#### = %𝕣'𝕣 = 𝑃 + 𝑡𝑃𝑄
*****⃗ ,
Line segment: 𝑃𝑄
Lines and Planes
↱ 𝑥 = 𝑥- + 𝑡𝑎, 𝑦 = 𝑦- + 𝑡𝑏, 𝑧 = 𝑧- + 𝑡𝑐 And…
=>=?
@
=
A>A?
B
=
Cartesian Equation: Given 𝕣- = (𝑥- , 𝑦- , 𝑧-), 𝕕 = (𝑎, 𝑏, 𝑐) Then: line {𝕣|𝕣 = 𝕣- + 𝑡𝕕} = {(𝑥, 𝑦, 𝑧)|(𝑥, 𝑦, 𝑧) = (𝑥- , 𝑦- , 𝑧- ) + 𝑡(𝑎, 𝑏, 𝑐)}
C>C?
(symmetric form)
D
**********⃗
'S
? T? ×𝕕'
#####
Distance (point → line): Given point 𝑄- , line 𝑙 = 𝑙𝕕 (𝑃- ) Then 𝑑(𝑄- , 𝑙) = |𝑄
-𝑃| =
|𝕕|
And… ↲
|𝕕 ⋅𝕕 |
Angle: If 𝑙F = 𝑙𝕕F(𝕣- ), 𝑙G = 𝑙𝕕G (𝕣- ) Then: 𝑎𝑛𝑔(𝑙F , 𝑙G ) = cos >F |𝕕 N||𝕕P |
Line: 𝑙𝕕(𝕣- ) = {𝕣|𝕣 = 𝕣- + 𝑡𝕕}
N
*****⃗ + 𝑢𝑃𝑅
*****⃗ ,
Plane: 𝑃𝑄𝑅 = %𝕣'𝕣 = 𝑃 + 𝑡𝑃𝑄
*****⃗ '{𝑃, 𝑄} ⊆ 𝑃,
Direction Vectors: 𝑑𝑖𝑟 𝑝 = %𝑃𝑄
P
Normal Vectors: ↲
Parameterized Vector to Cartesian: If 𝕣- ∈ ℝa , 𝑘 = 𝕟 ⋅ 𝕣- , then {𝕣|𝕣 = 𝕣- + 𝑡𝕕 + 𝑢𝕖} = {𝕣 ∈ ℝa| det[𝕕|𝕖|𝕣 − 𝕣- ] = (𝕕 × 𝕖) ⋅ (𝕣 − 𝕣-) = 0} = {(𝑥, 𝑦, 𝑧) ∈ ℝa |𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑘}
k𝕟
, then {(𝑥, 𝑦, 𝑧)|𝑎𝑥 + 𝑏𝑦 + 𝑐𝑧 = 𝑘} = {𝕣|𝕟 ⋅ 𝕣 = 𝕟 ⋅ 𝕣- } = {𝕣|𝕣 = 𝕣- + 𝑡𝕕 + 𝑢𝕖}
Distance (point → plane): Given 𝕟 ∈ ℝl , 𝑄- ∈ ℝl If 𝑃- ∈ ℝl and ↲
↱ 𝑛𝑜𝑟 𝑝 = {𝕟|𝑛 ⊥ 𝕕}
Cartesian to Parameterized Vector: If 𝑘 ∈ ℝ, 𝕣- =
𝕟⋅𝕟
|𝕟⋅(T?>S? )|
|𝕟⋅T?>𝕟S? |
#### | = '𝑝𝑟𝑜𝑗𝕟 𝑃
*********⃗
*********⃗ 𝕟
↱ if 𝑝 = {𝕣 ∈ ℝl |𝑛 ⋅ (𝕣 − 𝑃- ) = 0 ⇔ 𝕟 ⋅ 𝕣 = 𝕟 ⋅ 𝑃- } then 𝑑(𝑝, 𝑄-) = |𝑃𝑄
=
- 𝑄- ' = |𝑃- 𝑄- ⋅ |𝕟| =
|𝕟|
|𝕟|
(in cartesian): If 𝑘 ∈ ℝ, 𝑝 = {𝕣 ∈ ℝl |𝕟 ⋅ 𝕣 = 𝑘} then 𝑑(𝑝, 𝑄- ) =
l
l
↱ Also if 𝑝 = %o𝑥, 𝑦(, 𝑧)p ∈ ℝ '𝑎𝑥 + 𝑏𝑦(+𝑐𝑧) = 𝑘, and {𝑎, 𝑏(, 𝑐), 𝑘} ⊆ ℝ and o𝑎, 𝑏(, 𝑐)p ≠ o0,0(, 0)p and 𝑄- = o𝑥- , 𝑦- (, 𝑧-)p ∈ ℝ then 𝑑(𝑝, 𝑄-) =
𝕟
********⃗
↱ if 𝑃F = {𝕣|𝕟 ⋅ (𝕣 − 𝑃F) = 0} = {𝕣|𝕟 ⋅ 𝕣 = 𝑘F }, 𝑃G = {𝕣|𝕟 ⋅ (𝕣 − 𝑃G ) = 0} = {𝕣|𝕟 ⋅ 𝕣 = 𝑘G} then 𝑑(𝑃F , 𝑃G) = '𝑝𝑟𝑜𝑗𝕟********⃗
𝑃F 𝑃G' = t𝑃
F 𝑃G ⋅ |𝕟|t =
l
∗
|(SP>SN )⋅𝕟|
|𝕟|
l
|@=?rBA?(rDC? )>k|
=
=
|𝕟|
↲
Distance (parallel planes or lines): ↲
s@PrBP(rD P )
|𝕟⋅SP>𝕟⋅SN|
|kP>kN|
|𝕟|
|𝕟⋅T?>k|
|𝕟 ⋅𝕟 |
Planes: 𝑎𝑛𝑔(𝑃F , 𝑃G) = cos >F |𝕟 N||𝕟P |
|𝕟|
N
P
Limits and Continuity
Given 𝕣: ℝ → ℝ , 𝑆 ⊆ ℝ, 𝑎 ∈ 𝑆 , 𝕃 ∈ ℝ
𝕣 is continuous at 𝑎 relative to 𝑆 iff (i) 𝕣 is definted at 𝑎, (ii) as 𝑡 →y 𝑎, 𝕣(𝑡) → 𝕣(𝑎)
𝕣 is continuous on 𝑆 iff ∀𝑡 ∈ 𝑆 𝕣 is cont. at 𝑡 rel to 𝑆
𝕣(•)>𝕣(•?)
