Calc 4 Final

polar coordinate

(r, θ) r: length of the line segment, distance from origin θ: angle from the origin to the line segment r^2 = x^2+ y^2 x= r cos θ y= r sin θ tanθ = y/x

cylindrical coordinates

(r, θ, z) Type 1: x= r cos θ y = r sin θ leave z as if dV = r dz dr dθ -r is the radius from the axis -θ is in terms of x-y plane in counterclockwise z is height

Type 2 region

(x simple) y is bounded by two constants x is bounded by two functions in terms of y (on graph its horizontal) (integrate with respect to x first )

Type 1 Region

(y simple) y is two functions in terms of x x is bounded by two constants (on graph its vertical) (integrate in terms of y first)

changing to polar coordinates in a double integral

*dont forget the r drdθ

triple integrals

*fubinis theorem still holds true

fundamental theorem of line integrals

*implies that the line integral is independent of path --here you find the potential function and then sub in the values of

Finding a limit

- For polynomials and rationals just plug it in (just watch out for division by zero) If 0/0 is found then... 1. Conjugate trick: multiply by the reciprocal ( BIG 1) *use when the denominator is the difference of square roots * 2. Approach method: because the value is contingent on the direction lim DNE *IF TWO DIRECTIONS YIELD THE SAME NUMBER, YOU HAVE NOT SHOWN THE LIM DOES EXIST 3. Squeeze Theorem

domain no no's

-(-) under square root -1/0 -ln (0) -ln (negative)

div(curl(F))=

0 IF the vector field on R3 is continuous with 2nd order partials

use greens theorem to prove area

1 = dq/dx - dp/dy 3 options: one of the two functions P or Q can equal 0 or you split it evenly -1/2 & 1/2

How to find absolute points

1. find all critical points 4. Determine different edges on the boundary (L1, L2, L3) 2. find location of the extreme values along the boundary 5. Plug into original function and see if there is a min or max on the specific segment (if so add it to the CP list) 6. compute F @ each point and compare

Finding the Line Integral efficiently

1. is F a gradient/ independent/ conservative IF YES then.... - if a closed and simple curve then the line integral will equal 0 -if not closed and simple then use the Fundamental Theorem of Line Integrals ***Find your potential function and plug in the values of your r(t) IF NO then ..... you must parametrize C as r(t) and do the line integral as usual

graphing a linear function

1. move all variables to one side and the constants to the other 2. to find the intercepts set two variables equal to 0 and solve for the other 3. gives you a plane

How to line integral of a vector field

1. parametrize (this is your r(t) values and find the range of t 2. find the derivative of r(t) 3. plug in parametrized values into the given function 4. set up and solve the integral (f*dr)

How to line integral

1. set up r(t) by parametrizing the function given (also find the bounds of t) 2. find the derivative of r(t) by deriving the x and y values 3. find the magnitude of r'(t) by doing the square root of the sum of the squares 4. plug in the parametrized values of x and y into the integral 5. put it all together 6. integrate

lagrange multipliers and extrema

1. take the derivative of the function to find the critical points 2. then set up the lagrange multipliers equations (derive both f and g and set them equal to eachother with the lagrange multiplier) 3. Add all the critical points found in #1 and #2 to a list 4. Find the value of all critical points from the function.

(cos θ) ^2

1/2 ( 1 + cos 2θ)

(sin θ)^2

1/2 (1 - cos 2θ)

sin 2θ

2 sinθcos θ

Curl F

<Ry - Qz, Pz - Rx, Qx - Py> HOW TO CALCULATE: del x F del = d/dx, d/dy, d/dz F = P, Q, R (add - before J component)

piecewise smooth curve

C is the union of a finite number of smooth curves for which you can integrate to sum up to the entire line integral

Fundamental theorem of line integrals

Cannot use without the potential function

Finding out what the critical points actually are

D= fxx( fyy) - f(xy)^2 -D>0 & fxx > 0 , local min -D>0 & fxx< 0, local max -D< 0 , saddle point -D=0 NO INFO ORRRRR use lagrange multipliers

Directional Derivatives

Duf (x0,y0) = lim f(x0+ha, y0+hb) - f(x0,y0) / h Duf (x,y) = fx(x,y) a +fy(x,y)b Duf (x,y) = ∇ f(x,y) * u (gradient times the component of the unit vector) DOT PRODUCT *If F is a diff function of x and y then f has a directional derivative in the direction of any unit vector u

