ππ¦ ππ‘ ππ₯ ππ‘ π2 π¦ ππ¦ ′ ) ππ₯ ππ₯ ππ‘ ( ο· ππ¦ ο· π = 2π ∫πΌ π₯√( ππ‘ ) + ( ππ‘ ) , if rotating about the π¦ − ππ₯ππ , put π¦ instead of π₯ if rotating about the π₯ − ππ₯ππ ο· to find the horizontal tangent to a parametric curve, ππ‘ = 0, for vertical, ππ‘ π·ππΈ ο· if there is a value of π‘ that makes ππ₯ both zero and π·ππΈ, then use lim ππ₯ , ο· ππ¦ ο· Symmetry test: ππππ’π‘ ? −ππ₯ππ ) ο· sin(π ± πΌ ) = π πππ cos πΌ ± π πππΌ cos π ο· π 2 = π₯ 2 + π¦ 2 , π = tan−1 (π₯ ) ο· π΄= ο· In graph, π = π is the same as π = −π ο· πΏ = ∫π √π 2 + (ππ) ππ ο· βββββ βββββ π΄πΆ = −πΆπ΄ ο· Slope ≡ ππ’π ο· π ⋅ π = |π |2 ο· Angle b\w two vectors is cos(π) = |π||π| ο· To get a vector which is parallel to a vector, get the unit vector. ο· Two vectors are parallel if and only if the ratio of their components are equal, π =< π, π, π >, πβ =< π, π, π >, π = π = π ο· Volume of parallelepiped is π = |π ⋅ (πβ × π)| ο· π ⋅ (πβ + π) = π ⋅ πβ + π ⋅ π ο· Symmetric equations are ο· Distance b\w two parallel planes is π· = ο· Two variable limits,: ππ₯ = , ππ₯ 2 π½ = ππ₯ 2 ππ¦ 2 ππ¦ ππ₯ ππ¦ ππ¦ t−? π′ sin π+π cos π = π′ cos π−π sin π ππ₯ π¦ 1 2 π ∫π π 2 ππ ππ 2 π π ππ π β πβ ⋅π π π₯−π₯0 π = π¦−π¦0 π = π π π§−π§0 π |π1 −π2 | |π| Direct substitution ο Split the limit ο Algebraic(conjugate) ο u substitution ο Squeeze ο 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,,,)ο calculate the limit of this new function. ο· π§ is always considered as a function of π₯ and π¦ unless it is declared otherwise. ο· π2 π§ ππ¦ππ₯ means first differentiate z with respect to x then with respect to y. ο· Equation of tangent plane π§ − π§0 = ππ₯ ⋅ (π₯ − π₯0 ) + ππ¦ ⋅ (π¦ − π¦0 ), ο· Equation of linearization near π₯0 and π¦0 is z from the tangent plane equation π§ = πΏ(π₯, π¦) = π§0 + ππ₯ ⋅ (π₯ − π₯0 ) + ππ¦ ⋅ (π¦ − π¦0 ) (π§0 ππ π‘βπ π£πππ’π ππ π§ ππ‘ π₯0 πππ π¦0 , ππ π(π₯0 , π¦0 )) ο· π is differentiable at (π, π) if and only if ππ₯ and ππ¦ are continuous at (π, π) ο· ππ§ = ππ₯ ππ₯ + ππ¦ ππ¦ ο· Partial derivative of multiple variables ππ‘ = ππ₯ ⋅ ππ‘ + ππ¦ ⋅ ππ‘ , and it is better to draw a tree to know ο· ππ¦ ο· ∇π (π₯, π¦, π§) =< ππ₯ , ππ¦ , ππ§ > ο· π·π’ π (π₯0 , π¦0 ) = ∇π (π₯0 , π¦0 ) ⋅ π’ = |∇π(π₯0 , π¦0 )| cos π, π€βπππ π‘βππ‘π ππ π‘βπ ππππ π\π€ π‘βπ π’πππ‘ π£πππ‘ππ πππ ∇π ο· In Directional derivative, π’ must be a unit vector, if you were given a ordinary vector, you must convert it to its unit ο· The value of maximum directional derivative is |∇π(π₯, π¦, π§)| and the minimum −|∇π(π₯, π¦, π§)| ο· The maximum directional derivative occurs at many points, at points where π’ has the same direction as ∇π ο· The gradient vector is the normal of the tangent plane at some point ο· Extreme values ≡ Local max or min ο· Critical points occur at either when {ππ₯ = ππ¦ = 0} or {ππ₯ ππ ππ¦ π·ππΈ} ο· ππ§ ππ§ ππ§ ππ₯ −πΉ ππ§ ππ§ ππ₯ ππ§ ππ¦ πΉ = πΉ π₯ , ππ πΉ (π₯, π¦) = 0, πππ ππ₯ = − πΉπ₯ , ππ π¦, ππ πΉ (π₯, π¦, π§) = 0 π¦ π§