# This - r z Directional Derivatives Double Integrals Let...

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Directional Derivatives Let z=f(x,y) be a fuction, (a,b) ap point in the domain (a valid input point) and ˆ u a unit vector (2D). The Directional Derivative is then the derivative at the point (a,b) in the direction of ˆ u or: D ~ u f ( a, b ) = ˆ u · r f ( a, b ) This will return a scalar . 4-D version: D ~ u f ( a, b, c ) = ˆ u · r f ( a, b, c ) Tangent Planes let F(x,y,z) = k be a surface and P = ( x 0 , y 0 , z 0 ) be a point on that surface. Equation of a Tangent Plane: r F ( x 0 , y 0 , z 0 ) · < x - x 0 , y - y 0 , z - z 0 > Approximations let z = f ( x, y ) be a di erentiable function total di erential of f = dz dz = r f · < dx, dy > This is the approximate change in z The actual change in z is the di erence in z values: Δ z = z - z 1 Maxima and Minima Internal Points 1. Take the Partial Derivatives with respect to X and Y ( f x and f y ) (Can use gradient) 2. Set derivatives equal to 0 and use to solve system of equations for x and y 3. Plug back into original equation for z. Use Second Derivative Test for whether points are local max, min, or saddle Second Partial Derivative Test 1. Find all (x,y) points such that 2. Let D = f xx ( x, y ) f yy ( x, y ) - f 2 xy ( x, y ) IF (a) D > 0 AND f xx < 0 , f(x,y) is local max value (b) D > 0 AND f xx ( x, y ) > 0 f(x,y) is local min value (c) D < 0, (x,y,f(x,y)) is a saddle point (d) D = 0, test is inconclusive 3. Determine if any boundary point gives min or max. Typically, we have to parametrize boundary and then reduce to a Calc 1 type of min/max problem to solve. The following only apply only if a boundary is given 1. check the corner points 2. Check each line (0 x 5 would give x=0 and x=5 ) On Bounded Equations, this is the global min and max...second derivative test is not needed. Lagrange Multipliers Given a function f(x,y) with a constraint g(x,y), solve the following system of equations to find the max and min points on the constraint (NOTE: may need to also find internal points.): r f = λ r g g ( x, y ) = 0( orkifgiven ) Double Integrals With Respect to the xy-axis, if taking an integral, R R dydx is cutting in vertical rectangles, R R dxdy is cutting in horizontal rectangles Polar Coordinates When using polar coordinates, dA = rdrd Surface Area of a Curve let z = f(x,y) be continuous over S (a closed Region in 2D domain) Then the surface area of z = f(x,y) over S is: SA = R R S q f 2 x + f 2 y + 1 dA Triple Integrals R R R s f ( x, y, z ) dv = R a 2 a 1 R φ 2 ( x ) φ 1 ( x ) R 2 ( x,y ) 1 ( x,y ) f ( x, y, z ) dzdydx Note: dv can be exchanged for dxdydz in any order, but you must then choose your limits of integration according to that order Jacobian Method R R G f ( g ( u, v ) , h ( u, v )) | J ( u, v ) | dudv = R R R f ( x, y ) dxdy J ( u, v ) = @ x @ u @ x @ v @ y @ u @ y @ v Common Jacobians: Rect. to Cylindrical: r Rect. to Spherical: 2 sin( φ ) Vector Fields let f ( x, y, z ) be a scalar field and ~ F ( x, y, z ) = M ( x, y, z ) ˆ i + N ( x, y, z ) ˆ j + P ( x, y, z ) ˆ k be a vector field, Grandient of f = r f = < @ f @ x , @ f @ y , @ f @ z > Divergence of ~ F : r · ~ F = @ M @ x + @ N @ y + @ P @ z Curl of ~ F : r ⇥ ~ F = ˆ i ˆ j ˆ k @ @ x @ @ y @ @ z M N P Line Integrals C given by x = x ( t ) , y = y ( t ) , t 2 [ a, b ] R c f ( x, y ) ds = R b a f ( x ( t ) , y ( t )) ds where ds = q ( dx dt ) 2 + ( dy dt ) 2 dt or q 1 + ( dy dx ) 2 dx or q 1 + ( dx dy ) 2 dy To evaluate a Line Integral, ·
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