Let 𝑐 ∈ ℝ, 𝑓: ℝ → ℝ, 𝕔 ∈
Vector Valued
𝕣(𝜏) − 𝕣(𝑡)
Given 𝕣 = (𝑓… )l : ℝ → ℝl
Properties of vector
• (𝑡)
𝕣 is differentiable at 𝑡- relative to 𝑆 iff 𝑡- ∈ 𝐷𝕣 , and lim
∈ ℝl
𝕣
=:
lim
•
•>•?
•→€ •?
ℝl , 𝐴 ∈ ℝˆ,l
Functions:
then 𝕣• = (𝑓… )l
valued derivatives:
‚→•
𝜏−𝑡
(𝑐𝕧)• = 𝑐𝕧′
(𝕔 ⋅ 𝕧)• = 𝕔 ⋅ 𝕧′
(𝑓𝕧)• = 𝑓 • 𝕧 + 𝑓𝕧′
(𝕦 ⋅ 𝕧)• = 𝕦• ⋅ 𝕧 + 𝕦 ⋅ 𝕧′
If 𝑛 ∈ {2,3}, (𝕔 × 𝕧)• = 𝕔 × 𝕧• and (𝕦 × 𝕧)• = (𝕦• × 𝕧) + (𝕦 × 𝕧• )
(𝕦 ∘ 𝑓)• = (𝕦• ∘ 𝑓)𝑓′
(𝕦/𝑓)′ = (𝑓𝕦• − 𝑓 • 𝕦)/𝑓 G
l
𝑓(𝑎 + ℎ, 𝑏) − 𝑓(𝑎, 𝑏)
B
B
B
(𝕦 ∘ 𝑓)•(𝑡) = 𝕦•o𝑓(𝑡)p𝑓′(𝑡)
(𝐴𝕦)• = 𝐴𝕦′
Vector valued integral: ∫@ 𝕣 = ‘∫@ 𝑓… ’
Fundamental Theorem: ∫@ 𝕣 = [ℝ]B@ = ℝ(𝑏) − ℝ(𝑎)
𝐷F (𝑓)(𝑎, 𝑏) = lim
“→ℎ
𝕣•
›𝕋
B
Unit Tangent: 𝕋 = |𝕣•|
Arc Length
Curvature
Curvature: 𝜅 = t t
Let 𝕣: 𝐼 → ℝl , 𝑎 ∈ 𝐼 if 𝐼 is an interval, 𝕣 is 𝐶 F on 𝐼, {𝑎, 𝑏} ⊆ 𝐼, then arc length for 𝕣 relative to 𝑎 is 𝑆𝕣,@ st 𝑠𝕣,@ = ∫•˜@|𝕣′(𝑡)| 𝑑𝑡
𝕋•
Unit Normal: ℕ = |𝕋ž|
𝕣ž
Note: if 𝕣 is smooth, then 𝕋 = |𝕣ž | =
Binormal Vector (n=3): 𝔹 = 𝕋 × ℕ
Motion in Space
𝑣 = |𝕧| = |𝕣• |
𝕧 = 𝕣′
Multivariable Limits and Continuity
𝕒 = 𝕧• = 𝕣••
↱ 𝐷k 𝑓 = 𝐷𝕖± 𝑓 = lim
•
•→-
where 𝕖k = (𝛿…k )l
›𝕋
𝕋ž
›œ
𝕣
›œ
Note: if 𝕣 is a parameterization wrt arc length then 𝕋 = 𝕣′ and 𝜅 = |𝕋•| = |𝕣•• |
and 𝜅 = t t = t ž t
•
𝑎𝕋 = 𝕒 ⋅ 𝕋, 𝑎ℕ = 𝕒 ⋅ ℕ so 𝑎𝕋 = 𝑣′ and 𝑎ℕ = 𝑣 G 𝜅
By fund. Thm. 𝕧(𝑡) = 𝕧(𝑡-) +
𝑑𝑢
And 𝕣(𝑡) = 𝕣(𝑡- ) + ∫¤˜• 𝕧(𝑢) 𝑑𝑢
?
