→0
h
= 0.
fxx (0, 0) =
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lim
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Multivariable Calculus
AY 2005/2006 Sem 1
fy (0, h) − fy (0, 0)
h→0
h
0−0
= lim
h→0
h
= 0.
fyy (0, 0) =
lim
fy (h, 0) − fy (0, 0)
h→0
h
0−0
= lim
h→0
h
= 0.
fyx (0, 0) =
lim
fx (0, h) − fx (0, 0)
h→0
h
0−0
= lim
h→0
h
= 0.
fxy (0, 0) =
lim
(iv) For all real h, let the proposition P (n) be the following:
P (n) :
∂nf
(h, 0) = 0, n ∈ N.
∂xn
P (1) is true as illustrated in (iii).
Assume that P (k) is true for k ∈ N. Consider P (k + 1):
∂ k+1 f
(h, 0) = lim
t→0
∂xk+1
∂k f
(h
∂xk
+ t, 0) −
t
∂k f
(h, 0)
∂xk
0−0
t→0
t
= 0.
= lim
Since P (1) is true and P (k) ⇒ P (k + 1), by Mathematical Induction, P (n) is true for all
n
natural numbers n. Therefore, ∂∂xnf (0, 0) = 0.
(v) No. It suffices to show that fxxy does not exist at (0, 0).
Working out the expression of fxx ,
fx (x, y) =
=
fxx (x, y) =
=
=
(x2 + y 2 )(2xy 2 ) − (x2 y 2 )(2x)
(x2 + y 2 )2
2xy 4
(x2 + y 2 )2
(x2 + y 2 )2 (2y 4 ) − (2xy 4 )(2)(x2 + y 2 )(2x)
(x2 + y 2 )4
2x2 y 4 + 2y 6 − 8x2 y 4
(x2 + y 2 )3
2y 4 (y 2 − 3x2 )
.
(x2 + y 2 )3
For any h 6= 0
2h4 (h2 )
(h2 )3
= 2.
fxx (0, h) =
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AY 2005/2006 Sem 1
Therefore,
fxx (0, h) − fxx (0, 0)
h
2
= lim .
h→0 h
fxxy (0, 0) =
lim
h→0
And hence fxxy does not exist at (0, 0).
Question 3
(i) Let g(x, y, z) = x2 + y 2 + z 2 − h, where 0 ≤ h ≤ 1. ∇f = h2x + y + z, 2y + x + z, 2z + x + yi,
λ∇g = h2xλ, 2yλ, 2zλi. By the Method of Lagrange Multiplier, ∇f = λ∇g and g(x, y, z) = 0,
resulting in the following system of equations:

2x + y + z = 2xλ;



x + 2y + z = 2yλ;
x + y + 2z = 2zλ;


