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𝑑𝑦
𝑑𝑑
𝑑π‘₯
𝑑𝑑
𝑑2 𝑦
𝑑𝑦 ′
)
𝑑π‘₯
𝑑π‘₯
𝑑𝑑
(
ο‚·
𝑑𝑦
ο‚·
𝑆 = 2πœ‹ ∫𝛼 π‘₯√( 𝑑𝑑 ) + ( 𝑑𝑑 ) , if rotating about the 𝑦 − π‘Žπ‘₯𝑖𝑠, put 𝑦 instead of π‘₯ if rotating about the π‘₯ − π‘Žπ‘₯𝑖𝑠
ο‚·
to find the horizontal tangent to a parametric curve, 𝑑𝑑 = 0, for vertical, 𝑑𝑑 𝐷𝑁𝐸
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if there is a value of 𝑑 that makes 𝑑π‘₯ both zero and 𝐷𝑁𝐸, then use lim 𝑑π‘₯ ,
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𝑑𝑦
ο‚·
Symmetry test: π‘Žπ‘π‘œπ‘’π‘‘ ? −π‘Žπ‘₯𝑖𝑠 )
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sin(πœƒ ± 𝛼 ) = π‘ π‘–π‘›πœƒ cos 𝛼 ± 𝑠𝑖𝑛𝛼 cos πœƒ
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π‘Ÿ 2 = π‘₯ 2 + 𝑦 2 , πœƒ = tan−1 (π‘₯ )
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𝐴=
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In graph, π‘Ÿ = π‘Ž is the same as π‘Ÿ = −π‘Ž
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𝐿 = ∫π‘Ž √π‘Ÿ 2 + (π‘‘πœƒ) π‘‘πœƒ
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βƒ—βƒ—βƒ—βƒ—βƒ—
βƒ—βƒ—βƒ—βƒ—βƒ—
𝐴𝐢 = −𝐢𝐴
ο‚·
Slope ≡ π‘Ÿπ‘’π‘›
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π‘Ž ⋅ π‘Ž = |π‘Ž |2
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Angle b\w two vectors is cos(πœƒ) = |π‘Ž||𝑏|
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To get a vector which is parallel to a vector, get the unit vector.
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Two vectors are parallel if and only if the ratio of their components are equal, π‘Ž =< π‘Ž, 𝑏, 𝑐 >, 𝑏⃗ =< 𝑑, 𝑒, 𝑓 >, 𝑑 = 𝑒 = 𝑓
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Volume of parallelepiped is 𝑉 = |π‘Ž ⋅ (𝑏⃗ × π‘)|
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π‘Ž ⋅ (𝑏⃗ + 𝑐) = π‘Ž ⋅ 𝑏⃗ + π‘Ž ⋅ 𝑐
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Symmetric equations are
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Distance b\w two parallel planes is 𝐷 =
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Two variable limits,:
𝑑π‘₯
=
,
𝑑π‘₯ 2
𝛽
=
𝑑π‘₯ 2
𝑑𝑦 2
𝑑𝑦
𝑑π‘₯
𝑑𝑦
𝑑𝑦
t−?
π‘Ÿ′ sin πœƒ+π‘Ÿ cos πœƒ
= π‘Ÿ′ cos πœƒ−π‘Ÿ sin πœƒ
𝑑π‘₯
𝑦
1
2
𝑏
∫π‘Ž π‘Ÿ 2 π‘‘πœƒ
π‘‘π‘Ÿ 2
𝑏
𝑅𝑖𝑠𝑒
βƒ—
π‘Žβƒ— ⋅𝑏
π‘Ž
π‘₯−π‘₯0
π‘Ž
=
𝑦−𝑦0
𝑏
=
𝑏
𝑐
𝑧−𝑧0
𝑐
|𝑑1 −𝑑2 |
|𝑛|
Direct substitution οƒ  Split the limit οƒ Algebraic(conjugate) οƒ u substitution οƒ Squeeze οƒ Polar . type of the function depends;
you should look at it and see which is best.
ο‚· Path testing for two variable limits is ONLY for checking EXISTANCE, NOT FOR CALCULATING THE VALUE of the limit.
