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MA 261 - Fall 2009
Study Guide # 1
1. Vectors in R
2
and R
3
(a) ~v = 〈a, b, c〉 = a
i + b
j + c
k; vector addition and subtraction geometrically using paral-
lelograms spanned by ~u and ~v; length or magnitude of ~v = 〈a, b, c〉, |~v| =
a
2
2
2 ;
directed vector from P 0
(x 0
, y 0
, z 0
) to P 1
(x 1
, y 1
, z 1
) given by ~v = P 0
P
1
= P
1
− P
0
〈x 1
− x 0
, y 1
− y 0
, z 1
− z 0
(b) Dot (or inner) product of ~a = 〈a 1
, a 2
, a 3
〉 and
b = 〈b 1
, b 2
, b 3
〉: ~a ·
b = a 1
b 1
b 2
b 3
properties of dot product; useful identity: ~a · ~a = |~a|
2
; angle between two vectors ~a and
b:
cos θ =
~a ·
b
|~a| |
b|
; ~a ⊥
b if and only if ~a ·
b = 0; the vector in R
2 with length r with angle θ is
~v = 〈r cos θ, r sin θ〉 :
x
y
0
r θ
(c) Projection of
b along ~a: proj ~a
b =
~a ·
b
|~a|
~a
|~a|
; Work =
F ·
D.
b
proj
a
proj
a b
b
a
a
b
(d) Cross product (only for vectors in R
3
):
~a ×
b =
i
j
k
a 1
a 2
a 3
b 1
b 2
b 3
a 2
a 3
b 2
b 3
i −
a 1
a 3
b 1
b 3
j +
a 1
a 2
b 1
b 2
k
properties of cross products; ~a ×
b is perpendicular (orthogonal or normal) to both ~a and
b; area of parallelogram spanned by ~a and
b is A = |~a ×
b|:
b
a
the area of the triangle spanned is A =
1
2
|~a ×
b|:
b
a
Volume of the parallelopiped spanned by ~a,
b,~c is V = |~a · (
b × ~c)|:
b
a
c
2. Equation of a line L through P
0
(x 0
, y 0
, z 0
) with direction vector
d = 〈a, b, c〉:
Vector Form: ~r(t) = 〈x 0
, y 0
, z 0
〉 + t
d.
(x ,y ,z )
0 0 0
d
Parametric Form:
x = x 0 +^ a t
y = y 0
z = z 0
Symmetric Form:
x − x 0
a
y − y 0
b
z − z 0
c
. (If say b = 0, then
x − x 0
a
z − z 0
c
, y = y 0
3. Equation of the plane through the point P
0
(x 0
, y 0
, z 0
) and perpendicular to the vector ~n = 〈a, b, c〉
(~n is a normal vector to the plane) is 〈(x − x 0 ), (y − y 0 ), (z − z 0 )〉 · ~n = 0; Sketching planes
(consider x, y, z intercepts).
n
(x ,y ,z ) 0 0 0
4. Quadric surfaces (can sketch them by considering various traces, i.e., curves resulting from the
intersection of the surface with planes x = k, y = k and/or z = k); some generic equations have
the form:
(a) Ellipsoid:
x
2
a
2
y
2
b
2
z
2
c
2
(b) Elliptic Paraboloid:
z
c
x
2
a
2
y
2
b
2
(c) Hyperbolic Paraboloid (Saddle):
z
c
x
2
a
2
y
2
b
2
(d) Cone:
z
2
c
2
x
2
a
2
y
2
b
2
(e) Hyperboloid of One Sheet:
x
2
a
2
y
2
b
2
z
2
c
2
(f) Hyperboloid of Two Sheets: −
x
2
a
2
y
2
b
2
z
2
c
2
13. CHAIN RULE; different forms of the Chain Rule: Form 1, Form 2; CHAIN RULE (Gen-
eral Form): Tree diagrams. For example:
(a) If z = f (x, y) and
x = x(t)
y = y(t)
, then
df
dt
∂f
∂x
dx
dt
∂f
∂y
dy
dt
x y
dx
dy
dt
dt
z=f(x,y)
t
t
x
y
f
f
(b) If z = f (x, y) and
x = x(s, t)
y = y(s, t)
, then
∂f
∂s
∂f
∂x
∂x
∂s
∂f
∂y
∂y
∂s
and
∂f
∂t
∂f
∂x
∂x
∂t
∂f
∂y
∂y
∂t
x y
f
x
z=f(x,y)
s t s t
x
s
x
t
y
s t
y
f
y
etc.....
14. Implicit Differentiation:
Part I: If F (x, y) = 0 defines y as function of x (i.e., y = y(x)), then to compute
dy
dx
differentiate both sides of the equation F (x, y) = 0 w.r.t. x and solve for
dy
dx
If F (x, y, z) = 0 defines z as function of x and y (i.e. z = z(x, y)) , then to compute
∂z
∂x
differentiate the equation F (x, y, z) = 0 w.r.t. x (hold y fixed) and solve for
∂z
∂x
. For
∂z
∂y
differentiate the equation F (x, y, z) = 0 w.r.t. y (hold x fixed) and solve for
∂z
∂y
Part II: If F (x, y) = 0 defines y as function of x =⇒
dy
dx
∂F
∂x
∂F
∂y
while if F (x, y, z) = 0 defines z as function of x and y =⇒
∂z
∂x
∂F
∂x
∂F
∂z
and
∂z
∂y
∂F
∂y
∂F
∂z
15. Gradient vector for f (x, y): ∇f (x, y) =
∂f
∂x
∂f
∂y
, properties of gradients; gradient points in
direction of maximum rate of increase of f ; ∇f (x 0
, y 0
) ⊥ level curve f (x, y) = C and, in the case
of 3 variables, ∇f (x 0 , y 0 , z 0 ) ⊥ level surface f (x, y, z) = C:
0
(x ,y )
x
y
f(x,y,z)=C
x
y
(x ,y ,z ) 0 0 0
n =
0
n =
z
f(x ,y ) 0 0
f(x,y)=C
f(x ,y ,z ) 0 0 0
16. Directional derivative of f (x, y) at (x
0
, y 0
) in the direction ~u : D ~u
f (x 0
, y 0
) = ∇f (x 0
, y 0
) · ~u,
where ~u must be a unit vector; tangent planes to level surfaces f (x, y, z) = C (a normal vector
at (x 0
, y 0
, z 0
) is ~n = ∇f (x 0
, y 0
, z 0
17. Relative/local extrema; critical points (points where ∇f =
0 or ∇f does not exist).
nd
Derivatives Test: Suppose the 2
nd
partials of f (x, y) are continuous in a disk with center (a, b)
and ∇f (a, b) =
- Let D =
fxx fxy
f yy
f yx
(a,b)
(a) If D > 0 and f xx
(a, b) > 0 =⇒ f (a, b) is a local minimum value.
(b) If D > 0 and f xx
(a, b) < 0 =⇒ f (a, b) is a local maximum value.
(c) If D < 0 =⇒ f (a, b) is a not a local min or local max value. So (a, b) is a saddle point of f.
If D = 0 (or if ∇f (a, b) does not exist or f has more than 2 variables) the test gives no information
and you need to do something else to determine if a relative extremum exists.
19. Absolute extrema; Max-Min Problems.
