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The University of British Columbia
Final Examination – 12 December 2006
Mathematics 217
Multivariable and Vector Calculus
Closed book examination Time: 150 minutes
Special Instructions: To receive full credit, all answers must be supported with clear and correct derivations.
No calculators, notes, or other aids are allowed. A formula sheet is provided with the test.
[12] 1. A laser fired from the origin strikes the point P (1, 1 , 3) on the mirrored surface
z = 6 − (x − 2)
2
− 2(y − 2)
2
Find the point where the reflected beam strikes the plane z = 6.
Hint: The component of the incident beam direction that is normal to the mirror gets reversed by reflection;
the component parallel to the mirror is unchanged.
[13] 2. Astronaut Alpha patrols Sector Zero, the plane region x > 0, monitoring Cosmic Disorder (CD). The
true CD density at point (x, y) is given by a function Alpha does not know, namely,
f (x, y) = x
2
ye
−x
2 − 2 y
2
However, Alpha’s ship carries instruments that measure f (x, y) and ∇f (x, y) when it is at (x, y).
(a) Find and classify the critical points of f in Sector Zero.
(b) Alpha flies a mission where the ship’s coordinates at time t are given by
x = cos(t), y = 2 sin(t), t ≥ 0.
As Alpha passes through the point P (
1 2
3), does the on-board CD detector indicate that CD is
increasing or decreasing? At what rate?
(c) What direction should Alpha fly from P to maximize the instantaneous rate of increase in CD?
What is the angle between this direction and the line from P to the point of maximum CD?
(Note: A calculator-ready numerical expression for the cosine of the requested angle is fully
acceptable.)
Continued on page 2
12 December 2006 Mathematics 217 Page 2 of 3 pages
[13] 3. Given a > 0, consider the triangle D whose vertices are (0, 0), (0, a), (2a, a). Define a surface S by
z = 1 + 3x + 2y
2
(a) Find the plane tangent to S at the point where x = a and y = a.
(b) Find the area of the part of the tangent plane that lies above D. Call this area β(a).
(c) Find the area of the part of the surface S that lies above D. Call this area γ(a).
(d) [OPTIONAL BONUS QUESTION] Prove: lim
a→ 0 +
γ(a)
β(a)
[12] 4. Evaluate I =
0
∫ √y
0
ye
27 x−x
3
dx dy +
9
√ 18 −y
0
ye
27 x−x
3
dx dy.
[13] 5. Let C denote the closed loop in which the cylinder x
2
+ y
2
= 2ax meets the plane z = y. Given
F = y
2
i + tan
− 1
z j + (1 + x
2
) k, a > 0 ,
find the work done by F acting around C. Orient C counterclockwise when viewed from above.
[12] 6. Let F(x, y, z) =
y
2
cos z
i +
− xy
2
sin z
k.
(a) Prove that F is not conservative.
(b) Find a scalar field Q = Q(x, y, z) such that G is conservative, where
G(x, y, z) = F(x, y, z) + Q(x, y, z)j.
(c) Find W =
C
F • dr, given C: x = cos t, y = sin t, z = t, 0 ≤ t ≤ π/2.
[13] 7. Evaluate the flux I =
S
F • dS in each of the situations below.
(a) S is the boundary surface for the solid cylinder E =
(x, y, z) : x
2
+ y
2
≤ a
2
, 0 ≤ z ≤ H
, and
F(x, y, z) =
−yze
xyz
, xze
xyz
, xy
2
e
z
+ z
2
(b) S =
(x, y, z) : 0 ≤ z = 9 − x
2
− y
2
and F = 〈xy, yz, xz〉. (Use upward orientation on S.)
