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54 Cards in this Set
- Front
- Back
Projection: Acceleration |
a(t) = oi-gjWith a, i, j as vector quantities and g being the acceleration of objects due to gravity |
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Projection: Velocity |
v(t) = v₀cosθi+ (-gt +v₀sinθ)j |
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Projection: Position |
r(t) = (v₀cosθ)ti+(-1/2gt²+(v₀sinθ)t+h₀)jWith h₀ being some initial height, i and j being the horizontal and vertical components of projection |
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Unit tangent vector function |
T(t) = r'(t) / ‖r'(t)‖ a tangent line in direction of function divided by the magnitude of tangent line |
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Arc length of a function (function length) |
S = ∫‖r'(t)‖ dtfrom a to b. Alternate form: S = [ ∫√x'(t)² + y'(t)² + z'(t)² ] dt |
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Angle between two vectors |
θ=arccos(V*W)/(‖V‖*‖W‖)
Where V and W are vectors |
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Dot Product |
V*W= ‖V‖*‖W‖Cosθ |
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Dot Product to prove to vectors are perpendicular |
V*W = 0 Cos=0 θ=π/2 |
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Projection of V onto W |
ProjV on W=( V*W / V*V)*V |
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Projection of W onto V |
ProjW on V=( V*W / W*W)*W |
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Equation of a line |
Need points on the line, and a vector in the direction of the line. x= x₀+tA, y=y₀+tB, z=z₀+tC |
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Parallel Vectors |
VXW=0, V and W are parallel vectors |
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Equation of a plane |
Need a point on the plane, and a normal vector to the plane Normal vector between points is a cross product of common points. |
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Normal component of acceleration |
an(t) =[ √‖a(t)‖² - at² ]a(t) being acceleration functionat being the tangential component of acceleration |
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Tangential component of acceleration |
at(t) = a(t) · T'(t) |
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Curvature |
K(t) = ‖r'(t) X r''(t)‖ / ‖r'(t)‖³ |
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Unit Normal Vector |
N(t) = B(t) X T(t) alternative form: N(t) = T'(t) / ‖T(t)‖ |
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To find the distance travelled by a particle |
Arc length: ∫a to b of ‖v(t)‖dt |
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Gradient Vector |
∨f(c,d) = fx(c,d)i+fy(c,d)j Max rate of change: ‖∨f(c,d)‖ Min. rate of change: -‖∨f(c,d)‖ No change: orthogonal to ∨f(c,d) Need: fx at point, fy at point |
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Directional derivative Rate of Change |
Duf(c,d) where ‖u‖=1: = fx(c,d)α+fy(c,d)β u=αi+βj Duf(c,d) = ∨f(c,d)·u (where ∨ is the gradient vector) Gives us the rate of change in z as we move from (c,d) in the direction of u. Need: unit vector, a point, fx at that point, fy at that point, or gradient vector |
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unit vector |
Find V, divide V by its magnitude |
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Local Linearization |
F(x,y) ≈ f(c, d) + fx(c, d)(x-c) + fy(c, d)(y-d) used to approximate a function using a rate of change, also called local linearization. Need: function evaluated at points.fx evaluated at points, fy evaluated at points |
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Total Differential |
dz = fx(x, y)dx + fy(x, y) dy ≈ ∆z Approximate the change in z. similar to local linearization without the function value. Need: fx evaluated at points, fy evaluated at points, the change in points in x and y (three points total) |
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Taylor Polynomial (of degree 2) |
f(x, y) ≈ f(c, d) +fx(c, d)(x-c)+ fy(c, d)(y-d) + 1/2![(fxx(c, d)(x-c)²+2fxy(c,d)(x-c)(y-d)+fyy(c,d)(y-d)²]+ 1/3![(fxxx(c,d)(x-c)³+3fxxy(c,d)(x-c)²(y-d)+3fxyy(c,d)(x-c)(y-d)²+fyyy(c,d)(y-d)³] You need: fx, fy, fxx, fxy, fyy, fxxx, fxxy, fxyy, fyyy Patterns: first (x-c) matches degree of denominator ! second coefficiant matches degree of denominator !, with (x-c) degree in this being the coefficient -1For 1/3! the fxxy and fxyy both have coefficient 3, one with (x-c)² and the second with (y-d)²the last term (y-d) is the degree of the ! coefficiant. |
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Implicit Diff. |
dz/dx = -fx/fx dz/dy = -fy/fz dy/dx= -fx/fy Note: d means greek del symbol |
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Equation of a tangent plane in an area |