#####
#####
Closed Ball: 𝐵#¦ (𝑃- ) = {𝑃||𝑃
Sphere: 𝑆¦ (𝑃- ) = {𝑃||𝑃
- 𝑃 | ≤ 𝑟}
- 𝑃 | = 𝑟}
#####
Open Ball: 𝐵¦ (𝑃- ) = {𝑃||𝑃
- 𝑃 | < 𝑟}
As 𝕩 → 𝕩- , 𝕗(𝕩) → 𝕃 means that ↲
Given: 𝑓: ℝl → ℝˆ , unit 𝕦 ∈ ℝl , 𝑘 ∈ ℕl then Directional Derivative (𝐷𝕦 𝑓)(𝕩) = : lim
↱ ∀𝜖 > 0, ∃𝛿 > 0 ∀𝕩 ∈ 𝑆 [0 < |𝕩 − 𝕩- | < 𝛿 ⇒ 𝕗(𝕩) ∈ 𝐷 and |𝕗(𝕩) − 𝕃 < 𝜖]
°(𝕩r•𝕖±)>°(𝕩)
›𝕣
𝒅𝒔
•
∫¤˜•? 𝕒(𝑢)
°(𝕩r•𝕦)>°(𝕩)
•
•→-
and Partial Derivative ↲
Let 𝑓: ℝl → ℝˆ , if 𝑛 = 1, then for any 𝕩 𝕗 is differentiable at 𝕩 ⇔ ∃ a unique 𝕒 ∈ ℝˆ 𝑠𝑡 as 𝕦 → 𝕩, ↲
Differentiability and Total Derivative
|𝕗(𝕦) − 𝕗(𝕩) − 𝕒(𝕦 − 𝕩)|
'𝕗(𝕦)>𝕗(𝕩)>³⋮¸¹ 𝕗(𝕩)⋮¶(𝕦>𝕩)'
Total Derivative: 𝐴 = ³⋮ 𝐷µ 𝕗(𝕩) ⋮¶ i.e. [𝐷F 𝕗 ⋮ 𝐷G 𝕗 ⋮ ⋯ ⋮ 𝐷l 𝕗]
Linearization
𝕗 is differentiable at 𝕩 ⇔ lim
= 0 ⇒ 𝕗• (𝕩) = ³⋮ 𝐷µ 𝕗(𝕩) ⋮¶
→0
|𝕦>𝕩|
𝕦→𝕩
|𝕦 − 𝕩|
If f is differentiable at 𝕩 then its linearization at 𝕩- is 𝐿𝕗,𝕩(𝕩) = 𝕗(𝕩- ) + 𝕗′(𝕩- )(𝕩 − 𝕩- )
If all partials 𝐷µ 𝑓 are continuous at 𝕩, then f is differentiable at 𝕩
If f is differentiable at 𝕩, then f is continuous at 𝕩
l
(𝐷𝕦 𝑓)(𝕩) = 𝑓 • (𝕩)𝕦 = (∇𝑓)(𝕩) ⋅ 𝕦
Properties of Total Derivative & Chain Rule
If 𝑓(𝕩) = 𝑐, then 𝑓 • (𝕩) = [0]l
(𝑐𝕘)•(𝕩) = 𝑐𝕘′(𝕩)
(𝐴𝕘)•(𝕩) = 𝐴𝕘′(𝕩)
Gradient ∇𝑓(𝕩) = 𝑓 • (𝕩)¼ = o𝐷… 𝑓(𝕩)p
(𝕗 + 𝕘)•(𝕩) = 𝕗• (𝕩) + 𝕘′(𝕩)
(𝑢𝕘)•(𝕩) = 𝑢(𝕩)𝕘•(𝕩) + 𝕘(𝕩)𝑢′(𝕩)
(𝕗/𝑢)′(𝕩) = o𝑢(𝕩)𝕗• (𝕩) − 𝕗(𝕩)𝑢• (𝕩)p/(𝑢(𝕩)G )
Then (𝕗 ∘ 𝕘)• (𝕩) = 𝕗• o𝕘(𝕩)p𝕘′(𝕩)
Chain Rule let 𝕗: ℝˆ → ℝk , 𝕘: ℝl → ℝˆ
𝑙 is tangent to 𝐶 at 𝕣- iff: ∃ 𝜎⃗ st 𝜎⃗(𝑡- ) = 𝕣- and 𝜎⃗ •(𝑡- ) ≠ 0 and 𝑙 = 𝑙*¿*⃗ ž(•? )(𝕣- )
Tangent Lines and Planes
Let 𝐶 = {(𝑥, 𝑦)|𝑦 = 𝑓(𝑥)}, f be diff’ble at 𝑥- , then 𝑙 is tangent to C at (𝑥- , 𝑦-) ⇔ 𝑦- = 𝑓(𝑥- ) and ↲
↱ 𝑙 = 𝐿°,=? = {(𝑥, 𝑦)|𝑦 = 𝑓(𝑥-) + 𝑓 •(𝑥- )(𝑥 − 𝑥- )}
OR if 𝑥 = 𝑔(𝑦) then 𝑙 is tangent to C ⇔ 𝑥- = 𝑔(𝑦- ) and 𝑙 = 𝐿Ä,A? = {(𝑥, 𝑦)|𝑥 = 𝑔(𝑦-) + 𝑔• (𝑦-)(𝑦 − 𝑦- )}
The plane 𝑝 is tangent to 𝑆 𝑎𝑡 𝕣- iff ↲
↱ ∃ 𝜙: ℝG → 𝑆 and ∃ 𝕥- ∈ 𝑑𝑜𝑚𝜙 st 𝜙(𝕥- ) = 𝕣- and 𝑝 = {𝕣|𝕣 = 𝕣- + 𝑡𝐷F (𝜙)(𝕥- ) + 𝑢𝐷G (𝜙)(𝕥- )
Maximum and Minimum Extrema
Local max at (𝑎, 𝑏): 𝑓(𝑥, 𝑦) ≤ 𝑓(𝑎, 𝑏) when (𝑥, 𝑦) is near (𝑎, 𝑏)
Local Min at (𝑎, 𝑏): 𝑓(𝑥, 𝑦) ≥ 𝑓(𝑎, 𝑏) when (𝑥, 𝑦) is near (𝑎, 𝑏)
If 𝑓 has a local min or max at (𝑎, 𝑏) then 𝐷F 𝑓(𝑎, 𝑏) = 0 and 𝐷G 𝑓(𝑎, 𝑏) = 0