(c) S is the two-part surface with bottom z =
x
2
+ y
2
and top z =
a
2
− x
2
− y
2
, and
F =
xy
2
, yz
2
, zx
2
Continued on page 3
MATH 217 FORMULAS FOR FINAL EXAMINATION, 12 DECEMBER 2006
VECTOR IDENTITIES
For u = u 1 i + u 2 j + u 3 k, v = v 1 i + v 2 j + v 3 k, w = w 1 i + w 2 j + w 3 k,
u • v = u 1 v 1 + u 2 v 2 + u 3 v 3 = |u| |v| cos(θ), 0 ≤ θ ≤ π u×v =
i j k
u 1 u 2 u 3
v 1 v 2 v 3
= (u 2 v 3 −u 3 v 2 )i+(u 3 v 1 −u 1 v 3 )j+(u 1 v 2 −u 2 v 1 )k
Length of u: |u| =
u • u =
u^21 + u^22 + u^23 Angle between u and v: θ = cos
− 1
u • v
|u||v|
, 0 ≤ θ ≤ π
Triple product identities: u • (v × w) = v • (w × u) = w • (u × v) u × (v × w) = (u • w)v − (u • v)w
DISTANCES AND PROJECTIONS
Point (x 0 , y 0 , z 0 ) to plane Ax + By + Cz = D: s =
|Ax 0 + By 0 + Cz 0 − D| √ A^2 + B^2 + C^2
F = proju(F) + orthu(F)
Point r 0 = (x 0 , y 0 , z 0 ) to line r = r 1 + tv: s =
|(r 0 − r 1 ) × v|
|v|
proju(F) =
F • u
u • u
u
Line r = r 1 + tv 1 to line r = r 2 + tv 2 : s =
|(r 2 − r 1 ) • (v 1 × v 2 )|
|v 1 × v 2 |
orthu(F) = F − proju(F) =
(u • u)F − (F • u)u
u • u
VECTOR-VALUED FUNCTIONS OF ONE VARIABLE
d
dt
(λ(t)u(t)) = λ
′ (t)u(t) + λ(t)u
′ (t)
d
dt
(u(t) • v(t)) = u
′ (t) • v(t) + u(t) • v
′ (t)
d
dt
(u(t) × v(t)) = u
′ (t) × v(t) + u(t) × v
′ (t)
d
dt
(u(λ(t))) = λ
′ (t)u
′ (λ(t))
d
dt
|u(t)| =
u(t) • u
′ (t)
|u(t)|
, u(t) 6 = 0
Position r = r(t) gives velocity v(t) = r
′ (t), speed v(t) = |v(t)| , acceleration a(t) = v
′ (t) = r
′′ (t) =
dv
dt
T +
v
2
ρ
N; T̂ =
v
|v|
ds = v(t) dt = |v(t)| dt =
dr
dt
∣ dt^ =^ |dr|^ dr^ =^
dr
dt
dt = v(t) dt =
dx
dt
dy
dt
dz
dt
dt = 〈dx, dy, dz〉 dt
APPROXIMATIONS
Differentiability test for scalar field f at a: 0 = lim x→a
E(x)
|x − a|
, where E(x) = f (x) − f (a) − ∇f (a) • (x − a)
Tangent Hyperplane for G(x) = 0 at a: 0 = ∇G(a) • (x − a) (a line in R
2 ; a plane in R
3 ; a hyperplane in R
n )
Linearization of f around a: f (x) ≈ L(x) for x ≈ a, where L(x) = f (a) + (x − a) • ∇f (a)
Differentials (case x ∈ R
3 ): df =
∂f
∂x
dx +
∂f
∂y
dy +
∂f
∂z
dz = ∇f • dr ∆f ≈
∂f
∂x
∆x +
∂f
∂y
∆y +
∂f
∂z
∆z = ∇f • ∆r
Quadratic Approx, for (x, y) ∈ R
2 near (a, b): f (x, y) ≈ Q(x, y) = f (a, b) + f 1 (a, b)(x − a) + f 2 (a, b)(y − b)
2
[
f 11 (a, b)(x − a)
