{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

FinalFormulaSheet

# FinalFormulaSheet - Formula Sheet(Final Exam The length of...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}