If (𝑎, 𝑏) is crit pt of f, let 𝐷 = 𝑓== (𝑎, 𝑏)𝑓AA (𝑎, 𝑏) − ³𝑓=A (𝑎, 𝑏)G¶ ↲
↱ then if 𝐷 > 0 and 𝑓== (𝑎, 𝑏) > 0 then 𝑓(𝑎, 𝑏) is a local min or if 𝐷 > 0 and 𝑓== (𝑎, 𝑏) < 0 then 𝑓(𝑎, 𝑏) is a local max or if 𝐷 < 0 then 𝑓(𝑎, 𝑏) is neither local min or max (saddle point)
To find extrema: ↲
↱ 1) Find values of 𝑓 at the critical points of 𝑓 in 𝐷. 2) Find the extreme values of 𝑓 on the boundary of 𝐷. 3) The largest values from steps 1 and 2 is the absolute max, the smallest is the absolute min
Lagrange Multipliers
To find the max and min values of 𝑓(𝑥, 𝑦, 𝑧) subject to the constraint 𝑔(𝑥, 𝑦, 𝑧) = 𝑘 [assuming they exist and ∇𝑔 ≠ 𝟎 on the surface of 𝑔(𝑥, 𝑦, 𝑧) = 𝑘]:
a) find all values of 𝑥, 𝑦, 𝑧, and ↲
↱ 𝜆 such that ∇f(x, y, z) = λ∇g(x, y, z) and 𝑔(𝑥, 𝑦, 𝑧) = 𝑘
b) Evaluate 𝑓 at all points (𝑥, 𝑦, 𝑧) that result from a). The largest is the max of 𝑓, the smallest is the min of 𝑓
This produces a system of eq’s: ↲
Union of regions: ∬¸ 𝑓(𝑥, 𝑦)𝑑𝐴 = ∬¸ 𝑓(𝑥, 𝑦)𝑑𝐴 + ∬¸ 𝑓(𝑥, 𝑦)𝑑𝐴
Area: ∬¸ 1 𝑑𝐴 = 𝐴(𝐷)
↱ 𝐷F 𝑓 = 𝜆𝐷F 𝑔, 𝐷G 𝑓 = 𝜆𝐷G 𝑔, 𝐷a 𝑓 = 𝜆𝐷a 𝑔, 𝑔(𝑥, 𝑧, 𝑦) = 𝑘
Multiple Integrals
N
P
↱
l
lim ∑Ø…˜F ∑ˆ
µ˜F ∑k˜F 𝑓(𝑥… , 𝑦µ
Ø,ˆ,l→Ù
B ÄP(=) ¤P (=,A)
∭Ü 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫@ ∫ÄN(=) ∫¤N(=,A) 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧 𝑑𝑦 𝑑𝑥
=P
=P
³[𝑚]¶= = [𝑚(𝑥)]=˜=
= 𝑚(𝑥G ) − 𝑚(𝑥F )
N
f over box B: ∭× 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 =
Type I:
Physics
œ
, 𝑧k )Δ𝑉
› “ (A) ¤ (=,A)
Type II: ∭Ü 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫D ∫“ P(A) ∫¤ P(=,A) 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑧 𝑑𝑥 𝑑𝑦
N
N
Mass: ∫× 𝑓 𝑑𝑚 = ∑à
𝑚
𝑓(𝑥
⃗) and 𝑚[𝑆] = ∫× 𝑑𝑚 = ∑𝑵
k˜F k
𝒌˜𝟏 𝒎𝒌
N
N=2: ‘∬(=,A)∈y 𝑥 𝑑𝑚(𝑥, 𝑦) , ∬(=,A)∈y 𝑦 𝑑𝑚(𝑥, 𝑦)’, called (𝑀ç [𝑆], 𝑀è [𝑆])
Center of m: 𝑐𝑚[𝑆] = 𝑀æ [𝑆]/𝑚[𝑠] = ∫¦⃗∈y 𝑟⃗𝑑𝑚(𝑟⃗) / ∫¦⃗∈y 𝑑𝑚(𝑟⃗)
For 𝜌: 𝑐𝑚[𝑆] =
ëì [y]
ˆ[y]
=
∫*î⃗∈ï ¦⃗í(¦⃗)›¦⃗
∫*î⃗∈ï í(¦⃗)›¦⃗
Centroid: 𝑆̅ =
Moment of Inertia (about): 𝐼æ = 𝐼A + 𝐼= = ∬¸
(𝑥 G
∫î*⃗∈ï ¦⃗›¦⃗
+𝑦
G)
G
𝜌(𝑥, 𝑦) 𝑑𝐴
B
General regions:
¤ (=,A)
Volume: 𝑉(𝐸) = ∭Ü 𝑑𝑉 i.e. ∭Ü 𝑑𝑉 = ∬¸ Þ∫¤ P(=,A) 𝑑𝑧ß 𝑑𝐴
N
1st m-moment about origin: 𝑀æ [𝑆] = ∑à
⃗k = ∫¦⃗∈y 𝑟⃗ 𝑑𝑚(𝑟⃗) if defined.
k˜F 𝑚k 𝑟