2
2
- 2f 12 (a, b)(x − a)(y − b)
]
SECOND DERIVATIVE TEST FOR (a, b) WHERE ∇f (a, b) = (0, 0)
H(x, y) =
[
f 11 (x, y) f 12 (x, y)
f 21 (x, y) f 22 (x, y)
]
H(a, b) =
[
A B
B D
]
∆ = det(H(a, b)) = AD − B
2
∆ < 0 =⇒ saddle ∆ > 0 , A > 0 =⇒ loc min ∆ > 0 , A < 0 =⇒ loc max
DERIVATIVES
∇ = i
∂x
∂y
∂z
∂x
∂y
∂z
F(x, y, z) = F 1 (x, y, z) i + F 2 (x, y, z) j + F 3 (x, y, z) k
∇φ(x, y, z) = grad φ(x, y, z) =
∂φ
∂x
i +
∂φ
∂y
j +
∂φ
∂z
k ∇ • F(x, y, z) = div F(x, y, z) =
∂F 1
∂x
∂F 2
∂y
∂F 3
∂z
∇ × F(x, y, z) = curl F(x, y, z) =
i j k
∂ ∂x
∂y
∂z
F 1 F 2 F 3
∂F 3
∂y
∂F 2
∂z
i +
∂F 1
∂z
∂F 3
∂x
j +
∂F 2
∂x
∂F 1
∂y
k
∇(φψ) = φ∇ψ + ψ∇φ ∇ • (F × G) = (∇ × F) • G − F • (∇ × G)
∇ • (φF) = (∇φ) • F + φ (∇ • F) ∇ × (F × G) = F(∇ • G) − G (∇ • F) − (F •∇ )G + (G •∇ )F
∇ × (φF) = (∇φ) × F + φ (∇ × F) ∇(F • G) = F × (∇ × G) + G × (∇ × F) + (F •∇ )G + (G •∇ )F
∇ × (∇φ) = 0 (curl grad = 0 ) ∇ • (∇ × F) = 0 (div curl = 0)
2 φ(x, y, z) = ∇ • ∇φ(x, y, z) = div grad φ =
2 φ
∂x^2
2 φ
∂y^2
2 φ
∂z^2
∇ × (∇ × F) = ∇(∇ • F) − ∇
2 F (curl curl = grad div − laplacian)
File “fecs”, version of 11 December 2006, page 1. Typeset at 22:34 December 11, 2006.
SURFACE NORMALS AND AREA ELEMENTS
r = r(u, v) (parametrized surface): n =
∂r
∂u
×
∂r
∂v
; N̂ = ±
n
|n|
dS = ±
∂r
∂u
×
∂r
∂v
du dv
G(x, y, z) = 0 (smooth level surface): n = ∇G(x, y, z); N̂ = ±
n
|n|
dS = ±
∇G(x, y, z)
|∂G/∂z|
dx dy
dS =
n
|n • k|
dx dy =
n
|n • j|
dx dz =
n
|n • i|
dy dz for other projections dS = N̂ dS; dS = |dS|
POLAR AND CYLINDRICAL COORDINATES
Transformation: x = r cos θ, y = r sin θ, z = z Position vector: r = r cos θ i + r sin θ j + zk
Volume element: dV = r dr dθ dz Surface area element (on r = a): dS = a dθ dz
Surface area element (on z = 0): dS = r dr dθ
SPHERICAL COORDINATES
Transformation: x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ Position vector: r = ρ sin φ cos θ i + ρ sin φ sin θ j + ρ cos φk
Volume element: dV = ρ
2 sin φ dρ dφ dθ Surface area element (on ρ = a): dS = a
2 sin φ dθ dφ
INTEGRATING DERIVATIVES: THE FUNDAMENTAL THEOREM OF CALCULUS
∫ (^) b
a
f
′ (t) dt = f (b) − f (a) (the one-dimensional Fundamental Theorem)
C
∇φ • dr = φ
r(b)
− φ
r(a)
, if C is the curve r = r(t), a ≤ t ≤ b F(x, y, z) = P (x, y, z)i+Q(x, y, z)j+R(x, y, z)k
