𝐕𝐞𝐜𝐭𝐨𝐫 𝐚𝐥𝐠𝐞𝐛𝐫𝐚:
𝟐
̂)
‖𝐮‖ = √(𝐱𝐢̂)𝟐 + (𝐲𝐣̂)𝟐 + (𝐳𝐤
𝐮 ∙ 𝐯 = ‖𝐮‖‖𝐯‖ cos θ
‖𝐮‖ = √𝐮 ∙ 𝐮
𝐮
Unit vector: 𝐮
̂=
‖𝐮‖
𝐮∙𝐯= 𝐯∙𝐮
(𝛂𝐮) ∙ 𝐯 = 𝛂(𝐮 ∙ 𝐯)
(𝐮 + 𝐯) ∙ 𝐰 = 𝐮 ∙ 𝐰 + 𝐯 ∙ 𝐰
̂ ∙𝐤
̂ =𝟏
𝐢̂ ∙ 𝐢̂ = 𝐣̂ ∙ 𝐣̂ = 𝐤
̂ = 𝐢̂ ∙ 𝐤
̂ =𝟎
𝐢̂ ∙ 𝐣̂ = 𝐣̂ ∙ 𝐤
(α𝐮 + β𝐯) ∙ 𝐰 = α(𝐮 ∙ 𝐰) + β(𝐯 ∙ 𝐰)
Magnitude of 𝐮 × 𝐯 = ‖𝐮‖‖𝐯‖ sin θ𝐞̂
𝐮 × 𝐯 = −(𝐯 × 𝐮)
̂ , 𝐣̂ × 𝐢̂ = −𝐤
̂
𝐢̂ × 𝐣̂ = 𝐤
(𝛂𝐮) × 𝐯 = 𝛂(𝐮 × 𝐯)
(𝐮 + 𝐯) × 𝐰 = 𝐮 × 𝐰 + 𝐯 × 𝐰
‖𝐮 × 𝐯‖ = area of 𝐮, 𝐯 parallelogram
𝐌 ≡ 𝐑 × 𝐅, Moment 𝐌 of 𝐅 about 𝐏
𝟐
‖𝐮 × 𝐯‖𝟐 = ‖𝐮‖𝟐 ‖𝐯‖𝟐 − (𝐮 ∙ 𝐯)
̂ ×𝐤
̂ =𝟎
𝐢̂ × 𝐢̂ = 𝐣̂ × 𝐣̂ = 𝐤
̂
𝐢̂
𝐣̂
𝐤
̂
𝐮 × 𝐯 = |𝑢𝑥 𝑢𝑦 𝑢𝑧 | = (𝑢2 𝑣3 − 𝑢3 𝑣2 )𝐢̂ + (𝑢3 𝑣1 − 𝑢1 𝑣3 )𝐣̂ + (𝑢2 𝑣1 − 𝑢2 𝑣1 )𝐤
𝑣𝑥 𝑣𝑧 𝑣𝑧
𝐮 × 𝐯 = 𝑢𝐢 𝑣𝐣 𝜀𝐢𝐣𝐤 𝑒𝐤 𝜀𝐢𝐣𝐤 is the permutation tensor
𝑒𝐢 × 𝑒𝐣 = 𝜀𝐢𝐣𝐤 𝑒𝐤
̂ = ‖𝐑 𝟎 ‖,
𝐑 ∙𝐧
R is any point plane, ‖𝐑 𝟎 ‖ is shortest distance to plane
Plane: 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑑, (𝑎𝑖̂ + 𝑏𝑗̂ + 𝑐𝑘̂ ) is normal vector (not unit normal)
|d|
Shortest distance D =
√𝑎2 + 𝑏2 + 𝑐 2
|𝐮 ∙ 𝐯 × 𝐰| = volume of 𝐮, 𝐯, 𝐰 parallelepiped
𝑢1 𝑢2 𝑢3
𝐮 ∙ 𝐯 × 𝐰 = | 𝑣1 𝑣2 𝑣3 |
𝑤1 𝑤2 𝑤3
𝐮 ∙ 𝐯 × 𝐰 = 𝐮 × 𝐯 ∙ 𝐰 (if = 0, 𝐮, 𝐯, 𝐰 are LD, in plane)
𝐮 × (𝐯 × 𝐰) = (𝐮 ∙ 𝐰)𝐯 − (𝐮 ∙ 𝐯) × 𝐰
(𝐮 ∙ 𝐯 × 𝐰)′ = 𝐮′ ∙ 𝐯 × 𝐰 + 𝐮 ∙ 𝐯′ × 𝐰 + 𝐮 ∙ 𝐯 × 𝐰′
[(𝐮 × (𝐯 × 𝐰)]′ = (𝐮′ × (𝐯 × 𝐰) + (𝐮 × (𝐯 ′ × 𝐰) + (𝐮 × (𝐯 × 𝐰′)
𝐮 ∙ 𝐮′
‖𝐮‖ =
‖𝐮‖
𝐮𝐢 𝛅𝐢𝐣 = 𝐮𝐣
𝐮 ∙ 𝐯 = 𝐮𝐢 𝐞𝐢 ∙ 𝐯𝐣 𝐞𝐣 = 𝐮𝐢 𝐯𝐣 𝐞𝐢 ∙ 𝐞𝐣 = 𝐮𝐢 𝐯𝐣 𝛅𝐢𝐣
𝐮 ∙ 𝐯 = 𝐮𝐓 𝐯
𝐂𝐚𝐫𝐭𝐞𝐬𝐢𝐨𝐧:
̂
𝐑 = 𝑥𝐢̂ + 𝑦𝐣̂ + 𝑧𝐤
̂
𝐯(𝑡) = 𝑥′𝐢̂ + 𝑦′𝐣̂ + 𝑧′𝐤
̂
𝐚(𝑡) = 𝑥′′𝐢̂ + 𝑦′′𝐣̂ + 𝑧′′𝐤
𝑑𝑦 𝑑𝑧 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑥 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑑𝐴 = { 𝑑𝑥 𝑑𝑧 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑦 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑑𝑦 𝑑𝑥 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑧 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
∂
∂
∂
̂
Vector differential operator (Gradient): 𝛁 ≡ 𝐢̂
+ 𝐣̂
+𝐤
∂𝑥
∂𝑦
∂𝑧
∂𝑣𝑥 ∂𝑣𝑦 ∂𝑣𝑧
Div 𝐯 = 𝛁 ∙ 𝐯 =
+
+
∂𝑥
∂𝑦
∂𝑧
∂
∂
∂
𝐯 ∙ 𝛁 = 𝑣𝑥
+ 𝑣𝑦
+ 𝑣𝑧
∂𝑥
∂𝑦
∂𝑧
∂𝑢
∂𝑢
∂𝑢
̂,
Grad 𝑢 ≡ 𝛁𝑢 =
𝐢̂ +
𝐣̂ +
𝐤
scalar field 𝑢 to vector field
∂𝑥
∂𝑦
∂𝑧
̂
𝐢̂
𝐣̂
𝐤
∂
∂
∂
Curl 𝐯 ≡ 𝛁 × 𝐯 = ||
||
∂𝑥 ∂𝑦 ∂𝑧
𝑣𝑥 𝑣𝑦 𝑣𝑧
𝜕𝑣𝑧 𝜕𝑣𝑦
𝜕𝑣𝑧 𝜕𝑣𝑥
=(
−
−
) 𝐣̂
) 𝐢̂ − (
𝜕𝑦
𝜕𝑧
𝜕𝑥
𝜕𝑧
𝜕𝑣𝑦 𝜕𝑣𝑥
̂ , vector field 𝐯 to vector field
+(
−
)𝐤
𝜕𝑥
𝜕𝑦
𝐏𝐨𝐥𝐚𝐫: 𝐞̂𝐫 (θ), 𝐞̂𝛉 (θ)
𝐑 = 𝑟𝒆̂𝒓
𝐯(t) = 𝑟 ′ 𝐞̂𝐫 + 𝑟θ′ 𝐞̂𝛉
𝐚(𝑡) = (𝑟 ′′ − 𝑟θ′2 )𝐞̂𝐫 + (𝑟𝜃 ′′ + 2𝑟 ′ 𝜃 ′ ) 𝐞̂𝛉
𝜕𝐞̂𝐫
𝜕 𝐞̂𝛉
= 𝐞̂𝛉 ,
= −𝐞̂𝐫
𝜕𝜃
𝜕𝜃
𝐢̂ = cos 𝜃𝐞̂𝐫 − sin 𝜃 𝐞̂𝛉 ,
𝐣̂ = sin 𝜃𝐞̂𝐫 + cos 𝜃 𝐞̂𝛉
𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃
̂
𝐂𝐲𝐥𝐢𝐧𝐝𝐫𝐢𝐜𝐚𝐥: 𝐞̂𝐫 (θ), 𝐞̂𝛉 (θ), 𝐞̂𝐳 = 𝐤
𝐑 = 𝑟𝐞̂𝐫 + 𝑧𝐞̂𝐳 ,
𝑑𝐑 = 𝑑𝑟𝐞̂𝐫 + 𝑟𝑑𝜃𝐞̂𝛉 + 𝑑𝑧𝐞̂𝐳