n=3: ‘∭(=,A,C)∈y 𝑥 𝑑𝑚(𝑥, 𝑦, 𝑧) , ∭(=,A,C)∈y 𝑦 𝑑𝑚(𝑥, 𝑦, 𝑧) , ∭(=,A,C)∈y 𝑧 𝑑𝑚(𝑥, 𝑦, 𝑧)’, called (𝑀çé [𝑆], 𝑀éè [𝑆], 𝑀èç [𝑆])
n=2: (𝑀ç [𝑆]/𝑚[𝑆], 𝑀è [𝑆]/𝑚[𝑆])
n=2: 𝑆̅ = 𝑐𝐴[𝑆] =
∫î*⃗∈ï ›¦⃗
›
Fubini’s Theorem: if 𝐵 = [𝑎, 𝑏] × [𝑐, 𝑑] × [𝑟, 𝑠] then, ∭× 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫¦ ∫D ∫@ 𝑓(𝑥, 𝑦, 𝑧) 𝑑𝑥 𝑑𝑥 𝑑𝑧
∬(ñ,ò)∈ï(=,A)›(=A)
∬(ñ,ò)∈ï ›(=A)
=ó
about point 𝑃- : 𝑀𝐼 S? [𝑆] =
G
ô[y]
∫y 𝑟÷G? 𝑑𝑚
,
ô[y]
õ
n=3: 𝑆̅ = 𝑐𝑉[𝑆] =
about line ℓ: 𝑀𝐼ℓ [𝑆] =
G
𝜌 continuous on S: ∫y 𝑑𝑚 = ∫y 𝜌 = ∫¦⃗∈y 𝜌(𝑟⃗)𝑑𝑟⃗
∭(ñ,ò,ö)∈ï(=,A,C) ›(=AC)
i.e.: 𝑥̅ =
∭(ñ,ò,ö)∈ï ›(=AC)
∫y 𝑟ℓG 𝑑𝑚
about origin: 𝑀𝐼æ [𝑆] =
G
G
ëò
ˆ
∫y 𝑟æG 𝑑𝑚
=
F
ˆ
∬¸ 𝑥 𝜌(𝑥, 𝑦) 𝑑𝐴
= ∫¦⃗∈y|𝑟⃗ − 0|G 𝑑𝑚(𝑟⃗)
n=3: 𝑀𝐼è = ∭(=,A,C)∈y 𝑦 + 𝑧 𝑑𝑚(𝑥𝑦𝑧) , 𝑀𝐼ç = ∭(=,A,C)∈y 𝑧 + 𝑥 𝑑𝑚(𝑥𝑦𝑧) , 𝑀𝐼é = ∭(=,A,C)∈y 𝑥 G + 𝑦 G 𝑑𝑚(𝑥𝑦𝑧)
wrt x-axis: 𝑚𝑦# G = 𝐼= , wrt y-axis: 𝑚𝑥̅̅ G = 𝐼A
n=2: 𝑀𝐼è [𝑆] = ∬(=,A)∈y 𝑦 𝑑𝑚(𝑥, 𝑦) , 𝑀𝐼ç [𝑆] = ∬(=,A)∈y 𝑥 𝑑𝑚(𝑥, 𝑦) , 𝑀𝐼æ = 𝑀𝐼è [𝑆] + 𝑀𝐼ç [𝑆]
n=3: 𝑀𝐼è + 𝑀𝐼ç + 𝑀𝐼é = ∭(=,A,C)∈y 2𝑥 G + 2𝑦 G + 2𝑧 G 𝑑𝑚(𝑥𝑦𝑧) = 2𝑀𝐼æ
n=3: (𝑀çé [𝑆]/𝑚[𝑆], 𝑀éè [𝑆]/𝑚[𝑆], 𝑀èç [𝑆]/𝑚[𝑆])
∬(ñ,ò)∈ï = ›(=A) ∬(ñ,ò)∈ï A ›(=A)
G
Radius of Gyration: 𝑟ˆ,ℓ[𝑆] = s𝑀𝐼ℓ [𝑆]/𝑚[𝑆]
If 𝑆 = 𝕋(𝐸FFF ) then 𝑉[𝑆] = ∭(¤,ü,ý)∈Ü |𝑑𝑒𝑡𝕋• (𝑢, 𝑣, 𝑤)|𝑑(𝑢𝑣𝑤)
Linear Transformations:
For any 𝐴 ∈ ℝˆ,l , 𝑇ô : ℝl → ℝˆ , 𝑇ô (𝑢
*⃗) = 𝐴𝑢
*⃗ ∀𝑢
*⃗ ∈ ℝl is a linear transformation from ℝl → ℝˆ
NNN
G
Expand/Contract/Invert:
Diagonal reflection:
Horizontal Shear:
Vertical Shear:
Given: %𝑃- , 𝑎⃗, 𝑏*⃗ , ⊆ ℝ , 𝐵FF = [0,1] × [0,1] and
area𝑇ô (𝐵) =
Then: areaD = 'det³𝑎⃗'𝑏*⃗¶' = '𝑎⃗ × 𝑏*⃗ '
𝑟 0
0 1
1 𝑠
1 0
|𝑑𝑒𝑡𝐴|(𝑎𝑟𝑒𝑎𝐵)
𝐴F = Þ
ß
𝐴G = Þ
ß
𝐴a = Þ
ß
𝐴þ = Þ
ß
𝐷 = %𝑃- + 𝑢𝑎⃗ + 𝑣𝑏*⃗ '(𝑢, 𝑣) ∈ 𝐵FF,
0 𝑠
1 0
0 1
𝑟 1
l
[𝑎… , 𝑏… ], 𝐴 ∈ ℝl,l , 𝑇 = 𝑇ô and 1: ℝl → {1}
Given: 𝐵 = 𝑥…˜F
Then: 𝑣𝑜𝑙𝐷 = 'det³𝑎⃗'𝑏*⃗'𝑐⃗¶' = '𝑎⃗ ⋅ o𝑏*⃗ × 𝑐⃗p'
Given: 𝐵FFF = [0,1]a , 𝐷 = {𝑃- + 𝑢𝑎⃗ + 𝑣𝑏*⃗ + 𝑤𝑐⃗|(𝑢, 𝑣, 𝑤) ∈ 𝐵FFF
Then: ∫¼(×) 1 = |𝑑𝑒𝑡𝐴| ∫× 1 = ∫×|𝑑𝑒𝑡𝐴|1 = ∫×|𝑑𝑒𝑡𝑇′|
n=2: = ∬(¤,ü)∈×|𝑑𝑒𝑡𝑇 • (𝑢, 𝑣)|𝑑(𝑢𝑣)
n=3: = ∭(¤,ü,ý)∈×|𝑑𝑒𝑡𝑇 • (𝑢, 𝑣, 𝑤)|𝑑(𝑢𝑣𝑤)
Polar Transformation: 𝑇÷ (𝑟, 𝜃) = 𝑟(𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃) = 𝑇÷ (𝑟 • , 𝜃 • )
Polar →Cartesian: (𝑥, 𝑦) = 𝑇÷ (𝑟, 𝜃) ⇔ [𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃] ⇒ [𝑟 G = 𝑥 G + 𝑦 G , 𝑥 ≠ 0 ⇒ 𝑡𝑎𝑛𝜃 =
