University:
Eastern Kentucky University
Course:
MAT 254 | Calculus III
Academic year:

2020
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370

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1
Author:

rezancuk2j

Derivatives D x e x = ex Dx sin(x) = cos(x) Dx cos(x) =  sin(x) Dx tan(x) = sec2 (x) Dx cot(x) =  csc2 (x) Dx sec(x) = sec(x) tan(x) Dx csc(x) =  csc(x) cot(x) Dx sin1 = p 1 , x 2 [1, 1] 2 1x Dx cos1 = p 1 Dx tan1 = Dx sec1 = , x 2 [1, 1] 1x2 1 , ⇡  x  ⇡2 1+x2 2 p1 , |x| > 1 |x| x2 1 Dx sinh(x) = cosh(x) Dx cosh(x) = sinh(x) Dx tanh(x) = sech2 (x) Dx coth(x) = csch2 (x) Dx sech(x) = sech(x) tanh(x) Dx csch(x) = csch(x) coth(x) Dx sinh1 = p 21 x +1 1 , x > 1 x2 1 1 1 ~ v =< v1 , v2 , v3 > ~ u·~ v = ~0 means the two vectors are Perpendicular ✓ is the angle between them. ~ u·~ v = ||~ u|| ||~ v || cos(✓ ) ~ u·~ v = u1 v1 + u2 v 2 + u3 v 3 NOTE: û · v̂ = cos(✓ ) 2 u|| = u ~·~ u ||~ ~ u·~ v = 0 when ? Angle Between u ~ and ~ v: u ~ ·~ v ✓ = cos1 ( ||~ u|| ||~ v || ) U-Substitution Let u = f (x) (can be more than one variable). (x) Determine: du = fdx dx and solve for dx. Then, if a definite integral, substitute the bounds for u = f (x) at each bounds Solve the integral using u. Integration by R R Parts udv = uv  vdu Fns and Identities p Directional Derivatives Double Integrals Let z=f(x,y) be a fuction, (a,b) ap point in the domain (a valid input point) and û a unit vector (2D). The Directional Derivative is then the derivative at the point (a,b) in the direction of û or: Du ~ f (a, b) = û · rf (a, b) This will return a scalar. 4-D version: Du ~ f (a, b, c) = û · rf (a, b, c) With Respect to the xy-axis, if taking an integral, R R R R dydx is cutting in vertical rectangles, dxdy is cutting in horizontal rectangles Tangent Planes Surface Area of a Curve let F(x,y,z) = k be a surface and P = (x0 , y0 , z0 ) be a point on that surface. Equation of a Tangent Plane: rF (x0 , y0 , z0 )· < x  x0 , y  y0 , z  z0 > Approximations let z = f (x, y) be a differentiable function total differential of f = dz dz = rf · < dx, dy > This is the approximate change in z The actual change in z is the difference in z values: ∆z = z  z1 Polar Coordinates When using polar coordinates, dA = rdrd✓ 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: R R q SA = fx2 + f 2y + 1dA S Triple Integrals R R R s f (x, y, z)dv = R a R  (x) R (x,y) 2 2 2 f (x, y, z)dzdydx a1 1 (x) 1 (x,y) Note: dv can be exchanged for dxdydz in any order, but you must then choose your limits of integration according to that order Maxima and Minima Jacobian Method Internal Points 1. Take the Partial Derivatives with respect to X and Y (fx and fy ) (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 R R R RG f (g (u, v), h(u, v))|J(u, v)|dudv = R f (x, y)dxdy    @x @x  @v  J(u, v) =  @u  Second Partial Derivative Test 1. Find all (x,y) points such that rf (x, y) =0~ 2 (x, y) 2. Let D = fxx (x, y)fyy (x, y)  fxy IF (a) D > 0 AND fxx < 0, f(x,y) is local max value (b) D > 0 AND fxx (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.): rf =  rg g(x, y) = 0(orkif given) @y @u @y @v Common Jacobians: Rect. to Cylindrical: r Rect. to Spherical: ⇢2 sin() Vector Fields let f (x, y, z) be a scalar field and ~ F (x, y, z) = M (x, y, z)ˆi + N (x, y, z)jˆ+ P (x, y, z)k̂ be a vector field, @f @f > Grandient of f = rf =< @f @x , @y , @z ~: Divergence of F ~ = @M + @N + @P r· F @z @x @y ~: Curl of F   ˆk ˆ ĵ  i  @ @ ~ =@  r⇥F  @x @y @z M  N P Line Integrals C given by x = x(t), y = y(t), t 2 [a, b] R R = ab f (x(t), y(t))ds c f (x, y)ds q dy 2 2 where ds = ( dx dt ) + ( dt ) dt q dy 2 1 + ( dx ) dx or q 2 or 1 + ( dx dy ) dy To evaluate a Line Integral, · get