𝐯(𝑡) = 𝑟 ′ 𝐞̂𝐫 + 𝑟𝜃 ′ 𝐞̂𝛉 + 𝑧 ′ 𝐞̂𝐳
𝐚(𝑡) = (𝑟 ′′ − 𝑟𝜃 ′2 )𝐞̂𝐫 + (𝑟𝜃 ′′ + 2𝑟 ′ 𝜃 ′ )𝐞̂𝛉 + 𝑧 ′′ 𝐞̂𝐳
𝜕𝐞̂𝐫
𝜕 𝐞̂𝛉
= 𝐞̂𝛉 ,
= −𝐞̂𝐫
𝜕𝜃
𝜕𝜃
𝐢̂ = cos 𝜃𝐞̂𝐫 − sin 𝜃 𝐞̂𝛉 ,
𝐣̂ = sin 𝜃𝐞̂𝐫 + cos 𝜃 𝐞̂𝛉
𝐞̂𝐫 × 𝐞̂𝐳 = −𝐞̂𝛉 ,
𝐞̂𝛉 × 𝐞̂𝐳 = 𝐞̂𝐫 ,
𝐞̂𝐫 × 𝐞̂𝛉 = 𝐞̂𝐳
If taking the cross product, set it up as 𝐞̂𝐫 → 𝐞̂𝛉 → 𝐞̂𝐳 , L to R
𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 , 𝑧 = 𝑧
E.g. for a cone 𝑟, 𝜃, and 𝑧 are not
𝑟 𝑑𝜃𝑑𝑧 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑟 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑑𝐴 = { 𝑑𝑟 𝑑𝑧 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜃 𝑠𝑢𝑟𝑓𝑎𝑐𝑒) constant. For a cylinder, r is constant.
𝑟 𝑑𝑟 𝑑𝜃 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑧 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑑𝑉 = 𝑟 𝑑𝑟 𝑑𝜃 𝑑𝑧
𝜕𝑢
1 𝜕𝑢
𝜕𝑢
𝛁𝑢 =
𝐞̂ +
𝐞̂ +
𝐞̂
𝜕𝑟 𝐫 𝑟 𝜕𝜃 𝛉 𝜕𝑧 𝐳
2
2
𝜕 𝑢 1 𝜕𝑢 1 𝜕 𝑢 𝜕 2 𝑢
𝛁𝟐 𝑢 = 2 +
+
+
𝜕𝑟
𝑟 𝜕𝑟 𝑟 2 𝜕𝜃 2 𝜕𝑧 2
1 𝜕
1 𝜕
𝜕
𝛁 ∙𝐯 =
(𝑟 𝑣𝑟 ) +
𝑣 +
𝑣
𝑟 𝜕𝑟
𝑟 𝜕𝜃 𝜃 𝜕𝑧 𝑧
1 𝜕𝑣𝑧 𝜕𝑣𝜃
𝜕𝑣𝑟 𝜕𝑣𝑧
1 𝜕(𝑟𝑣𝜃 ) 𝜕𝑣𝑟
𝛁×𝐯 =(
−
) 𝐞̂𝐫 + (
−
) 𝐞̂ + (
−
) 𝐞̂
𝑟 𝜕𝜃
𝜕𝑧
𝜕𝑧
𝜕𝑟 𝛉 𝑟
𝜕𝑟
𝜕𝜃 𝐳
𝐒𝐩𝐡𝐞𝐫𝐢𝐜𝐚𝐥: 𝐞̂𝛒 (𝜙, 𝜃),
𝐞̂𝛟 (𝜙, 𝜃),
𝐞̂𝛉 (𝜃)
𝜕𝐞̂𝛒
𝜕𝐞̂𝛒
𝜕𝐞̂𝛒
= 0,
= 𝐞̂𝛟 ,
= sin 𝜙 𝐞̂𝛉
𝜕𝜌
𝜕𝜙
𝜕𝜃
𝜕𝐞̂𝛟
𝜕𝐞̂𝛟
𝜕𝐞̂𝛟
= 0,
= −𝐞̂𝛒,
= cos 𝜙 𝐞̂𝜃
𝜕𝜌
𝜕𝜙
𝜕𝜃
̂
̂
̂
𝜕𝐞𝛉
𝜕𝐞𝛉
𝜕𝐞𝛉
= 0,
= 0,
= − sin ϕ 𝐞̂𝛒 − cos 𝜙 𝐞̂𝜙
𝜕𝜌
𝜕𝜙
𝜕𝜃
𝐑 = 𝜌𝑒̂𝜌 ,
𝑑𝐑 = 𝑑𝜌𝐞̂𝛒 + 𝜌𝑑𝜙𝐞̂𝛟 + 𝜌 sin 𝜙 𝑑𝜃𝐞̂𝛉
𝐯(𝑡) = 𝜌 ′ 𝐞̂𝛒 + 𝜌𝜙 ′ 𝐞̂𝛟 + 𝜌𝜃 ′ sin 𝜙 𝐞̂𝛉
𝐚(𝑡) = (𝜌 ′′ − 𝜌𝜙 ′2 − 𝜌𝜃 ′2 sin2 𝜙)𝐞̂𝛒 + (𝜌𝜙 ′′ + 2𝜌 ′ 𝜙 ′
− 𝜌𝜃 ′2 sin 𝜙 cos 𝜙)𝐞̂𝛟 + (𝜌𝜃 ′′ sin 𝜙 + 2𝜌 ′ 𝜙′ sin 𝜙
+ 2𝜌𝜃 ′ 𝜙 ′ cos 𝜙) 𝐞̂𝛉
2 |sin
𝜌
𝜙|𝑑𝜙 𝑑𝜃 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜌 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝑑𝐴 = { 𝜌|sin 𝜙|𝑑𝜌 𝑑𝜃 (𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜙 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
(𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝜃 𝑠𝑢𝑟𝑓𝑎𝑐𝑒)
𝜌 𝑑𝜌 𝑑𝜙
𝑑𝑉 = 𝜌 2 |sin 𝜙|𝑑𝜌 𝑑𝜙 𝑑𝜃
𝐞̂𝛉 × 𝐞̂𝐫 = 𝐞̂𝛟 ,
𝐞̂𝛟 × 𝐞̂𝐫 = −𝐞̂𝛉 ,
𝐞̂𝛟 × 𝐞̂𝛉 = 𝐞̂𝐫
̂
𝐞̂𝛒 = sin 𝜙 (cos 𝜃 𝐢̂ + sin 𝜃 𝐣̂) + cos 𝜙 𝐤
̂
𝐞̂𝛟 = cos 𝜙 (cos 𝜃 𝐢̂ + sin 𝜃 𝐣̂) − sin 𝜙 𝐤
𝐞̂𝛉 = − sin 𝜃 𝐢̂ + cos 𝜃 𝐣̂
𝐢̂ = sin 𝜙 cos 𝜃 𝐞̂𝛒 + cos 𝜙 cos 𝜃 𝐞̂𝛟 − sin 𝜃 𝐞̂𝛉
𝐣̂ = sin 𝜙 sin 𝜃 𝐞̂𝛒 + cos 𝜙 sin 𝜃 𝐞̂𝛟 + cos 𝜃 𝐞̂𝛉
̂ = cos 𝜙 𝐞̂𝛒 − sin 𝜙 𝐞̂𝛟
𝐤
𝑥 = 𝜌 sin 𝜙 cos 𝜃
𝑦 = 𝜌 sin 𝜙 sin 𝜃
𝑧 = 𝜌 cos 𝜙
𝜕𝑢
1 𝜕𝑢
1 𝜕𝑢
𝛁𝑢 =
𝐞̂ +
𝐞̂ +
𝐞̂
𝜕𝜌 𝛒 ρ 𝜕𝜙 𝛟 ρ sin ϕ 𝜕𝜃 𝛉
1 𝜕
𝜕𝑢
1 𝜕
𝜕𝑢
1 𝜕2𝑢
𝛁𝟐 𝑢 = 2 [ (𝜌 2 ) +
(sin 𝜙 ) + 2
]
𝜌 𝜕𝜌
𝜕𝜌
sin 𝜙 𝜕𝜙
𝜕𝜙
sin 𝜙 𝜕𝜃 2
1 𝜕
1
𝜕
1 𝜕 𝑣𝜃
𝛁 ∙ 𝐯 = 2 (𝜌 2 𝑣𝜌 ) +
(𝑣 sin 𝜙) +
𝜌 𝜕𝜌
𝜌sin 𝜙 𝜕𝜙 𝜙
𝜌sin 𝜙 𝜕𝜃
𝜕𝑣𝜙
1
𝜕
1
1 𝜕𝑣𝜌 𝜕(𝜌𝑣𝜃 )
𝛁 ×𝐯 =
−
( (𝑣 sin 𝜙) −
) 𝐞̂𝛒 + (
) 𝐞̂𝛟
𝜌 sin 𝜙 𝜕𝜙 𝜃
𝜕𝜃
𝜌 sin 𝜙 𝜕𝜃
𝜕𝜌
1 𝜕(𝜌𝑣𝜙 ) 𝜕𝑣𝜌