Maximizing the Directional Derivative

Duf = ∇ f(x,y) * u = | ∇ f(x,y) | | u | cos θ θ = 0 because then cos θ = 1 when the gradient and unit vector are parallel (minimizing: θ = 180 to get θ = -1 ) (ROC (directional derivative) = 0: θ = 90 perpendicular ( gradient and unit vector) to each other)

div (F) doesnt equal 0

F is not the curl of some vector field

implicit differentiation

F(x,y) = 0 defines y implicitly as a differentiable function y = f(x) F(x,y) =0 derivative in terms of x of F(x,y) = 0 means that dy/dx = dF/dx * dx/dx + dF/dy * dy/dx

Max/Min

Fermat's Theorem: if f has a local max/min @ (a,b) and the partials exist, fx = 0 fy = 0 [@ given point] CRITICAL POINTS: could be max, min or DNE

lagrange multipliers

Find extreme values when the domain is constrained to some subset of the plane DEF: find all values of x,y.z and λ such that ∇ f(x,y) = λ * ∇ g(x,y) and g(x,y,z) = k *Evaluate each point to see which is the extreme values

finding the potential function

First check if they are gradients of each other so (F is conservative) P = df/dx Q = df/dy R = df/dz Partially integrate one in terms of the variable it was partially derived in = f Then partially derive the f you found in terms of another variable. Use that derivative and equal it to the derivatives given in the equation. Keep going...

Riemann Sum

First set i = 0 and let j's run through then you increase i by 1 and let j's run through

Continuity

Function f is continuous @ (x0,y0) if lim f(x,y) = f(x0, y0) 1. (x0,y0) is in the domain of f 2. lim @ that point exists 3. 1=2 *POLYNOMIALS AND RATIONALS ARE CONTINUOUS WHEREVER THEY ARE DEFINED

gradient and tangent plane

GRADIENT IS NORMAL TO THE TANGENT PLANE OF A LEVEL SURFACE

Curl = 0

If f is a scalar function of 3 variables that has continuous 2nd order partials then the curl is 0 F is conservative

Fubini's Theorem

If f is cont on R = [a,b] x [c,d] then when taking a double integral the order in which you integrate doesn't matter

Integral theorem

If f is cont over R then it is integrable over R

Definition of a limit

If for every ϵ > 0, there exists δ > 0 such that for (x,y) in the domain of f, 0 < [(x-x0)^2 - (y-y0)^2] ^1/2 < δ implies | f(x,y) - L | < ϵ HOW TO: 1. start off with let ϵ > 0 2. Write the definition plugging in values -δ for distance part - ϵ for |f(x)| 3. Manipulate to somehow make a relationship between ϵ and δ 4. Write it all out .... Let ϵ > 0. Choose δ = ..... If 0 < ........ < δ then 0 < ....... < (δ in terms of ϵ) 5. Work your way back to ϵ

find the points on the sphere closest to the point (a,b,c)

Lagrange multipliers: f(x,y) = distance formula g(x,y) = function given Set up... ∇ f(x,y) = λ * ∇ g(x,y) Solve for λ Then solve for varibles ORRRR find normal vector (coefficients from equation) and find the point and write out their parametrized equations Sub into the function and solve for t then solve for variables

normal line to S at point

Normal line to S @ P= (x0,y0,z0) is the line thru P normal to the tangent plane x= x0+ tFx(x0,y0,z0) y= y0 + tFy(x0,y0,z0) z= z0 + tFz(x0,y0,z0)

vector field

R2: F(x,y) = P (x,y) i + Q (x,y) j = <P, Q> R3: F(x,y,z) = P(x,y,z) i + Q(x,y,z) j + R(x,y,z) k gives direction and magnitude *tangent to circles centered at the origin *dot the vector field with position vector you get zero

Green's Theorem

Suppose C is a piecewise smooth simple closed curve that is the boundary of a region R in the plane and oriented so that the region is on the left as we move around the curve. Suppose *F* = P*i* + Q*j* is a smooth vector field on an open region containing R and C. Then, ∫ (over C) *F* • d*r*= ∫ (over R) (∂Q/∂x - ∂P/∂y)dA If not moving counterclockwise add a negative dq/dx-dp/dy = curl F * k hat therefore the integral of that equals the line integral of f * dr

curl doesn't = 0

THIS IMPLIES THAT F IS NOT CONSERVATIVE thus it is not the gradient

Second Derivative Test

Tells you what the critical point found actually is.... D= fxx( fyy) - f(xy)^2 -D>0 & fxx > 0 , local min -D>0 & fxx< 0, local max -D< 0 , saddle point -D=0 NO INFO Saddle point is when it is neither a max or a min