a paramaterized version of the line (usually in terms of t, though in exclusive terms of x or y is ok) · evaluate for the derivatives needed (usually dy, dx, and/or dt) · plug in to original equation to get in terms of the independant variable · solve integral Projection of ~ u onto v ~: ~ ·~ v prv~ ~ u=( u v 2 )~ v || ||~ Cross Product ~ u⇥~ v Produces a Vector (Geometrically, the cross product is the area of a paralellogram with sides ||~ u|| and ||~ v ||) ~ u =< u1 , u2 , u3 > ~ v =< v1 , v2 , v3 >   ˆj k̂   î u2 u3  ~ u⇥~ v =  u1  v1 v2 v3  Surfaces Ellipsoid x2 a2 + y2 b2 + z2 c2 =1 Hyperboloid of One Sheet x2 a2 2 2 + yb2  zc2 = 1 (Major Axis: z because it follows - ) Equation of a Plane (x0 , y0 , z0 ) is a point on the plane and < A, B, C > is a normal vector Hyperboloid of Two Sheets 2 2  x 2  y2 = 1 b a (Major Axis: Z because it is the one not subtracted) z2 c2 A(x  x0 ) + B(y  y0 ) + C(z  z0 ) = 0 < A, B, C > · < xx0 , y y0 , zz0 >= 0 Ax + By + Cz = D where D = Ax0 + By0 + Cz0 Equation of a line A line requires a Direction Vector ~ u =< u1 , u2 , u3 > and a point (x1 , y1 , z1 ) then, a parameterization of a line could be: x = u1 t + x1 y = u2 t + y 1 z = u3 t + z1 Distance from a Point to a Plane The distance from a point (x0 , y0 , z0 ) to a plane Ax+By+Cz=D can be expressed by the formula: 0 +By0 +C z0 D| d = |Axp 2 2 2 A +B +C Elliptic Paraboloid 2 2 z = x 2 + y2 b a (Major Axis: z because it is the variable NOT squared) x2 +y 2 +z 2 Spherical to Cylindrical r = ⇢ sin() ✓=✓ z = ⇢ cos() Cylindrical to Spherical p ⇢ = r2 + z 2 ✓=✓ cos() = p z 2+ 2 Work ˆ (force) Let F~ = Mˆi + jˆ + k M = M (x, y, z ), N = N (x, y, z), P = P (x, y, z) (Literally)d~ r = dxî + dy ĵ + dzk̂ R Work w = c F~ · d~ r (Work done by moving a particle over curve C with forceF~) Independence of Path Fund Thm of Line Integrals C is curve given by ~ r (t), t 2 [a, b]; ~ r 0(t) exists. If f (~ r ) is continuously differentiable on an open set containing R C, then c rf (~ r ) · d~ r = f (~ b)  f (~ a) Equivalent Conditions ~ F (~ r ) continuous on open connected set D. Then, (a)F~ = rf for some fn f. (if ~ F is conservative) R , (b) c F~ (~ r ) · d~ r isindep.ofpathinD R r ) · d~ r = 0 for all closed paths , (c) c F~ (~ in D. Conservation Theorem ~ F = M î + N ĵ + Pkˆ continuously differentiable on open, simply connected set D. ~ F conservative , r ⇥ F~ = ~0 ~ =~ (in 2D r ⇥ F 0 if f M y = N x ) Green’s Theorem (method of changing line integral for double integral - Use for Flux and Circulation across 2D curve and line integrals over a closed H R R boundary) Mdy  N dx = (Mx + Ny )dxdy H R RR Mdx + N dy = R (Nx  My )dxdy Let: ·R be a region in xy-plane ·C is simple, closed curve enclosing R (w/ paramerization ~ r (t)) ~ (x, y) = M (x, y )iˆ+ N (x, y)ˆj be ·F continuously differentiable over R[C. Form 1: Flux Across Boundary ~ n = unit normal vector to C H R R ~ ·~ n= r ·F~dA c F R H R R , Mdy  N dx = (Mx + Ny )dxdy R Form 2: Circulation Along Boundary H R R ~ r= r ⇥RF~R · ûdA c F R H · d~ , Mdx + N dy = (Nx  My )dxdy R AreaH of R A = ( 1 ydx + 21xdy) 2 Gauss’ Divergence Thm (3D Analog of Green’s Theorem - Use for Flux over a 3D surface) Let: ~ (x, y, z) be vector field continuously ·F differentiable in solid S ·S is a 3D solid ·@S boundary of S (A Surface) ·n̂unit outer normal to @S Then, R R R R R ~ ~ dV r· F @S F (x, y, z) · n̂dS = S (dV = dxdydz) For @3 f , @x@ 2 y @3 f @x@ 2 y , work right to left in the denominator The Gradient of a function in 2 variables is rf =< fx , fy > The Gradient of a function in 3 variables is rf =< fx , fy , fz > Chain Rule(s) Take the Partial derivative with respect to the first-order variables of the