FinalFormulaSheet - Formula Sheet (Final Exam) The length...

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Unformatted text preview: Formula Sheet (Final Exam) The length of r = x, y, z is r = x2 + y 2 + z 2 The distance from r0 = (x0 , y0 , z0 ) to r1 = (x1 , y1 , z1 ) is (x1 − x0 )2 + (y1 − y0 )2 + (z1 − z0 )2 The line through the point r0 = (x0 , y0 , z0 ) with direction vector v = (a, b, c) is given by r(t) = r0 + t v = (at + x0 , bt + y0 , ct + z0 ) The plane through the point r0 = (x0 , y0 , z0 ) with normal vector n = (a, b, c) is given by r · n = 0, or a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. a·b= a b cos θ and a × b = a For a regular curve r(t), b sin θ . r ′ ( t) ; T ( t) = r ′ ( t) The curvature is given by κ(t) = N(t) = T ′ ( t) T ′ ( t) ; B ( t) = T ( t) × N ( t) r ′ (t) × r ′′ (t) T ′ ( t) = r ′ ( t) r ′ ( t) 3 If a particle is moving along the trajectory r(t), then v(t) = r ′ (t), v (t) = v(t) and a(t) = r ′′ (t). Also, a(t) = aT (t)T(t) + aN (t)N(t) where a T ( t) = v ′ ( t) = a ( t) · v ( t) , v ( t) aN (t) = κ(t)v (t)2 = a(t) 2 − | a T ( t) | 2 The tangent plane to a surface z = f (x, y ) at a point (a, b, f (a, b)) is given by z = f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b). ˆ ˆ The derivative of a function f in the direction u at the point P is given by Du f = ∇f (P ) · u ˆ Cylindrical Coordinates: x = r cos θ , y = r sin θ , z = z ; x2 + y 2 = r 2 , tan θ = y/x, Spherical Coordinates: x = ρ cos θ sin φ, y = ρ sin θ sin φ, z = ρ cos φ and x2 + y 2 + z 2 = ρ2 , tan θ = y/x, cos φ = z/ρ The Second Derivative Test for Functions of Two Variables: Let f (x, y ) be a twice differentiable function, whose second partial derivatives are continuous. If (a, b) is a critical point for f (∇f (a, b) = 0), the discriminant is D(a, b) = fxx (a, b) · fyy (a, b) − [fxy (a, b)]2 . If D(a, b) < 0, then (a, b) is a saddle point for f (x, y ). If D(a, b) = 0, then the test is inconclusive. If D(a, b) > 0 and fxx (a, b) > 0, then f (a, b) is a local minimum for f (x, y ). If D(a, b) > 0 and fxx (a, b) < 0, then f (a, b) is a local maximum for f (x, y ). Lagrange Multipliers: If a differentiable function f (x, y ) has a local maximum or minimum at (a, b) on a constraint curve of the form g (x, y ) = c (where g is differentiable) and if ∇g (a, b) = 0, then there is a constant λ so that ∇f (a, b) = λ∇g (a, b). 1 Jacobians and Change of Variables. For polar coordinates, dx dy = r dr dθ . In general, suppose that under the 1–1 differentiable transformation x = x(u, v ) and y = y (u, v ) that the region R in the xy -plane corresponds to the region S in the uv -plane. Then ∂x ∂x ∂ (x, y ) ∂ (x,y du dv where ∂ (u,v) = det ∂u ∂v . f (x, y ) dx dy = f (x, y ) ∂y ∂y ) ∂ (u, v ) R S ∂u ∂v The 3-D analogue (between coordinates u, v, w and x, y, z ) amounts to the change-of∂ (x,y,z ∂ (x,y,z variables relation dx dy dz = ∂ (u,v,w) du dv dw where ∂ (u,v,w) is the determinant of ) ) the analogous 3 × 3 matrix of partial derivatives. For cylindrical coordinates, dx dy dz = r dr dθ dz . For spherical coordinates, dx dy dz = ρ2 sin(φ) dρ dφ dθ . Criterion for the Existence of a Potential Function. If F = (F1 , F2 ) is a vector field in a simply connected plane region, F is a gradient field if and only if ∂F1 /∂y = ∂F2 /∂x. If F = (F1 , F2 , F3 ) is a vector field in a simply connected region in 3-space, F is a gradient vector field if and only if the three cross partials agree: for all i and j , ∂Fi /∂xj = ∂Fj /∂xi . Line integrals. Let r(t) parametrize a path C (t0 ≤ t ≤ t1 ). Then C f (x, y, z )ds = t1 t1 f (r(t))(ds/dt)dt. If F is a vector field, then C F(x, y, z ) · ds = t=t0 F(r(t)) · (dr/dt)dt t=t0 If F = ∇U , C F · ds = U (r(t1 )) − U (r(t0 )). Surface Integrals: If a surface S is parametrized by u, v in a domain D then n = Tu × Tv , ∂ (x, y, z ) ∂ (x, y, z ) where Tu = and Tv = , and f (x, y, z ) dS = f (u, v ) n du dv . ∂u ∂v S D The surface area is S dS . If S is defined by z = z (x, y ) then n = 1 + (∂z/∂x)2 + (∂z/∂y )2 . If F is a vector field dened on an oriented surface S then F · dS = F · n du dv . S D ∂ ∂ ∂ curl ∇ × F and div ∇ · F: Use ∇ = ( ∂x , ∂y , ∂z ). For example ∇ · F = Green’s Theorem. ∂ Q ∂P − ∂x ∂y P dx+Q dy = ∂D Stoke’s Theorem. D F · ds = ∂S S Divergence Theorem. + ∂F2 ∂x + ∂F3 . ∂x D is a domain in R2 , F = (P, Q) a vector field. (∇ × F) · dS, where S and its boundary ∂S have compatible orientations. F · dS = S dx dy ∂F1 ∂x R 2 (∇ · F) dx dy dz , where S is the boundary of a region R in 3-space. ...
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This note was uploaded on 11/16/2011 for the course MULTIVARIA 251 taught by Professor Staff during the Fall '11 term at Rutgers.

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