D
∂Q
∂x
∂P
∂y
dA =
∂D
F • dr =
∂D
P (x, y) dx + Q(x, y) dy, where ∂D is the positively oriented boundary of D (Green’s Theorem)
S
∇×F• N̂ dS =
∂S
F•dr =
∂S
P (x, y, z) dx+Q(x, y, z) dy +R(x, y, z) dz, where ∂S is the oriented boundary of S (Stokes’s Theorem)
E
∇ • F dV =
∂E
F • N̂ dS, where ∂E is the closed boundary of E, with outward unit normal N (Divergence Theorem)̂
AVERAGE VALUES FOR FUNCTION f ON CURVE C, FUNCTION g ON SURFACE S, FUNCTION h ON SOLID E
f =
C
f ds
∫
C
1 ds
g =
S
g dS
∫∫
S
1 dS
h =
E
h(x, y, z) dV
∫∫∫
E
1 dV
SINGLE INTEGRALS
x sin(bx) dx =
sin(bx)
b^2
x cos(bx)
b
x cos(bx) dx =
cos(bx)
b^2
x sin(bx)
b
xe
ax dx =
e
ax
a^2
(ax − 1)
e
ax sin(bx) dx =
eax^ (a sin bx − b cos bx)
a^2 + b^2
e
ax cos(bx) dx =
eax^ (a cos bx + b sin bx)
a^2 + b^2
dx √ a
2 − x
2
= sin
− 1
x
a
sec
2 x dx = tan x
sin
2 x dx =
x
2
sin 2x
cos
2 x dx =
x
2
sin 2x + C
∫
tan x dx = ln | sec x|
sin
3 x dx =
cos
3 x − cos x
cos
3 x dx = sin x −
sin
3 x
tan
2 x dx = tan x − x
dx
a^2 + x^2
a
tan
− 1 x
a
(a > 0)
dx
a^2 − x^2
2 a
ln
x + a
x − a
, (a > 0)
dx
(a^2 − x^2 )^3 /^2
x
a^2
a^2 − x^2
dx √ a^2 − x^2
= sin
− 1
x
a
(a > 0)
a^2 − x^2 dx =
x
2
a^2 − x^2 +
a
2
sin
− 1
x
a
dx
(x^2 ± a^2 )^3 /^2
±x
a
2
x
2 ± a
2
dx √ x
2 ± a
2
= ln
∣x +
x
2 ± a
2
x
2 ± a
2 dx =
x
x
2 ± a
2 ±
a^2
ln
∣x +
x
2 ± a
2
∫ (^) π/ 2
0
sin x dx =
∫ (^) π/ 2
0
cos x dx = 1
∫ (^) π/ 2
0
sin
2 x dx =
∫ (^) π/ 2
0
cos
2 x dx =
π
∫ (^) π/ 2
0
sin
3 x dx =
∫ (^) π/ 2
0
cos
3 x dx =
∫ (^) π/ 2
0
sin
4 x dx =
∫ (^) π/ 2
0
cos
4 x dx =
3 π
16
∫ (^) π/ 2
0
sin
5 x dx =
∫ (^) π/ 2
0
cos
5 x dx =
∫ (^) π/ 2
0
sin
6 x dx =
∫ (^) π/ 2
0
cos
6 x dx =
5 π
32
TRIGONOMETRIC IDENTITIES
sin
2 x + cos^2 x = 1 sin(−x) = − sin x cos(−x) = cos x
sec
2 x = 1 + tan
2 x csc
2 x = 1 + cot
2 x tan(x ± y) =
tan x ± tan y
1 ∓ tan x tan y
sin(x ± y) = sin x cos y ± cos x sin y cos(x ± y) = cos x cos y ∓ sin x sin y sin
π
2
= 1 = cos(0)
sin
2 x =
1 − cos 2x
cos^2 x =
1 + cos 2x
sin
π
= cos
π
sin(0) = 0 = cos
π
2
sin
π
6
= cos
π
3
sin
π
3
= cos
π
6
Adapted from R. A. Adams, Calculus, A Complete Course, Addison-Wesley, 2003.
File “fecs”, version of 11 December 2006, page 2. Typeset at 22:34 December 11, 2006.