function times the partial (or normal) derivative of the first-order variable to the ultimate variable you are looking for summed with the same process for other first-order variables this makes sense for. Example: let x = x(s,t), y = y(t) and z = z(x,y). z then has first partial derivative: @z and @z @x @y x has the partial derivatives: @x and @x @t @s and y has the derivative: dy dt Hyperbolic Paraboloid (Major Axis: Z axis because it is not squared) z= y2 b2  x2 a2 Coord Sys Conv Cylindrical to Rectangular x = r cos(✓ ) y = r sin(✓) z=z Rectangular to Cylindrical p r = x2 + y 2 y tan(✓) = x z=z Spherical to Rectangular x = ⇢ sin() cos(✓ ) y = ⇢ sin() sin(✓ ) z = ⇢ cos() Rectangular to Spherical p ⇢ = x2 + y 2 + z 2 y tan(✓) = x z cos() = p fxyy = Gradients ~ u⇥~ v = ~0 means the vectors are paralell Lines and Planes Given z=f(x,y), the partial derivative of z with respect to x is: @f (x,y) fx (x, y) = zx = @z @x = @x likewise for partial with respect to y: @f (x,y) fy (x, y) = zy = @z @y @y = Notation For fxyy , work ”inside to outside” fx then fxy , then fxyy In this case (with z containing x and y as well as x and y both containing s and @z = @z @x t), the chain rule for @z is @s @x @s @s The chain rule for @z @t is @z = @z @x + @z dy @t @x @t @y dt Note: the use of ”d” instead of ”@” with the function of only one independent variable Limits and Continuity Elliptic Cone (Major Axis: Z axis because it’s the only one being subtracted) x2 a2 + y2 b2  z2 c2 =0 Cylinder 1 of the variables is missing OR (x  a)2 + (y  b2 ) = c (Major Axis is missing variable) Partial Derivatives Partial Derivatives are simply holding all other variables constant (and act like constants for the derivative) and only taking the derivative with respect to a given variable. Surface Integrals Let ·R be closed, bounded region in xy-plane ·f be a fn with first order partial derivatives on R ·G be a surface over R given by z = f (x, y) ·g (x, y, z) = g (x, y, f (x, y)) is cont. on R Then, R R R RG g(x, y, z)dS = (x, y))dS R g(x, y, fq where dS = fx2 + fy2 + 1dydx ~ Flux R R of F across G ~ · ndS = F R RG R [Mfx  N fy + P ]dxdy where: ~ (x, y, z) = ·F M (x, y, z)ˆi + N (x, y, z)jˆ+ P (x, y, z)k̂ ·G is surface f(x,y)=z ·~ n is upward unit normal on G. ·f(x,y) has continuous 1st order partial derivatives Unit Circle (cos, sin) Limits in 2 or more variables Limits taken over a vectorized limit just evaluate separately for each component of the limit. Strategies to show limit exists 1. Plug in Numbers, Everything is Fine 2. Algebraic Manipulation -factoring/dividing out -use trig identites 3. Change to polar coords if(x, y) ! (0, 0) , r ! 0 Strategies to show limit DNE 1. Show limit is different if approached from different paths (x=y, x = y 2 , etc.) 2. Switch to Polar coords and show the limit DNE. Continunity A fn, z = f (x, y), is continuous at (a,b) if f (a, b) = lim(x,y)!(a,b) f (x, y) Which means: 1. The limit exists 2. The fn value is defined 3. They are the same value Other Information p p pa b a = b Where p a Cone is defined as z = a(x2 + y 2 ), In SphericalqCoordinates, 1 a  = cos ( 1+a ) Right Circular Cylinder: V = ⇡r 2 h, SA = ⇡ r 2 + 2⇡rh limn!inf (1 + m )pn = emp n Law of Cosines: a2 = b2 + c2  2bc(cos(✓ )) Stokes Theorem Let: ·S be a 3D surface ·F~(x, y, z) = M (x, y, z)î + N (x, y, z)ĵ + P (x, y, z)l̂ ·M,N,P have continuous 1st order partial derivatives ·C is piece-wise smooth, simple, closed, curve, positively oriented ·Tˆ is unit tangent vector to C. Then, H R R ~ · T̂ dS = ~ ) · n̂dS = (r ⇥ F F c s R R F~) · ~ ndxdy R (r ⇥ Remember: H ~ · T~ ds = R (Mdx + N dy + P dz) F c

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