Exam 1Exam 2Exam 3Exam 418.02 Final Exam Formula Sheet Exam 1 Dot product AB⋅rr = ||||ABrr cos θ AB⋅rr= (a1b1 + a2b2) Cross product AB×rr= ||||ABrr sin θ ˆnArea of parallelogram is equal to half the length of the × prod Matrix arithmetic 1. Addition (A + B) add corresponding elements must be same shape 2. Multiplication (AB = C) multiply each element of row i by column j, then add 3. Inversion (A-1) A → M (minors) → C (cofactors) → J (adjoint) → divide by |A|-1 M: just determinant of the minors C: checkerboard signs J: flip around top left – bottom right diagonal One or many solutions If |A| is nonzero, the matrix has exactly one solution. If |A| = 0, the matrix has either 0 or ∞ solutions. Lines in parametric With P0 (x0, y0, z0) and v (a, b, c): r x = at + x0 y = bt + y0 z = ct + z0 Planes in parametric With P0 (x0, y0, z0) and v (a, b, c): r a(x-x0) + b(y-y0) + c(z-z0) = 0 because · v = 0 NrrGiven P1, P2, P3: Find 12 13,PP PP Take × prod to get NrGiven surface ax + by + cz, = a + b + c Nrˆiˆjˆk More parametric v(t) = dxdt + ˆiˆdyjdt + ˆdzkdt18.02 Final Exam Formula Sheet Intersection of curve/surface: sub curve eqn into surface Angle between two planes: Find 12,NN Take · prod Product rule for · and × ()dABdt⋅rr = dB dAABdt dt⋅+⋅rrrr ()dA B dB dAABdt dt dt×=× +×rrrrrr Exam 2 Tangent plane 00()(zzzz xx yyxy∂∂−= − + −∂∂0) 000(,,) 0 0 0 0 0 0[( ) ( ) ( )] ( ) ( ) ( )xyzffffxx yy zz xx yy zzxyz∂∂∂∇−+−+−=−+−+∂∂ ∂− Normal vector if z = f(x,y), ˆˆˆzzijxy∂∂+−∂∂k 000(,,)xyzf∇ Max/min Find partials, set to zero (critical points) d = fxxfyy – fxy2 d – saddle d + fxx + min fxx – max Chain rule z = f(u,v) zfdufdvxudx vdx∂∂ ∂=+∂∂ ∂ u = u(x) v = v(x) Gradient ˆˆfffijxy∂∂∇= +∂∂r magnitude ⇒ steepness, always points uphill18.02 Final Exam Formula Sheet Directional derivative ||RfR∇⋅rrr dzzsds∆≈ ∆ Lagrange multipliers f(x,y), restraint g(x,y) (, , ) (, ) (, )Lxy fxy xyλλ=− take,,LLLxyλ∂∂∂∂∂∂, set equal to zero, solve system Exam 3 Double integrals (, )bdacfx y dydx∫∫ Do inner ∫ as usual, treat outer variable as constant Do outer Polar coordinates (, )rfrrdrdθθθ∫∫ Integration applications Mass: RdAδ∫∫ Average of f over R: (, )RfxydAarea∫∫ Center of mass: RxdAxmassδ=∫∫ Moment of inertia: 2()axisRdist dAδ∫∫ Work: line integrals CCFdr Mdx Ndy⋅= +∫∫rr Paramaterize x and y as functions of t ⇒ 10ttdx dyMNddt dt+∫t) Gradient fields For a gradient field, ()(CFdr fend fstart⋅= −∫rrGradient field test: MNyx∂∂=∂∂ To find the potential function, integrate Mdx and Ndy. Green’s theorem18.02 Final Exam Formula Sheet CRNMMdx Ndy dAxy∂∂+= −∂∂∫∫∫ If Nx – My = 1, CFdr area⋅=∫rr Flux CRMNNdx Mdy dAxy∂∂−+ = +∂∂∫∫∫ Exam 4 Triple integrals 2211() (,)() (,)(, ,) (, ,)bgx hxyDagxhxyfxyz f xyzdzdydx=∫∫∫ ∫ ∫ ∫ get outer and middle variables from shadow Cylindrical coordinates (, , )rzfr z rdzdrdθθθ∫∫∫ cossinyrxrzzθθ=== Spherical coordinates 2(,,) sinfdddθϕρρϕθ ρ ϕ ρ ϕ θ∫∫∫ sin cossin sincosxyzρϕθρϕθρθ=== Applications of 3D integrals Mass: ∫∫∫D δ dV Average of f over D: ∫∫∫D f(x,y,z) dV volume D Center of mass: ∫∫∫D x δ dV Moment of inertia: ∫∫∫D (distance from axis)2 δ dV mass Gravitational attraction: G ∫∫∫D δ cos φ sin φ dV Surface integrals Find ∫∫S F · n ds. 1. Inspection. F · n is constant. ∫∫S F · n ds = (F · n)(area s) 2. Cylindrical/spherical coords. ∫z∫θ F · n (radius) dzdθ ncyl = x i + y j radius ∫θ∫φ F · n (radius)2 sin φ dφdθ nsph = x i + y j + z k radius 3. Rectangular case. S is z = f(x,y).18.02 Final Exam Formula Sheet ∫∫S F · n ds = ∫∫shadow F · (-fxi – fyj + k) dA Divergence theorem S is closed surface oriented outward §§ F · ds = ∫∫∫D (div F) dV *Flux div F = Mx + Ny + Pz = ∇ · F Stokes’ theorem §F · dr = ∫∫S ∇ × F ds *Work Properties of div, curl, grad 1. div(curl F) = 0 ∇ · (∇ × F) = 0 2. curl(gradient field) = 0 gradient field test: ∇ × F =
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