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Multivariable calculus, 2012-10-30.

Per-Sverre Svendsen, Tel.035 - 16 76 15/0709 - 398 526.

Minimum requirements: Grade 5: 27p, 4: 21p, 3: 15p.You are allowed to use one unmarked, ordinary (non-mathematical) dictionary of your choice.No additional written material is allowed. Also, no calculators or any other electronic equipment.

1. Calculate the directional derivative of   f (x, y) = 

8 − x2 − 2y2

at the point  P  = (2, 1) and in the direction  v   = (1, 1).   (2p)

2. The equation   ez + z − x2 − y2 = 1 implicitly defines a function   z  =  f (x, y)for which  f (0, 0) = 0. Show that

(a) (0, 0) is a critical point,   (2p)

(b)   f (0, 0) is a local minimum value.   (2p)

3. Calculate

 

D

(x + y) ex2−y2 dx dy,   D   = {(x, y) ∈   R2 | 0 ≤ x + y ≤ 1,   0 ≤ x− y ≤ 1}.   (3p)

4. Calculate

 

A

y x2 + y2

dx dy,   A   = {(x, y) ∈   R2 | x2 + y2 ≤ y }.   (3p)

5. Find the absolute minimum and maximum values of   g(x, y) =   x2 +   y2 −   2x

on the set ∆ = {(x, y

) ∈  R

2

|x− 2 ≤

y ≤ −

x + 2

, x≥ 0}  .   (5p)

6. We are given a function   f (x ,y ,z) =   xy + z.

(a) Find the absolute minimum and maximum values of  f subject to the constraint   x2 + y2 + z2 = 9.   (5p)

(b) Calculate the surface integral

 

S 

f (x ,y ,z) dS ,

where   S   = {(x ,y ,z) ∈   R3 | x2 + y2 + z2 = 9, x ≥ 0, y ≥ 0, z ≥ 0}.   (3p)

7. Calculate the  average  value of   g(x ,y ,z) =  z

x2 + y2 + 1 given by

g   =  1

V  K 

 

K 

g(x ,y ,z) dx dy dz, where  V  K  =

 

K 

dx dy dz, and

K   = {(x ,y ,z) ∈   R3 | x2 + y2 ≤ 1,   0 ≤ z ≤ 

2− x2 − y2}.   (5p)

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Reference formulas and equations in Multivariable Calculus

Trigonometry and Logarithms

sin(x± y) = sin x cos y ± cos x sin y   sin x± sin y = 2 sin x ± y2

  cos x ∓ y2

cos(x± y) = cos x cos y ∓ sin x sin y   cos x− cos y = −2sin x + y2

  sin x − y

2

tan(x± y) =   tan x ± tan y1∓ tan x tan y   cos x + cos y = 2cos

 x + y

2  cos

 x − y2

cot(x± y) =   cot x cot y ∓ 1± cot x + cot y   2sin x sin y = cos(x− y)− cos(x + y)

sin2x = 2sin x cos x   2cos x cos y = cos(x− y) + cos(x + y)

cos2x = cos2 x− sin2 x = 2cos2 x− 1 = 1− 2sin2 x   2sin x cos y  = sin(x− y) + sin(x + y)

ln x + ln y = ln xy   ln x

−ln y = ln

 x

yln xa = a ln x   (x, y > 0)

Standard limits

limx→0+

xα loga x = 0 (a > 1,  α > 0) limx→∞

ax

xα  = ∞   (a > 1)

limx→0

sin x

x  = 1 lim

x→∞

xα

loga x = ∞ (a > 1,  α > 0)

limx→0

ln(1 + x)

x  = 1 lim

n→∞

an

n!  = 0

limx→0

ex − 1

x

  = 1

Basic derivatives

f (x)   f (x)

xa axa−1

ax ax ln a

ln |x|   1x

sin x   cos x

cos x   − sin x

tan x   1 + tan2 x =  1

cos2 x

arcsin x  1√ 

1− x2arccos x   −   1√ 

1− x2arctan x

  1

1 + x2

lnx + x2 + α

1√ x2 + α

12x√ x

2

+ α +  α

2  lnx +√ x

2

+ α  

x2

+ α

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Taylor Series

Taylor’s formula for a function   f   :   R→ R

f (a + h) =   f (a) +   f (a) h   +  f (a)