A
𝑑𝑒𝑡𝑇 • (𝑟, 𝜃) = det(x, y)• = det(𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)′ = 𝑟 → (𝑟 𝑑𝑟 𝑑𝜃)
Cylindrical transformation:
=
(𝑥, 𝑦, 𝑧) = 𝑇é (𝑟, 𝜃, 𝑧) ⇔ (𝑥, 𝑦) = 𝑇÷ (𝑟, 𝜃) ⇔ (𝑥, 𝑦) = 𝑟(𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃), 𝑧 = 𝑧
𝑇C (𝑟, 𝜃, 𝑧) = (𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃, 𝑧)
𝑇é (𝑟, 𝜃, 𝑧) = 𝑇é , 𝜃 , 𝑧 ⇔ [𝑟(𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃) = 𝑟 (𝑐𝑜𝑠𝜃 , 𝑠𝑖𝑛𝜃 ), and z=z’]
Converting:
𝑇é (𝑟, 𝜃, 𝑧) = 𝑇æ (𝜌, 𝜙, 𝜃) ⇔ [(𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃) = (𝜌𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 𝜌𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃), 𝑧 = 𝜌𝑐𝑜𝑠𝜙] ⇒ [𝜌G = 𝑟 G + 𝑧 G , 𝑟 ≠ 0 ⇒ 𝑐𝑜𝑡𝜙 = 𝑧/𝑟
𝑑𝑒𝑡𝑇 • (𝑟, 𝜃, 𝑧) = 𝑟𝑐𝑜𝑠 G 𝜃 + 𝑟𝑠𝑖𝑛G 𝜃 = 𝑟 → (𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃)
Cyl→Rec: 𝑥 = 𝑟𝑐𝑜𝑠𝜃, 𝑦 = 𝑟𝑠𝑖𝑛𝜃, 𝑧 = 𝑧
Rec→Cyl: 𝑟^2 = 𝑥^2 + 𝑦^2, 𝑡𝑎𝑛𝜃 = 𝑦/𝑥, 𝑧 = 𝑧
Triple Cylindrical: 𝐸 = {(𝑥, 𝑦, 𝑧)|(𝑥, 𝑦) ∈ 𝐷, 𝑢F (𝑥, 𝑦) ≤ 𝑧 ≤ 𝑢G(𝑥, 𝑦), 𝐷 = {(𝑟, 𝜃)|𝛼 ≤ 𝛽, ℎF (𝜃) ≤ 𝑟 ≤ ℎG(𝜃)
(𝑟 •
&
•
•)
•
¤ (¦Dæœ$,¦œ…l$)
“
Then: ∭Ü 𝑓(𝑥, 𝑦, 𝑧)𝑑𝑉 = ∫' ∫“ P(%) ∫¤ P(¦Dæœ$,¦œ…l$) 𝑓(𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃, 𝑧) 𝑟 𝑑𝑧 𝑑𝑟 𝑑𝜃
N(%)
N
•
•
Spherical Transformation:
𝜌 ≥ 0, 0 ≤ 𝜃 ≤ 2𝜋, 0 ≤ 𝜙 ≤ 𝜋
Sph→Rec: 𝑟 = 𝜌𝑠𝑖𝑛𝜙, 𝑥 = 𝜌𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 𝑦 = 𝜌𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃, ↲
(𝑥, 𝑦, 𝑧) = 𝑇æ (𝜌, 𝜙, 𝜃) ⇔ (𝑥, 𝑦, 𝑧) = 𝑇é (𝜌𝑠𝑖𝑛𝜙, 𝜃, 𝜌𝑐𝑜𝑠𝜙)
Rec→Sph: 𝜌G = 𝑥 G + 𝑦 G + 𝑧 G , 𝑡𝑎𝑛𝜃 = 𝑦/𝑥, 𝑐𝑜𝑠𝜙 = 𝑧/𝜌
𝑇æ (𝜌, 𝜙, 𝜃) = 𝑇é (𝜌𝑠𝑖𝑛𝜙, 𝜃, 𝜌𝑐𝑜𝑠𝜙) = (𝜌𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 𝜌𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃, 𝜌𝑐𝑜𝑠𝜙)
B
•˜B
•˜B
Where arc length 𝐿(𝐶) = ∫)⃗ 𝑑𝑠
𝜎 smth[a,b]⇔ 𝜎 𝐶 F , 𝜎 • ≠ 0
Line Integral: ∫) 𝑓 𝑑𝑠 = ∫@ 𝑓(𝜎⃗(𝑡))||𝜎⃗ • (𝑡)||𝑑𝑡
For each 𝑗 ∈ ℕl , ∫)⃗ 𝑓 𝑑𝜄µ = ∫•˜@ 𝑓o𝜎(𝑡)p 𝜎µ• (𝑡) 𝑑𝑡
And ∫)⃗ 𝐹⃗ ⋅ 𝑑𝐼⃗ = ∫•˜@ 𝐹⃗ o𝜎⃗(𝑡)p ⋅ 𝜎⃗ ′(𝑡) 𝑑𝑡
For piecewise-smooth curves: ∫) 𝑓(𝑥, 𝑦)𝑑𝑠 = ∫) 𝑓(𝑥, 𝑦)𝑑𝑠 + ∫) 𝑓(𝑥, 𝑦)𝑑𝑠 + ⋯ + ∫ _𝐶l 𝑓(𝑥, 𝑦)𝑑𝑠
To parameterize line segment 𝕣 → 𝕣F , 𝕣(𝑡) = (1 − 𝑡)𝕣- + 𝑡𝕣F , 0 ≤ 𝑡 ≤ 1
show F conservative:
𝑧 = 𝜌𝑐𝑜𝑠𝜙
N
P
So ∫) 𝐹 ⋅ 𝑑𝑟 = ∫) ∇𝑓 ⋅ 𝑑𝑟 = 𝑓(𝑏) − 𝑓(𝑎)