Level Curves/ Surfaces

The curves with equations f(x,y) / f(x,y,z) = k, where k is a constant. level curves: R2 level surfaces: R3

Region types (2D)

Type 1: y Type 2: x

Region types (3D)

Type 1: z sandwiched Type 2: x Type 3: y

when in a closed curve the line integral

the line integral takes the subscript c but not the line integral equals f(r(b)) - f(r(a))

conservative vector field

vector field is a conservative vector field if it is the gradient os some scalar function that is F = gradient of F for some f called a potential function

changing the bounds on double integral

when the bounds are in terms of another variable and you want to switch the order of integration you must draw the graphs of the bounds and change it

parametrize circle

x = cos t y = sin t 0 < t < 2pi

parametrize a line

x = x0 + t (x1 - x0) y = y0 + t (y1 - y0)

spherical for a unit sphere

x^2 + y^2 + z^2 = 1 p =1 0< ⌀ < pi 0 < θ < 2pi

gradient vector ∇

[fx, fy] *points normal to the level curve that runs through p the rate of greatest increase to find direction find the unit vector GREATEST DECREASE: opposite signs of the greatest decrease NO CHANGE: vector perpendicular to gradient dot product must be equal to 0

simple curve

a curve that does not intersect itself

simple region

both x and y (unit disk is simple)

Conservative vector field

conservative if partial dp/dy = partial dQ/dx

Tangent Plane

consist of all possible tangent lines to S @ P most closely resembles S near P P = (x0, y0, z0) S - surface EQUATION: z-z0 = fx (x0,y0) (x-x0) +fy (x0,y0) (y-y0) TO APPROXIMATE Use equation of tangent line and plug in value that you are approximating if fx and fy are not continuous at p the tangent plane may not b a good approximation If it is a good approximation then the function is differentiable

divergence

div F = del * F

3D check for it to be conservative

dp/dy = dq/dx dq/dz = dr/dy dp/dz=dr/dx

Implicit Function Theorem

dy/dx = -Fx/Fy = (-dF/dx)/ (dF/dy)

Chain Rule

dz/dt = dz/dx*dx/dt + dfzdy*dy/dt *use tree diagram

Clairaut's Theorem

fxy=fyx

simply connected region

if every simple closed curve encloses only points that are in D

Extreme value theorem

if f is cont on a closed bounded set D within R^2 then f attains an abs. max and abs. min @ some point in D -Extreme values will occur either @ critical points or boundary points

Differentiable

if fx and fy exists near (x0,y0) and are continous @ (x0,y0), then f is differentiable @ (x0,y0) PLUS THE PARTIAL DERIVATIVES EXIST AT THAT POINT

area

integral of 1 dA = A(D)

independence of path

line integral is independent of path if and only if the line integral of F dot dr is equal to zero -if the line integral is independent then F is conservative on D that is F = gradient F for some f

line integrals

must parametrize parametric equations: x= x(t) y = y(t) a < t < b def: If f is defined on a smooth curve C then the line integral of f along C is *MULTIPLY BY ARC LENGTH

second derivative leibniz notation

opposite read right to left to know which to differentiate first

tangent plane to the level surface

plane that passes through p and has normal vector gradient. EQUATION: Fx (x0,y0,z0)(x-x0) + Fy (x0,y0,z0)(y-y0)+ Fz (x0,y0,z0)(z-z0) = 0

derivative of tan

sec ^2

positive orientation

simple closed curve C that refers to the single counterclockwise traversal of C * region D is on the left

spherical coordinates

simplifies the evaluation of triple integrals over regions bounded by spheres or cones coordinate: (p, θ, ⌀) -p is distance from the origin - θ is the usual polar angle - ⌀ is the clockwise angle from z axis *useful when there is symmetry about a point MEMORIZE: x^2 + y^2 + z^2 = p^2 z = p cos ⌀ y = p sin ⌀ sin θ x = p sin ⌀ cos θ

finding the bounds for triple integrals

solve for z in terms of the other two variables -set equal to zero to solve for the next variable and so on ?????? *graphing could help

what happens if its not type 1 or type 2?

split the integral up into two regions. D = D1 + D2

closed curve

terminal point coincides with its initial point r(b) = r(a) line integral is independent of path if and only if the line integral of F dot dr is equal to zero.

gradient implies

the gradient of f means f is conservative which means that f is path independent

line integral of vector fields

the integral of f dot with r'(t) dt (this is work)