2  h2 + · · ·   =

∞n=0

f (n)(a)

n!  hn

Table of particular expansions (a = 0, h → x)

1.  1

1− x   =∞k=0

xk = 1 +   x   + x2 +   x3 + · · ·   (−1 < x

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Double Integrals

General substitution

Assume a one-to-one mapping between a region  D  in the  xy-plane and a region  Duv   in the  uv -plane

x   =   x(u, v)

y   =   y(u, v)⇔

u   =   u(x, y)

v   =   v(x, y)

Then

 

D

f (x, y) dxdy   =

 

Duv

f (x(u, v), y(u, v))∂ (x, y)

∂ (u, v)

dudv,

with  ∂ (x, y)

∂ (u, v)  =

xu   xv

yu   yv

= 0.

Polar coordinates

x   =   r cos θ

y   =   r sin θ

∂ (x, y)

∂ (r, θ)  =   r,

 

D

f (x, y) dxdy   =

 

Drθ

f (r cos θ, r sin θ) rdr dθ

Triple Integrals

General substitution

As above assume a one-to-one mapping between points (x ,y ,z) in ∆ and (u ,v ,w) in ∆uvw.

 

∆

f (x ,y ,z) dxdydz   =

 

∆uvw

f (x(u ,v ,w), y(u ,v ,w), z(u ,v ,w)) ∂ (x ,y ,z)

∂ (u ,v ,w)

dudvdw,

with  ∂ (x ,y ,z)

∂ (u ,v ,w) = 0

Spherical coordinates

x   =   ρ sin φ cos θ

y   =   ρ sin φ sin θ

z   =   ρ cos φ

∂ (x ,y ,z)

∂ (ρ,φ,θ)  =   ρ2 sin φ

 

∆

f (x ,y ,z) dxdydz   =

 

∆ρφθ

f (ρ sin φ cos θ, ρ sin φ sin θ, ρ cos φ) ρ2 sin φ dρ dφ dθ

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Line Integrals

Tangent line integral

Given a parametrized curve   C  :   r(t) = (x(t), y(t), z(t)), a ≤ t ≤ b  and a vector field   F    = (P,Q,R).

 C 

F 

 · dr

  =

b

 a

(P,Q,R) · (x

(t), y

(t), z

(t)) dt   =

b

 a

P x

(t) +   Q y

(t) +   R z

(t)

dt

Line integral with respect to arc length

Assume a curve  C  as above and a function  f   :   R3 → R. 

C 

f (x ,y ,z) ds   =

b a

f (x(t), y(t), z(t)) 

(x(t))2 + (y(t))2 + (z(t))2 dt,   (ds =  d|r|)

Green’s theorem

Given a plane, closed, positively oriented curve  C   that encloses a region  D  and a field  F    = (P, Q).

 

C 

F  · dr   = 

D

∂Q∂x −   ∂P 

∂y

dxdy

Surface Integrals

General parametrized surface

S  :   r   =   r(u, v) = (x(u, v), y(u, v), z(u, v)),   (u, v) ∈   D. 

S 

f (x ,y ,z) dS   =

 

D

f (r(u, v))∂ r

∂u ×  ∂ r

∂v

dudv

Parametrization of (parts of) the surface   x2 + y2 + z2 = R2 (R > 0).

S  :  r   = (x ,y ,z) = (R sin φ cos θ, R sin φ sin θ, R cos φ),   (φ, θ) ∈ D. 

S 

f (x ,y ,z) dS   =

 

D

f (r(φ, θ))∂ r

∂φ ×  ∂ r

∂θ

dφ dθ   = 

D

f (r(φ, θ)) R2 sin φ dφ dθ.

Function graph   z  =  h(x, y).

S  :   r   = (x ,y ,z) = (x ,y ,h(x, y)),   (x, y) ∈   S xy. 

S 

f (x ,y ,z) dS   =

 

S xy

f (x ,y ,h(x, y)) 

1 + h2x + h2y dxdy

Gauss’ theorem

Given a space region  K  with (closed) surface boundary  S  and a field  F .n is the  outer  unit normal vector of the surface.

 

S 

F  · n dS   = 

K 

∇ · F  dx dy dz.