F is Conservative iff 𝐹 = ∇𝑓
n=2: 𝐹(𝑥, 𝑦) = (𝑃, 𝑄), 𝐷F 𝑄(𝑥, 𝑦) must = 𝐷G 𝑃(𝑥, 𝑦)
n=3: 𝐹(𝑥, 𝑦, 𝑧) = (𝑃, 𝑄, 𝑅), CurlF=0 (𝜕𝑅/𝜕𝑦 = 𝜕𝑄/𝜕𝑧, 𝜕𝑅/𝜕𝑥 = 𝜕𝑃/𝜕𝑧, 𝑒𝑡𝑐)
B
Green’s Theorem: ∮) 𝐹 ⋅ 𝑑𝑟 = ∮ 𝑃𝑑𝑥 + 𝑄𝑑𝑦 = ∬ [𝜕𝑄/𝜕𝑥 − 𝜕𝑃/𝜕𝑦] 𝑑𝐴
F is vector function, C given by 𝕣(𝑡), 𝑎 ≤ 𝑡 ≤ 𝑏, 𝕋(𝑥, 𝑦, 𝑧) unit tangent vector on C: ∫) 𝔽 ⋅ 𝑑𝕣 = ∫) 𝔽 ⋅ 𝕋 𝑑𝑠 = ∫@ 𝔽o𝕣(𝑡)p ⋅ 𝕣• (𝑡) 𝑑𝑡
More generally (given 𝔽 = (𝐹F , 𝐹G ), 𝐹F and 𝐹G real-valued and 𝐶 F ): ∫1¸ 𝔽 ⋅ 𝑑𝕀 = ∬¸ 𝑐𝑢𝑟𝑙𝐹 𝑑𝐴 = ∬¸ ∇ × 𝔽 𝑑𝐴 so (∫1¸2(𝐹F 𝑑𝜄F , 𝐹G 𝑑𝜄G) = ∬¸(𝐷F 𝐹G − 𝐷G 𝐹F )𝑑𝐴)
Vector to scalar ∫) 𝔽 ⋅ 𝑑𝕣 = ∫) 𝑃 𝑑𝑥 + 𝑄 𝑑𝑦 + 𝑅 𝑑𝑧, where 𝔽 = (𝑃, 𝑄, 𝑅) and d𝕣 = o𝑥 • (𝑡), 𝑦 •(𝑡), 𝑧 •(𝑡)p
Divergence: ∇=
1
1=
𝚤̂ +
1
1A
𝚥̂ +
1
1C
𝑘6 , 𝐹(𝑥, 𝑦, 𝑧) = 𝑃𝚤̂ + 𝑄𝚥̂ + 𝑅𝑘6 , 𝑫𝒊𝒗𝑭 = ∇ ⋅ 𝐹
n=3: ∇ × 𝐹 = (𝐷G 𝐹a − 𝐷a 𝐹G , 𝐷a 𝐹F − 𝐷F 𝐹a , 𝐷F 𝐹G − 𝐷G 𝐹F ) = (0,0,0)
And flux ∫𝕊 𝔽 ⋅ 𝑑𝔸 = ∬(•,¤)∈¸ 𝔽 ‘𝜙*⃗ (𝑡, 𝑢)’ ⋅ (𝐷F × 𝐷G )o𝜙*⃗p(𝑡, 𝑢) 𝑑(𝑡𝑢)
e.g. ∫) 𝑦 𝑑𝑥 + 𝑧 𝑑𝑦 + 𝑥 𝑑𝑧 = ∫) 𝔽 ⋅ 𝑑𝕣 where 𝔽(𝑥, 𝑦, 𝑧) = (𝑦, 𝑧, 𝑥)
Curl: (+) ccw rotation, (-) cw rotation, (0) no rotation
Surface Integrals:
Examples
n=2: ∇ × 𝐹 = (𝐷F , 𝐷G) × (𝐹F , 𝐹G ) = 𝐷F 𝐹G − 𝐷G 𝐹F = 0
Curl F: ∇ × 𝐹
Given 𝜙*⃗: 𝐷 (𝐶 F and onto) → 𝑆 ⊆ 𝐸, 𝕊 = (𝑆, 𝜙*⃗) then ∫𝕊 𝑓 𝑑𝜎 = ∬(•,¤)∈¸ 𝑓 ‘𝜙*⃗(𝑡, 𝑢)’ '(𝐷F × 𝐷G )o𝜙*⃗p(𝑡, 𝑢)' 𝑑(𝑡𝑢)
Find a simp. Cart. Eqtn. of plane p tangent to 𝑆 = {𝑥 − 𝑦 G + 𝑒 C = 𝑒} at (1, −1,1)
𝑓(𝑥, 𝑦, 𝑧) = 𝑥 − 𝑦 G + 𝑒 C ↲
C
↱ ∇𝐹(𝑥, 𝑦, 𝑧) = (1, −2𝑦, 𝑒 ) and 𝕣- = (1, −1,1)
𝑃 = %(𝑥, 𝑦, 𝑧)'∇𝐹(1, −1,1) ⋅ o(𝑥, 𝑦, 𝑧) − (1, −1,1)p = 0, ⇔ 𝑥 + 2𝑦 + 𝑒𝑧 = 𝑒 − 1
Determine at which of these pts f is locally max or min or has saddle pt ↲
∇𝑓(𝑥, 𝑦) = (𝑓F , 𝑓G ) = (4𝑥 a − 4𝑥, 4𝑦 − 8) and 𝑓FF = 12𝑥 G − 4, 𝑓GG = 4, 𝑓FG = 𝑓GF = 0
a) 4𝑥 a − 4𝑥 = 0 ⇔ 4𝑥(𝑥 G − 1) = 0 ⇔ 𝑥 = 0 𝑜𝑟 𝑥 = ±1 b) 4𝑦 − 8 = 0 ⇔ 𝑦 = 2 ↲
𝑓(𝑥, 𝑦) = 𝑥 þ − 2𝑥 G + 2𝑦 G − 8𝑦
So pts = {(0,2), (1,2), (−1,2)}
For 𝕣 = (0,2), 𝑓FF = −4 < 0, 𝐷 = −16 < 0 ⇒Saddle
For 𝕣 = (1,2), 𝑓FF = 8 > 0, 𝐷 = 32 > 0 ⇒local min
For 𝕣 = (−1,2), 𝑓FF = 8 > 0, 𝐷 = 32 > 0 ⇒local min
B/G G
B/G G
Compute ∮𝝏𝑫 𝔽 ⋅ 𝒅𝕀
Given 𝐷 = {(𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)|1 ≤ 𝑟 ≤ 2, 0 ≤ 𝜃 ≤ 𝜋/2 and 𝔽(𝑥, 𝑦) = (𝑐𝑜𝑠𝑥 − 3𝑦 G , 1 − 𝑒 œ…lA )
Greens Thm ⇒ ∮ 𝔽 ⋅ 𝑑𝕀 = ∫ ∫ [𝐷F 𝐹G − 𝐷G 𝐹F ]𝑑𝐴 = ∫ ∫ 6𝑟 G 𝑠𝑖𝑛𝜃 𝑑𝑟 𝑑𝜃 = 14