 2
x + y 2 + z 2 = h.
By adding the first three equations, we obtain λ = 2. Hence, by substituting λ back into the
first three equations, it is easy to find out that x = y = z. Hence, for 0 ≤ h ≤ 1, the critical
points hold the following x, y and z values:
x2 = y 2 = z 2 =
h
.
3
By substituting this back into f (x, y, z):
f (x, y, z) = 2h.
And hence, the maximum value of f is 2 and the minimum value of f is 0.
(ii) We have,
Z
Z
f (x, y, z) ds =
C
=
=
=
=
=
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1
p
f (cos t, sin t, t) (− sin t)2 + (cos t)2 + 1 dt
0
√ Z 1
2
cos2 t + sin2 t + t2 + sin t cos t + t sin t + t cos t dt
0
√ Z 1
1
2
1 + t2 + sin 2t + t(sin t + cos t) dt
2
0
1
Z
√
1 3 1
2 t + t − cos 2t + t(sin t − cos t) − sin t − cos t dt
3
4
0
1
√
1
1
2 t + t3 − cos 2t + t(sin t − cos t) + sin t + cos t
3
4
0
√
7
1
2
− cos 2 + 2 sin 1 .
12 4
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Multivariable Calculus
AY 2005/2006 Sem 1
Question 4
(a) Let D be the domain of the unit circle around the origin. By Green’s Theorem,
Z
Z
(x2 + y 2 ) dx + (y 3 + x) dy
F · dr =
C
C
x
=
(1 − 2y) dA
D
Z
2π
Z
1
r(1 − 2r sin θ) dr dθ
=
0
Z
=
0
2π
=
=
=
=
1
r − 2r2 sin θ dr dθ
1
Z 2π 1 2 2 3
r − r sin θ dθ
2
3
0
0
Z 2π
1 2
− sin θ dθ
2 3
0
2π
1
2
θ + cos θ
2
3
0
2 2
π+ −
3 3
π.
0
=
Z
0
(b) Let G = hP, Q, Ri. Hence,
curl G =
∂R ∂Q ∂P
∂R ∂Q ∂P
−
,
−
,
−
∂y
∂z ∂z
∂x ∂x
∂y
.
If F = curl G, then by observation with the given F , a suitable G would be the following:
1 2
2 3 2
G(x, y, z) = xz , x y, y z .
2
2
Let C be the unit circle with the anticlockwise orientation, which is also the bounding curve of
surface S. C has equation r(t) = hcos t, sin t, 0i, where 0 ≤ t ≤ 2π. Since S is negatively oriented,
by Stoke’s Theorem,
x
x
F · dS =
curl G · dS
S
S
Z
= −
G · dr
C
Z 2π
G(cos t, sin t, 0) · r0 (t) dt
0
Z 2π 3
2
−
0, cos t sin t, 0 · h− sin t, cos t, 0i dt
2
0
Z 2π
3
−
cos3 t sin t dt
2 0
2π
3
1
4
=
− cos t
2
4
0
0.
= −
=
=
=
=
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AY 2005/2006 Sem 1
(c) This writer believes there is a typographic error in this question and hence will not provide a model
answer for it.
Question 5
R
H
(a) (1. ⇒
2.)
F
·
dr
is
independent
of
path
⇒
C
C F · dr = 0, and hence for any positive number
R
B, | C F · dr| < B whenever C is closed.
(2.
i.e. (NOT 1.) ⇒
shall be shown instead.
H (NOT 2.)
R ⇒ 1.) The equivalent contrapositive,
H
C F · dr = a, where a > 0. Let the
C F · dr is dependent on path ⇒ C F · dr 6= 0. Suppose
closed curve CM , where M is a positive
integer, be RC looped upon itself exactly M times. For all
R
B > 0, let M = d Ba e. Therefore, | CM F · dr| = M | C F · dr| = aM = ad Ba e ≥ B. This represents
the negation of (2.)
Hence, statement 1. and statement 2. are equivalent.
(b) We have,
curl F (x, y, z) = hcos z − cos z, cos x − cos x, cos y − cos yi
= h0, 0, 0i.
Therefore, F is conservative.
Question 6
(a) By spherical coordinates, B = {(ρ, θ, φ) | 0 ≤ ρ ≤ 1, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}. Hence,
y
Z
3
(x2 +y 2 +z 2 ) 2
e
dV
π
Z
2π
Z
1
=
0
B
0
Z
=
3
2) 2
ρ2 sin φ e(ρ
dρ dθ dφ
0
π
Z
2π
sin φ dφ
dθ
0
0
1 3
=
[2π] eρ
3
1
1
= (2)(2π)
e−
3
3
4π
=
(e − 1).
3
[− cos φ]π0
Z
1
2
ρ e
ρ3
dρ
0
1
0
(b) Figure 1 below shows one symmetric half of a slice of the visualization in the xz-plane for positive
x and positive z, with the sphere appearing as a semicircle and the cone appearing as a slanted
line. O is the origin, C is the center of the sphere and T is the touching point, also the tangent, of
the sphere with the cone on this slice.
π
π
It can be seen that ∠T OC =
since the cone has slant of magnitude 1, ∠CT O =
since
4
2
T is a tangent point of the sphere and the cone and CT = 1 as the sphere has unit radius.
1 1
Hence, OT = 1, and hence T has xz-coordinates √ , √ . By Pythagoras’ Theorem, C has
2 2
√
xz-coordinates (0, 2).
√
2 + (z − 2)2 = 1. In particular, the
The line has equations z = x, while the circle
has
equation
x
√
√
lower half of the circle has equation z = 2 − 1 − x2 .
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AY05-06 Sem 1 Qn 6b Diagram.jpg
The revolution of the shaded area about the z-axis is the region X. Since region X has unit density,
the mass of solid X is simply the volume of region X. By the volume of revolution using cylindrical
shells,
√1
2
Z
Mass of X =
p
√
2πx( 2 − 1 − x2 − x) dx
0
Z
= 2π
√1
2
√
2x − x2 − x
p
1 − x2 ) dx
0
=
=
=
=
"√
# √1
2
3
2 2 x3 1
2π
x −
+ (1 − x2 ) 2
2
3
3
!#0
"√
√
√
2
2 1
2
−
+
−1
2π
4
12
3
4
"√
#
2 1
2π
−
4
3
√
π
(3 2 − 4).
6
By rotational symmetry of region X about the z-axis, the x-coordinate and y-coordinate of the
center of mass of X are 0. The z-coordinate of the center of mass is simply the weighted average
of the middle of the heights of each cylindrical shell considered above. Hence,
R
z-coordinate of center of mass of X =
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1
√
2
0
√ √
√
√
2 +x
2πx( 2 − 1 − x2 − x) 2− 1−x
dx
2
√
.
π
6 (3 2 − 4)
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Multivariable Calculus
AY 2005/2006 Sem 1
Computing the integral in the numerator of the fraction,
√
√
p
√
2
−
1 − x2 + x
2πx( 2 − 1 − x2 − x)
dx
2
0
Z √1
p
√
2
x(( 2 − 1 − x2 )2 − x2 ) dx
= π
Z
√1
2
0
Z
= π
√1
2
√ p
x(2 − 2 2 1 − x2 + 1 − x2 − x2 ) dx
√1
2
√ p
3x − 2x3 − 2 2x 1 − x2 dx
0
Z
= π
0
# √1
√
2
3
3 2 1 4 2 2
π
x − x +
(1 − x2 ) 2
2
2
3
0
"
√ #
3 1 1 2 2
π
− + −
4 8 3
3
"
√ #
23 2 2
π
−
24
3
√
π
(23 − 16 2).
24
"
=
=
=
=
Hence, the z-coordinate of the center of mass of X is
√
π
− 16 2)
24 (23√
π
6 (3 2 − 4)
=
√
23 − 16 2
√
.
4(3 2 − 4)
And therefore, the center of mass of X has coordinates
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√ !
23 − 16 2
√
0, 0,
.
4(3 2 − 4)
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