ο‚· To check a path, change the function based on the path’s equation(substitute x,,,)οƒ  calculate the limit of this new function.
ο‚· 𝑧 is always considered as a function of π‘₯ and 𝑦 unless it is declared otherwise.
ο‚·
πœ•2 𝑧
πœ•π‘¦πœ•π‘₯
means first differentiate z with respect to x then with respect to y.
ο‚· Equation of tangent plane 𝑧 − 𝑧0 = 𝑓π‘₯ ⋅ (π‘₯ − π‘₯0 ) + 𝑓𝑦 ⋅ (𝑦 − 𝑦0 ),
ο‚· Equation of linearization near π‘₯0 and 𝑦0 is z from the tangent plane equation 𝑧 = 𝐿(π‘₯, 𝑦) = 𝑧0 + 𝑓π‘₯ ⋅ (π‘₯ − π‘₯0 ) + 𝑓𝑦 ⋅ (𝑦 − 𝑦0 )
(𝑧0 𝑖𝑠 π‘‘β„Žπ‘’ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑧 π‘Žπ‘‘ π‘₯0 π‘Žπ‘›π‘‘ 𝑦0 , π‘œπ‘Ÿ 𝑓(π‘₯0 , 𝑦0 ))
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𝑓 is differentiable at (π‘Ž, 𝑏) if and only if 𝑓π‘₯ and 𝑓𝑦 are continuous at (π‘Ž, 𝑏)
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𝑑𝑧 = πœ•π‘₯ 𝑑π‘₯ + πœ•π‘¦ 𝑑𝑦
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Partial derivative of multiple variables πœ•π‘‘ = πœ•π‘₯ ⋅ 𝑑𝑑 + πœ•π‘¦ ⋅ 𝑑𝑑 , and it is better to draw a tree to know
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𝑑𝑦
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∇𝑓 (π‘₯, 𝑦, 𝑧) =< 𝑓π‘₯ , 𝑓𝑦 , 𝑓𝑧 >
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𝐷𝑒 𝑓 (π‘₯0 , 𝑦0 ) = ∇𝑓 (π‘₯0 , 𝑦0 ) ⋅ 𝑒 = |∇𝑓(π‘₯0 , 𝑦0 )| cos πœƒ, π‘€β„Žπ‘’π‘Ÿπ‘’ π‘‘β„Žπ‘’π‘‘π‘Ž 𝑖𝑠 π‘‘β„Žπ‘’ π‘Žπ‘›π‘™π‘’ 𝑏\𝑀 π‘‘β„Žπ‘’ 𝑒𝑛𝑖𝑑 π‘£π‘’π‘π‘‘π‘œπ‘Ÿ π‘Žπ‘›π‘‘ ∇𝑓
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In Directional derivative, 𝑒 must be a unit vector, if you were given a ordinary vector, you must convert it to its unit
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The value of maximum directional derivative is |∇𝑓(π‘₯, 𝑦, 𝑧)| and the minimum −|∇𝑓(π‘₯, 𝑦, 𝑧)|
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The maximum directional derivative occurs at many points, at points where 𝑒 has the same direction as ∇𝑓
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The gradient vector is the normal of the tangent plane at some point
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Extreme values ≡ Local max or min
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Critical points occur at either when {𝑓π‘₯ = 𝑓𝑦 = 0} or {𝑓π‘₯ π‘œπ‘Ÿ 𝑓𝑦 𝐷𝑁𝐸}
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πœ•π‘§
πœ•π‘§
πœ•π‘§
𝑑π‘₯
−𝐹
πœ•π‘§
πœ•π‘§
𝑑π‘₯
πœ•π‘§
𝑑𝑦
𝐹
= 𝐹 π‘₯ , 𝑖𝑓 𝐹 (π‘₯, 𝑦) = 0, π‘Žπ‘›π‘‘ πœ•π‘₯ = − 𝐹π‘₯ , π‘œπ‘Ÿ 𝑦, 𝑖𝑓 𝐹 (π‘₯, 𝑦, 𝑧) = 0
𝑦
𝑧
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