+ (
−
) 𝐞̂
𝜌
𝜕𝜌
𝜕𝜙 𝛉
Curves and line integrals
𝜏
Arc legth 𝑠(𝜏) = ∫ √𝐑′ (𝑡) ∙ 𝐑′ (𝑡)𝑑𝑡
𝜏0
𝑏
Line Integral: ∫ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑠 = ∫ 𝑓(𝑥(𝜏), 𝑦(𝜏), 𝑧(𝜏))√𝐑′(𝜏) ∙ 𝐑′ (𝜏)𝑑𝜏
𝐶
𝑎
Note: R is the vector that traces out the curve. For example, if the curve is a
semicircle in the quadrant I and II, then 𝐑(𝛉) = 𝑟𝐞̂𝐫 = cos θ 𝐢̂ + sin θ 𝐣̂, where
0≤θ≤ 𝝅. And in this case, τ =θ.
𝐏𝐚𝐫𝐚𝐦𝐞𝐭𝐞𝐫𝐢𝐳𝐚𝐭𝐢𝐨𝐧 𝐨𝐟 𝐚 𝐬𝐭𝐫𝐚𝐢𝐠𝐡𝐭 𝐥𝐢𝐧𝐞:
x = x1 + (x2 − x1 )τ,
y = y1 + (y2 − y1 )τ,
z = z1 + (z2 − z1 )τ,
(0 ≤ τ < 1)
Parameterization: The goal is to solve for your curve or surface in terms of a
variable that suits your preferred coordinate system. If your curve is an intersection
of two surfaces, then solve for the intersection just like you would a system of
equations (Gauss elimination, etc.), but before you solve, chose a parameterization
that makes sense for the surfaces in question. I.e. if you have a plane intersecting a
cylinder, chose θ as the parameter (from 0 to 2π), make the appropriate substitutions
𝑥 = 𝑟 cos 𝜃 , 𝑦 = 𝑟 sin 𝜃 for the cylinder and then solve for x,y, and z in terms of
the new parameter.
𝐍𝐞𝐭 𝐰𝐨𝐫𝐤 𝐝𝐨𝐧𝐞 𝐭𝐫𝐚𝐯𝐞𝐫𝐬𝐢𝐧𝐠 𝐜𝐮𝐫𝐯𝐞 𝐂: ∫ 𝐯 ∙ 𝑑𝐑 = ∫ 𝑓𝑑𝑆,
𝐶
𝐶
Where v is a force vector field and R is the position vector to some reference point.
∫ 𝐯 ∙ 𝑑𝐑 = ∫ (𝑣𝑥 (𝑥, 𝑦, 𝑧)𝑑𝑥 + 𝑣𝑦 (𝑥, 𝑦, 𝑧)𝑑𝑦 + 𝑣𝑧 (𝑥, 𝑦, 𝑧)𝑑𝑧),
𝐶
𝐶
Note 1: The dot turns this from two vectors to a scalar.
Note 2: If the curve is not continuous, need to break up the integral.
̂ ∙ 𝛁 × 𝐯𝑑𝐴,
𝐒𝐭𝐨𝐤𝐞′ 𝐬 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: ∮ 𝐯 ∙ 𝑑𝐑 = ∮ 𝐧
𝑆
𝐶
̂ follows the right hand rule.
Note 1: If the surface is not closed, the 𝐧
Note 2: The line integral must be closed.
∂Q ∂P
𝐆𝐫𝐞𝐞𝐧𝐞′ 𝐬𝐓𝐡𝐞𝐨𝐫𝐞𝐦: ∫
−
𝑑𝐴 = ∮ 𝑃𝑑𝑥 + 𝑄𝑑𝑦
∂y
𝑆 ∂x
𝐶
Note 1: Vector field: 𝐯 = 𝑃(𝑥, 𝑦)𝐢̂ + 𝑄(𝑥, 𝑦)𝐣̂
Note 2: Edge of S must be piecewise, smooth, simple, closed, oriented CC.
𝜕𝑣
𝐆𝐫𝐞𝐞𝐧𝐞′ 𝐬 𝟏𝐬𝐭 𝐢𝐝𝐞𝐧𝐭𝐢𝐭𝐲: ∫ (𝛁𝑢 ∙ 𝛁𝑣 + 𝑢𝛁2 𝑣)𝑑𝑉 = ∫ 𝑢
𝑑𝐴
𝜕𝑛
𝑉
𝑆
𝜕𝑣
𝜕𝑢
𝐆𝐫𝐞𝐞𝐧𝐞′ 𝐬 𝟐𝐧𝐝 𝐢𝐝𝐞𝐧𝐭𝐢𝐭𝐲: ∫ (𝑢𝛁𝟐 𝑣 ∙ 𝛁𝑣 − 𝑣𝛁2 𝑢)𝑑𝑉 = ∫ (𝑢
− 𝑣 ) 𝑑𝐴
𝜕𝑛
𝜕𝑛
𝑉
𝑆
𝜕𝑣
̂ ∙ 𝛁𝑣
𝐍𝐨𝐭𝐞 𝟏: 𝑢
= 𝑢𝐧
𝜕𝑛
𝐍𝐨𝐭𝐞 𝟐: 𝑢 and 𝑣 are scalar fields
̂ ∙ 𝐯 𝑑𝐴
𝐃𝐢𝐯𝐞𝐫𝐠𝐞𝐧𝐜𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: ∫ ∇ ∙ 𝐯 𝑑𝑉 = ∫ 𝐧
𝜈
𝑆
̂𝑢𝑑𝐴 = ∫ 𝛁𝑢𝑑𝑉 , ∫ 𝐧
̂ × 𝐯𝒅𝑨 = ∫ 𝛁 × 𝐯𝑑𝑉
∫𝐧
𝑆
𝑉
𝑆
𝑉
̂ is the unit normal vector to the surface.
Where: 𝐯 is a vector field, and 𝐧
If you have several discontinuous surfaces forming one piecewise smooth
surface, you need to integrate each one seperately, and then add them.
The unit normal vectors always point outward from the surface.