For a plane we need: a point, a normal vector Normal Vector: ∨f(c,d,m)use (x-c) (y-d) and (z-m) for the equation set equal to zero and each part times the normal vector |
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Finding a normal line |
We need a point and a normal vector, (same work as tangent plane, however we write the equation for a line differently)as x=, y=, z=, |
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Given explicitly |
as z = f(x,y) Convert by subtracting z over, as fx(c,d)i+fy(c,d)j-1k |
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Local min/max second derivative test |
1.) f(x,y) critical points: (c,d) in the domain, where fx=0 and fy=o or where they DNE 2.)define D(xy) as: fxx(fyy)-(fxy)² 3.) if D(c,d) > 0 and fxx or fyy(c,d)>0, local min: concave up 4.) if D(c,d) >0 and fxx or fyy(c,d)<0, local max: concave down 5.) if D(c,d)< 0: saddle point: concave up/down |
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Global min/max |
1.) check critical points, evaluate f(x) at these points 2.) check end points, evaluate f(x) at these points 3.) evaluate boundaries by replacing equation with x= y= z= values, taking the derivative, find its =0 values, and evaluating the function at these values. 4.) Choose the largest and smallest matching x, y, and z values for the max and min. |
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Solving min/max with Lagrange multiplier λ |
fx(x,y) = λgx(xy) : system of equations with 3 fy(xy) = λgy(xy) : unknowns, substitution works on g(xy) = c : min/max occurs at solutions :substitute solutions back into z=f(xy) for the min and max :solve for an x or y for substitution, may need to solve for λ and take BOTH routes of complete system solving and plugging into f(xy). |
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Polar conversions |
x²+y²=r² or r=√x²+y²sinθ = y/r or y=rsinθcosθ = x/r or x=rcosθtanθ = y/x or θ = arctan(y/x) +π if x<0 |
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Double integral for center of mass in a lamina in the shape of R with δ(x,y) |
(∫∫xδ(x,y)dA / ∫∫δ(x,y)dA , ∫∫yδ(x,y)dA / ∫∫δ(x,y)dA) |
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Double integral for average value of f(x,y) on R |
∫∫f(x,y)dA / ∫∫ 1dA |
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Double integral for moment of inertia |
Ix = ∫∫y²δ(x,y)dA : rotated lamina about x Iy = ∫∫x²δ(x,y)dA : rotated lamina about y Io = ∫∫x²+y²δ(x,y)dA : rotated lamina about origin |
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Double integral for surface area over a region |
SA = ∫∫ √fx(x,y)²+fy(x,y)²+1 dA fx and fy are partials Surface area problems use 2 graphs. One is the three dimensional representation of the region. One is the projection of that region to determine boundary limits. |
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Triple integral moment of inertia |
Ix = ∫∫∫(y²+z²)p(x,y,z)dV : about x Iy = ∫∫∫(x²+z²)p(x,y,z)dV : about y Iz = ∫∫∫(x²+y²)p(x,y,z)dV : about z Io = ∫∫∫(x²+y²+z²)p(x,y,z)dV : about origin where dV is the order of integration of three variables where p is the given density function |
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Common spherical functions to evaluate in triple integrals |
common functions easy to evaluate in spherical: Sphere: x²+y²+z²=R² or p=R Cone: z=√x²+y² or ∅ = π/4z=√3x²+3y² or ∅ = π/6z= 0 or ∅= π/2z= √x²/3 + y²/3 or ∅=π/3 |
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Spherical triple integrals |
∫∫∫f(psin∅cosθ, psin∅sinθ, pcos∅)p²sin∅dpd∅dθ inner bounds: from solid closest to origin to solid furthest from origin in terms of p middle bounds: rotate your arm about the region, this is the angle from the z-axis as ∅outer bounds: how far you go around z axis as θ note: r is always positive in the bounds. if origin is included in region, bound is 0 |
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Cylinderical triple integrals |
∫∫∫f(rcosθ,rsinθ,z)rdzdrdθ inner bounds: function closes to origin in terms of rmiddle bounds: rotational arm r where you enter and exit the region outer bounds: range of θ values |
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Line integrals for arc length |
∫over c, f(x,y,z)ds 1.) replace x, y, and z with their parametric equivalent: x > i, y> j, z> k, (replace x,y,z with t) 2.) Multiply by the square root of (di/dt)² + (dj/dt)²+(dk/dt)² 3.) Simplify, bounds can be given as an inequality, or found by graphing and setting equations equal to each other to find intercepts |
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Line integrals for work. |
∫over c, F*dr where F, is the vector field function 1.) replace x,y,z with their parametric t equivalent. Multiply by the parametric equivalent's partial with respect to t, dt. 2.) Simplify, bounds can be given as a inequality or found by graphing and setting equations equal to each other to find intercepts |