1¸
Compute 𝒎[𝑺] = ∫𝑺 𝜹 𝒅𝝈 = ∫𝑺 𝜹|𝒅𝔸| Given 𝑆 = 𝑆F (0) = {𝕣||𝕣| = 1} and 𝛿(𝑥, 𝑦, 𝑧) = 1 + 𝑧 G
-
F
-
F
𝑆 = 𝜙*⃗(𝐸) and 𝜙*⃗(𝜙, 𝜃) = (𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜙, 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜙) = 𝕋- (1, 𝜙, 𝜃) and 𝜙*⃗ is 𝐶 F and 1-to-1 on 𝐸 ° ↲
GB B
|(𝐷F × 𝐷G )|o𝜙*⃗p(𝜙, 𝜃) = |(𝑐𝑜𝑠𝜙𝑐𝑜𝑠𝜃, 𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜃, −𝑠𝑖𝑛𝜙) × (−𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃, 𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 0)| = 𝑠𝑖𝑛𝜙 so 𝒎[𝑺] = ∫y 𝛿 𝑑𝜎 = ∬(G,$)∈Ü 𝛿 ‘𝜙*⃗ (𝜙, 𝜃)’ '(𝐷F × 𝐷G )o𝜙*⃗p(𝜙, 𝜃)' 𝑑(𝜙, 𝜃) = ∫$˜- ∫G˜-(1 + cos G 𝜙)𝑠𝑖𝑛𝜙 𝑑𝜙 𝑑𝜃
Find flux ∫𝕊 𝔽 ⋅ 𝒅𝔸 given 𝔽(𝑥, 𝑦, 𝑧) = (0,0,3) same surface as above ∫𝕊 𝔽 ⋅ 𝑑𝔸 = ∬(G,$)∈Ü 𝔽 ‘𝜙*⃗(𝜙, 𝜃)’ ⋅ (𝐷F × 𝐷G )o𝜙*⃗p(𝜙, 𝜃)𝑑(𝜙𝜃) = ∬(G,$)∈Ü 𝔽(𝑠𝑖𝑛𝜙𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜙𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜙) ⋅ (sinG 𝜙 𝑐𝑜𝑠𝜃, sinG 𝜙 𝑠𝑖𝑛𝜃, 𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜙)
GB
B
=∫$˜- ∫G˜- 3𝑐𝑜𝑠𝜙𝑠𝑖𝑛𝜙 𝑑𝜙 𝑑𝜃 = 0
Compute the volume 𝑽[𝑬𝟏𝟏𝟏 ] given 𝐸FFF = {(𝑢, 𝑣, 𝑤)|0 ≤ 𝑢 ≤ 1, 0 ≤ 𝑣 ≤ 𝑢, 0 ≤ 𝑤 ≤ 𝑣} 𝑽[𝑬𝟏𝟏𝟏] = ∭(¤,ü,ý)∈Ü
*****⃗ + 𝑣𝐵𝐶
*****⃗ + 𝑤𝐶𝐷
*****⃗ |(𝑢, 𝑣, 𝑤) ∈ 𝐸FFF, 𝑽[𝑺] = ∭
*****⃗ ⋮ *****⃗
*****⃗ ¶' 𝑑(𝑢𝑣𝑤) = F |−1(−1) + 1(2)| = F
Compute 𝑽[𝑺] given 𝑆 = Δ𝐴𝐵𝐶𝐷 = %𝐴 + 𝑢𝐴𝐵
'det³𝐴𝐵
𝐵𝐶 ⋮ 𝐶𝐷
(¤,ü,ý)∈Ü
J
NNN
G
NNN
F
¤
ü
𝑑(𝑢𝑣𝑤) = ∫¤˜- ∫ü˜- ∫ý˜- 𝑑𝑤 𝑑𝑣 𝑑𝑢 = 1/6
Express 𝑫𝟐 (𝒉)(𝒙, 𝒓, 𝜽) in terms of ↲
𝑫𝟏 𝑭, 𝑫𝟐 𝑭, 𝑫𝟑 𝑭, 𝒙, 𝒓, 𝜽 given 𝕋(𝑥, 𝑟, 𝜃) = (𝑥, 𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃) and ℎ = 𝑓 ∘ 𝕋 𝑫𝟐 (𝒉) = ℎ• (𝑥, 𝑟, 𝜃)(0,1,0) = 𝑓 •o𝕋(𝑥, 𝑟, 𝜃)p𝕋•(𝑥, 𝑟, 𝜃)(0,1,0) = 𝑓 • o𝕋(𝑥, 𝑟, 𝜃)p𝐷G(𝕋)(𝑥, 𝑟, 𝜃) = 𝑓 • (𝑥, 𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)(0, 𝑐𝑜𝑠𝜃, 𝑠𝑖𝑛𝜃) ↲
= 𝐷G (𝑓)(𝑥, 𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)𝑐𝑜𝑠𝜃 + 𝐷a (𝑓)(𝑥, 𝑟𝑐𝑜𝑠𝜃, 𝑟𝑠𝑖𝑛𝜃)𝑠𝑖𝑛𝜃
Compute 𝛁𝒇(𝒙, 𝒚) given 𝑓(𝑥, 𝑦) = 2𝑥𝑠𝑖𝑛𝑦 𝛁𝒇(𝒙, 𝒚) = (2𝑠𝑖𝑛𝑦, 2𝑥𝑐𝑜𝑠𝑦) At 𝕣𝟎 = (𝟏, 𝝅/𝟔) compute direction 𝕦𝟎 of greatest rate of change ↲
B
B
B
J
J
J
𝕦𝟎 = ∇𝑓(1, 𝜋/6) / |∇𝑓(1, 𝜋/6)| = (2𝑠𝑖𝑛(𝜋/6),2𝑐𝑜𝑠(𝜋/6))/ t‘2 sin ‘ ’ , 2 cos ‘ ’’t = (1, √3)/2 Compute 𝑫𝕦𝟎 (𝒇)(𝕣𝟎 ) = ∇𝑓(𝕣-) ⋅ 𝕦- = t∇𝑓 ‘1, ’t = '1, √3' = 2