∇𝑔
̂=±
𝐧
, Where 𝑔 = 𝑔(𝑥, 𝑦, 𝑧), or 𝑔(𝜌, 𝜙, 𝜃), 𝑒𝑡𝑐. = a surface
‖∇𝑔‖
Example: We have the equation for a paraboloid: 𝑥 2 + 𝑦 2 = 𝑧. First, get all the
variables to one side, so: 𝑥 2 + 𝑦 2 − 𝑧 = 0. This is now 𝑔(𝑥, 𝑦, 𝑧). The gradient of g
is now the normal to the surface. This is a “level surface”. If we want to find the
normal to a “level curve”, then we set x,y,or z to a constant and the take the
gradient, e.g.: 12 + 𝑦 2 − 𝑧 = 0. This is now our 𝑔(𝑥, 𝑦, 𝑧).
Div curl 𝐯 = 𝛁 ∙ 𝛁 × 𝐯 = 0
curl grad 𝑢 = 𝛁 × 𝛁𝑢 = 0
𝛁 ∙ 𝛁 = 𝛁𝟐
∂2
∂2
∂2
𝛁𝟐 ≡ 2 + 2 + 2
∂𝑥
∂𝑦
∂𝑧
𝛁𝐱 = 𝐢̂, 𝛁𝐲 = 𝐣̂
𝛁 ∙ (𝛼𝐮 + 𝛽𝐯) = 𝛼𝛁 ∙ 𝐮 + 𝛽𝛁 ∙ 𝐯
𝛁(𝛼𝑢 + 𝛽𝑣) = 𝛼𝛁𝑢 + 𝛽𝛁𝑣
𝛁 × (𝛼𝐮 + 𝛽𝒗) = 𝛼𝛁 × 𝐮 + 𝛽𝛁 × 𝐯
𝛁 ∙ (𝑢𝐯) = 𝛁𝑢 ∙ 𝐯 + 𝑢𝛁 ∙ 𝐯
𝛁 × (𝑢𝐯) = 𝛁𝑢 × 𝐯 + 𝑢𝛁 × 𝐯
𝛁 ∙ (𝑢𝛁𝑣) = 𝛁𝑢 ∙ 𝛁𝑣 + 𝑢𝛁 ∙ 𝛁𝑣
𝛁 ∙ (𝐮 × 𝐯) = 𝐯 ∙ 𝛁 × 𝐮 − 𝐮 ∙ 𝛁 × 𝐯
𝛁 × (𝐮 × 𝐯) = 𝐮𝛁 ∙ 𝐯 − 𝐯𝛁 ∙ 𝐮 + (𝐯 ∙ 𝛁)𝐮 − (𝐮 ∙ 𝛁)𝐯
𝛁(𝐮 ∙ 𝐯) = (𝐮 ∙ 𝛁)𝐯 + (𝐯 ∙ 𝛁)𝐮 + 𝐮 × (𝛁 × 𝐯) + 𝐯 × 𝛁 × 𝐮)
𝜕𝑣
̂ ∙ 𝛁𝑣 = 𝑢
𝑢𝐧
𝜕𝑛
̂)
(𝐮 ∙ 𝛁)𝐯 = (𝐮 ∙ 𝛁)(𝑣𝑥 𝐢̂ + 𝑣𝑦 𝐣̂ + 𝑣𝑧 𝐤
𝜕𝑣𝑦
𝜕𝑣𝑥
𝜕𝑣𝑧
̂
= (𝑢𝑥
+ 𝑢𝑦
+ 𝑢𝑧
) 𝒊̂ + (𝑒𝑡𝑐. )𝒋̂ + (𝑒𝑡𝑐. )𝒌
𝜕
𝜕
𝜕
Laplace equation: 𝛁𝟐 𝛗 = 𝛁 ∙ 𝛁𝛗 = 0
Poisson equation: 𝛁𝟐 𝛗 = 𝐹
∂𝛗
Diffusion equation: 𝛁𝟐 𝛗 =
∂t
∂2 𝛗
2 𝟐
Wave equation: c 𝛁 𝛗 = 2
∂t
Level Curve: The function parameters that yield a specified “z”.
̂ ∙ 𝐯𝑑𝐴
∫𝐧
div 𝐯(𝑃) ≡ lim ( 𝑆
)
𝐵→0
𝑉
Trig Identities:
1 = cos2 𝜃 + sin2 𝜃
1
csc 𝜃 =
sin 𝜃
1
sec 𝜃 =
cos 𝜃
sin 2𝜃 = 2 sin 𝜃 cos 𝜃
cos 2𝜃 = cos2 𝜃 − sin2 𝜃
sin θ
tan θ =
cos θ
cos θ
cot 𝜃 =
sin θ
𝑏
𝑎
𝑐
=
=
= 2𝑅
sin 𝛽
sin 𝛼 sin 𝜃
where 𝑅 is the radius of the triangle′s circumference.
We can derive all the others from these
sin(𝛼 ± 𝛽) = sin 𝛼 cos 𝛽 ± cos 𝛼 sin 𝛽
cos(𝛼 ± 𝛽) = cos 𝛼 cos 𝛽 ∓ sin 𝛼 sin 𝛽 two and sin2 𝜃 + cos 2 𝜃 = 1
Law of cosines: 𝑐 2 = 𝑎2 + 𝑏2 − 2𝑎𝑏 cos 𝛾
Length of any one side of a triangle cannot exceed the sum of the lengths of the
other two sides.
𝒂+𝒃+𝒄
A of triangle a, b, c = √𝒔(𝒔 − 𝒂)(𝒔 − 𝒃)(𝒔 − 𝒄), where 𝑠 =
𝟐
Distance between 𝐱 and 𝐱 ′ : d(𝐱, 𝐱 ′ ) = √(𝑥1 − 𝑥1′ )2 + ⋯
𝜕 𝜕𝑓
( )
𝜕𝑦 𝜕𝑥
𝑑𝐹 𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑓 𝜕𝑦
Chain rule:
=
+
+
; 𝐹 = 𝑓(𝑥(𝑡), 𝑦(𝑡))
𝑑𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑡 𝜕𝑦 𝜕𝑡
𝑑 𝑏(𝑡)
Leibniz rule:
∫ 𝑓(𝑥, 𝑡)𝑑𝑥
𝑑𝑡 𝑎(𝑡)
𝑓𝑥 𝑦 =
𝑏(𝑡)
𝜕
𝑓(𝑥, 𝑡)𝑑𝑥 + 𝑏′ (𝑡)𝑓(𝑏(𝑡), 𝑡) − 𝑎′ (𝑡)𝑓(𝑎(𝑡), 𝑡)
𝜕𝑡
𝑎(𝑡)
𝜕𝑦1
𝜕𝑦1
⋯
𝜕𝑥
𝜕𝑥
1
𝑛
∂(𝑦1 , … , 𝑦𝑚 )
Jacobian Matrix:
= ⋮
⋱
⋮
𝜕(𝑥1 , … , 𝑥𝑛 ) 𝜕𝑦
𝜕𝑦𝑚
𝑚
⋯
𝜕𝑥1
𝜕𝑥𝑛
Div 𝐯 ≡ 𝛁 ∙ 𝐯,
vector field 𝐯 to scalar field
Physical significance of curl: if 𝐯 is a fluid velocity field, then 𝛁
× 𝐯 at any point P is twice the angular velocity of the fluid at P.
If Curl 𝐯 = 0, we have irrotation field
=∫
Change of Variables Example:
Geometry:
Plane General Equation: 𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧
= 𝑑, where 𝑎, 𝑏, 𝑐 makes a unit normal vector.
𝑑
Distance from origin: 𝐷 =
√𝑎2 + 𝑏2 + 𝑐 2
Given 3 points: (𝑥1, 𝑦1, 𝑧1), (𝑥2, 𝑦2, 𝑧2), (𝑥3, 𝑦3, 𝑧3)
𝑥1 𝑦1 𝑧1
1 𝑦1 𝑧1
𝑥1 𝑦1 1
𝑥1 1 𝑧1
𝑎 = |1 𝑦2 𝑧2 | , 𝑏 = |𝑥2 1 𝑧2 | 𝑐 = |𝑥2 𝑦2 1| 𝑑 = − |𝑥2 𝑦2 𝑧2 |
𝑥3 𝑦3 𝑧3
𝑥3 1 𝑧3
1 𝑦3 𝑧3
𝑥3 𝑦3 1
𝐒𝐩𝐡𝐞𝐫𝐞: Equation for a sphere, centered at 𝑥0 , 𝑦0 , 𝑧0 , radius 𝑟: (𝑥 − 𝑥0 )2
+ (𝑦 − 𝑦0 )2 + (𝑧 − 𝑧0 )2 = 𝑟 2
Surface area of sphere: 𝐴 = 4𝜋𝑟 2
4
Volume of a sphere: 𝑉 = 𝜋𝑟 3
3
𝐂𝐲𝐥𝐢𝐧𝐝𝐞𝐫: 𝑎𝑥 2 + 𝑏𝑦 2 = 𝑟, where 𝑟 is the radius.