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Line integral for flux. |
∫over c, F*nds where F is the vector flow function, nds is the normal vector 1.) Replace function as idy-jdx 2.) Replace x,y,z with their t parametric equivalent. 3.) Replace dy as j with respect to t,dt partial 4.) Replace dx as i with respect to t,dt partial 5) Simplify, bounds are given as an inequality or solved for by graphing |
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Solving the Fundamental Theorem of Line Integrals |
1.) Solve each i, j, and k part of F as a separate integral with respect to their corresponding x,y,z part. i>x, j>y, z>k. As dx, dy, dz. Add an uncertainty considering what variable(s) was lost in integration 2.) Combine equations into one, expressions that are common are used once. 3.) Plug in given bounds or found bounds, subtract first bound from the second. |
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Solving Green's Theorem |
1.) graph the parametric region. rcosti+rsintj is a circle of radius 1. Piece wise functions are graphed and connected together as one. 2.)Given a piece wise function, find the equation of the line (usually the last line) on top, bottom, right, or left etc. for a dydx, or dxdy boundary 3.) Find dN/dx-dM/dy. 4.) Set up integral, boundaries are given or found from graph, order of integration is found from graph 5.) Solve |
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Green's Theorem |
Let C be a simple closed curve orientated counterclockwise that encloses the region R. Let M(x,y) and N(x,y) have continuous partials on R. then ∫over c closed curve, M(x,y)dx+N(x,y)dy= ∫∫over R, (dn/dx-dm/dy)dA |
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Surface Area over a surface given explicitly |
∫∫over R, f(x,y,g(x,y)√fx²+fy²+1 dA |
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Solving Surface Area over a surface given explicitly |
1.) graph a 3d graph of the portion and constrictions 2.) graph a 2d graph of the region 3.) If needed, change coordinates to an easier system 4.) Find fx, fy in the portion region equation. Square each partial and add 1. 5.) Replace z with a function of x and y and multiple by partial square roots +1 6.) Simplify, look for common variables to pull out, look for "1's" to cancel, sub in an easier coordinate system if needed 7.) boundaries are given or found from 3d graph, and 2d graph. |
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Surface Area over a surface given implicitly |
∫a to b ∫c to d, f(x(u,v)), y(u,v), z(u,v)*||ruXrv||dvdu |
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Solving Surface Area over a surface given implicitly |
1.) find ru, and rv. partials of r with respect to u or v. 2.) find ruXrv 3.) find the magnitude of ruXrv by taking the square root of each part squared 4.) Replace function with equal u,v values given from r 5.) multiple the magnitude and the u,v functioning simplify. 6.) boundaries are given or found through graphing |
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Surface S and flux given explicitly |
∫∫ over R, [-gx(x,y)*M(x,y,g(x,y))-gy(x,y)*N(x,y,g(x,y))+P(x,y,g(x,y))]dA 1.) Make function variables only x and y 2.) Nds is the partials of g(x,y) with respect to x and y, +k 3.) Graph the region to find boundaries 4.) multiply parts with respect to i, j, and k together (Nds multiplied with the function) 5.) Simplify integral with bounds and order of integration dependent on graph |
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Surface S and flux given implicitly |
∫a to b ∫c to d, F(x(u,v),y(u,v),z(u,v))*(ruXrv)dvdu 1.) find ru, rv, and cross them in a matrix 2.) make function a function of u and v by subbing equal i, j, k parts 3) Multiply ruxrv i,j,k parts with function i,j,k parts and simplify 4.) bounds are given or solved for by graphing |
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Use the Divergence Theorem to evaluate a surface integral as a triple integral in either (spherical, cylindrical, or rectangular) coordinates |
∫∫∫over Q, ∨*FdV 1.) Graph a 3d graph of the region 2.) graph a 2d graph of the region 3.) Take partials of each i,j,k component and simplify 4.) Decide if a coordinate change is needed to simplify solving 5.) Boundaries are given or solved in graphing |
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Stoke's Theorem |
∫∫over S, (∨XF)*NdS 1.) If C is given piecewise, find the plane they all lie in. 2.) graph a 3d graph of the C region 3.) graph a 2d graph of the C region in the XY place 4.) Solve ∨XF as a matrix with partials dx, dy, dz and i, j, k parts of function given 5.) Find Nds as partials of the z= function given or found 6.) Multiple Nds and ∨XF by their i, j, k components 7.) solve as boundries are found or given, and order of integration is found from graph |