If a and b are 1, then it is a circular cylinder.
𝑥 2 𝑦2
𝐄𝐥𝐥𝐢𝐩𝐬𝐞: 2 + 2 = 1, where a is semiminor axis, b is semimajor axis
𝑎
𝑏
𝐏𝐚𝐫𝐚𝐛𝐨𝐥𝐨𝐢𝐝: 𝑧 = 𝑥 2 + 𝑦 2 (elliptic),
𝑧 = 𝑥 2 − 𝑦 2 (hyperbolic)
2
𝐏𝐚𝐫𝐚𝐛𝐨𝐥𝐚 (𝐞𝐱𝐚𝐦𝐩𝐥𝐞): 𝑧 = 𝑦 + 1, note that y
= r ir we revolve the parabola about z.
𝐀𝐧𝐨𝐭𝐡𝐞𝐫 𝐩𝐚𝐫𝐚𝐛𝐨𝐥𝐚: 𝑥 2 − 𝑦 2 = 1
𝐂𝐢𝐫𝐜𝐥𝐞: 𝑥 2 + 𝑦 2 = 𝑟 2
Parameterization of curves, surfaces, and volumes:
̂
𝐑(𝜏) = 𝑥(𝜏)𝐢̂ + 𝑦(𝜏)𝐣̂ + 𝑧(𝜏)𝐤
̂
𝐑(𝑢, 𝑣) = 𝑥(𝑢, 𝑣)𝐢̂ + 𝑦(𝑢, 𝑣)𝐣̂ + 𝑧(𝑢, 𝑣)𝐤
̂
𝐑(𝑢, 𝑣, 𝑤) = 𝑥(𝑢, 𝑣, 𝑤)𝐢̂ + 𝑦(𝑢, 𝑣, 𝑤)𝐣̂ + 𝑧(𝑢, 𝑣, 𝑤)𝐤
𝑑𝑠 = √𝐑′(𝜏) ∙ 𝐑′ (𝜏)𝑑𝜏
𝑑𝐴 = ‖𝐑 u × 𝐑 v ‖𝑑𝑢𝑑𝑣
𝑑𝑉 = |𝐑 u ∙ 𝐑 v × 𝐑 w |𝑑𝑢𝑑𝑣𝑑𝑤
𝐑𝑢 × 𝐑𝑣
∂𝐑
̂=
𝐧
, 𝐑(𝑢, 𝑣)is parameterized surface, 𝐑 𝑢 =
‖𝐑 𝑢 × 𝐑 𝐯 ‖
∂u
Computationally, we can express these in terms of the components x,y,z of R:
𝑑𝑠 = √𝑥 ′2 + 𝑦 ′2 + 𝑧 ′2 𝑑𝜏
𝑑𝐴 = √𝐸𝐺 − 𝐹 2 𝑑𝑢 𝑑𝑣, where: 𝐸 = 𝑥𝑢2 + 𝑦𝑢2 + 𝑧𝑢2 ,
𝐺 = 𝑥𝑣2 + 𝑦𝑣2 + 𝑧𝑣2 ,
𝐹 = 𝑥𝑢 𝑥𝑣 + 𝑦𝑢 𝑦𝑣 + 𝑧𝑢 𝑧𝑣
Notes: Find x,y, and z in terms of the two parameters u and v. z will be in terms of u
and v, e.g. 𝑓(𝑥, 𝑦). Then take the appropriate derivatives. The limits of integration
are over the new parameters u and v.
∂(𝑥, 𝑦, 𝑧)
𝑑𝑉 =
𝑑𝑢𝑑𝑣𝑑𝑤,
note the Jacobian.
𝜕(𝑢, 𝑣, 𝑤)
Special cases of the above:
∂(𝑥, 𝑦)
Case 1: surface is flat and in the xy plane: 𝑑𝐴 =
𝑑𝑢𝑑𝑣
𝜕(𝑢, 𝑣)
Case 2: surface is known of the form 𝑧 = 𝑓(𝑥, 𝑦): 𝑑𝐴 = √1 + 𝑓𝑥2 + 𝑓𝑦2 𝑑𝑥𝑑𝑦
When we integrate Case 2, the limits of integration are defined by the region in the
xy plane under the surface.
𝐓𝐚𝐧𝐠𝐞𝐧𝐭 𝐩𝐥𝐚𝐧𝐞 at 𝑥𝑝 , 𝑦𝑝 , 𝑧𝑝 on a surface: 𝑓𝑥 (𝑥𝑝 , 𝑦𝑝 , 𝑧𝑝 )(𝑥 − 𝑥𝑝 )
+ 𝑓𝑦 (𝑥𝑝 , 𝑦𝑝 , 𝑧𝑝 )(𝑦 − 𝑦𝑝 ) + 𝑓𝑧 (𝑥𝑝 , 𝑦𝑝 , 𝑧𝑝 )(𝑧 − 𝑧𝑝 ) = 0
Note: 𝑓(𝑥, 𝑦, 𝑧) is the function, e.g. if our equation is 𝑥 2 + 𝑦 2 = 𝑧, then
𝑓(𝑥, 𝑦, 𝑧) = x 2 + y 2 − z. And 𝑓𝑥 is
̂)
(𝑓𝑥 𝐢̂+𝑓𝑦 𝐣̂+𝑓𝑧 𝐤
√𝑓𝑥2 +𝑓𝑦2 +𝑓𝑧2
𝜕𝑓
𝜕𝑥
. 𝐧̂ =
, where 𝑓𝑥 , 𝑓𝑦 , 𝑓𝑧 are evaluated at the point (𝑥𝑝 , 𝑦𝑝 , 𝑧𝑝 )on 𝑆.
ℜ
𝐴 = ∬ √1 + 𝑓𝑥2 + 𝑓𝑦2 𝑑𝑥𝑑𝑦 , where 𝐴 is the area of the surface and
ℜ
ℜ is the region in the x, y plane under 𝑆. We only use this equation if we can
write the surface as z=f(x,y).
y2 x2(y)
𝐴 = ∬ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∫
∫ 𝑓(𝑥, 𝑦)𝑑𝑥𝑑𝑦
𝑦1 𝑥1(𝑦)
𝑧2 y2(z) x2(y,z)
𝑉 = ∭ 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫ ∫
ℜ
𝑘=1
Note sizes of 𝐀𝐁𝐂: 𝑚 × 𝑛 times 𝑛 × 𝑝 = 𝑚 × 𝑝
𝐀𝐁 ≠ 𝐁𝐀
𝐀𝐀 … 𝐀 ≡ 𝐀𝐩 , 𝐀𝐩 𝐀𝐪 = 𝐀𝐩+𝐪 , (𝐀𝐩 )𝐪 = 𝐀𝐩𝐪
If 𝐀𝐁 = 𝐀𝐂, does 𝐧𝐨𝐭 imply that 𝐁 = 𝐂
(𝐀𝐁)𝟐 ≠ 𝐀𝟐 𝐁𝟐 : (𝐀𝐁)𝟐 = 𝐀𝐁𝐀𝐁, 𝐀𝟐 𝐁𝟐 = 𝐀𝐀𝐁𝐁
Transpose: switch rows for columns: (𝐀𝐓)𝐓 = 𝐀
(𝐀 + 𝐁)𝐓 = 𝐀𝐓 + 𝐁 𝐓
(α𝐀)𝐓 = α𝐀𝐓
(𝐀𝐁)𝐓 = 𝐁 𝐓 𝐀𝐓, (𝐀𝐁𝐂𝐃)𝐓 = 𝐃𝐓 𝐂𝐓 𝐁 𝐓 𝐀𝐓
x and y are column vectors:
𝐱 ∙ 𝐲 = 𝐱𝐓𝐲
If 𝐀𝐓 = 𝐀, then 𝐀 is symmetric.
If 𝐀𝐓 = −𝐀, then 𝐀 is antisymmetric.
1
1
Decompose: 𝐀 = (𝐀 + 𝐀𝐓 ) + (𝐀 − 𝐀𝐓 )
2
2
= 𝐀𝟏 + 𝐀𝟐 , where 𝐀 is square, 𝐀𝟏 is symmetric, and 𝐀𝟐 is antisymmetric.
𝐀𝟐 𝐈 does not imply that 𝐀 = ±𝐈
𝐀𝐱 = 𝐜,
𝐱 = 𝐀−𝟏 𝐜
𝐀−𝟏 𝐀 = 𝐀 𝐀−𝟏 = 𝐈
𝐀𝐁 = 𝐀[𝐂𝟏 … 𝐂𝐧 ] = [𝐀𝐂𝟏 … 𝐀𝐂𝐧 ] If 𝐁 is paritioned into n columns.
n
∫
𝑧1 𝑦1(𝑧) 𝑥1(𝑦,𝑧)
𝐌𝐚𝐭𝐫𝐢𝐜𝐞𝐬
𝐀 + 𝐁 = 𝐁 + 𝐀, α(β𝐀) = (αβ)𝐀
where 𝐀jk = (−1)j+k 𝐌jk
det𝐀 = ∑ 𝑎𝑗𝑘 𝐀𝑗𝑘 ,
k=1
Notes: Fix j, Find the M’s (little determinants), carry out the summation.
Properties of determinants:
1) If a row or column is modified by adding 𝛼 times another row, then detA does
not change.
2) If any two rows are changed then detA=-detB.
3) If A is triangular, then detA is the product of the diagonal.
4) If a row or column is 0 then the det is 0.
5) If a row or a column is a linear combo of other rows or columns then the det =
0
6) det(α𝐀) = αdet𝐀
7) det(𝐀T ) = det𝐀
8) det(𝐀 + 𝐁) ≠ det(𝐀) + det(𝐁)
9) det(𝐀𝐁) = det(𝐀) det(𝐁)
10) If any two rows are equal, det=0.
11) If det ≠ 0, then Ax=c has a unique solution.
12) If det=0 then A is singular.
13) If m < 𝑛 then system is underdetermined.
14) If m > 𝑛 then system is overdetermined.
15) Underdetermined and 2 unknowns 2 parameter family of solutions and
solutions lie in a plane. 1 parameter family and solutions like on a line, etc.
16) Inconsistent system: 0= -15 for a solution (for example).
17) Scaling a row or column scales the determinant.
𝐀𝟏 ⋯
0
A=[ ⋮
⋱
⋮ ] , det𝐀 = (det 𝐀𝟏 )(… )(det 𝐀𝐦 )
0
⋯ 𝐀𝐦
𝑛
𝑛
∑
∑
𝑖=1
𝑛
𝑐𝑖𝑗 = ∑
𝑗=1
𝑛
(∑
𝑖=1
𝑐𝑖𝑗 )
𝑗=1
⋯ 𝐴𝑛1
⋱
⋮ ],
⋯ 𝐴𝑛𝑛
1𝑛
adj𝐀 is the 𝐭𝐫𝐚𝐧𝐬𝐩𝐨𝐬𝐞 of the cofactor matrix.
Recall that the cofactor is: 𝐀jk = (−1)j+k 𝐌jk. Remember to take the Transpose of
Ajk to get the adjA
Another way to find 𝐀−𝟏 : Augment A|I, then use elementary ops to get 𝐈|𝐀−𝟏
𝟏
(𝐀𝐓 )−𝟏 = (𝐀−𝟏 )𝐓 ,
𝐈𝐧𝐯𝐞𝐫𝐬𝐞𝐬: (𝐀𝐁)−𝟏 = 𝐁 −𝟏 𝐀−𝟏 ,
det(𝐀−𝟏 ) =
det 𝐀
(𝐈 − 𝐀)−𝟏 = 𝐈 + 𝐀 + 𝐀𝟐 + ⋯ + 𝐀𝐩−𝟏 ,
where p is the power that yields 𝐀𝐩 = 0 (Nilpotent).
If A is invertible, then AB=AC implies that B=C, BA=CA implies that B=C, AB=0
implies that B=0.
𝐀−𝟏 =
𝐴 = ∬‖𝐑 u × 𝐑 v ‖𝑑𝑢𝑑𝑣, 𝑑𝐴 = ‖𝐑 u × 𝐑 v ‖𝑑𝑢𝑑𝑣
ℜ
𝑛
𝐀𝐁 = 𝐂 = {𝑐𝑖𝑗 } = {∑ 𝑎𝑖𝑘 𝑏𝑘𝑗 } ; (1 ≤ 𝑖 ≤ 𝑚, 1 ≤ 𝑗 ≤ 𝑝)
𝑓(𝑥, 𝑦, 𝑧)𝑑𝑥𝑑𝑦𝑑𝑧
−𝟏
𝑨
1
1 𝐴11
adj 𝐀 =
[ ⋮
det 𝐀
det 𝐀 𝐴
𝐀−𝟏 𝟏
=[ ⋮
0
⋯
⋱
⋯
0
⋮
]
𝐀−𝟏 𝐦
Theorem: If A is 𝑛 × 𝑛 and det 𝐀 ≠ 0, then 𝐀𝐱 =
𝐜 admits the unique solution 𝐱 = 𝐀−𝟏 𝐜.
1
𝑎 𝑏 −1
𝑑 −𝑏
] =[
]
𝑐 𝑑
−𝑐 𝑎 𝑎𝑑 − 𝑏𝑐
Cramer’s Rule: If Ax=c where A is invertible, then each component 𝑥𝑖 of x may
be computed as the ratio of two determinants; the denominator is det A, and the
numerator is det A but with the ith column replaced by c.
1
0 0 𝑢11 𝑢12 𝑢13
𝐀 = 𝐋𝐔 = [𝑙21 1 0] [ 0 𝑢22 𝑢23 ]
0 𝑢33
𝑙31 𝑙32 1 0
Notes: Decompose A into LU (e.g. start with u11 = a11, then u11 l21=a21, etc.), then
solve Ly=c for y, then solve Ux=y for x.
Ill conditioned: Small changes in matrix elements yield large changes in det,
𝐀−1 = [
det 𝐀
inverse, etc. To test if 𝑛 𝑛
∑𝑗 ∑𝑖 𝑎𝑗𝑖
≪ 1 then the matrix is ill conditioned.
Vector Spaces:
N − space: ℝ𝒏 = (𝑎1 , 𝑎2 , … , 𝑎𝑛 ) This means we have a vector with “n” dimesions.
Subspace: If a subset T of a vector space is itself a vector space (with the same
definitions as S for vector addition u+v, scalar multiplication au, zero vector 0, and
negative vector –u), then T is a subspace of S.
Dot product, norm, and angle for n-space
𝑛
𝐮 ∙ 𝐯 ≡ 𝑢1 𝑣1 + ⋯ + 𝑢𝑛 𝑣𝑛 = ∑ 𝑢𝑗 𝑣𝑗
𝑗=1
We can “weight” each component:
𝑛
𝐮 ∙ 𝐯 ≡ 𝑤1 𝑢1 𝑣1 + ⋯ + 𝑤𝑛 𝑢𝑛 𝑣𝑛 = ∑ 𝑤𝑗 𝑢𝑗 𝑣𝑗
𝑗=1
𝐮∙𝐯
𝜃 = cos −1 (
)
‖𝐮‖‖𝐯‖
|𝐮 ∙ 𝐯| ≤ ‖𝐮‖‖𝐯‖
Orthonormal: a set of vectors where each vector is normalized and each vector in
one set is orthogonal to every other vector in the other set. This can be described as
1, 𝑖 = 𝑗
the Kronecker delta = 𝐮𝐢 ∙ 𝐯𝐣 = 𝛿𝑖 𝑗 = {
0, 𝑖 ≠ 𝑗
Norms:
𝑛
 𝐮 = 𝛼1 𝐞𝟏 + ⋯ + 𝛼𝑘 𝐞𝐤 . A set of e’s is a basis for u
iff it is LI and span S.
 Orthogonal basis are preferred. Given orthog. basis
vectors: {𝐞𝟏 , … 𝐞𝐤 }, suppose we wish to expand a
given u in terms of these, then
𝐮∙𝐞
𝐮 = 𝛼1 𝐞𝟏 + ⋯ + 𝛼𝑘 𝐞𝐤 , where 𝛼1 = ( 𝟏 ) 𝐞𝟏 ,
u
e1
𝐞𝟏 ∙𝐞𝟏
e2
Orthogonalization process: Given k LI vectors 𝐯𝟏 , … , 𝐯𝐤 we can get k ON vectors
𝐞̂𝟏 , … , 𝐞̂𝐤 in span{𝐯𝟏 , … 𝐯𝐤 }by:
𝐞̂𝟏 =
𝐯𝟏
,
‖𝐯𝟏 ‖
𝐞̂𝟐 =
𝐯𝟐 − (𝐯𝟐 ∙ 𝐞̂𝟏 )𝐞̂𝟏
‖𝐯𝟐 − (𝐯𝟐 ∙ 𝐞̂𝟏 )𝐞̂𝟏 ‖
𝐣−𝟏
,
𝐞̂𝒋 =
𝐯𝐣 − ∑𝐢=𝟏(𝐯𝐣 ∙ 𝐞̂𝐢 )𝐞̂𝐢
𝐣−𝟏
‖𝐯𝐣 − ∑𝐢=𝟏(𝐯𝐣 ∙ 𝐞̂𝐢 )𝐞̂𝐢 ‖
Dimensions: The dimension of a vector space is the greatest number of LI vectors
in that vector space. If a vector space contains only a zero vector, the dimension is
0. Dimension relates to bases: The number of basis vectors equals the dimension
(because the bases are LI). The dim of a space will be no greater than n (ℜ𝑛 ).
Linear Independence: A set of vectors is LD if at least one of them can be
expressed as a linear combination of the others. Example: (1,0), (1,1), and (5,4)are
LD (𝐮𝟑 = 𝐮𝟏 + 𝟒𝐮𝟐 ).
A set of vectors is LD iff there exist scalars, not all zero, such that 𝛼1 𝐮𝟏 + ⋯ +
𝛼𝑘 𝐮𝐤 = 0. To solve for the scalars: 1) Set up a system of equations: 𝐀𝛂 = 𝟎. Where
A is the matrix of the vectors. 2) Use elementary row operations to reduce to “row
echelon form” or as close to it. The non-zero rows are LI.

A set containing the zero vector is LD.

Every orthogonal set of nonzero vectors is LI.
Best Approximations:
If u is any vector with ‖𝐮‖ = √𝐮 ∙ 𝐮
𝐍
𝐮 ≈ ∑(𝐮 ∙ 𝐞̂𝐣 )𝐞̂𝐣
𝑗=1
Where we are given u and an orthonormal basis set {𝐞̂𝟏 , … , 𝐞̂𝐍 }. The more basis
vectors we use, the closer our approximation will be (up to the number of bases
equal to the dimension of u.) The error of our approximation is:
𝑁
‖𝐄‖2 = ‖𝐮‖2 − ∑ 𝛼𝑗2
Taxicab norm: ‖𝐮‖ = ∑|𝑢𝑗 |
𝑗=1
𝑗=1
‖𝐮‖ ≡ √∑𝑛𝑗=1 𝑤𝑗 𝑢𝑗2, if we are using weights.
𝑛
‖𝐮‖ ≡ √∑ 𝑢𝑗2
𝑗=1
If vector space consists of functions, say, 𝑢(𝑥), 𝑣(𝑥) then the inner product is: 𝐮 ∙
𝐛
𝐯 = 〈𝑢(𝑥), 𝑣(𝑥)〉 ≡ ∫𝐚 𝑢(𝑥)𝑣(𝑥)𝑑𝑥, where a and b are the bounds of the function.
1
Note that the norm can be found for a vector ‖𝐮‖ = ‖𝑢(𝑥)‖ = √∫0 𝑢2 (𝑥)𝑑𝑥
Span: The set of all linear combinations of the vectors in a vector space is called
the span. E.g. for a vector space u1 , … , uk , the span is α1 u1 , … , αk uk and is denoted
as span{u1 , … , uk }. The span is a subspace of S. The span has to include the origin.
The above example shows a few linear combinations of the vectors u and v; the
span of this vector space would include all the linear combinations (a plane).
 To discover if two spans are equal, say span1 {𝐮𝟏 , 𝐮𝟐 }and span2 {𝐯𝟏 , 𝐯𝟐 }, write
the equation (𝛼1 𝐮𝟏 + α2 𝐮𝟐 ) = 𝐰, and (𝛼1 𝐯𝟏 + 𝛼2 𝐯𝟐 ) = 𝐰 and solve for
𝛼1 and 𝛼2 in terms of w. Compare the w vectors. E.g. (for a ℜ3 vectors):
𝑢11 𝑢21
𝑤1
𝛼1
[𝑢12 𝑢22 ] [𝛼 ] = [𝑤2 ]. In this case we have a non-square matrix, so reduce using
2
𝑢13 𝑢23
𝑤3
elementary operations so that the last row of A is zero (keeping track of the
operations on w). This gives us an equation only in terms of w, e.g. 0 = 𝑐𝑤1 +
𝑑𝑤2 + 𝑒𝑤3
Bases:
Where: 𝛼𝑗 = 𝐮 ∙ 𝐞̂𝐣
Row-echelon form:
1) In each row not made up entirely of zeros, the first nonzero element is a 1.
2) In any two consecutive rows not made up entirely of zeros, the leading 1 in the
lower row is to the right of the leading 1 in the upper row.
3) If a column contains a leading 1, every other element in that column is a zero.
4) All rows made up entirely of zeros are grouped together at the bottom of the
matrix.
Rank:
1) A matrix (maybe not square) is of rank r if it contains at least one r × r
submatrix with nonzero determinant but no square submatrix larger than r × r
with nonzero determinant. You can swap rows and columns to find these
submatrices. The zero matrix is of rank 0.
2) Elementary row operations do not alter the rank.
3) # of LI row vectors = # of LI column vectors = rank.
Elementary operations:
1) Operate on the augmented matrix (glue c onto A).
2) Addition of a multiple of one row to another.
3) Multiplication of a row by a nonzero constant.
4) Interchange of two rows.
Terminology
1) Consistent: one or more solutions
2) Unique: only one solution
3) Non-unique: more than one solution
4) Inconsistent: No solutions.
5) If m<n: Consistent or inconsistent. If consistent no unique solution exists (p
parameter family, where 𝑛 − 𝑚 ≤ 𝑝 ≤ 𝑛.
6) If m>n: Consistent or inconsistent. Can have unique or non-unique solution. (p
parameter family, where 1 ≤ 𝑝 ≤ 𝑛).
Eigenvalue Problem
Decompose any function into its even and odd parts:
𝑓(𝑥) =
𝑓(𝑥)+𝑓(−𝑥)
2
+
𝑓(𝑥)−𝑓(−𝑥)
∞
FS 𝑓 = 𝑎0 + ∑ (𝑎𝑛 cos
𝑛=1
2
𝑛𝜋𝑥
𝑛𝜋𝑥
+ 𝑏𝑛 sin
)
𝑙
𝑙
𝑙
If v is an eigenvector of A, then v lies on the vector Av. In other words, if v is an
eigenvector of A, then Av is the same as some constant times A, e.g. 𝜆𝑣.
(𝐀 − λ𝐈)𝐱 = 𝟎, Characteristic equation: det(𝐀 − λ𝐈) = 0
To find eigenvalues, solve the characteristic equation for 𝜆. To find the eigenspaces,
1) find the eigenvalues 𝜆, 2) for each 𝜆, plug back into (𝐀 − λ𝐈)𝐱 = 𝟎, and solve for
x. This will give us an eigenvector for each eigenvalue. 3) We should end up with
at least one arbitrary solution (0=0) for each eigenvector. This will give us our
arbitrary constants (𝛼, 𝛽, 𝛾, 𝑒𝑡𝑐. ) that when multiplied by the eigenvector gives us
our eigenspace.
1
𝑎0 = ∫ 𝑓(𝑥)𝑑𝑥 ,
2𝑙
−𝑙
𝑙
𝐴
𝐴
Even function: 𝑏𝑛 = 0, 𝑓(−𝑥) = 𝑓(𝑥), ∫−𝐴 𝑓(𝑥)𝑑𝑥 = 2 ∫0 𝑓(𝑥)𝑑𝑥
Odd function: 𝑎𝑛 = 𝑏𝑛 = 0, 𝑓(−𝑥) = −𝑓(𝑥),
𝐴
∫−𝐴 𝑓(𝑥)𝑑𝑥
=0
𝑙
𝑎𝑛 =
1
𝑛𝜋𝑥
∫ 𝑓(𝑥) cos
𝑑𝑥
𝑙
𝑙
−𝑙
1
𝑛𝜋𝑥
𝑏𝑛 = ∫ 𝑓(𝑥) sin
𝑑𝑥
𝑙
𝑙
−𝑙
𝑓 is 2𝑙 periodic, e. g. if the period is 2π, 𝑙 = 𝜋.
Elementary integral formulas:
𝑙
0, 𝑚 ≠ 𝑛
𝑚𝜋𝑥
𝑛𝜋𝑥
∫ cos
cos
𝑑𝑥 { 𝑙, 𝑚 = 𝑛 ≠ 0
𝑙
𝑙
2𝑙, 𝑚 = 𝑛 = 0
−𝑙
𝑙
∫ sin
Symmetric Matrices:
 If A is symmetric, then all of its eigenvalues are real.
 If an 𝜆 of a symmetric matrix A is of multiplicity k, then the eigenspace
corresponding to 𝜆 is of dimension k.
 If A is symmetric, then eigenvectors corresponding to distinct eigenvalues are
orthogonal.
 If an 𝑛 × 𝑛 matrix A is symmetric, then its eigenvectors provide an orthogonal
basis for n-space.
 If A is symmetric, then
𝐞T 𝐀𝐞
𝜆= 𝐓
𝐞 𝐞
Where e is an eigenvector of A.
Diagonalization: 𝐐−𝟏 𝐀𝐐 = 𝐃
Especially useful when solving a system of differential equations. Given a system
𝐀𝐱 = 𝐱′, our goal is to solve for x by 1) Find the Q and D matrices. Q has the
eigenvectors for rows, D has the eigenvalues in the diagonal (the order of the
eigenvalues matches the order of the eigenvectors in Q). 2) Write 𝐱̃′ = 𝐃𝐱̃. 3) This
gives you uncoupled equations, so you can do things like: 𝑥̃′ = 𝜆𝑥̃ → 𝑥̃ = 𝐶e𝜆𝑡 . 4)
Now that you have 𝐱̃, you can solve for x, by 𝐱 = 𝐐𝐱̃.
 Every symmetric matrix is diagonalizable.
 If an 𝑛 × 𝑛 matrix has n distinct eigenvalues, then it is diagonalizable. (It may be
diagonalizable anyway—does not read iff).
 If an 𝑛 × 𝑛 matrix has eigenvalues, then the corresponding eigenvectors are LI.
 A is diagonalizable iff it has n LI eigenvectors.
 𝐀𝐦 = 𝐐𝐃𝐦 𝐐−𝟏
Quadratic Form/Cononical Form: A quadratic form is said to be canonical if all
mixed terms (such as 𝑥1 𝑥2 ) are absent.
Example: reduce 𝑓(𝑥1 , 𝑥2 ) = 𝑎11 𝑥12 + 𝑎22 𝑥22 + (𝑎12 + 𝑎21 )𝑥1 𝑥2 to canonical form.
𝑎11 𝑎12
1) Identify the A matrix: 𝐀 = [𝑎
𝑎22 ], chose 𝑎12 , 𝑎21 to be equal so that the
21
matrix is symmetrical. 2) Find the eigenvalues. 3) Plug in to get the canonical form:
𝑓(𝑥̃1 , 𝑥̃2 ) = 𝜆1 𝑥̃12 + 𝜆2 𝑥̃22 . 4) Find the connection between 𝑥̃and 𝑥 by 𝐱 = 𝐐𝐱̃
where Q is the eigenvector matrix from A.
Tensors:
𝐀𝐢𝐣 = 𝐮⨂𝐯 = 𝐮𝐯 𝐓 = 𝐮𝐢 𝐯𝐣
𝐀𝐢𝐣𝐤𝐥 = 𝐮⨂𝐯⨂𝐰⨂𝐱 = 𝐮𝐢 𝐯𝐣 𝐰𝐤 𝐱𝐥
𝐈𝟏 (𝐀) = 𝐀𝐢𝐢 = trace(𝐀) = 1st Invariant
1
𝐈𝟐 (𝐀) = (𝐀𝐢𝐢 𝐀𝐣𝐣 − 𝐀𝐣𝐢 𝐀𝐢𝐣 ) = trace(𝐀) = 2nd Invariant
2
1st invariant of stress is hydrostatic pressure.
Eigenvalues of stress and strain tensors are principal stresses and strains.
Eigenvectors of stresses and strains are the directions.
𝐈𝟑 (𝐀) = 𝐀𝟏𝐢 𝐀𝟐𝐣 𝐀𝟑𝐤 𝜀𝐢𝐣𝐤 = 𝐝𝐞𝐭(𝐀)=3rd Invariant
To find eigenvalues solve for the roots of:
𝛌𝟑 − 𝐈𝟏 𝛌𝟐 + 𝐈𝟐 𝛌 − 𝐈𝟑 = 0
𝟏 𝛛𝐮𝐢 𝛛𝐮𝐣 𝛛𝐮𝐤 𝛛𝐮𝐤
𝐄𝐢𝐣 = ( ′ + ′ + ′
)
𝟐 𝛛𝐱𝐣 𝛛𝐱𝐢 𝛛𝐱𝐢 𝛛𝐱𝐣′
= Finite strain tensor (as opposed to the small strain tensor).
Where: 𝐱 ′ is the reference position.
Fourier Series:
= 𝑓𝑒 (𝑥) + 𝑓𝑜 (𝑥)
−𝑙
𝑙
∫ cos
−𝑙
0, 𝑚 ≠ 𝑛
𝑚𝜋𝑥
𝑛𝜋𝑥
sin
𝑑𝑥 {𝑙, 𝑚 = 𝑛 ≠ 0
𝑙
𝑙
𝑚𝜋𝑥
𝑛𝜋𝑥
sin
𝑑𝑥 = 0, for all 𝑚, 𝑛
𝑙
𝑙
Integration by parts: ∫ 𝑥 2 sin 𝑥 𝑑𝑥 = 𝑢𝑣 − ∫ 𝑣𝑑𝑢 , where 𝑣 ′ = sin 𝑥 , 𝑢 = 𝑥 2
Questions:
If we have 4 vectors of Rank 3, are they guaranteed to be LD? It seems they are if
we have more vectors than the rank…
Integral Table:
∫ sin2 𝑎𝜃𝑑𝜃 =
𝜃 sin 2𝑎𝜃
−
+𝐶
2
4𝑎
∫ ln 𝑥 𝑑𝑥 = 𝑥 ln 𝑥 − 𝑥 + 𝐶
∫ 𝑎 𝑥 𝑑𝑥 =
𝑎𝑥
+𝐶
ln 𝑎
1
∫ 𝑑𝑥 = ln 𝑥 + 𝐶
𝑥
𝑑𝑥
𝑥
∫
= sin−1 + 𝐶
𝑎
√𝑎2 − 𝑥 2
∫ 𝑒 𝑥 𝑑𝑥 = 𝑒 𝑥 + 𝐶
𝑑 𝑥
𝑢 = 𝑢 𝑥 ln u
𝑑𝑥
‖𝐫(t + h) −𝐫(t)‖
𝑑𝑠
= lim
𝑑𝑡 ℎ→0
ℎ
𝜏
1
1
2
∫ √1 + 4𝑡 𝑑𝑡 = 𝜏√1 + 4𝜏 2 + ln(2𝜏 + √1 + 4𝜏 2 )
2
4
0
𝐴
∫
𝑑𝑥 = −𝐴 ln(𝐵 − 𝑥) + 𝐶
𝐵−𝑥
u substitution for simplifying integrals: 1) Substitute a single variable (u) for a hardto-integrate portion (x). 2) Find du. 3) Rework limits of integration.
Power rule: